The level of exposition is quite uneven throughout these notes. After all, they aren't meant to be a polished exposition but rather notes I take for myself. In some places I build concepts from first principles, while in others I gloss over details and focus only on the main results.

Sometime during the second half of the month, I also developed an open-source tool called Wander Console on a whim. It lets anyone with a website host a decentralised web console that recommends interesting websites from the 'small web' of independent, personal websites. Check my console here: wander/.

Although the initial version was ready after just about 1.5 hours of development during a break I was taking from studying algebraic graph theory, the subsequent warm reception on Hacker News and a growing community around it, along with the resulting feature requests and bug fixes, ended up taking more time than I had anticipated, at the expense of my algebraic graph theory studies. With a full-time job, it becomes difficult to find time for both open source development and mathematical studies. But eventually, I managed to return to my studies while making Wander Console improvements only occasionally during breaks from my studies.

Contents

1. Group Theory
   1. Permutation
   2. Group Homomorphism
   3. Group Homomorphism Preserves Identity
   4. Group Homomorphism Preserves Inverses
   5. Image of a Group Homomorphism
   6. Group Monomorphism
      1. Standard Proof
      2. Alternate Proof
   7. Permutation Representation
   8. Group Action
      1. Why Right Action?
      2. Example 1
      3. Example 2
   9. Group Actions Induce Permutations
   10. Group Actions Determine Permutation Representations
   11. Permutation Representations Determine Group Actions
   12. Bijection Between Group Actions and Permutation Representations
   13. Orbits
   14. Stabilisers
   15. Orbit-Stabiliser Theorem
   16. Faithful Actions
   17. Semiregular Actions
   18. Transitive Actions
   19. Conjugacy
       1. Conjugation as Group Action
       2. Right Conjugation vs Left Conjugation
   20. Conjugate Subgroups
   21. Conjugacy of Stabilisers
2. Algebraic Graph Theory
   1. Stabiliser Index
   2. Strongly Connected Directed Graph
   3. Shunting
   4. Automorphisms Preserve Successor Relation
   5. Test of \( s \)-arc Transitivity
   6. Moore Graphs
   7. Generalised Polygons
3. Computing
   1. Select Between Lines, Inclusive
   2. Select Between Lines, Exclusive
   3. Signing and Verification with SSH Key
   4. Block IP Address with nftables
   5. Debian Logrotate Setup

Group Theory

Permutation

A permutation of a set \( X \) is a bijection \( X \to X . \)

For example, take \( X = \{ 1, 2, 3, 4, 5, 6 \} \) and define the map \[ \pi : X \to X; \; x \mapsto 1 + ((x + 1) \bmod 6). \] This maps \begin{align*} 1 &\mapsto 3, \\ 2 &\mapsto 4, \\ 3 &\mapsto 5, \\ 4 &\mapsto 6, \\ 5 &\mapsto 1, \\ 6 &\mapsto 2. \end{align*}

We can describe permutations more succinctly using cycle notation. The cycle notation of a permutation \( \pi \) consists of one or more sequences written next to each other such that the sequences are pairwise disjoint and \( \alpha \) maps each element in a sequence to the next element on its right. If the sequence is finite, then \( \alpha \) maps the final element back to the first one. Any element that does not appear in any sequence is mapped to itself. For example the cycle notation for the above permutation is \( (1 3 5)(2 4 6). \)

Group Homomorphism

A map \( \phi : G \to H \) from a group \( (G, \ast) \) to a group \( (H, \cdot) \) is a group homomorphism if, for all \( x, y \in G, \) \[ \phi(x \ast y) = \phi(x) \cdot \phi(y). \] We say that a group homomorphism is a map between groups that preserves the group operation. In other words, a group homomorphism sends products in \( G \) to products in \( H . \) For example, consider the groups \( (\mathbb{Z}, +) \) and \( (\mathbb{Z}_3, +). \) Then the map \[ \phi : \mathbb{Z} \to \mathbb{Z}_3; \; n \mapsto n \bmod 3 \] is a group homomorphism because \[ \phi(x + y) = (x + y) \bmod 3 = (x \bmod 3) + (y \bmod 3) = \phi(x) + \phi(y) \] for all \( x, y \in \mathbb{Z}. \) As another example, consider the groups \( (\mathbb{R}_{\gt 0}, \times) \) and \( (\mathbb{R}, +). \) Then the map \[ \log : \mathbb{R}_{\gt 0} \to \mathbb{R} \] is a group homomorphism because \[ \log(m \times n) = \log m + \log n. \] Note that a group homomorphism preserves the identity element. For example, \( 1 \) is the identity element of \( (\mathbb{R}_{\gt 0}, \times) \) and \( 0 \) is the identity element of \( (\mathbb{R}, +) \) and indeed \( \log 1 = 0. \) Also, a group homomorphism preserves inverses. Indeed \( \log m^{-1} = -\log m \) for all \( m \in \mathbb{R}_{\gt 0}. \) These observations are proved in the next two sections.

Group Homomorphism Preserves Identity

Let \( \phi : G \to H \) be a group homomorphism from \( (G, \ast) \) to \( (H, \cdot). \) Let \( e_1 \) be the identity in \( G \) and let \( e_2 \) be the identity in \( H. \) Then \( \phi(e_1) = e_2. \) The proof is straightforward. Note first that \[ \phi(e_1) \cdot \phi(e_1) = \phi(e_1 \ast e_1) = \phi(e_1) \] Multiplying both sides on the right by \( \phi(e_1)^{-1}, \) we get \[ (\phi(e_1) \cdot \phi(e_1)) \cdot \phi(e_1)^{-1} = \phi(e_1) \cdot \phi(e_1)^{-1}. \] Using the associative and inverse properties of groups, we can simplify both sides to get \[ \phi(e_1) = e_2. \]

Group Homomorphism Preserves Inverses

Let \( \phi : G \to H \) be a group homomorphism from \( (G, \ast) \) to \( (H, \cdot). \) Let \( e_1 \) be the identity in \( G \) and let \( e_2 \) be the identity in \( H. \) Then for all \( x \in G, \) \(\phi(x^{-1}) = (\phi(x))^{-1}. \) The proof of this is straightforward too. Note that \[ \phi(x) \cdot \phi(x^{-1}) = \phi(x \ast x^{-1}) = \phi(e_1) = e_2. \] Thus \( \phi(x^{-1}) \) is an inverse of \( \phi(x), \) so \[ \phi(x^{-1}) = (\phi(x))^{-1}. \] The image of the inverse of an element is the inverse of the image of that element.

Image of a Group Homomorphism

Let \( \phi : G \to H \) be a group homomorphism. Then the image of the \( \phi, \) denoted \[ \phi(G) = \{ \phi(x) : x \in G \} \] is a subgroup of \( H. \) We will prove this now.

Let \( a, b \in \phi(G). \) Then \( a = \phi(x) \) and \( b = \phi(y) \) for some \( x, y \in G. \) Now \( ab = \phi(x)\phi(y) = \phi(xy) \in \phi(G). \) Therefore \( \phi(G) \) satisfies the closure property.

Let \( e_1 \) and \( e_2 \) be the identities in \( G \) and \( H \) respectively. Since a group homomorphism preserves the identity, \( \phi(e_1) = e_2. \) Hence the identity of \( H \) lies in \( \phi(G). \)

Finally, let \( a \in \phi(G). \) Then \( a = \phi(x) \) for some \( x \in G. \) Then \( a^{-1} = \phi(x)^{-1} = \phi(x^{-1}) \in \phi(G). \) Therefore \( \phi(G) \) satisfies the inverse property as well. Therefore \( \phi(G) \) is a subgroup of \( H. \)

Group Monomorphism

A map \( \phi : G \to H \) from a group \( (G, \ast) \) to a group \( (H, \cdot) \) is a group monomorphism if \( \phi \) is a homomorphism and is injective. In other words, a homomorphism \( \phi \) is called a monomorphism if, for all \( x, y \in G, \) \[ \phi(x) = \phi(y) \implies x = y. \] Let \( e_1 \) be the identity element of \( G \) and let \( e_2 \) be the identity element of \( H. \) A useful result in group theory states that a homomorphism \( \phi : G \to H \) is a monomorphism if and only if its kernel is trivial, i.e. \[ \ker(\phi) = \{ x \in G : \phi(x) = e_2 \} = \{ e_1 \}. \] Let us prove this now.

Standard Proof

Suppose \( \phi : G \to H \) is a monomorphism. Since a homomorphism preserves the identity element, we have \( \phi(e_1) = e_2. \) Therefore \[ e_1 \in \ker(\phi). \] Let \( x \in \ker(\phi). \) Then \( \phi(x) = e_2 = \phi(e_1). \) Since \( \phi \) is injective, \( x = e_1. \) Therefore \[ \ker(\phi) = \{ e_1 \}. \] Conversely, suppose \( \ker(\phi) = \{ e_1 \}. \) Let \( x, y \in G \) such that \( \phi(x) = \phi(y). \) Then \[ \phi(x \ast y^{-1}) = \phi(x) \cdot \phi(y^{-1}) = \phi(x) \cdot (\phi(y))^{-1} = \phi(y) \cdot (\phi(y))^{-1} = e_2. \] Hence \[ x \ast y^{-1} \in \ker(\phi) = \{ e_1 \}, \] so \[ x \ast y^{-1} = e_1. \] Multiplying both sides on the right by \( y, \) we obtain \[ x = y. \] This completes the proof.

Alternate Proof

Here I briefly discuss an alternate way to think about the above proof. The above proof is how most texts usually present these arguments. In particular, the proof of injectivity typically proceeds by showing that equal images imply equal preimages. It's a standard proof technique. When I think about these proofs, however, the contrapositive argument feels more intuitive to me. I prefer to think about how unequal preimages must have unequal images. Mathematically, there is no difference at all but the contrapositive argument has always felt the most natural to me. Let me briefly describe how this proof runs in my mind when I think about it more intuitively.

Suppose \( \phi \) is a monomophorism. Since a homomorphism preserves the identity element, clearly \( \phi(e_1) = e_2. \) Since \( \phi \) is injective, it cannot map two distinct elements of \( G \) to \( e_2. \) Thus \( e_1 \) is the only element of \( G \) that \( \phi \) maps to \( e_2 \) which means \( \ker(\phi) = \{ e_1 \}. \)

To prove the converse, suppose \( \ker(\phi) = \{ e_1 \}. \) Consider distinct elements \( x, y \in G. \) Since \( x \ne y, \) we have \( x \ast y^{-1} \ne e_1. \) Therefore \( x \ast y^{-1} \notin \ker(\phi). \) Thus \( \phi(x \ast y^{-1}) \ne e_2. \) Since \( \phi \) is a homomorphism, \[ \phi(x \ast y^{-1}) = \phi(x) \cdot \phi(y^{-1}) = \phi(x) \cdot \phi(y)^{-1}. \] Therefore \( \phi(x) \cdot \phi(y)^{-1} \ne e_2 \) which implies \[ \phi(x) \ne \phi(y). \] This proves that \( \ker(\phi) = \{ e_1 \} \) implies that \( \phi \) is injective and thus a monomorphism.

Permutation Representation

Let \( G \) be a group and \( X \) a set. Then a homomorphism \[ \phi : G \to \operatorname{Sym}(X) \] is called a permutation representation of \( G \) on \( X . \) The homomorphism \( \phi \) maps each \( g \in G \) to a permutation of \( X. \) We say that each \( g \in G \) induces a permutation of \( X. \)

For example, let \( G = (\mathbb{Z}_3, +) \) and \( X = \{ 0, 1, 2, 3, 4, 5 \}. \) Define the map \( \phi : G \to \operatorname{Sym}(X) \) by \begin{align*} \phi(0) &= (), \\ \phi(1) &= (024)(135), \\ \phi(2) &= (042)(153). \end{align*} It is easy to verify that this is a homomorphism. Here is one way to verify it: \begin{align*} \phi(0)\phi(1) &= ()(024)(135) = (024)(135) = \phi(0 + 1), \\ \phi(0)\phi(2) &= ()(042)(153) = (042)(153) = \phi(0 + 2), \\ &\;\,\vdots \\ \phi(2)\phi(1) &= (042)(153)(024)(135) = () = \phi(0) = \phi(2 + 1), \\ \phi(2)\phi(2) &= (042)(153)(042)(153) = (024)(135) = \phi(1) = \phi(2 + 2). \end{align*} We will meet this homomorphism again in the form of group action \( \alpha \) in the next section.

Group Action

Let \( G \) be a group with identity element \( e. \) Let \( X \) be a set. A right action of \( G \) on \( X \) is a map \[ \alpha : X \times G \to X \] such that \begin{align*} \alpha(x, e) &= x, \\ \alpha(\alpha(x, g), h) &= \alpha(x, gh) \end{align*} for all \( x \in X \) and all \( g, h \in G. \) The two conditions above are called the identity and compatibility properties of the group action respectively. Note that in a right action, the product \( gh \) is applied left to right: \( g \) acts first and then \( h \) acts. If we denote \( \alpha(x, g) \) as \( x^g, \) then the notation for the two conditions can be simplified to \( x^e = x \) and \( (x^g)^h = x^{gh} \) for all \( g, h \in G. \)

Why Right Action?

We discuss right group actions here instead of left group actions because we want to use the notation \( \alpha(x, g) = x^g, \) which is quite convenient while studying permutations and graph automorphisms. It is perfectly possible to use left group actions to study permutations as well. However, we lose the benefit of the convenient \( x^g \) notation. In a left group action, the compatibility property is \( \alpha(g, \alpha(h, x)) = \alpha(gh, x) , \) so if we were to use the notation \( \alpha(g, x) = x^g, \) the compatibility property would look like \( (x^h)^g = x^{gh}. \) This reverses the order of exponents which can be confusing. Right group actions avoid this notational inconvenience.

Example 1

Let \( G = \mathbb{Z}_3 \) be the group under addition modulo \( 3 . \) Let \( X = \{ 0, 1, 2, 3, 4, 5 \}. \) Define an action \( \alpha \) of \( G \) on \( X \) by \[ \alpha(x, g) = x^g = (x + 2g) \bmod 6. \] Each \( g \in G \) acts as a permutation of \( X. \) For example, the element \( 0 \in \mathbb{Z}_3 \) acts as the identity permutation. The element \( 1 \in \mathbb{Z}_3 \) acts as the permutation \( (0 2 4)(1 3 5). \) The element \( 2 \in \mathbb{Z}_3 \) acts as the permutation \( (0 4 2)(1 5 3). \) The following table shows how each \( g \in G \) permutes \( X. \) \[ \begin{array}{c|ccc} x_{\downarrow} \backslash g_{\rightarrow} & 0 & 1 & 2 \\ \hline 0 & 0 & 2 & 4 \\ 1 & 1 & 3 & 5 \\ 2 & 2 & 4 & 0 \\ 3 & 3 & 5 & 1 \\ 4 & 4 & 0 & 2 \\ 5 & 5 & 1 & 3 \\ \end{array} \] From the table we see that each \( g \in G \) permutes the elements of \( \{ 0, 2, 4 \} \) among themselves. Similarly, the elements of \( \{ 1, 3, 5 \} \) are permuted among themselves. These sets \( \{0, 2, 4 \} \) and \( \{ 1, 3, 5 \} \) are called the orbits of the action. The concept of orbits is formally introduced in its own section further below.

Example 2

Now let \( G = \mathbb{Z}_6 \) be the group under addition modulo \( 6. \) Let \( X = \{ 0, 1, \dots, 8 \}. \) Define an action \( \beta \) of \( G \) on \( X \) by \[ \beta(x, g) = x^g = (x + 3g) \bmod 9. \] Now the table for the action looks like this: \[ \begin{array}{c|cccccc} x_{\downarrow} \backslash g_{\rightarrow} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 0 & 0 & 3 & 6 & 0 & 3 & 6 \\ 1 & 1 & 4 & 7 & 1 & 4 & 7 \\ 2 & 2 & 5 & 8 & 2 & 5 & 8 \\ 3 & 3 & 6 & 0 & 3 & 6 & 0 \\ 4 & 4 & 7 & 1 & 4 & 7 & 1 \\ 5 & 5 & 8 & 2 & 5 & 8 & 2 \\ 6 & 6 & 0 & 3 & 6 & 0 & 3 \\ 7 & 7 & 1 & 4 & 7 & 1 & 4 \\ 8 & 8 & 2 & 5 & 8 & 2 & 5 \end{array} \] This action splits \( X \) into three orbits \( \{ 0, 3, 6 \}, \) \( \{ 1, 4, 7 \} \) and \( \{ 2, 5, 8 \}. \)

Group Actions Induce Permutations

Earlier, we saw an example of a group action and observed that each element of the group acts as a permutation. That was not merely a coincidence. It is indeed a general property of group actions. Whenever a group \( G \) acts on a set \( X, \) each element \( g \in G \) determines a bijection \( X \to X. \) In other words, every element of \( G \) acts as a permutation of \( X. \) Let us see why this must be the case.

Consider the group action \( \alpha : X \times G \to X. \) Fix \( g \in G \) and let \( x \) vary over \( X \) to obtain the map \[ \alpha_g : X \to X; \; x \mapsto \alpha(x, g). \] We show that \( \alpha_g \) is a bijection. First we prove injectivity. Let \( e \) be the identity element of \( G. \) Let \( x, y \in X. \) Then \begin{align*} \alpha_g(x) = \alpha_g(y) & \implies \alpha(x, g) = \alpha(y, g) \\ & \implies \alpha(\alpha(x, g), g^{-1}) = \alpha(\alpha(y, g), g^{-1}) \\ & \implies \alpha(x, gg^{-1}) = \alpha(y, gg^{-1}) \\ & \implies \alpha(x, e) = \alpha(y, e) \\ & \implies x = y. \end{align*} The \( x^g \) notation allows us to write the above proof more conveniently as follows: \begin{align*} \alpha_g(x) = \alpha_g(y) & \implies \alpha(x, g) = \alpha(y, g) \\ & \implies (x^g)^{g^{-1}} = (y^g)^{g^{-1}} \\ & \implies x^{g g^{-1}} = y^{g g^{-1}} \\ & \implies x^e = y^e \\ & \implies x = y. \end{align*} This completes the proof of injectivity. Now we prove surjectivity. Let \( y \in X. \) Take \( x = \alpha(y, g^{-1}). \) Then \[ \alpha_g(x) = \alpha(x, g) = \alpha(\alpha(y, g^{-1}), g) = \alpha(y, g^{-1} g) = \alpha(y, e) = y. \] Again, if we write \( x = y^{g^{-1}}, \) the above step can be written more succinctly as \[ \alpha_g(x) = x^g = (y^{g^{-1}})^g = y^{(g^{-1} g)} = y^e = y. \] Thus every element \( y \in X \) has a preimage in \( X \) under \( \alpha_g. \) Hence \( \alpha_g \) is surjective. Since we have already shown that \( \alpha_g \) is injective, we now conclude that \( \alpha_g \) is bijective. Therefore \( \alpha_g \) is a permutation of \( X. \) Stated symbolically, \[ \alpha_g \in \operatorname{Sym}(X). \] Note that \[ \alpha_g(x) = \alpha(x, g) = x^g. \] Thus both \( \alpha_g(x) \) and \( x^g \) serve as convenient shorthands for \( \alpha(x, g). \)

Group Actions Determine Permutation Representations

We have seen that each group element \( g \in G \) induces (acts as) a permutation of \( X. \) Precisely speaking, each \( g \in G \) determines a permutation \( \alpha_g \) of \( X. \) Now define a map \[ \phi: G \to \operatorname{Sym}(X); \; g \mapsto \alpha_g. \] We now show that this map is a homomorphism. This means that we want to show that \( \phi(gh) = \phi(g) \phi(h). \) Since \( \phi(g), \phi(h) \in \operatorname{Sym}(X), \) the right-hand side is a product of permutations of \( X. \) We first define the product of two permutations \( \pi, \rho : X \to X \) by \[ \pi \rho : X \to X; \; x \mapsto \rho(\pi(x)). \] In other words, \( \pi \rho = \rho \circ \pi. \) Now \begin{align*} \phi(gh)(x) & = \alpha_{gh}(x) \\ & = \alpha(x, gh) \\ & = \alpha(\alpha(x, g), h) \\ & = \alpha_h(\alpha_g(x)) \\ & = (\alpha_h \circ \alpha_g)(x) \\ & = (\alpha_g \alpha_h)(x) \\ & = (\phi(g) \phi(h))(x). \end{align*} Since the above equality holds for all \( x \in X, \) we conclude that \[ \phi(gh) = \phi(g) \phi(h). \] Hence \( \phi \) is a group homomorphism from \( G \) to \( \operatorname{Sym}(X). \) Therefore \( \phi \) is a permutation representation of \( G \) on \( X. \) It maps each group element \( g \in G \) to a permutation \( \alpha_g \in \operatorname{Sym}(X). \)

Note the multiple levels of abstraction here. The group action \( \alpha : X \times G \to X \) determines a permutation representation \( \phi : G \to \operatorname{Sym}(X). \) Each element \( g \in G \) together with the group action \( \alpha \) determines a permutation \( \alpha_g : X \to X. \)

Also note that \( \phi(g)(x) = \alpha_g(x) = \alpha(g, x) = x^g. \) In fact, \( \phi(g) = \alpha_g. \)

Permutation Representations Determine Group Actions

Consider a permutation representation \( \phi : G \to \operatorname{Sym}(X). \) Define a map \[ \alpha : X \times G \to X; \; (x, g) \mapsto \phi(g)(x). \] First we verify the identity property of group actions. Since \( \phi \) is a homomorphism, it preserves the identity element. Therefore \( \phi(e) \) is the identity permutation. Hence \[ \alpha(x, e) = \phi(e)(x) = x \] Now we verify the compatibility property of the action. For all \( g, h \in G \) and \( x \in X, \) we have \begin{align*} \alpha(\alpha(x, g), h) & = \alpha(\phi(g)(x), h) \\ & = \phi(h)(\phi(g)(x)) \\ & = (\phi(h) \circ \phi(g))(x) \\ & = (\phi(g)\phi(h))(x) \\ & = \phi(gh)(x) \\ & = \alpha(x, gh). \end{align*} This completes the proof of the fact that every permutation representation determines a group action.

Bijection Between Group Actions and Permutation Representations

There is a bijection between the group actions \( \alpha : X \times G \to X \) and permutation representations \( \phi : G \to \operatorname{Sym}(X). \) We now show that these two constructions are inverses of each other.

Given a right action \( \alpha : X \times G \to X, \) define \[ \phi_{\alpha} : G \to \operatorname{Sym}(X) \quad \text{by} \quad \phi_{\alpha}(g)(x) = \alpha(x, g). \] Given a permutation representation \( \phi : G \to \operatorname{Sym}(X), \) define \[ \alpha_{\phi} : X \times G \to X \quad \text{by} \quad \alpha_{\phi}(x, g) = \phi(g)(x). \] We now show that these two constructions undo each other. Take an arbitrary group action \( \alpha : X \times G \to X \) and construct the corresponding permutation representation \( \phi_{\alpha}. \) Then take this permutation representation and construct the group action \( \alpha_{\phi_{\alpha}}. \) But \[ \alpha_{\phi_{\alpha}}(x, g) = \phi_{\alpha}(g)(x) = \alpha(x, g). \] Therefore \( \alpha_{\phi_{\alpha}} = \alpha. \) Similarly, starting with the permutation representation \( \phi, \) we get \[ \phi_{\alpha_{\phi}}(g)(x) = \alpha_{\phi}(x, g) = \phi(g)(x). \] Therefore \( \phi_{\alpha_{\phi}} = \phi. \) Hence there is a bijection between group actions \( \alpha : X \times G \to X \) and permutation representations: \( \phi : G \to \operatorname{Sym}(X) . \) In fact, a group action and the corresponding permutation representation contain the same information, namely how the elements \( g \in G \) acts as permutations of \( X. \) For this reason, many advanced texts do not make any distinction between the group action and its permutation representation. They often use them interchangeably even though technically they have different domains.

Orbits

Let \( G \) act on a set \( X. \) For an element \( x \in X, \) the orbit of \( x \) under the action of \( G \) is the set of all elements of \( X \) that can be reached from \( x \) by the action of elements of \( G. \) Symbolically, the orbit of \( x \) is the set \[ x^G = \{ x^g : g \in G \}. \] In other words, the orbit of \( x \) contains every element of \( X \) that \( x \) can be moved to by the group action. If \( y \in x^G, \) then there exists some \( g \in G \) such that \( y = x^g . \)

The orbits of a group action partition the set \( X. \) That is, every element of \( X \) lies in exactly one orbit and two orbits are either identical or disjoint. Thus the group action decomposes the set \( X \) into disjoint subsets (the orbits), each consisting of elements that can be transformed into one another by the action of \( G. \)

Stabilisers

Let \( G \) be a group acting on a set \( X. \) For an element \( x \in X, \) the stabiliser of \( x \) is the set \[ G_x = \{ g \in G : x^g = x \}. \] The stabiliser \( G_x \) consists of all elements of \( G \) that fix the element \( x. \) The stabiliser \( G_x \) is a subgroup of \( G. \) Indeed, the identity element \( e \in G \) satisfies \( x^e = x, \) so \( e \in G_x. \) If \( g, h \in G_x, \) then \( x^{gh} = (x^g)^h = x. \) If \( g \in G_x, \) then \( x^{g^{-1}} = (x^g)^{g^{-1}} = x. \)

Intuitively, the stabiliser measures how much symmetry of the group action leaves the element \( x \) unchanged. The larger the stabiliser, the more elements of \( G \) fix \( x. \)

Orbit-Stabiliser Theorem

Let \( G \) be a group acting on a set \( X. \) The orbit-stabiliser theorem states that for any \( x \in X, \) \[ \lvert G_x \rvert \cdot \lvert x^G \rvert = \lvert G \rvert. \] Stated differently, the index of the stabiliser \( G_x \) in the group \( G \) is given by \[ [ G : G_x ] = \lvert G_x \backslash G \rvert = \lvert G \rvert / \lvert G_x \rvert = \lvert x^G \rvert. \] There is a bijection between the right cosets of \( G_x \) and the elements of \( x^G. \) Demonstrating this bijection proves the above equation. We will work with right cosets of \( G_x. \) Define \[ \phi : G_x \backslash G \to x^G; \; G_x g \mapsto x^g. \] We want to show that \( \phi \) is a bijection. But first we need to show that \( \phi \) is well defined. A coset \( G_x g \in G_x \backslash G \) can also be written as \[ G_x g = G_x h \] for some \( h \in G. \) If \( x^g \ne x^h, \) then \( \phi \) would not be well defined, since \( \phi \) must assign each coset in \( G_x \backslash G \) to exactly one element in the orbit \( x^G \) in order to be a function. This can be shown using the following equivalences: \begin{align*} G_x g = G_x h & \iff hg^{-1} \in G_x \\ & \iff x = x^{h g^{-1}} \\ & \iff x^g = x^h \\ \end{align*} This proves two things at once. The fact that \[ G_x g = G_x h \implies x^g = x^h \] proves that when the same coset is written using two different representatives, the image does not change. Therefore \( \phi \) is well defined. Further \[ x^g = x^h \implies G_x g = G_x h \] proves that \( \phi \) is injective. To show that \( \phi \) is surjective, let \( y \in x^G. \) Then \( y = x^g \) for some \( g \in G. \) Since \( \phi(G_x g) = x^g, \) we get \[ \phi(G_x g) = y. \] Thus every element of \( x^G \) is the image of some right coset \( G_x g \) under \( \phi. \) This completes the proof of a bijection between the right cosets of \( G_x \) and the elements of \( x^G. \) Therefore \( \lvert G_x \backslash G \rvert = \lvert x^G \rvert \) and hence \( \lvert G \rvert / \lvert G_x \rvert = \lvert x^G \rvert , \) which establishes the orbit-stabiliser theorem.

Faithful Actions

Let \( G \) act on a set \( X. \) The action is called faithful if distinct elements of \( G \) induce distinct permutations of \( X. \) In other words, the only element of \( G \) that acts as the identity permutation of \( X \) is the identity element \( e \in G. \) Symbolically, the action is faithful if \[ g \ne e \implies \exists x \in X, \; x^g \ne x. \] Equivalently, \[ \forall x \in X, \; x^g = x \implies g = e. \] The action is faithful if the only element of \( G \) that fixes every element of \( X \) is the identity, i.e. \[ \bigcap_{x \in X} G_x = \{ e \}. \] Recall that every group action determines a permutation representation \( \phi : G \to \operatorname{Sym}(X). \) From this point of view, the action is faithful precisely when the permutation representation is faithful, that is, when the homomorphism \( \phi \) is injective (or equivalently when \( \ker(\phi) = \{ e \} \)). In other words, the action is faithful if and only if the associated homomorphism \( \phi \) is a monomorphism.

Semiregular Actions

A group action of \( G \) on \( X \) is called semiregular if no non-identity element of \( G \) fixes any element of \( X. \) In other words, whenever \( g \ne e, \) the permutation of \( X \) induced by \( g \) moves every element of \( X. \) Symbolically, \[ g \ne e \implies \forall x \in X, \; x^g \ne x. \] Equivalently, \[ \exists x \in X, \; x^g = x \implies g = e. \] The action is semiregular if \[ \forall x \in X, \; G_x = \{ e \}. \] This is a stronger property than faithfulness. Faithfulness only guarantees that when \( g \ne e, \) the element \( g \) moves at least one element of \( X. \) But semiregularity guarantees that when \( g \ne e, \) the element \( g \) moves every element of \( X . \) Therefore every semiregular action is faithful, but not every faithful action is semiregular.

Transitive Actions

Let \( G \) act on a set \( X. \) The action is called transitive if there is only one orbit. In other words, the action is transitive if every element of \( X \) can be reached from any other element by the action of some element of \( G. \) Symbolically, the action is transitive if \[ \forall x, y \in X \; \exists g \in G, \; x^g = y. \] Equivalently, the action is transitive if \[ x^G = X \] for some (and hence every) \( x \in X. \)

Conjugacy

Let \( G \) be a group. Let \( x, g \in G. \) The element \[ g^{-1} x g \] is called a conjugate of \( x \) by \( g. \) Any element \( y \in G \) that can be written as \( g^{-1} x g \) for some \( g \in G \) is said to be a conjugate of \( x. \) The conjugacy class of \( x \) in \( G \) is the set \[ x^G = \{ g^{-1} x g : g \in G \}. \] In other words, the conjugacy class of \( x \) is the set of all elements of \( G \) that are conjugate to \( x. \) At first, reusing the orbit notation \( x^G \) for the conjugacy class may seem like an abuse of notation. However, we will see in the next section that the conjugacy class is precisely the orbit of \( x \) under the action of \( G \) on itself by conjugation. Thus \( x^G \) is in fact a natural and accurate notation for the conjugacy class.

Conjugation as Group Action

Conjugation can be seen as an action of a group on itself. Define the map \[ \alpha : G \times G \to G; \; (x, g) \mapsto g^{-1} x g. \] Note that \[ \alpha(x, e) = e^{-1} x e = x \] and \[ \alpha(\alpha(x, g), h) = h^{-1} (g^{-1} x g) h = (gh)^{-1} x (gh) = \alpha(x, gh). \] Therefore \( \alpha \) satisfies the two defining properties of a right group action. The conjugacy class \( x^G \) is precisely the orbit of \( x \) under the conjugation action. Therefore the orbits of the conjugation action of \( G \) on itself are the conjugacy classes of \( G. \)

Right Conjugation vs Left Conjugation

We observed above that the conjugation action is a right action of a group on itself. Let \( x, g \in G \) and let \[ y = g^{-1} x g. \] Now let \( h = g^{-1}. \) Then we can write the above equation as \[ y = h x h^{-1}. \] According to the previous section, \( y \) is the conjugate of \( x \) by \( g. \) However, many texts call \( y \) the conjugate of \( x \) by \( h. \) Both are valid perspectives. In both perspectives, \( x \) and \( y \) are conjugates of each other. Precisely,

• In the first perspective, we have \( y = g^{-1} x g \) and we say that \( y \) is a conjugate of \( x \) by \( g. \) A corollary is that \( x \) is a conjugate of \( y \) by \( g^{-1}. \)
• In the second perspective, we have \( y = h x h^{-1} \) and we say that \( y \) is a conjugate of \( x \) by \( h. \) A corollary is that \( x \) is a conjugate of \( y \) by \( h^{-1}. \)

Although in both perspectives, \( x \) and \( y \) are conjugates of each other, the group element by which one is conjugated to the other is different. This leads to different group actions as well.

When we say that \( y = g^{-1} x g \) is a conjugate of \( x \) by \( g, \) the group action \[ \alpha : G \times G \to G; \; (x, g) \mapsto g^{-1} x g. \] is a right group action as demonstrated in the previous section. But when we say that \( y = h x h^{-1} \) is a conjugate of \( x \) by \( h, \) the conjugation action is no longer a right group action because the compatibility property is violated: \[ \alpha(\alpha(x, g), h) = h ( g x g^{-1} ) h^{-1} = (hg) x (hg)^{-1} = \alpha(x, hg). \] We get \( \alpha(x, hg) \) instead of the required \( \alpha(x, gh) . \) So with the second perspective, the group action is no longer a right action. Instead it is a left action since \[ \alpha(g, \alpha(h, x)) = g (h x h^{-1}) g^{-1} = (gh) x (gh)^{-1} = \alpha(gh, x). \] In this post we will work only with the first perspective because we will use right actions throughout.

Conjugate Subgroups

Let \( G \) be a group. Let \( H \le G. \) Define \[ g^{-1} H g = \{ g^{-1} h g : h \in H \}. \] We say that \( g^{-1} H g \) is a conjugate of \( H \) by \( g. \)

Conjugacy of Stabilisers

Let \( G \) be a group acting on a set \( X. \) Let \( x \in X \) and \( g \in G. \) Then \[ g^{-1} G_x g = G_{x^g}. \] That is, \( G_{x^g} \) is a conjugate of \( G_x \) by \( g. \) This result can be summarised as follows: stabilisers of elements in the same orbit are conjugate. Or more explicitly: the stabiliser of \( x^g \) is a conjugate of the stabiliser of \( x \) by \( g. \) The proof is straightforward. Let \( h \in G. \) Then \begin{align*} h \in g^{-1} G_x g & \iff g^{-1} (g h g^{-1}) g \in g^{-1} G_x g \\ & \iff g h g^{-1} \in G_x \\ & \iff x^{g h g^{-1}} = x \\ & \iff (x^g)^h = x^g \\ & \iff h \in G_{x^g}. \end{align*} Therefore \( g^{-1} G_x g = G_{x^g}. \)

Algebraic Graph Theory

Stabiliser Index

In a vertex-transitive graph \( \Gamma, \) for any \( x \in V(\Gamma) \) and all \( y \in V(\Gamma), \) there exists \( g \in G \) such that \( x^g = y. \) Therefore \( x^G = V(\Gamma). \) Thus by the orbit-stabiliser theorem, \[ [ G : G_x ] = \lvert G_x \backslash G \rvert = \lvert x^G \rvert = \lvert V(\Gamma) \rvert. \]

Strongly Connected Directed Graph

A path in a directed graph \( \Gamma \) is a sequence of vertices \( v_0, \dots, v_r \) of distinct vertices such that \( (v_{i - 1}, v_i) \) is an arc of \( \Gamma \) for \( i = 1, \dots, r . \)

A directed graph is strongly connected if for every ordered pair of vertices \( (u, v) \) there is a path from \( u \) to \( v . \)

Shunting

Let \( \alpha = ( \alpha_0, \dots, \alpha_s ) \) and \( \beta = ( \beta_0, \dots, \beta_s ) \) be two \( s \)-arcs in a graph \( \Gamma. \) We say that \( \beta \) is a successor of \( \alpha \) if \( \beta_i = \alpha_{i + 1} \) for \( 0 \le i \le s - 1. \) We also say that \( \alpha \) can be shunted onto \( \beta. \)

In section 4.2 of Godsil and Royle, there is a rather technical setup which first defines \( X^{(s)} \) as the directed graph with the \( s \)-arcs of a graph \( X \) as its vertices such that \( (\alpha, \beta) \) is an arc of \( X^{(s)} \) if and only if \( \alpha \) can be shunted onto \( \beta \) in \( X. \) Then it goes on to show that if \( X \) is a connected graph with a minimum degree two and \( X \) is not a cycle, then \( X^{(s)} \) is strongly connected for all \( s \ge 0. \)

That is a very technical way of saying that in a connected graph \( X \) that is not a cycle and has a minimum degree two, any \( s \)-arc \( \alpha \) can be sent to any \( s \)-arc \( \beta \) by repeated shunting. The proof is quite technical too and pretty long, so I'll omit it here.

Automorphisms Preserve Successor Relation

We will obtain a nifty result here that will prove to be very useful in the next section. Let \( S(\gamma) \) denote the set of all successors of the \( s \)-arc \( \gamma \) of a graph. Let \( g \) be an automorphism of the graph. Then \[ \delta \in S(\gamma) \iff \delta^g \in S(\gamma^g). \] This follows directly from the fact that automorphisms preserve adjacency, so they must preserve successor relation as well. A corollary of this is that for an automorphism \( h, \) we have \[ \delta^{h^{-1}} \in S(\gamma) \iff \delta \in S(\gamma^h). \] This is the form that will be useful soon.

Test of \( s \)-arc Transitivity

The results in the previous two sections lead to a remarkably simple proof of the fact that the Petersen graph is \( 3 \)-arc transitive. Let us see how.

Let \( P \) be the Petersen graph whose vertices are the \( 2 \)-subsets of \( \{ 1, 2, 3, 4, 5 \} \) with adjacency given by disjointness of the \( 2 \)-subsets. Then \( \operatorname{Aut}(P) \cong S_5 \) since any permutation of \( \{ 1, 2, 3, 4, 5 \} \) induces a permutation of the vertices that preserves disjointness and hence adjacency. We will use the shorthand \( ab \) to represent each vertex \( \{ a, b \} \) of \( P. \) Consider the \( 3 \)-arc \[ \alpha = (12, 34, 15, 23). \] It has exactly two successors, namely \[ \beta_1 = (34, 15, 23, 14), \quad \beta_2 = (34, 15, 23, 45). \] Let \( g_1 = (13)(245) \) and \( g_2 = (13524). \) Then \begin{align*} \alpha^{g_1} & = (12, 34, 15, 23)^{(13)(245)} = (34, 15, 23, 14) = \beta_1, \\ \alpha^{g_2} & = (12, 34, 15, 23)^{(13524)} = (34, 51, 23, 45) = \beta_2. \end{align*} Let \( H = \langle g_1, g_2 \rangle \le \operatorname{Aut}(P). \) Consider an \( s \)-arc \( \alpha^h \) for some \( h \in H. \) Let \( \delta \in S(\alpha^h). \) Then by the result in the previous section, we get \[ \delta^{h^{-1}} \in S(\alpha) = \{ \beta_1, \beta_2 \} = \{ \alpha^{g_1}, \alpha^{g_2} \}. \] Therefore \[ \delta \in \{ \alpha^{g_1 h}, \alpha^{g_2 h} \}. \] Thus \[ \delta \in \alpha^{H}. \] We started with an \( s \)-arc \( \alpha^h \in \alpha^H \) and showed that its successors \( \delta \) also lie in \( \alpha^H. \) Thus the orbit \( \alpha^H \) is closed under taking successors.

Now by the shunting result discussed previously, \( \alpha \) can be sent to any \( 3 \)-arc of \( P \) by repeated shunting. Therefore all \( 3 \)-arcs of \( P \) belong to \( \alpha^H. \) Therefore the automorphisms in \( H \) can send any \( 3 \)-arc of \( P \) to any other thus making \( P \) \( 3 \)-arc transitive.

Moore Graphs

Graphs with diameter \( d \) and girth \( 2d + 1 \) are known as Moore graphs.

There are an infinite number of Moore graphs with diameter \( 1 \) since the complete graphs \( K_n, \) where \( n \ge 3, \) have diameter \( 1 \) and girth \( 3. \)

There are three known Moore graphs of diameter \( 2. \) They are \( C_5, \) \( J(5, 2, 0) \) also known as the Petersen graph and the Hoffman-Singleton graph. They are respectively \( 2 \)-regular, \( 3 \)-regular and \( 7 \)-regular. There is a famous result that proves that a Moore graph must be \( 2 \)-regular, \( 3 \)-regular, \( 7 \)-regular or \( 57 \)-regular. It is unknown currently whether a \( 57 \)-regular Moore graph of diameter \( 2 \) exists.

There are infinitely many Moore graphs of diameter \( d \ge 3 \) because the odd cycles \( C_{2d + 1} \) are \( 2 \)-regular graphs with diameter \( d \) and girth \( 2d + 1 \) for all \( d \ge 1. \) However, there are no \( k \)-regular Moore graphs for diameter \( d \ge 3 \) when \( k \ge 3. \)

Generalised Polygons

Bipartite graphs with diameter \( d \) and girth \( 2d \) are known as generalised polygons. This is easy to understand. If we take a classical \( d \)-gon and create the incidence graph of its vertices and edges, then the incidence graph is the cycle \( C_{2d} \) which has diameter \( d \) and girth \( 2d. \)

The converse is not always true. For example, the Heawood graph which has diameter \( d = 3 \) and girth \( 2d = 6. \) It is the incidence graph of Fano plane, which is a projective plane rather than a classical \( d \)-gon.

Although a generalised polygon is not always the incidence graph of a classical polygon, the idea behind the definition comes from a simple observation. If we take a classical \( d \)-gon and form the incidence graph of its vertices and edges, we obtain the cycle \( C_{2d}. \) This graph is bipartite and has diameter \( d \) and girth \( 2d. \) The definition of a generalised polygon abstracts these properties. Any bipartite graph with diameter \( d \) and girth \( 2d \) is called a generalised polygon, even when it is not the incidence graph of a classical \( d \)-gon. In this way the definition allows much richer graphs than simple cycles.

Computing

Select Between Lines, Inclusive

Select text between two lines, including both lines:

sed '/pattern1/,/pattern2/!d'

sed -n '/pattern1/,/pattern2/p'

Here are some examples:

$ printf 'A\nB\nC\nD\nE\nF\nG\nH\n' | sed '/C/,/F/!d'
C
D
E
F
$ printf 'A\nB\nC\nD\nE\nF\nG\nH\n' | sed -n '/C/,/F/p'
C
D
E
F

Select Between Lines, Exclusive

Select text between two lines, excluding both lines:

sed '/pattern1/,/pattern2/!d; //d'

Here is an example usage:

$ printf 'A\nB\nC\nD\nE\nF\nG\nH\n' | sed '/C/,/F/!d; //d'
D
E

The negated command !d deletes everything not matched by the 2-address range /C/,/F/, i.e. it deletes everything before the line matching /C/ as well as everything after the line matching /F/. So we are left with only the lines from C to F, inclusive. Finally, // (the empty regular expression) reuses the most recently used regular expression. So when /C/,/F/ matches C, the command //d also matches C and deletes it. Similarly, F is deleted too. That's how we are left with the lines between C and F, exclusive.

Here are some excerpts from POSIX.1-2024 that help understand the !d and //d commands better:

A function can be preceded by a '!' character, in which case the function shall be applied if the addresses do not select the pattern space. Zero or more <blank> characters shall be accepted before the '!' character. It is unspecified whether <blank> characters can follow the '!' character, and conforming applications shall not follow the '!' character with <blank> characters.

If an RE is empty (that is, no pattern is specified) sed shall behave as if the last RE used in the last command applied (either as an address or as part of a substitute command) was specified.

Signing and Verification with SSH Key

Here are some minimal commands to demonstrate how we can sign some text using SSH key and then later verify it.

ssh-keygen -t ed25519 -f key
echo hello > hello.txt
ssh-keygen -Y sign -f key.pub -n file hello.txt
echo "jdoe $(cat key.pub)" > allowed.txt
ssh-keygen -Y verify -f allowed.txt -I jdoe -n file -s hello.txt.sig < hello.txt

Here are some examples that demonstrate what the outputs and signature file look like:

$ ssh-keygen -Y sign -f key.pub -n file hello.txt
Signing file hello.txt
Write signature to hello.txt.sig

$ cat hello.txt.sig
-----BEGIN SSH SIGNATURE-----
U1NIU0lHAAAAAQAAADMAAAALc3NoLWVkMjU1MTkAAAAgAwP6RnmFVrZO0m/nRIHyvr2S19
itsKegj9p/BZKqP1sAAAAEZmlsZQAAAAAAAAAGc2hhNTEyAAAAUwAAAAtzc2gtZWQyNTUx
OQAAAEB8ylqjCLgInF8DvROnLSm1UUWd0VuLPesI+1NhMrV9BjH5lf0w20kHunJW3qRIjw
Jfs9+q/e47KdlR8wBQaHYD
-----END SSH SIGNATURE-----

$ ssh-keygen -Y verify -f allowed.txt -I jdoe -n file -s hello.txt.sig < hello.txt
Good "file" signature for jdoe with ED25519 key SHA256:9ZJuUJNMy1UXo3AlQy8L7baD3LOfEbgQ30ELIt+8wWc

Block IP Address with nftables

Here is a sequence of commands to create an nftables rule from scratch to block an IP address:

$ sudo nft list ruleset
$ sudo nft add table inet filter
$ sudo nft list ruleset
table inet filter {
}
$ sudo nft add chain inet filter input { type filter hook input priority 0 \; }
$ sudo nft list ruleset
table inet filter {
        chain input {
                type filter hook input priority filter; policy accept;
        }
}
$ sudo nft add rule inet filter input ip saddr 172.236.0.216 drop
$ sudo nft list ruleset
table inet filter {
        chain input {
                type filter hook input priority filter; policy accept;
                ip saddr 172.236.0.216 drop
        }
}

Here is how to undo the above setup step by step:

$ sudo nft -a list ruleset
table inet filter { # handle 1
        chain input { # handle 1
                type filter hook input priority filter; policy accept;
                ip saddr 172.236.0.216 drop # handle 2
        }
}
$ sudo nft delete rule inet filter input handle 2
$ sudo nft list ruleset
table inet filter {
        chain input {
                type filter hook input priority filter; policy accept;
        }
}
$ sudo nft delete chain inet filter input
$ sudo nft list ruleset
table inet filter {
}
$ sudo nft delete table inet filter
$ sudo nft list ruleset
$

Finally, the following command deletes all rules, chains and tables. It wipes the entire ruleset, so use it with care.

$ sudo nft flush ruleset
$ sudo nft list ruleset
$

All outputs above were obtained using nftables v1.1.3 on Debian 13.2 (Trixie).

Debian Logrotate Setup

Observed on Debian 11.5 (Bullseye) that logrotate is set up on it via systemd. Here are some outputs that show what the setup is like:

$ sudo systemctl status logrotate.service
● logrotate.service - Rotate log files
     Loaded: loaded (/lib/systemd/system/logrotate.service; static)
     Active: inactive (dead) since Mon 2026-03-30 00:00:17 UTC; 19h ago
TriggeredBy: ● logrotate.timer
       Docs: man:logrotate(8)
             man:logrotate.conf(5)
    Process: 2148235 ExecStart=/usr/sbin/logrotate /etc/logrotate.conf (code=exited, status=0/SUCCESS)
   Main PID: 2148235 (code=exited, status=0/SUCCESS)
        CPU: 574ms

Mar 30 00:00:16 spweb systemd[1]: Starting Rotate log files...
Mar 30 00:00:17 spweb systemd[1]: logrotate.service: Succeeded.
Mar 30 00:00:17 spweb systemd[1]: Finished Rotate log files.
$ sudo systemctl status logrotate.timer
● logrotate.timer - Daily rotation of log files
     Loaded: loaded (/lib/systemd/system/logrotate.timer; enabled; vendor preset: enabled)
     Active: active (waiting) since Mon 2026-01-19 19:19:34 UTC; 2 months 9 days ago
    Trigger: Tue 2026-03-31 00:00:00 UTC; 4h 7min left
   Triggers: ● logrotate.service
       Docs: man:logrotate(8)
             man:logrotate.conf(5)

Warning: journal has been rotated since unit was started, output may be incomplete.
$ sudo systemctl list-timers logrotate
NEXT                        LEFT         LAST                        PASSED  UNIT            ACTIVATES
Tue 2026-03-31 00:00:00 UTC 4h 7min left Mon 2026-03-30 00:00:16 UTC 19h ago logrotate.timer logrotate.service

1 timers listed.
Pass --all to see loaded but inactive timers, too.
$ head /lib/systemd/system/logrotate.service
[Unit]
Description=Rotate log files
Documentation=man:logrotate(8) man:logrotate.conf(5)
RequiresMountsFor=/var/log
ConditionACPower=true

[Service]
Type=oneshot
ExecStart=/usr/sbin/logrotate /etc/logrotate.conf

$ cat /lib/systemd/system/logrotate.timer
[Unit]
Description=Daily rotation of log files
Documentation=man:logrotate(8) man:logrotate.conf(5)

[Timer]
OnCalendar=daily
AccuracySec=1h
Persistent=true

[Install]
WantedBy=timers.target
$ grep -vE '^#|^$' /etc/logrotate.conf
weekly
rotate 4
create
include /etc/logrotate.d
$ ls -l /etc/logrotate.d/
total 40
-rw-r--r-- 1 root root 120 Aug 21  2022 alternatives
-rw-r--r-- 1 root root 173 Jun 10  2021 apt
-rw-r--r-- 1 root root 130 Oct 14  2019 btmp
-rw-r--r-- 1 root root  82 May 26  2018 certbot
-rw-r--r-- 1 root root 112 Aug 21  2022 dpkg
-rw-r--r-- 1 root root 128 May  4  2021 exim4-base
-rw-r--r-- 1 root root 108 May  4  2021 exim4-paniclog
-rw-r--r-- 1 root root 329 May 29  2021 nginx
-rw-r--r-- 1 root root 374 May 20  2022 rsyslog
lrwxrwxrwx 1 root root  28 Mar 17 01:52 susam -> /opt/susam.net/etc/logrotate
-rw-r--r-- 1 root root 145 Oct 14  2019 wtmp

To force log rotation right now, execute:

sudo systemctl start logrotate.service

Read on website | #notes | #mathematics | #linux | #technology

Most of last month's notes grew out of my reading of Algebraic Graph Theory by Godsil and Royle. I am still exploring and learning this subject. This month, however, I dove into another book with the same title but this book is written by Norman Biggs. As a result, many of the notes that follow are drawn from Biggs's treatment of the topic.

Since I already had a good understanding of the subject from the earlier book, I decided to skip the first fourteen chapters of the new book. I began with Chapter 15, which discusses automorphisms of graphs and then moved on to the following chapters on graph symmetries. My main reason for picking up Biggs's book was to understand Tutte's well known result that any \( s \)-arc-transitive finite cubic graph must satisfy \( s \le 5. \) While I did not reach that chapter this month, I made substantial progress with the book. I hope to work through the proof of Tutte's theorem next month.

Contents

1. Degree of Vertices in an Orbit
2. Regular Non-Vertex-Transitive Graphs
3. Vertex-Transitive But Not Edge-Transitive
4. Edge-Transitivex But Not Vertex-Transitive
5. Bipartiteness as a Necessary Condition
6. Graph with an Automorphism Group
7. Permutation Groups Need Not Be Automorphism Groups
8. Symmetric Graphs

Degree of Vertices in an Orbit

If two vertices of a graph belong to the same orbit, then they have the same degree. In other words, for a graph \( X, \) if \( x, y \in V(X) \) and there is an automorphism \( \alpha \) such that \( \alpha(x) = y, \) then \( \deg(x) = \deg(y). \)

The proof is quite straightforward. Let \begin{align*} N(x) &= \{ v_1, \dots, v_r \}, \\ N(y) &= \{ w_1, \dots, w_s \} \end{align*} represent the neighbours of \( x \) and \( y \) respectively. Therefore we have \[ x \sim v_1, \; \dots, \; x \sim v_r. \] Since an automorphism preserves adjacency, we get \[ \alpha(x) \sim \alpha(v_1), \; \dots, \; \alpha(x) \sim \alpha(v_r). \] Substituting \( \alpha(x) = y, \) we get \[ y \sim \alpha(v_1), \; \dots, \; y \sim \alpha(v_r). \] Thus \[ \alpha(N(x)) = \{ \alpha(v_1), \; \dots, \; \alpha(v_r) \} \subseteq N(y). \] A similar argument works in reverse as well. By the definition of automorphism, if \( \alpha \) is an automorphism, so is \( \alpha^{-1}. \) From the definition of \( N(y) \) above, we have \[ y \sim w_1, \; \dots, \; y \sim w_s. \] Therefore \[ \alpha^{-1}(y) \sim \alpha^{-1}(w_1), \; \dots, \; \alpha^{-1}(y) \sim \alpha^{-1}(w_s). \] This is equivalent to \[ x \sim \alpha^{-1}(w_1), \; \dots, \; x \sim \alpha^{-1}(w_s). \] Thus \[ \alpha^{-1}(N(y)) = \{ \alpha^{-1}(w_1), \; \dots, \; \alpha^{-1}(w_s) \} \subseteq N(x) \] This can be rewritten as \[ \{ \alpha^{-1}(w_1), \; \dots, \; \alpha^{-1}(w_s) \} \subseteq \{ v_1, \dots, v_r \}. \] Therefore \[ N(y) = \{ w_1, \dots, w_s \} \subseteq \{ \alpha(v_1), \dots, \alpha(v_r) \} = \alpha(N(x)). \] We have shown that \( \alpha(N(x)) \subseteq N(y) \) and \( N(y) \subseteq \alpha(N(x)). \) Thus \[ \alpha(N(x)) = N(y). \] Thus \[ \lvert N(y) \rvert = \lvert \alpha(N(x)) \rvert = r. \] Therefore both \( x \) and \( y \) have \( r \) neighbours each. Hence \( \deg(x) = \deg(y). \)

Regular Non-Vertex-Transitive Graphs

The Frucht graph and the Folkman graph are examples of graphs that are \( k \)-regular but not vertex-transitive. In fact, the Folkman graph is a semi-symmetric graph, i.e. it is regular and edge-transitive but not vertex-transitive.

Vertex-Transitive But Not Edge-Transitive

The circular ladder graph \( CL_3, \) i.e. the triangular prism graph, is vertex-transitive but not edge-transitive.

Every vertex has the same local structure. Every vertex has degree \( 3 \) and it lies on exactly one of the two triangles and it has exactly one 'vertical' edge connecting it to the corresponding edge on the other triangle. Any vertex can be sent to any other by an automorphism.

Since triangle edges are in a triangle and vertical edges are in no triangle, no automorphism can send a triangle edge to a vertical edge or vice versa. Therefore the graph is not edge-transitive.

Edge-Transitivex But Not Vertex-Transitive

The complete bipartite graphs \( K_{m,n} \) with \( m \ne n \) are edge-transitive but not vertex-transitive.

Every edge connects one vertex from the \( m \)-part to one vertex from the \( n \)-part. Any permutation of vertices inside the \( m \)-part preserves adjacency. Similarly, any permutation of vertices inside the \( n \)-part preserves adjacency.

Take two arbitrary edges \[ uv, \; u'v' \in E(K_{m,n}) \] where \( u, u' \) are vertices that lie in the \( m \)-part and \( v, v' \) are vertices that lie in the \( n \)-part. Permute vertices within the \( m \)-part to send \( u \) to \( u'. \) Similarly, permute vertices within the \( n \)-part to send \( v \) to \( v'. \) This gives an automorphism that sends the edge \( uv \) to \( u'v'. \) In this manner we can find an automorphism that sends any edge to any other. Therefore, \( K_{m,n} \) is edge-transitive.

However, \( K_{m,n} \) is not vertex-transitive since no automorphism can send a vertex in the \( m \)-part to a vertex in the \( n \)-part since the vertices in the \( m \)-part have degree \( n \) and the vertices in the \( n \)-part have degree \( m. \)

Bipartiteness as a Necessary Condition

If a connected graph is edge-transitive but not vertex-transitive, then it must be bipartite.

Graph with an Automorphism Group

In 1938, Frucht proved that for every finite abstract group \( G, \) there exists a graph whose automorphism group is isomorphic to \( G . \)

Remarkably, this result remains valid even when we restrict our attention to cubic graphs. That is, for every finite abstract group \( G, \) there exists a cubic graph whose automorphism group is isomorphic to \( G. \) Moreover, the result has been extended to graphs satisfying various additional graph-theoretical properties, such as \( k \)-connectivity, \( k \)-regularity and prescribed chromatic number.

Permutation Groups Need Not Be Automorphism Groups

Consider the following specialised version of the problem discussed in the previous section: Given a permutation group on a set \( X, \) must there exist a graph with vertex set \( X \) whose automorphism group is precisely that permutation group?

The answer is no. Consider the cyclic group \( C_3 \) acting on \( X = \{ a, b, c \}. \) There is no graph \( \Gamma \) with \( V(\Gamma) = X \) and \( \operatorname{Aut}(\Gamma) \cong C_3. \) If we take \( \Gamma = K_3, \) then \( C_3 \subset S_3 = \operatorname{Aut}(K_3) \) but \( C_3 \ne \operatorname{Aut}(K_3) . \)

Symmetric Graphs

It is interesting that while we study graph symmetry through concepts such as graph automorphisms, vertex-transitivity, edge-transitivity, etc. the name symmetric graph is reserved for graphs that are \( 1 \)-arc-transitive. A vertex-transitive graph or an edge-transitive graph need not be \(1\)-arc-transitive and therefore need not be symmetric.

However, every \( s \)-arc-transitive graph is \(1 \)-arc-transitive for \( s \ge 1. \) Consequently, every \( s \)-arc-transitive graph is symmetric. Moreover, every distance-transitive graph is also \( 1 \)-arc-transitive and hence symmetric.

Formally, we say that a graph \( \Gamma \) is \( 1 \)-arc-transitive (or equivalently, symmetric) if for all \( 1 \)-arcs \( uv \) and \( u'v' \) of \( \Gamma, \) there is an automorphism \( \alpha \in \operatorname{Aut}(\Gamma) \) such that \( \alpha(uv) = u'v'. \)

Stated in more basic terms, we can say that \( \Gamma \) is symmetric if for all \( u, v, u', v' \in V(\Gamma) \) satisfying \( u \sim v \) and \( u' \sim v', \) there exists \( \alpha \in \operatorname{Aut}(\Gamma) \) such that \( \alpha(u) = u' \) and \( \alpha(v) = v'. \)

Switching gears now, we say that \( \Gamma \) is distance-transitive if for all \( u, v, u', v' \in V(\Gamma) \) satisfying \( d(u, v) = d(u', v'), \) there exists \( \alpha \in \operatorname{Aut}(\Gamma) \) such that \( \alpha(u) = u' \) and \( \alpha(v) = v'. \) Since all \( 1 \)-arcs \( uv \) and \( u'v' \) satisfy \( d(u, v) = d(u', v') = 1, \) distance-transitivity implies that there is an automorphism that sends \( uv \) to \( u'v'. \) Therefore a distance-transitive graph is also \( 1 \)-arc-transitive.

To summarise, a graph must possess a certain degree of symmetry in order to be called symmetric. It turns out that merely having a non-trivial automorphism group is not sufficient. Even being vertex-transitive or edge-transitive is not enough for a graph to be called symmetric. The graph needs to be at least \( 1 \)-arc-transitive to be called symmetric.

Another interesting aspect of this terminology is that the property of being asymmetric is not the exact opposite of being symmetric. For example, a vertex-transitive graph need not be symmetric. However, that does not make it asymmetric. A graph is called asymmetric if it has no non-trivial automorphisms, i.e. its automorphism group contains only the identity permutation. Thus, if a graph has at least two vertices and is vertex-transitive, it must admit a non-trivial automorphism that maps one vertex to another. So while such a vertex-transitive may not be symmetric, it isn't asymmetric either.

Read on website | #notes | #mathematics

Contents

1. Cayley Graphs
2. Vertex-Transitive Graphs
3. Arc-Transitive Graphs
4. Bipartite Graphs and Cycle Parity
5. Tutte's Theorem
6. Tutte's 8-Cage
7. Linear Congruential Generator
8. Numbering Lines

Cayley Graphs

Let \( G \) be a group and let \( C \subseteq G \) such that \( C \) is closed under taking inverses and does not contain the identity, i.e. \[ \forall x \in C, \; x^{-1} \in C, \qquad e \notin C. \] Then the Cayley graph \( X(G, C) \) is the graph with the vertex set \( V(X(G, C)) \) and edge set \( E(X(G, C)) \) defined by \begin{align*} V(X(G, C)) &= G, \\ E(X(G, C)) &= \{ gh : hg^{-1} \in C \}. \end{align*} The set \( C \) is known as the connection set.

Vertex-Transitive Graphs

A graph \( X \) is vertex-transitive if its automorphism group acts transitively on its set of vertices \( V(X). \) Intuitively, this means that no vertex has a special role. We can 'move' the graph around so that any chosen vertex becomes any other vertex. In other words, all vertices are indistinguishable. The graph looks the same from each vertex.

The \( k \)-cube \( Q_k \) is vertex-transitive. So are the Cayley graphs \( X(G, C). \) However the path graph \( P_3 \) is not vertex-transitive since no automorphism can send the middle vertex of valency \( 2 \) to an end vertex of valency \( 1. \)

Arc-Transitive Graphs

The cube \( Q_3 \) is \( 2 \)-arc-transitive but not \( 3 \)-arc-transitive. In \( Q_3, \) a \( 3 \)-arc belonging to a \( 4 \)-cycle cannot be sent to a \( 3 \)-arc that does not belong to a \( 4 \)-cycle. This is easy to explain. The end vertices of a \( 3 \)-arc belonging to a \( 4 \)-cycle are adjacent but the end vertices of a \( 3 \)-arc not belonging to a \( 4 \)-cycle are not adjacent. Therefore, no automorphism can map the end vertices of the first \( 3 \)-arc to those of the second \( 3 \)-arc.

For intuition, imagine that a traveller stands on a vertex and chooses an edge to move along. They do this \( s \) times thereby walking along an arc of length \( s, \) also known as an \( s \)-arc. By the definition of \( s \)-arcs, the traveller is not allowed to backtrack from one vertex to the previous one immediately. In an \( s \)-arc-transitive graph, these arcs look the same no matter which vertex they start from or which edges they choose. In the cube, this is indeed true for \( s = 2. \) All arcs of length \( 2 \) are indistinguishable. No matter which arc of length \( 2 \) the traveller has walked along, the graph would look the same from their perspective at each vertex along the arc. However, this no longer holds good for arcs of length \( 3 \) since there are two distinct kinds of arcs of length \( 3. \) The first kind ends at a distance of \( 1 \) from the starting vertex of the arc (when the arc belongs to a \( 4 \)-cycle). The second kind ends at a distance \( 3 \) from the starting vertex of the arc (when the arc does not belong to a \( 4 \)-cycle). Therefore the cube is not \( 3 \)-arc-transitive.

Bipartite Graphs and Cycle Parity

A graph is bipartite if and only if it contains no cycles of odd length. Equivalently, every cycle in a bipartite graph has even length. Conversely, if every cycle in a graph has even length, then the graph is bipartite.

Tutte's Theorem

For any \( s \)-arc-transitive cubic graph, \( s \le 5. \) This was demonstrated by W. T. Tutte in 1947. A proof can be found in Chapter 18 of Algebraic Graph Theory by Norman Biggs.

In 1973, Richward Weiss established a more general theorem that proves that for any \( s \)-arc-transitive graph, \( s \le 7. \) The bound is weaker but it applies to all graphs rather than only to cubic ones.

Tutte's 8-Cage

The book Algebraic Graph Theory by Godsil and Royle offers the following two descriptions of Tutte's 8-cage on 30 vertices:

Take the cube and an additional vertex \( \infty. \) In each set of four parallel edges, join the midpoint of each pair of opposite edges by an edge, then join the midpoint of the two new edges by an edge, and finally join the midpoint of this edge to \( \infty. \)

Construct a bipartite graph \( T \) with the fifteen edges as one colour class and the fifteen \( 1 \)-factors as the other, where each edge is adjacent to the three \( 1 \)-factors that contain it.

It can be shown that both descriptions construct a cubic bipartite graph on \( 30 \) vertices of girth \( 8. \) It can be further shown that there is a unique cubic bipartite graph on \( 30 \) vertices with girth \( 8. \) As a result both descriptions above construct the same graph.

Linear Congruential Generator

Here is a simple linear congruential generator (LCG) implementation in JavaScript:

function srand (seed) {
  let x = seed
  return function () {
    x = (1664525 * x + 1013904223) % 4294967296
    return x
  }
}

Here is an example usage:

> const rand = srand(0)
undefined
> rand()
1013904223
> rand()
1196435762
> rand()
3519870697

Numbering Lines

Both BSD and GNU cat can number output lines with the -n option. For example:

$ printf 'foo\nbar\nbaz\n' | cat -n
     1  foo
     2  bar
     3  baz

However I have always used nl for this. For example:

$ printf 'foo\nbar\nbaz\n' | nl
     1  foo
     2  bar
     3  baz

While nl is specified in POSIX, the cat -n option is not.

Read on website | #notes | #mathematics | #programming | #javascript | #shell

Like most sensible people with a reasonable sense of priorities, I do not carry a ruler with me wherever I go. Nevertheless, I often find myself needing to measure something at short notice, usually in situations where a certain amount of inaccuracy is entirely forgivable. When I cannot easily fetch a ruler, I end up doing what many people do and reach for the next best thing, which for me is a sheet of A4 paper, available in abundant supply where I live.

From photocopying night-sky charts to serving as a scratch pad for working through mathematical proofs, A4 paper has been a trusted companion since my childhood days. I use it often. If I am carrying a bag, there is almost always some A4 paper inside: perhaps a printed research paper or a mathematical problem I have worked on recently and need to chew on a bit more during my next train ride.

Dimensions

The dimensions of A4 paper are the solution to a simple, elegant problem. Imagine designing a sheet of paper such that, when you cut it in half parallel to its shorter side, both halves have exactly the same aspect ratio as the original. In other words, if the shorter side has length \( x \) and the longer side has length \( y , \) then \[ \frac{y}{x} = \frac{x}{y / 2} \] which gives us \[ \frac{y}{x} = \sqrt{2}. \] Test it out. Suppose we have \( y/x = \sqrt{2}. \) We cut the paper in half parallel to the shorter side to get two halves, each with shorter side \( x' = y / 2 = x \sqrt{2} / 2 = x / \sqrt{2} \) and longer side \( y' = x. \) Then indeed \[ \frac{y'}{x'} = \frac{x}{x / \sqrt{2}} = \sqrt{2}. \] In fact, we can keep cutting the halves like this and we'll keep getting even smaller sheets with the aspect ratio \( \sqrt{2} \) intact. To summarise, when a sheet of paper has the aspect ratio \( \sqrt{2}, \) bisecting it parallel to the shorter side leaves us with two halves that preserve the aspect ratio. A4 paper has this property.

But what are the exact dimensions of A4 and why is it called A4? What does 4 mean here? Like most good answers, this one too begins by considering the numbers \( 0 \) and \( 1. \) Let me elaborate.

Let us say we want to make a sheet of paper that is \( 1 \, \mathrm{m}^2 \) in area and has the aspect-ratio-preserving property that we just discussed. What should its dimensions be? We want \[ xy = 1 \, \mathrm{m}^2 \] subject to the condition \[ \frac{y}{x} = \sqrt{2}. \] Solving these two equations gives us \[ x^2 = \frac{1}{\sqrt{2}} \, \mathrm{m}^2 \] from which we obtain \[ x = \frac{1}{\sqrt[4]{2}} \, \mathrm{m}, \quad y = \sqrt[4]{2} \, \mathrm{m}. \] Up to three decimal places, this amounts to \[ x = 0.841 \, \mathrm{m}, \quad y = 1.189 \, \mathrm{m}. \] These are the dimensions of A0 paper. They are precisely the dimensions specified by the ISO standard for it. It is quite large to scribble mathematical solutions on, unless your goal is to make a spectacle of yourself and cause your friends and family to reassess your sanity. So we need something smaller that allows us to work in peace, without inviting commentary or concerns from passersby. We take the A0 paper of size \[ 84.1 \, \mathrm{cm} \times 118.9 \, \mathrm{cm} \] and bisect it to get A1 paper of size \[ 59.4 \, \mathrm{cm} \times 84.1 \, \mathrm{cm}. \] Then we bisect it again to get A2 paper with dimensions \[ 42.0 \, \mathrm{cm} \times 59.4 \, \mathrm{cm}. \] And once again to get A3 paper with dimensions \[ 29.7 \, \mathrm{cm} \times 42.0 \, \mathrm{cm}. \] And then once again to get A4 paper with dimensions \[ 21.0 \, \mathrm{cm} \times 29.7 \, \mathrm{cm}. \] There we have it. The dimensions of A4 paper. These numbers are etched in my memory like the multiplication table of \( 1. \) We can keep going further to get A5, A6, etc. We could, in theory, go all the way up to A\( \infty. \) Hold on, I think I hear someone heckle. What's that? Oh, we can't go all the way to A\( \infty? \) Something about atoms, was it? Hmm. Security! Where's security? Ah yes, thank you, sir. Please show this gentleman out, would you?

Sorry for the interruption, ladies and gentlemen. Phew! That fellow! Atoms? Honestly. We, the mathematically inclined, are not particularly concerned with such trivial limitations. We drink our tea from doughnuts. We are not going to let the size of atoms dictate matters, now are we?

So I was saying that we can bisect our paper like this and go all the way to A\( \infty. \) That reminds me. Last night I was at a bar in Hoxton and I saw an infinite number of mathematicians walk in. The first one asked, "Sorry to bother you, but would it be possible to have a sheet of A0 paper? I just need something to scribble a few equations on." The second one asked, "If you happen to have one spare, could I please have an A1 sheet?" The third one said, "An A2 would be perfectly fine for me, thank you." Before the fourth one could ask, the bartender disappeared into the back for a moment and emerged with two sheets of A0 paper and said, "Right. That should do it. Do know your limits and split these between yourselves."

In general, a sheet of A\( n \) paper has the dimensions \[ 2^{-(2n + 1)/4} \, \mathrm{m} \times 2^{-(2n - 1)/4} \, \mathrm{m}. \] If we plug in \( n = 4, \) we indeed get the dimensions of A4 paper: \[ 0.210 \, \mathrm{m} \times 0.297 \, \mathrm{m}. \]

Measuring Stuff

Let us now return to the business of measuring things. As I mentioned earlier, the dimensions of A4 are lodged firmly into my memory. Getting hold of a sheet of A4 paper is rarely a challenge where I live. I have accumulated a number of A4 paper stories over the years. Let me share a recent one. I was hanging out with a few folks of the nerd variety one afternoon when the conversation drifted, as it sometimes does, to a nearby computer monitor that happened to be turned off. At some point, someone confidently declared that the screen in front of us was 27 inches. That sounded plausible but we wanted to confirm it. So I reached for my trusted measuring instrument: an A4 sheet of paper. What followed was neither fast, nor especially precise, but it was more than adequate for settling the matter at hand.

I lined up the longer edge of the A4 sheet with the width of the monitor. One length. Then I repositioned it and measured a second length. The screen was still sticking out slightly at the end. By eye, drawing on an entirely unjustified confidence built from years of measuring things that never needed measuring, I estimated the remaining bit at about \( 1 \, \mathrm{cm}. \) That gives us a width of \[ 29.7 \, \mathrm{cm} + 29.7 \, \mathrm{cm} + 1.0 \, \mathrm{cm} = 60.4 \, \mathrm{cm}. \] Let us round that down to \( 60 \, \mathrm{cm}. \) For the height, I switched to the shorter edge. One full \( 21 \, \mathrm{cm} \) fit easily. For the remainder, I folded the paper parallel to the shorter side, producing an A5-sized rectangle with dimensions \( 14.8 \, \mathrm{cm} \times 21.0 \, \mathrm{cm}. \) Using the \( 14.8 \, \mathrm{cm} \) edge, I discovered that it overshot the top of the screen slightly. Again, by eye, I estimated the excess at around \( 2 \, \mathrm{cm}. \) That gives us \[ 21.0 \, \mathrm{cm} + 14.8 \, \mathrm{cm} -2.0 \, \mathrm{cm} = 33.8 \, \mathrm{cm}. \] Let us round this up to \( 34 \, \mathrm{cm}. \) The ratio \( 60 / 34 \approx 1.76 \) is quite close to \( 16/9, \) a popular aspect ratio of modern displays. At this point the measurements were looking good. So far, the paper had not embarrassed itself. Invoking the wisdom of the Pythagoreans, we can now estimate the diagonal as \[ \sqrt{(60 \, \mathrm{cm})^2 + (34 \, \mathrm{cm})^2} \approx 68.9 \,\mathrm{cm}. \] Finally, there is the small matter of units. One inch is \( 2.54 \, \mathrm{cm}, \) another figure that has embedded itself in my head. Dividing \( 68.9 \) by \( 2.54 \) gives us roughly \( 27.2 \, \mathrm{in}. \) So yes. It was indeed a \( 27 \)-inch display. My elaborate exercise in showing off my A4 paper skills was now complete. Nobody said anything. A few people looked away in silence. I assumed they were reflecting. I am sure they were impressed deep down. Or perhaps... no, no. They were definitely impressed. I am sure.

Hold on. I think I hear another heckle. What is that? There are mobile phone apps that can measure things now? Really? Right. Security. Where's security?

Read on website | #absurd | #mathematics

Theorem. If \( \mathcal{T} \) is a set of transpositions, then the Cayley graph \( X(\operatorname{Sym}(n), \mathcal{T}) \) has no triangles.

Proof. Suppose the vertices \( a, b, c \in \operatorname{Sym}(n) \) form a triangle in the Cayley graph \( X(\operatorname{Sym}(n), \mathcal{T}). \) Since multiplication by \( a^{-1} \) is an automorphism of the Cayley graph (by the proof of Theorem 3.1.2 that comes earlier), the vertices \( e, ba^{-1}, ca^{-1} \) form a triangle too. Let us label them as \( e, b', c' \) respectively.

Now by the definition of a Cayley graph, for any two vertices \( a, b \in \operatorname{Sym}(n), \) we have \begin{align*} a \sim b & \iff ba^{-1} \in \mathcal{T} \\ & \iff ba^{-1} = g \\ & \iff b = ga \end{align*} for some \( g \in \mathcal{T}. \) Therefore \begin{align*} e \sim b' & \iff b' = ge = g, \\ e \sim c' & \iff c' = he = h, \\ b' \sim c' & \iff c' = lb' \end{align*} for some \( g, h, l \in \mathcal{T}. \) Note that the last equality gives \[ l =c'b'^{-1} = hg^{-1} = hg \in \mathcal{T} \] Therefore \( g, h, hg \in \mathcal{T}. \) However, this is impossible since the product of two transpositions is \( e, \) a \( 3 \)-cycle or a product of two disjoint transpositions. For example, \( (12)(12) = e, \) \( (12)(13) = (123) \) and \( (12)(24) = (12)(24). \) Therefore \( hg \) cannot be a transposition, i.e. \( hg \notin \mathcal{T}. \) This is a contradiction. Therefore the vertices \( a, b, c \) cannot form a triangle. We conclude that the Cayley graph \( X(\operatorname{Sym}(n), \mathcal{T}) \) has no triangles.

Read on website | #mathematics

for n in range(1, 101):
    if n % 15 == 0:
        print('FizzBuzz')
    elif n % 3 == 0:
        print('Fizz')
    elif n % 5 == 0:
        print('Buzz')
    else:
        print(n)

Here is the output: fizz-buzz.txt. Can we make the program more complicated? The words 'Fizz', 'Buzz' and 'FizzBuzz' repeat in a periodic manner throughout the sequence. What else is periodic? Trigonometric functions! Perhaps we can use trigonometric functions to encode all four rules of the sequence in a single closed-form expression. That is what we are going to explore in this article, for fun and no profit.

By the end, we will obtain a discrete Fourier series that can take any integer \( n \) and select the corresponding text to be printed. In fact, we will derive it using two different methods. First, we will follow a long-winded but hopefully enjoyable approach that relies on a basic understanding of complex exponentiation, geometric series and trigonometric functions. Then, we will obtain the same result through a direct application of the discrete Fourier transform.

Contents

• Definitions
  • Symbol Functions
  • Index Function
  • Fizz Buzz Sequence
• From Indicator Functions to Cosines
  • Indicator Functions
  • Complex Exponentials
  • Cosines
• Discrete Fourier Transform
  • One Period of Fizz Buzz
  • Fourier Coefficients
  • Inverse Transform
• Conclusion

Definitions

Before going any further, we establish a precise mathematical definition for the Fizz Buzz sequence. We begin by introducing a few functions that will help us define the Fizz Buzz sequence later.

Symbol Functions

We define a set of four functions \( \{ s_0, s_1, s_2, s_3 \} \) for integers \( n \) by: \begin{align*} s_0(n) &= n, \\ s_1(n) &= \mathtt{Fizz}, \\ s_2(n) &= \mathtt{Buzz}, \\ s_3(n) &= \mathtt{FizzBuzz}. \end{align*} We call these the symbol functions because they produce every term that appears in the Fizz Buzz sequence. The symbol function \( s_0 \) returns \( n \) itself. The functions \( s_1, \) \( s_2 \) and \( s_3 \) are constant functions that always return the literal words \( \mathtt{Fizz}, \) \( \mathtt{Buzz} \) and \( \mathtt{FizzBuzz} \) respectively, no matter what the value of \( n \) is.

Index Function

We define a function \( f(n) \) for integer \( n \) by \[ f(n) = \begin{cases} 1 & \text{if } 3 \mid n \text{ and } 5 \nmid n, \\ 2 & \text{if } 3 \nmid n \text{ and } 5 \mid n, \\ 3 & \text{if } 3 \mid n \text{ and } 5 \mid n, \\ 0 & \text{otherwise}. \end{cases} \] The notation \( m \mid n \) means that the integer \( m \) divides the integer \( n, \) i.e. \( n \) is a multiple of \( m. \) Equivalently, there exists an integer \( c \) such that \( n = cm . \) Similarly, \( m \nmid n \) means that \( m \) does not divide \( n, \) i.e. \( n \) is not a multiple of \( m. \)

This function covers all four conditions involved in choosing the \( n \)th item of the Fizz Buzz sequence. As we will soon see, this function tells us which of the four symbol functions produces the \( n \)th item of the Fizz Buzz sequence. For this reason, we call \( f(n) \) the index function.

Fizz Buzz Sequence

We now define the Fizz Buzz sequence as the sequence \[ (s_{f(n)}(n))_{n = 1}^{\infty} \] We can expand the first few terms of the sequence explicitly as follows: \begin{align*} (s_{f(n)}(n))_{n = 1}^{\infty} &= (s_{f(1)}(1), \; s_{f(2)}(2), \; s_{f(3)}(3), \; s_{f(4)}(4), \; s_{f(5)}(5), \; s_{f(6)}(6), \; s_{f(7)}(7), \; \dots) \\ &= (s_0(1), \; s_0(2), \; s_1(3), \; s_0(4), s_2(5), \; s_1(6), \; s_0(7), \; \dots) \\ &= (1, \; 2, \; \mathtt{Fizz}, \; 4, \; \mathtt{Buzz}, \; \mathtt{Fizz}, \; 7, \; \dots). \end{align*} Note how the function \( f(n) \) produces an index \( i \) which we then use to select the symbol function \( s_i(n) \) to produce the \( n \)th term of the sequence. This is precisely why we decided to call \( f(n) \) the index function while defining it in the previous section.

From Indicator Functions to Cosines

Here we discuss the first method of deriving our closed form expression, starting with indicator functions and rewriting them using complex exponentials and cosines.

Indicator Functions

Here is the index function \( f(n) \) from the previous section with its cases and conditions rearranged to make it easier to spot interesting patterns: \[ f(n) = \begin{cases} 0 & \text{if } 5 \nmid n \text{ and } 3 \nmid n, \\ 1 & \text{if } 5 \nmid n \text{ and } 3 \mid n, \\ 2 & \text{if } 5 \mid n \text{ and } 3 \nmid n, \\ 3 & \text{if } 5 \mid n \text{ and } 3 \mid n. \end{cases} \] This function helps us select another function \( s_{f(n)}(n) \) which in turn determines the \( n \)th term of the Fizz Buzz sequence. Our goal now is to replace this piecewise formula with a single closed-form expression. To do so, we first define indicator functions \( I_m(n) \) as follows: \[ I_m(n) = \begin{cases} 1 & \text{if } m \mid n, \\ 0 & \text{if } m \nmid n. \end{cases} \] The formula for \( f(n) \) can now be written as: \[ f(n) = \begin{cases} 0 & \text{if } I_5(n) = 0 \text{ and } I_3(n) = 0, \\ 1 & \text{if } I_5(n) = 0 \text{ and } I_3(n) = 1, \\ 2 & \text{if } I_5(n) = 1 \text{ and } I_3(n) = 0, \\ 3 & \text{if } I_5(n) = 1 \text{ and } I_3(n) = 1. \end{cases} \] Do you see a pattern? Here is the same function written as a table:

Do you see it now? If we treat the values in the first two columns as binary digits and the values in the third column as decimal numbers, then in each row the first two columns give the binary representation of the number in the third column. For example, \( 3_{10} = 11_2 \) and indeed in the last row of the table, we see the bits \( 1 \) and \( 1 \) in the first two columns and the number \( 3 \) in the last column. In other words, writing the binary digits \( I_5(n) \) and \( I_3(n) \) side by side gives us the binary representation of \( f(n). \) Therefore \[ f(n) = 2 \, I_5(n) + I_3(n). \] We can now write a small program to demonstrate this formula:

for n in range(1, 101):
    s = [n, 'Fizz', 'Buzz', 'FizzBuzz']
    i = (n % 3 == 0) + 2 * (n % 5 == 0)
    print(s[i])

We can make it even shorter at the cost of some clarity:

for n in range(1, 101):
    print([n, 'Fizz', 'Buzz', 'FizzBuzz'][(n % 3 == 0) + 2 * (n % 5 == 0)])

What we have obtained so far is pretty good. While there is no universal definition of a closed-form expression, I think most people would agree that the indicator functions as defined above are simple enough to be permitted in a closed-form expression.

Complex Exponentials

In the previous section, we obtained the formula \[ f(n) = I_3(n) + 2 \, I_5(n) \] which we then used as an index to look up the text to be printed. We also argued that this is a pretty good closed-form expression already.

However, in the interest of making things more complicated, we must ask ourselves: What if we are not allowed to use the indicator functions? What if we must adhere to the commonly accepted meaning of a closed-form expression which allows only finite combinations of basic operations such as addition, subtraction, multiplication, division, integer exponents and roots with integer index as well as functions such as exponentials, logarithms and trigonometric functions. It turns out that the above formula can be rewritten using only addition, multiplication, division and the cosine function. Let us begin the translation. Consider the sum \[ S_m(n) = \sum_{k = 0}^{m - 1} e^{i 2 \pi k n / m}, \] where \( i \) is the imaginary unit and \( n \) and \( m \) are integers. This is a geometric series in the complex plane with ratio \( r = e^{i 2 \pi n / m}. \) If \( n \) is a multiple of \( m , \) then \( n = cm \) for some integer \( c \) and we get \[ r = e^{i 2 \pi n / m} = e^{i 2 \pi c} = 1. \] Therefore, when \( n \) is a multiple of \( m, \) we get \[ S_m(n) = \sum_{k = 0}^{m - 1} e^{i 2 \pi k n / m} = \sum_{k = 0}^{m - 1} 1^k = m. \] If \( n \) is not a multiple of \( m, \) then \( r \ne 1 \) and the geometric series becomes \[ S_m(n) = \frac{r^m - 1}{r - 1} = \frac{e^{i 2 \pi n} - 1}{e^{i 2 \pi n / m} - 1} = 0. \] Therefore, \[ S_m(n) = \begin{cases} m & \text{if } m \mid n, \\ 0 & \text{if } m \nmid n. \end{cases} \] Dividing both sides by \( m, \) we get \[ \frac{S_m(n)}{m} = \begin{cases} 1 & \text{if } m \mid n, \\ 0 & \text{if } m \nmid n. \end{cases} \] But the right-hand side is \( I_m(n). \) Therefore \[ I_m(n) = \frac{S_m(n)}{m} = \frac{1}{m} \sum_{k = 0}^{m - 1} e^{i 2 \pi k n / m}. \]

Cosines

We begin with Euler's formula \[ e^{i x} = \cos x + i \sin x \] where \( x \) is a real number. From this formula, we get \[ e^{i x} + e^{-i x} = 2 \cos x. \] Therefore \begin{align*} I_3(n) &= \frac{1}{3} \sum_{k = 0}^2 e^{i 2 \pi k n / 3} \\ &= \frac{1}{3} \left( 1 + e^{i 2 \pi n / 3} + e^{i 4 \pi n / 3} \right) \\ &= \frac{1}{3} \left( 1 + e^{i 2 \pi n / 3} + e^{-i 2 \pi n / 3} \right) \\ &= \frac{1}{3} + \frac{2}{3} \cos \left( \frac{2 \pi n}{3} \right). \end{align*} The third equality above follows from the fact that \( e^{i 4 \pi n / 3} = e^{i 6 \pi n / 3} e^{-i 2 \pi n / 3} = e^{i 2 \pi n} e^{-i 2 \pi n/3} = e^{-i 2 \pi n / 3} \) when \( n \) is an integer.

The function above is defined for integer values of \( n \) but we can extend its formula to real \( x \) and plot it to observe its shape between integers. As expected, the function takes the value \( 1 \) whenever \( x \) is an integer multiple of \( 3 \) and \( 0 \) whenever \( x \) is an integer not divisible by \( 3. \)

Similarly, \begin{align*} I_5(n) &= \frac{1}{5} \sum_{k = 0}^4 e^{i 2 \pi k n / 5} \\ &= \frac{1}{5} \left( 1 + e^{i 2 \pi n / 5} + e^{i 4 \pi n / 5} + e^{i 6 \pi n / 5} + e^{i 8 \pi n / 5} \right) \\ &= \frac{1}{5} \left( 1 + e^{i 2 \pi n / 5} + e^{i 4 \pi n / 5} + e^{-i 4 \pi n / 5} + e^{-i 2 \pi n / 5} \right) \\ &= \frac{1}{5} + \frac{2}{5} \cos \left( \frac{2 \pi n}{5} \right) + \frac{2}{5} \cos \left( \frac{4 \pi n}{5} \right). \end{align*} Extending this expression to real values of \( x \) allows us to plot its shape as well. Once again, the function takes the value \( 1 \) at integer multiples of \( 5 \) and \( 0 \) at integers not divisible by \( 5. \)

Recall that we expressed \( f(n) \) as \[ f(n) = I_3(n) + 2 \, I_5(n). \] Substituting these trigonometric expressions yields \[ f(n) = \frac{1}{3} + \frac{2}{3} \cos \left( \frac{2 \pi n}{3} \right) + 2 \cdot \left( \frac{1}{5} + \frac{2}{5} \cos \left( \frac{2 \pi n}{5} \right) + \frac{2}{5} \cos \left( \frac{4 \pi n}{5} \right) \right). \] A straightforward simplification gives \[ f(n) = \frac{11}{15} + \frac{2}{3} \cos \left( \frac{2 \pi n}{3} \right) + \frac{4}{5} \cos \left( \frac{2 \pi n}{5} \right) + \frac{4}{5} \cos \left( \frac{4 \pi n}{5} \right). \] We can extend this expression to real \( x \) and plot it as well. The resulting curve takes the values \( 0, 1, 2 \) and \( 3 \) at integer points, as desired.

Now we can write our Python program as follows:

from math import cos, pi
for n in range(1, 101):
    s = [n, 'Fizz', 'Buzz', 'FizzBuzz']
    i = round(11 / 15 + (2 / 3) * cos(2 * pi * n / 3)
                      + (4 / 5) * cos(2 * pi * n / 5)
                      + (4 / 5) * cos(4 * pi * n / 5))
    print(s[i])

Discrete Fourier Transform

The keen-eyed might notice that the expression we obtained for \( f(n) \) is a discrete Fourier series. This is not surprising, since the output of a Fizz Buzz program depends only on \( n \bmod 15. \) Any function on a finite cyclic group can be written exactly as a finite Fourier expansion. In this section, we obtain \( f(n) \) using the discrete Fourier transform. It is worth mentioning that the calculations presented here are quite tedious to do by hand. Nevertheless, this section offers a glimpse of how such calculations are performed. By the end, we will arrive at exactly the same \( f(n) \) as before. There is nothing new to discover here. We simply obtain the same result by a more direct but more laborious method. If this doesn't sound interesting, you may safely skip the subsections that follow.

One Period of Fizz Buzz

We know that \( f(n) \) is a periodic function with period \( 15. \) To apply the discrete Fourier transform, we look at one complete period of the function using the values \( n = 0, 1, \dots, 14. \) Over this period, we have: \begin{array}{c|ccccccccccccccc} n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline f(n) & 3 & 0 & 0 & 1 & 0 & 2 & 1 & 0 & 0 & 1 & 2 & 0 & 1 & 0 & 0 \end{array} The discrete Fourier series of \( f(n) \) is \[ f(n) = \sum_{k = 0}^{14} c_k \, e^{i 2 \pi k n / 15} \] where the Fourier coefficients \( c_k \) are given by the discrete Fourier transform \[ c_k = \frac{1}{15} \sum_{n = 0}^{14} f(n) e^{-i 2 \pi k n / 15}. \] for \( k = 0, 1, \dots, 14. \) The formula for \( c_k \) is called the discrete Fourier transform (DFT). The formula for \( f(n) \) is called the inverse discrete Fourier transform (IDFT).

Fourier Coefficients

Let \( \omega = e^{-i 2 \pi / 15}. \) Then using the values of \( f(n) \) from the table above, the DFT becomes: \[ c_k = \frac{3 + \omega^{3k} + 2 \omega^{5k} + \omega^{6k} + \omega^{9k} + 2 \omega^{10k} + \omega^{12k}}{15}. \] Substituting \( k = 0, 1, 2, \dots, 14 \) into the above equation gives us the following Fourier coefficients: \begin{align*} c_{0} &= \frac{11}{15}, \\ c_{3} &= c_{6} = c_{9} = c_{12} = \frac{2}{5}, \\ c_{5} &= c_{10} = \frac{1}{3}, \\ c_{1} &= c_{2} = c_{4} = c_{7} = c_{8} = c_{11} = c_{13} = c_{14} = 0. \end{align*} Calculating these Fourier coefficients by hand can be rather tedious. In practice they are almost always calculated using numerical software, computer algebra systems or even simple code such as the example here: fizz-buzz-fourier.py.

Inverse Transform

Once the coefficients are known, we can substitute them into the inverse transform introduced earlier to obtain \begin{align*} f(n) &= \sum_{k = 0}^{14} c_k \, e^{i 2 \pi k n / 15} \\[1.5em] &= \frac{11}{15} + \frac{2}{5} \left( e^{i 2 \pi \cdot 3n / 15} + e^{i 2 \pi \cdot 6n / 15} + e^{i 2 \pi \cdot 9n / 15} + e^{i 2 \pi \cdot 12n / 15} \right) \\ & \phantom{=\frac{11}{15}} + \frac{1}{3} \left( e^{i 2 \pi \cdot 5n / 15} + e^{i 2 \pi \cdot 10n / 15} \right) \\[1em] &= \frac{11}{15} + \frac{2}{5} \left( e^{i 2 \pi \cdot 3n / 15} + e^{i 2 \pi \cdot 6n / 15} + e^{-i 2 \pi \cdot 6n / 15} + e^{-i 2 \pi \cdot 3n / 15} \right) \\ & \phantom{=\frac{11}{15}} + \frac{1}{3} \left( e^{i 2 \pi \cdot 5n / 15} + e^{-i 2 \pi \cdot 5n / 15} \right) \\[1em] &= \frac{11}{15} + \frac{2}{5} \left( 2 \cos \left( \frac{2 \pi n}{5} \right) + 2 \cos \left( \frac{4 \pi n}{5} \right) \right) \\ & \phantom{=\frac{11}{15}} + \frac{1}{3} \left( 2 \cos \left( \frac{2 \pi n}{3} \right) \right) \\[1em] &= \frac{11}{15} + \frac{4}{5} \cos \left( \frac{2 \pi n}{5} \right) + \frac{4}{5} \cos \left( \frac{4 \pi n}{5} \right) + \frac{2}{3} \cos \left( \frac{2 \pi n}{3} \right). \end{align*} This is exactly the same expression for \( f(n) \) we obtained in the previous section. We see that the Fizz Buzz index function \( f(n) \) can be expressed precisely using the machinery of Fourier analysis.

Conclusion

To summarise, we have defined the Fizz Buzz sequence as \[ (s_{f(n)}(n))_{n = 1}^{\infty} \] where \[ f(n) = \frac{11}{15} + \frac{2}{3} \cos \left( \frac{2 \pi n}{3} \right) + \frac{4}{5} \cos \left( \frac{2 \pi n}{5} \right) + \frac{4}{5} \cos \left( \frac{4 \pi n}{5} \right). \] and \( s_0(n) = n, \) \( s_1(n) = \mathtt{Fizz}, \) \( s_2(n) = \mathtt{Buzz} \) and \( s_3(n) = \mathtt{FizzBuzz}. \) A Python program to print the Fizz Buzz sequence based on this definition was presented earlier. That program can be written more succinctly as follows:

from math import cos, pi
for n in range(1, 101):
    print([n, 'Fizz', 'Buzz', 'FizzBuzz'][round(11 / 15 + (2 / 3) * cos(2 * pi * n / 3) + (4 / 5) * (cos(2 * pi * n / 5) + cos(4 * pi * n / 5)))])

We can also wrap this up nicely in a shell one-liner, in case you want to share it with your friends and family and surprise them:

python3 -c 'from math import cos, pi; [print([n, "Fizz", "Buzz", "FizzBuzz"][round(11/15 + (2/3) * cos(2*pi*n/3) + (4/5) * (cos(2*pi*n/5) + cos(4*pi*n/5)))]) for n in range(1, 101)]'

We have taken a simple counting game and turned it into a trigonometric construction consisting of a discrete Fourier series with three cosine terms and four coefficients. None of this makes Fizz Buzz any easier. Quite the contrary. But it does show that every \( \mathtt{Fizz} \) and \( \mathtt{Buzz} \) now owes its existence to a particular set of Fourier coefficients. We began with the modest goal of making this simple problem more complicated. I think it is safe to say that we did not fall short.

Read on website | #absurd | #python | #programming | #technology | #mathematics | #puzzle

I should mention the sections presented in that post are not in the same order in which we originally discussed them. The sections were edited and rearranged by Alex to improve the flow and avoid repetition of similar topics too close to each other.

This page preserves a copy of our discussion as edited by Alex, so I can keep an archived version on my website. In my copy, I have added a table of contents to make it easier to navigate to specific sections. The interview itself follows the table of contents. I hope you enjoy reading it.

Contents

1. Lisp and Other Things
2. Lisp, Emacs and Mathematics
3. Interests and Exploration
4. Computing for Fun
5. Computing Activities
6. Programming vs Domains
7. Old Functionality and New Problems
8. Designing for Composability
9. Small vs Large Functions
10. Domains and Projects
11. Double Spacing and Touch Typing
12. Approach to Learning
13. Managing Time and Distractions
14. Blogging
15. Forums
16. MathB Moderation Problems
17. Favourite Mathematics Textbooks
18. Mathematics and Computing

Our Conversation

Hi @susam, I primarily know you as a Lisper, what other things do you use?

Yes, I use Lisp extensively for my personal projects and much of what I do in my leisure is built on it. I ran a mathematics pastebin for close to thirteen years. It was quite popular on some IRC channels. The pastebin was written in Common Lisp. My personal website and blog are generated using a tiny static site generator written in Common Lisp. Over the years I have built several other personal tools in it as well.

I am an active Emacs Lisp programmer too. Many of my software tools are in fact Emacs Lisp functions that I invoke with convenient key sequences. They help me automate repetitive tasks as well as improve my text editing and task management experience.

I use plenty of other tools as well. In my early adulthood, I spent many years working with C, C++, Java and PHP. My first substantial open source contribution was to the Apache Nutch project which was in Java and one of my early original open source projects was Uncap, a C program to remap keys on Windows.

These days I use a lot of Python, along with some Go and Rust, but Lisp remains important to my personal work. I also enjoy writing small standalone tools directly in HTML and JavaScript, often with all the code in a single file in a readable, unminified form.

How did you first discover computing, then end up with Lisp, Emacs and mathematics?

I got introduced to computers through the Logo programming language as a kid. Using simple arithmetic, geometry, logic and code to manipulate a two-dimensional world had a lasting effect on me.

I still vividly remember how I ended up with Lisp. It was at an airport during a long layover in 2007. I wanted to use the time to learn something, so I booted my laptop running Debian GNU/Linux 4.0 (Etch) and then started GNU CLISP 2.41. In those days, Wi-Fi in airports was uncommon. Smartphones and mobile data were also uncommon. So it was fortunate that I had CLISP already installed on my system and my laptop was ready for learning Common Lisp. I had it installed because I had wanted to learn Common Lisp for some time. I was especially attracted by its simplicity, by the fact that the entire language can be built up from a very small set of special forms. I use SBCL these days, by the way.

I discovered Emacs through Common Lisp. Several sources recommended using the Superior Lisp Interaction Mode for Emacs (SLIME) for Common Lisp programming, so that's where I began. For many years I continued to use Vim as my primary editor, while relying on Emacs and SLIME for Lisp development. Over time, as I learnt more about Emacs itself, I grew fond of Emacs Lisp and eventually made Emacs my primary editor and computing environment.

I have loved mathematics since my childhood days. What has always fascinated me is how we can prove deep and complex facts using first principles and clear logical steps. That feeling of certainty and rigour is unlike anything else.

Over the years, my love for the subject has been rekindled many times. As a specific example, let me share how I got into number theory. One day I decided to learn the RSA cryptosystem. As I was working through the RSA paper, I stumbled upon the Euler totient function \( \varphi(n) \) which gives the number of positive integers not exceeding n that are relatively prime to n. The paper first states that \[ \varphi(p) = p - 1 \] for prime numbers \( p. \) That was obvious since \( p \) has no factors other than \( 1 \) and itself, so every integer from \( 1 \) up to \( p - 1 \) must be relatively prime to it. But then it presents \[ \varphi(pq) = \varphi(p) \cdot \varphi(q) = (p - 1)(q - 1) \] for primes \( p \) and \( q. \) That was not immediately obvious to me back then. After a few minutes of thinking, I managed to prove it from scratch. By the inclusion-exclusion principle, we count how many integers from \( 1 \) up to \( pq \) are not divisible by \(p \) or \( q. \) There are \( pq \) integers in total. Among them, there are \( q \) integers divisible by \( p \) and \( p \) integers divisible by \( q. \) So we need to subtract \( p + q \) from \(pq. \) But since one integer (\( pq \) itself) is counted in both groups, we add \( 1 \) back. Therefore \[ \varphi(pq) = pq - (p + q) + 1 = (p - 1)(q - 1). \] Next I could also obtain the general formula for \( \varphi(n) \) for an arbitrary positive integer \( n \) using the same idea. There are several other proofs too, but that is how I derived the general formula for \( \varphi(n) \) when I first encountered it. And just like that, I had begun to learn number theory!

You've said you prefer computing for fun. What is fun to you? Do you have an idea of what makes something fun or not?

For me, fun in computing began when I first learnt IBM/LCSI PC Logo when I was nine years old. I had very limited access to computers back then, perhaps only about two hours per month in the computer laboratory at my primary school. Most of my Logo programming happened with pen and paper at home. I would 'test' my programs by tracing the results on graph paper. Eventually I would get about thirty minutes of actual computer time in the lab to run them for real.

So back then, most of my computing happened without an actual computer. But even with that limited access to computers, a whole new world opened up for me: one that showed me the joy of computing and more importantly, the joy of sharing my little programs with my friends and teachers. One particular Logo program I still remember very well drew a house with animated dashed lines, where the dashes moved around the outline of the house. Everyone around me loved it, copied it and tweaked it to change the colours, alter the details and add their own little touches.

For me, fun in computing comes from such exploration and sharing. I enjoy asking 'what happens if' and then seeing where it leads me. My Emacs package devil-mode comes from such exploration. It came from asking, 'What happens if we avoid using the ctrl and meta modifier keys and use , (the comma key) or another suitable key as a leader key instead? And can we still have a non-modal editing experience?'

Sometimes computing for fun may mean crafting a minimal esoteric drawing language, making a small game or building a tool that solves an interesting problem elegantly. It is a bonus if the exploration results in something working well enough that I can share with others on the World Wide Web and others find it fun too.

How do you choose what to investigate? Which most interest you, with what commonalities?

For me, it has always been one exploration leading to another.

For example, I originally built MathB for my friends and myself who were going through a phase in our lives when we used to challenge each other with mathematical puzzles. This tool became a nice way to share solutions with each other. Its use spread from my friends to their friends and colleagues, then to schools and universities and eventually to IRC channels.

Similarly, I built TeXMe when I was learning neural networks and taking a lot of notes on the subject. I was not ready to share the notes online, but I did want to share them with my friends and colleagues who were also learning the same topic. Normally I would write my notes in LaTeX, compile them to PDF and share the PDF, but in this case, I wondered, what if I took some of the code from MathB and created a tool that would let me write plain Markdown (GFM) + LaTeX (MathJax) in a .html file and have the tool render the file as soon as it was opened in a web browser? That resulted in TeXMe, which has surprisingly become one of my most popular projects, receiving millions of hits in some months according to the CDN statistics.

Another example is Muboard, which is a bit like an interactive mathematics chalkboard. I built this when I was hosting an analytic number theory book club and I needed a way to type LaTeX snippets live on screen and see them immediately rendered. That made me wonder: what if I took TeXMe, made it interactive and gave it a chalkboard look-and-feel? That led to Muboard.

So we can see that sharing mathematical notes and snippets has been a recurring theme in several of my projects. But that is only a small fraction of my interests. I have a wide variety of interests in computing. I also engage in random explorations, like writing IRC clients (NIMB, Tzero), ray tracing (POV-Ray, Java ray tracer), writing Emacs guides (Emacs4CL, Emfy), developing small single-file HTML games (Andromeda Invaders, Guess My RGB), purely recreational programming (FXYT, may4.fs, self-printing machine code, prime number grid explorer) and so on. The list goes on. When it comes to hobby computing, I don't think I can pick just one domain and say it interests me the most. I have a lot of interests.

What is computing, to you?

Computing, to me, covers a wide range of activities: programming a computer, using a computer, understanding how it works, even building one. For example, I once built a tiny 16-bit CPU along with a small main memory that could hold only eight 16-bit instructions, using VHDL and a Xilinx CPLD kit. The design was based on the Mano CPU introduced in the book Computer System Architecture (3rd ed.) by M. Morris Mano. It was incredibly fun to enter instructions into the main memory, one at a time, by pushing DIP switches up and down and then watch the CPU I had built execute an entire program. For someone like me, who usually works with software at higher levels of abstraction, that was a thrilling experience!

Beyond such experiments, computing also includes more practical and concrete activities, such as installing and using my favourite Linux distribution (Debian), writing software tools in languages like Common Lisp, Emacs Lisp, Python and the shell command language or customising my Emacs environment to automate repetitive tasks.

To me, computing also includes the abstract stuff like spending time with abstract algebra and number theory and getting a deeper understanding of the results pertaining to groups, rings and fields, as well as numerous number-theoretic results. Browsing the On-Line Encyclopedia of Integer Sequences (OEIS), writing small programs to explore interesting sequences or just thinking about them is computing too. I think many of the interesting results in computer science have deep mathematical foundations. I believe much of computer science is really discrete mathematics in action.

And if we dive all the way down from the CPU to the level of transistors, we encounter continuous mathematics as well, with non-linear voltage-current relationships and analogue behaviour that make digital computing possible. It is fascinating how, as a relatively new species on this planet, we have managed to take sand and find a way to use continuous voltages and currents in electronic circuits built with silicon and convert them into the discrete operations of digital logic. We have machines that can simulate themselves!

To me, all of this is fun. To study and learn about these things, to think about them, to understand them better and to accomplish useful or amusing results with this knowledge is all part of the fun.

How do you view programming vs. domains?

I focus more on the domain than the tool. Most of the time it is a problem that catches my attention and then I explore it to understand the domain and arrive at a solution. The problem itself usually points me to one of the tools I already know.

For example, if it is about working with text files, I might write an Emacs Lisp function. If it involves checking large sets of numbers rapidly for patterns, I might choose C++ or Rust. But if I want to share interactive visualisations of those patterns with others, I might rewrite the solution in HTML and JavaScript, possibly with the use of the Canvas API, so that I can share the work as a self-contained file that others can execute easily within their web browsers. When I do that, I prefer to keep the HTML neat and readable, rather than bundled or minified, so that people who like to 'View Source' can copy, edit and customise the code themselves to immediately see their changes take effect.

Let me share a specific example. While working on a web-based game, I first used CanvasRenderingContext2D's fillText() to display text on the game canvas. However, dissatisfied with the text rendering quality, I began looking for IBM PC OEM fonts and similar retro fonts online. After downloading a few font packs, I wrote a little Python script to convert them to bitmaps (arrays of integers) and then used the bitmaps to draw text on the canvas using JavaScript, one cell at a time, to get pixel-perfect results! These tiny Python and JavaScript tools were good enough that I felt comfortable sharing them together as a tiny toolkit called PCFace. This toolkit offers JavaScript bitmap arrays and tiny JavaScript rendering functions, so that someone else who wants to display text on their game canvas using PC fonts and nothing but plain HTML and JavaScript can do so without having to solve the problem from scratch!

Has the rate of your making new Emacs functions has diminished over time (as if everything's covered) or do the widening domains lead to more? I'm curious how applicable old functionality is for new problems and how that impacts the APIs!

My rate of making new Emacs functions has definitely decreased. There are two reasons. One is that over the years my computing environment has converged into a comfortable, stable setup I am very happy with. The other is that at this stage of life I simply cannot afford the time to endlessly tinker with Emacs as I did in my younger days.

More generally, when it comes to APIs, I find that well-designed functionality tends to remain useful even when new problems appear. In Emacs, for example, many of my older functions continue to serve me well because they were written in a composable way. New problems can often be solved with small wrappers or combinations of existing functions. I think APIs that consist of functions that are simple, orthogonal and flexible age well. If each function in an API does one thing and does it well (the Unix philosophy), it will have long-lasting utility.

Of course, new domains and problems do require new functions and extensions to an API, but I think it is very important to not give in to the temptation of enhancing the existing functions by making them more complicated with optional parameters, keyword arguments, nested branches and so on. Personally, I have found that it is much better to implement new functions that are small, orthogonal and flexible, each doing one thing and doing it well.

What design methods or tips do you have, to increase composability?

For me, good design starts with good vocabulary. Clear vocabulary makes abstract notions concrete and gives collaborators a shared language to work with. For example, while working on a network events database many years ago, we collected data minute by minute from network devices. We decided to call each minute of data from a single device a 'nugget'. So if we had 15 minutes of data from 10 devices, that meant 150 nuggets.

Why 'nugget'? Because it was shorter and more convenient than repeatedly saying 'a minute of data from one device'. Why not something less fancy like 'chunk'? Because we reserved 'chunk' for subdivisions within a nugget. Perhaps there were better choices, but 'nugget' was the term we settled on and it quickly became shared terminology between the collaborators. Good terminology naturally carries over into code. With this vocabulary in place, function names like collect_nugget(), open_nugget(), parse_chunk(), index_chunk(), skip_chunk(), etc. immediately become meaningful to everyone involved.

Thinking about the vocabulary also ensures that we are thinking about the data, concepts and notions we are working with in a deliberate manner and that kind of thinking also helps when we design the architecture of software.

Too often I see collaborators on software projects jump straight into writing functions that take some input and produce some desired effect, with variable names and function names decided on the fly. To me, this feels backwards. I prefer the opposite approach. Define the terms first and let the code follow from them.

I also prefer developing software in a layered manner, where complex functionality is built from simpler, well-named building blocks. It is especially important to avoid layer violations, where one complex function invokes another complex function. That creates tight coupling between two complex functions. If one function changes in the future, we have to reason carefully about how it affects the other. Since both are already complex, the cognitive burden is high. A better approach, I think, is to identify the common functionality they share and factor that out into smaller, simpler functions.

To summarise, I like to develop software with a clear vocabulary, consistent use of that vocabulary, a layered design where complex functions are built from simpler ones and by avoiding layer violations. I am sure none of this is new to the Lobsters community. Some of these ideas also occur in domain-driven design (DDD). DDD defines the term ubiquitous language to mean, 'A language structured around the domain model and used by all team members within a bounded context to connect all the activities of the team with the software.' If I could call this approach of software development something, I would simply call it 'vocabulary-driven development' (VDD), though of course DDD is the more comprehensive concept.

Like I said, none of this is likely new to the Lobsters community. In particular, I suspect Forth programmers would find it too obvious. In Forth, it is very difficult to begin with a long, poorly thought-out monolithic word and then break it down into smaller ones later. The stack effects quickly become too hard to track mentally with that approach. The only viable way to develop software in Forth is to start with a small set of words that represent the important notions of the problem domain, test them immediately and then compose higher-level words from the lower-level ones. Forth naturally encourages a layered style of development, where the programmer thinks carefully about the domain, invents vocabulary and expresses complex ideas in terms of simpler ones, almost in a mathematical fashion. In my experience, this kind of deliberate design produces software that remains easy to understand and reason about even years after it was written.

Not enhancing existing functions but adding new small ones seems quite lovely, but how do you come back to such a codebase later with many tiny functions? At points, I've advocated for very large functions, particularly traumatized by Java-esque 1000 functions in 1000 files approaches. When you had time, would you often rearchitecture the conceptual space of all of those functions?

The famous quote from Alan J. Perlis comes to mind:

It is better to have 100 functions operate on one data structure than 10 functions on 10 data structures.

Personally, I enjoy working with a codebase that has thousands of functions, provided most of them are small, well-scoped and do one thing well. That said, I am not dogmatically opposed to large functions. It is always a matter of taste and judgement. Sometimes one large, cohesive function is clearer than a pile of tiny ones.

For example, when I worked on parser generators, I often found that lexers and finite state machines benefited from a single top-level function containing the full tokenisation logic or the full state transition logic in one place. That function could call smaller helpers for specific tasks, but we still need the overall switch-case or if-else or cond ladder somewhere. I think trying to split that ladder into smaller functions would only make the code harder to follow.

So while I lean towards small, composable functions, the real goal is to strike a balance that keeps code maintainable in the long run. Each function should be as small as it can reasonably be and no smaller.

Like you, I program as a tool to explore domains. Which do you know the most about?

For me too, the appeal of computer programming lies especially in how it lets me explore different domains. There are two kinds of domains in which I think I have gained good expertise. The first comes from years of developing software for businesses, which has included solving problems such as network events parsing, indexing and querying, packet decoding, developing parser generators, database session management and TLS certificate lifecycle management. The second comes from areas I pursue purely out of curiosity or for hobby computing. This is the kind I am going to focus on in our conversation.

Although computing and software are serious business today, for me, as for many others, computing is also a hobby.

Personal hobby projects often lead me down various rabbit holes and I end up learning new domains along the way. For example, although I am not a web developer, I learnt to build small, interactive single-page tools in plain HTML, CSS and JavaScript simply because I needed them for my hobby projects over and over again. An early example is QuickQWERTY, which I built to teach myself and my friends touch-typing on QWERTY keyboards. Another example is CFRS[], which I created because I wanted to make a total (non-Turing complete) drawing language that has turtle graphics like Logo but is absolutely minimal like P′′.

You use double spaces after periods which I'd only experienced from people who learned touch typing on typewriters, unexpected!

Yes, I do separate sentences by double spaces. It is interesting that you noticed this.

I once briefly learnt touch typing on typewriters as a kid, but those lessons did not stick with me. It was much later, when I used a Java applet-based touch typing tutor that I found online about two decades ago, that the lessons really stayed with me. Surprisingly, that application taught me to type with a single space between sentences. By the way, I disliked installing Java plugins into the web browser, so I wrote QuickQWERTY as a similar touch typing tutor in plain HTML and JavaScript for myself and my friends.

I learnt to use double spaces between sentences first with Vim and then later again with Emacs. For example, in Vim, the joinspaces option is on by default, so when we join sentences with the normal mode command J or format paragraphs with gqap, Vim inserts two spaces after full stops. We need to disable that behaviour with :set nojoinspaces if we want single spacing.

It is similar in Emacs. In Emacs, the delete-indentation command (M-^) and the fill-paragraph command (M-q) both insert two spaces between sentences by default. Single spacing can be enabled with (setq sentence-end-double-space nil).

Incidentally, I spend a good portion of the README for my Emacs quick-start DIY kit named Emfy discussing sentence spacing conventions under the section Single Space for Sentence Spacing. There I explain how to configure Emacs to use single spaces, although I use double spaces myself. That's because many new Emacs users prefer single spacing.

The defaults in Vim and Emacs made me adopt double spacing. The double spacing convention is also widespread across open source software. If we look at the Vim help pages, Emacs built-in documentation or the Unix and Linux man pages, double spacing is the norm. Even inline comments in traditional open source projects often use it. For example, see Vim's :h usr_01.txt, Emacs's (info "(emacs) Intro") or the comments in the GCC source code.

How do you approach learning a new domain?

When I take on a new domain, there is of course a lot of reading involved from articles, books and documentation. But as I read, I constantly try to test what I learn. Whenever I see a claim, I ask myself, 'If this claim were wrong, how could I demonstrate it?' Then I design a little experiment, perhaps write a snippet of code or run a command or work through a concrete example, with the goal of checking the claim in practice.

Now I am not genuinely hoping to prove a claim wrong. It is just a way to engage with the material. To illustrate, let me share an extremely simple and generic example without going into any particular domain. Suppose I learn that Boolean operations in Python short-circuit. I might write out several experimental snippets like the following:

def t(): print('t'); return True
def f(): print('f'); return False
f() or t() or f()

And then confirm that the results do indeed confirm short-circuit evaluation (f followed by t in this case).

At this point, one could say, 'Well, you just confirmed what the documentation already told you.' And that's true. But for me, the value lies in trying to test it for myself. Even if the claim holds, the act of checking forces me to see the idea in action. That not only reinforces the concept but also helps me build a much deeper intuition for it.

Sometimes these experiments also expose gaps in my own understanding. Suppose I didn't properly know what 'short-circuit' means. Then the results might contradict my expectations. That contradiction would push me to correct my misconception and that's where the real learning happens.

Occasionally, this process even uncovers subtleties I didn't expect. For example, while learning socket programming, I discovered that a client can successfully receive data using recv() even after calling shutdown(), contrary to what I had first inferred from the specifications. See my Stack Overflow post Why can recv() receive messages after the client has invoked shutdown()? for more details if you are curious.

Now this method cannot always be applied, especially if it is very expensive or unwieldy to do so. For example, if I am learning something in the finance domain, it is not always possible to perform an actual transaction. One can sometimes use simulation software, mock environments or sandbox systems to explore ideas safely. Still, it is worth noting that this method has its limitations.

In mathematics, though, I find this method highly effective. When I study a new branch of mathematics, I try to come up with examples and counterexamples to test what I am learning. Often, failing to find a counterexample helps me appreciate more deeply why a claim holds and why no counterexamples exist.

Do you have trouble not getting distracted with so much on your plate? I'm curious how you balance the time commitments of everything!

Indeed, it is very easy to get distracted. One thing that has helped over the years is the increase in responsibilities in other areas of my life. These days I also spend some of my free time studying mathematics textbooks. With growing responsibilities and the time I devote to mathematics, I now get at most a few hours each week for hobby computing. This automatically narrows down my options. I can explore perhaps one or at most two ideas in a month and that constraint makes me very deliberate about choosing my pursuits.

Many of the explorations do not evolve into something solid that I can share. They remain as little experimental code snippets or notes archived in a private repository. But once in a while, an exploration grows into something concrete and feels worth sharing on the Web. That becomes a short-term hobby project. I might work on it over a weekend if it is small or for a few weeks if it is more complex. When that happens, the goal of sharing the project helps me focus.

I try not to worry too much about making time. After all, this is just a hobby. Other areas of my life have higher priority. I also want to devote a good portion of my free time to learning more mathematics, which is another hobby I am passionate about. Whatever little spare time remains after attending to the higher-priority aspects of my life goes into my computing projects, usually a couple of hours a week, most of it on weekends.

How does blogging mix in? What's the development like of a single piece of curiosity through wrestling with the domain, learning and sharing it etc.?

Maintaining my personal website is another aspect of computing that I find very enjoyable. My website began as a loose collection of pages on a LAN site during my university days. Since then I have been adding pages to it to write about various topics that I find interesting. It acquired its blog shape and form much later when blogging became fashionable.

I usually write a new blog post when I feel like there is some piece of knowledge or some exploration that I want to archive in a persistent format. Now what the development of a post looks like depends very much on the post. So let me share two opposite examples to describe what the development of a single piece looks like.

One of my most frequently visited posts is Lisp in Vim. It started when I was hosting a Common Lisp programming club for beginners. Although I have always used Emacs and SLIME for Common Lisp programming myself, many in the club used Vim, so I decided to write a short guide on setting up something SLIME-like there. As a former long-time Vim user myself, I wanted to make the Lisp journey easier for Vim users too. I thought it would be a 30-minute exercise where I write up a README that explains how to install Slimv and how to set it up in Vim. But then I discovered a newer plugin called Vlime that also offered SLIME-like features in Vim! That detail sent me down a very deep rabbit hole. Now I needed to know how the two packages were different, what their strengths and weaknesses were, how routine operations were performed in both and so on. What was meant to be a short note turned into a nearly 10,000-word article. As I was comparing the two SLIME-like packages for Vim, I also found a few bugs in Slimv and contributed fixes for them (#87, #88, #89, #90). Writing this blog post turned into a month-long project!

At the opposite extreme is a post like Elliptical Python Programming. I stumbled upon Python's Ellipsis while reviewing someone's code. It immediately caught my attention. I wondered if, combined with some standard obfuscation techniques, one could write arbitrary Python programs that looked almost like Morse code. A few minutes of experimentation showed that a genuinely Morse code-like appearance was not possible, but something close could be achieved. So I wrote what I hope is a humorous post demonstrating that arbitrary Python programs can be written using a very restricted set of symbols, one of which is the ellipsis. It took me less than an hour to write this post. The final result doesn't look quite like Morse code as I had imagined, but it is quite amusing nevertheless!

What draws you to post and read online forums? How do you balance or allot time for reading technical articles, blogs etc.?

The exchange of ideas! Just as I enjoy sharing my own computing-related thoughts, ideas and projects, I also find joy in reading what others have to share.

Other areas of my life take precedence over hobby projects and hobby projects take precedence over technical forums.

After I've given time to the higher-priority parts of my life and to my own technical explorations, I use whatever spare time remains to read articles, follow technical discussions and occasionally add comments.

When you decided to stop with MathB due to moderation burdens, I offered to take over/help and you mentioned others had too. Did anyone end up forking it, to your knowledge?

I first thought of shutting down the MathB-based pastebin website in November 2019. The website had been running for seven years at that time. When I announced my thoughts to the IRC communities that would be affected, I received a lot of support and encouragement. A few members even volunteered to help me out with moderation. That support and encouragement kept me going for another six years. However, the volunteers eventually became busy with their own lives and moved on. After all, moderating user content for an open pastebin that anyone in the world can post to is a thankless and tiring activity. So most of the moderation activity fell back on me. Finally, in February 2025, I realised that I no longer want to spend time on this kind of work.

I developed MathB with a lot of passion for myself and my friends. I had no idea at the time that this little project would keep a corner of my mind occupied even during weekends and holidays. There was always a nagging worry. What if someone posted content that triggered compliance concerns and my server was taken offline while I was away? I no longer wanted that kind of burden in my life. So I finally decided to shut it down. I've written more about this in MathB.in Is Shutting Down.

To my knowledge, no one has forked it, but others have developed alternatives. Further, the Archive Team has archived all posts from the now-defunct MathB-based website. A member of the Archive Team reached out to me over IRC and we worked together for about a week to get everything successfully archived.

What're your favorite math textbooks?

I have several favourite mathematics books, but let me share three I remember especially fondly.

The first is Advanced Engineering Mathematics by Erwin Kreyszig. I don't often see this book recommended online, but for me it played a major role in broadening my horizons. I think I studied the 8th edition back in the early 2000s. It is a hefty book with over a thousand pages and I remember reading it cover to cover, solving every exercise problem along the way. It gave me a solid foundation in routine areas like differential equations, linear algebra, vector calculus and complex analysis. It also introduced me to Fourier transforms and Laplace transforms, which I found fascinating.

Of course, the Fourier transform has a wide range of applications in signal processing, communications, spectroscopy and more. But I want to focus on the fun and playful part. In the early 2000s, I was also learning to play the piano as a hobby. I used to record my amateur music compositions with Audacity by connecting my digital piano to my laptop with a line-in cable. It was great fun to plot the spectrum of my music on Audacity, apply high-pass and low-pass filters and observe how the Fourier transform of the audio changed and then hear the effect on the music. That kind of hands-on tinkering made Fourier analysis intuitive for me and I highly recommend it to anyone who enjoys both music and mathematics.

The second book is Introduction to Analytic Number Theory by Tom M. Apostol. As a child I was intrigued by the prime number theorem but lacked the mathematical maturity to understand its proof. Years later, as an adult, I finally taught myself the proof from Apostol's book. It was a fantastic journey that began with simple concepts like the Möbius function and Dirichlet products and ended with quite clever contour integrals that proved the theorem. The complex analysis I had learnt from Kreyszig turned out to be crucial for understanding those integrals. Along the way I gained a deeper understanding of the Riemann zeta function \( \zeta(s). \) The book discusses zero-free regions where \( \zeta(s) \) does not vanish, which I found especially fascinating. Results like \( \zeta(-1) = -1/12, \) which once seemed mysterious, became obvious after studying this book.

The third is Galois Theory by Ian Stewart. It introduced me to field extensions, field homomorphisms and solubility by radicals. I had long known that not all quintic equations are soluble by radicals, but I didn't know why. Stewart's book taught me exactly why. In particular, it demonstrated that the polynomial \( t^5 - 6t + 3 \) over the field of rational numbers is not soluble by radicals. This particular result, although fascinating, is just a small part of a much larger body of work, which is even more remarkable. To arrive at this result, the book takes us through a wonderful journey that includes the theory of polynomial rings, algebraic and transcendental field extensions, impossibility proofs for ruler-and-compass constructions, the Galois correspondence and much more.

One of the most rewarding aspects of reading books like these is how they open doors to new knowledge, including things I didn't even know that I didn't know.

How does the newer math jell with or inform past or present computing, compared to much older stuff?

I don't always think explicitly about how mathematics informs computing, past or present. Often the textbooks I pick feel very challenging to me, so much so that all my energy goes into simply mastering the material. It is arduous but enjoyable. I do it purely for the fun of learning without worrying about applications.

Of course, a good portion of pure mathematics probably has no real-world applications. As G. H. Hardy famously wrote in A Mathematician's Apology:

I have never done anything 'useful'. No discovery of mine has made or is likely to make, directly or indirectly, for good or ill, the least difference to the amenity of the world.

But there is no denying that some of it does find applications. Were Hardy alive today, he might be disappointed that number theory, his favourite field of 'useless' mathematics, is now a crucial part of modern cryptography. Electronic commerce wouldn't likely exist without it.

Similarly, it is amusing how something as abstract as abstract algebra finds very concrete applications in coding theory. Concepts such as polynomial rings, finite fields and cosets of subspaces in vector spaces over finite fields play a crucial role in error-correcting codes, without which modern data transmission and storage would not be possible.

On a more personal note, some simpler areas of mathematics have been directly useful in my own work. While solving problems for businesses, information entropy, combinatorics and probability theory were crucial when I worked on gesture-based authentication about one and a half decades ago.

Similarly, when I was developing Bloom filter-based indexing and querying for a network events database, again, probability theory was crucial in determining the parameters of the Bloom filters (such as the number of hash functions, bits per filter and elements per filter) to ensure that the false positive rate remained below a certain threshold. Subsequent testing with randomly sampled network events confirmed that the observed false positive rate matched the theoretical estimate quite well. It was very satisfying to see probability theory and the real world agreeing so closely.

Beyond these specific examples, studying mathematics also influences the way I think about problems. Embarking on journeys like analytic number theory or Galois theory is humbling. There are times when I struggle to understand a small paragraph of the book and it takes me several hours (or even days) to work out the arguments in detail with pen and paper (lots of it) before I really grok them. That experience of grappling with dense reasoning teaches humility and also makes me sceptical of complex, hand-wavy logic in day-to-day programming.

Several times I have seen code that bundles too many decisions into one block of logic, where it is not obvious whether it would behave correctly in all circumstances. Explanations may sometimes be offered about why it works for reasonable inputs, but the reasoning is often not watertight. The experience of working through mathematical proofs, writing my own, making mistakes and then correcting them has taught me that if the reasoning for correctness is not clear and rigorous, something could be wrong. In my experience, once such code sees real-world usage, a bug is nearly always found.

That's why I usually insist either on simplifying the logic or on demonstrating correctness in a clear, rigorous way. Sometimes this means doing a case-by-case analysis for different types of inputs or conditions and showing that the code behaves correctly in each case. There is also a bit of an art to reducing what seem like numerous or even infinitely many cases to a small, manageable set of cases by spotting structure, such as symmetries, invariants or natural partitions of the input space. Alternatively, one can look for a simpler argument that covers all cases. These are techniques we employ routinely in mathematics and I think that kind of thinking and reasoning is quite valuable in software development too.

Read on website | #programming | #technology | #mathematics

Read on website | #web | #mathematics | #technology

Read on website | #web | #programming | #mathematics | #technology

Contents

• Introduction
• Counting Placements as the Board Grows
  • Type A Squares
  • Type B Squares
  • Type C Squares
  • Type D Squares
  • Closed Form Expression
• Counting Placements for Each Square
  • Attacking Degrees of Squares
  • From Attacking Degrees to Counting Placements
  • Closed Form Expression
• Counting Placements From Minimal Attack Sections
  • Minimal Attack Sections
  • Closed Form Expression
• References

Introduction

A knight moves two squares in one direction, then one square perpendicular to it, forming an L-shaped path. If a piece occupies the destination square, the knight captures it. If two knights are placed such that each can capture the other in a single move, then we say the knights attack each other. We want to determine the number of ways to place two identical knights on an \( n \times n \) chessboard so that they attack each other.

The above illustration shows just one of several ways two knights can attack each other on a \( 3 \times 3 \) board. There are, in fact, a total of eight such placements, shown below.

Let \( f(n) \) denote the number of ways we can place two identical knights on an \( n \times n \) chessboard such that they attack each other, where \( n \ge 1. \)

A \( 1 \times 1 \) board has room for only one knight, so we define \( f(1) = 0. \) On a \( 2 \times 2 \) board, a knight cannot move two squares in any direction and therefore cannot attack. Therefore, \( f(2) = 0. \) To summarise, \[ f(1) = f(2) = 0. \] From the illustration above, we see that \( f(3) = 8. \) We want to find a closed-form expression for \( f(n). \)

We will analyse this problem from various perspectives. We begin with a couple of needlessly complicated approaches, followed by a simple and elegant solution. While I personally enjoy these long-winded explorations, if you prefer a more direct solution, please skip ahead to Counting Placements From Minimal Attack Sections.

Before we proceed, let us introduce the term mutually attacking knight placement to mean a placement of two knights on the chessboard such that they attack each other. Unless stated otherwise, the two knights are identical. This term will serve as a convenient shorthand for referring to such placements.

Counting Placements as the Board Grows

We now turn to the needlessly complicated solution promised in the previous section. We analyse the new mutually attacking knight placements introduced when an existing board is enlarged by adding a row and a column.

Let us define \[ \Delta f(n) = f(n) - f(n - 1) \] for \( n \ge 2, \) so that \( \Delta f(n) \) denotes the new mutually attacking knight placements introduced when an \( (n - 1) \times (n - 1) \) board is expanded to size \( n \times n \) by adding one row and one column.

For brevity, we will avoid restating the process of enlarging an \( (n - 1) \times (n - 1) \) board to an \( n \times n \) board by adding one row and one column whenever we refer to new placements. Instead, we use the term new placements on an \( n \times n \) board to refer to \( \Delta f(n). \) It is to be understood that these new placements are the mutually attacking knight placements introduced by enlarging the board from size \( (n - 1) \times (n - 1) \) to \( n \times n. \)

Without loss of generality, suppose the new row and column are added to the bottom and right respectively. We already know that \begin{align*} \Delta f(2) & = f(2) - f(1) = 0 - 0 = 0, \\ \Delta f(3) & = f(3) - f(2) = 8 - 0 = 8. \\ \end{align*} We will now find \( \Delta f(n) \) for \( n \ge 4. \) To do this, we first categorise the newly added squares due to board expansion, into four types, as illustrated below.

Here is a brief description of each square type:

• Type A squares are the three new corner squares.
• Type B squares are the two new squares adjacent to type A squares at the top and left edges.
• Type C squares are the new squares that are not adjacent to any type A square. If the new board has dimensions \( n \times n, \) where \( n \ge 4, \) then there are exactly \( 2n - 8 \) squares of type C.
• Type D squares are the two new squares adjacent to the bottom-right type A square.

We now calculate how many new mutually attacking knight placements are introduced by these additional squares as the board expands. We proceed with a case-by-case analysis for each square type.

Type A Squares

There are three squares of type A. If we place one knight on a type A square, there are two positions for the second knight such that the two knights attack each other.

Since there are three such squares, we get a total of \( 3 \times 2 = 6 \) new mutually attacking knight placements.

Type B Squares

There are two squares of type B. If we place one knight on a type B square, there are three positions for the second knight such that the two knights attack each other.

Since there are two such squares, we get a total of \( 2 \times 3 = 6 \) new mutually attacking knight placements.

Type C Squares

The number of type C squares depends on the board size. When we increase the size of a board from \( (n - 1) \times (n - 1) \) to \(n \times n, \) where \( n \ge 4, \) we add \( n^2 - (n - 1)^2 = 2n - 1 \) new squares. Among these, \( 3 \) are of type A, \( 2 \) are of type B and \( 2 \) are of type D. That gives us a total of \( 7 \) squares of type A, B or D. The remaining \( 2n - 1 - 7 = 2n - 8 \) squares are therefore of type C. Note that when the board size increases from \( 3 \times 3 \) to \( 4 \times 4, \) there are \( 2 \times 4 - 8 = 0 \) squares of type C.

However, for a board of size \( 5 \times 5 \) or greater, there is a positive number of type C squares since \( 2n - 8 \gt 0 \) if and only if \( n \gt 4. \)

If we place one knight on a type C square, there are four positions for the second knight such that the two knights attack each other.

Since there are \( 2n - 8 \) such squares, we get a total of \( 4(2n - 8) = 8(n - 4) \) new mutually attacking knight placements.

Type D Squares

There are two squares of type D. As with type B squares, placing one knight on a type D square yields three positions for the second knight such that the two knights attack each other. This gives \( 2 \times 3 = 6 \) potentially new placements.

However, unlike type B squares, not all of these placements are new. The two placements where one knight is on the right edge and the other on the bottom edge were already counted in a previous subsection.

For example, when we increase the board size from \( 3 \times 3 \) to \( 4 \times 4, \) both the placements described in the previous paragraph appear while analysing the placements with a knight on a type B square. More generally, for any board of size \( n \times n \) with \( n \ge 5, \) these placements occur while analysing the placements with a knight on a type C square. Therefore the total number of new mutually attacking knight placements is \( 2 \times 3 - 2 = 4. \)

Another way to describe this result is to observe that when one knight is placed on a type D square, only two positions for the second knight yield new mutually attacking knight placements. Since there are two type \( D \) squares, we get a total of \( 2 \times 2 = 4 \) new mutually attacking knight placements.

Closed Form Expression

If we add the number of new mutually attacking knight placements found in each of the cases above, we get \[ \Delta f(n) = 6 + 6 + 8(n - 4) + 4 = 8(n - 2) \] new mutually attacking knight placements as the board size increases from \( (n - 1) \times (n - 1) \) to \( n \times n, \) where \( n \ge 4. \) We already know that \( \Delta f(2) = 0 \) and \( \Delta f(3) = 8. \) Surprisingly, the above formula produces the correct values for those cases as well. Therefore, we can generalise this result as \[ \Delta f(n) = 8(n - 2) \] for all \( n \ge 2. \) We can now calculate \( f(n) \) for \( n \ge 1 \) as follows: \begin{align*} f(n) & = \sum_{k = 1}^n f(k) - \sum_{k = 1}^{n - 1} f(k) \\ & = \sum_{k = 1}^n f(k) - \sum_{k = 2}^n f(k - 1) \\ & = f(1) + \sum_{k = 2}^n (f(k) - f(k - 1)) \\ & = f(1) + \sum_{k = 2}^n \Delta f(k) \\ & = 0 + \sum_{k = 2}^n 8(k - 2) \\ & = 8 \sum_{k = 0}^{n - 2} k \\ & = \frac{8(n - 2)(n - 1)}{2} \\ & = 4(n - 1)(n - 2). \end{align*} To summarise, we now have a closed form expression for \( f(n). \) For all \( n \ge 1, \) we have \[ f(n) = 4(n - 1)(n - 2). \]

Counting Placements for Each Square

The previous section took a long-winded path to arrive at a closed form expression for \( f(n). \) In this section, we will reach the same result that is still a bit drawn out, but not quite as much as before.

This time, instead of looking only at the new squares created when the board grows, we consider every square on the board. To make the counting easier, we no longer treat the knights as identical. We first work with two distinct knights, count the mutually attacking knight placements and then divide the total by \( 2 \) to get the result for identical knights.

Attacking Degrees of Squares

Here, we introduce the term attacking degree of a square to mean the number of squares a knight can move to from that square in a single move. In other words, the attacking degree of a square is the number of squares that would be attacked if a knight were placed on it. For example, the corner squares have an attacking degree of \( 2. \)

Let us now label each square with its attacking degree. A \( 1 \times 1 \) board has only one square of attacking degree \( 0 \) since a knight placed on it has nothing to attack. Similarly, each square of a \( 2 \times 2 \) board has attacking degree \( 0 \) too.

On a \( 3 \times 3 \) board, all squares have attacking degree \( 2 \) except the centre square, whose attacking degree is \( 0. \) In other words, placing a knight on any square other than the middle one gives exactly two possible positions for the other knight so that they attack each other.

With eight such squares, we get \( 8 \times 2 = 16 \) mutually attacking knight placements when the two knights are distinct. If we divide this number by \( 2, \) we get \( 8 \) which is indeed the number of mutually attacking knight placements on a \( 3 \times 3 \) board when the two knights are identical. This matches the earlier result \( f(3) = 8. \)

From Attacking Degrees to Counting Placements

Let \( g(n) \) be the number of mutually attacking knight placements on an \( n \times n \) board when the knights are distinct. Then \( g(n) \) is simply the sum of the attacking degrees of all squares on the board. As before, let \( f(n) \) denote the number of mutually attacking knight placements on an \( n \times n \) board when the two knights are identical. We will now show that \( f(n) = g(n)/2. \)

Label all squares of the \( n \times n \) board as \( S_1, S_2, \dots, S_{n^2} \) in any fixed order. Label the two distinct knights as \( N_1 \) and \( N_2. \) We represent each mutually attacking knight placement as an ordered pair \( (S_i, S_j) \) if \( N_1 \) is on \( S_i \) and \( N_2 \) is on \( S_j, \) with the two knights attacking each other. Here \( 1 \le i, j \le n^2 \) and \( i \ne j. \)

Let \( M \) be the set of all mutually attacking knight placements for distinct knights on an \( n \times n \) board. Then \[ g(n) = \lvert M \rvert. \] If \( (S_i, S_j) \) is a mutually attacking knight placement of the distinct knights \( N_1 \) and \( N_2 \) for some \( i \) and \( j \) with \( 1 \le i, j \le n^2 \) and \( i \ne j, \) then \( (S_j, S_i) \) is also a mutually attacking knight placement, since swapping the positions of the two mutually attacking knights still yields a valid mutually attacking placement. Therefore \[ (S_i, S_j) \in M \iff (S_j, S_i) \in M. \] Each ordered placement \( (S_i, S_j) \) in \( M \) is thus paired with the ordered placement \( (S_j, S_i). \) When the knights are identical, the two arrangements are indistinguishable and count as one placement. Hence, the number of mutually attacking placements for identical knights is exactly half of the number for distinct knights, i.e. \[ f(n) = \frac{g(n)}{2}. \] The next subsection focuses on calculating \( g(n), \) from which \( f(n) \) follows immediately by the above formula.

Closed Form Expression

As noted in the previous section, the number of mutually attacking knight placements for two distinct knights on an \( n \times n \) board is simply the sum of attacking degrees of all squares on the board. If we label each square as discussed in the previous section and use the notation \( \deg(S_i) \) for the attacking degree of the square labelled \( S_i, \) where \( 1 \le i \le n^2, \) then \[ g(n) = \sum_{i=1}^{n^2} \deg(S_i). \] Recall that the attacking degree of a square is the number of squares a knight could attack if it were placed there. Earlier, we saw that on a \( 3 \times 3 \) board, all squares except the centre one have attacking degree \( 2, \) which gives \( g(3) = 8 \times 2 = 16 \) and \( f(3) = g(3)/2 = 8. \) Let us now write down the attacking degrees of all squares on a \( 4 \times 4 \) board.

From the above illustration we get \begin{align*} g(4) & = 4 \times 2 + 8 \times 3 + 4 \times 4 = 48, \\ f(4) & = g(4)/2 = 24. \end{align*} A more general pattern emerges if we consider a larger board, such as a \( 6 \times 6 \) board.

From this illustration, we get \begin{align*} g(6) & = 4 \times 2 + 8 \times 3 + 12 \times 4 + 8 \times 6 + 4 \times 8 = 160. \\ f(6) & = g(6)/2 = 80. \end{align*} Let us find a general formula now for \( n \ge 4. \) We introduce one more notation. Let \( D_k(n) \) denote the sum of the attacking degrees of all squares of attacking degree \( k \) on an \( n \times n \) board, i.e. \[ D_k(n) = \sum_{\mathclap{\deg(S_i) = k}} \deg(S_i). \] Since the only attacking degrees the squares can have are \( 2, 3, 4, 6 \) and \( 8, \) the sum of the attacking degrees of all squares can be written as \[ g(n) = D_2(n) + D_3(n) + D_4(n) + D_6(n) + D_8(n). \] There are exactly four squares of attacking degree \( 2. \) These are the corner ones. Therefore, \[ D_2(n) = 4 \times 2 = 8. \] The eight squares adjacent to the corner squares have attacking degree \( 3. \) Therefore, \[ D_3(n) = 8 \times 3 = 24. \] Let us define an inner corner square as one that shares a corner with a corner square but not an edge with it. There are four inner corner squares and each has attacking degree \( 4. \) Further, each row and column on the outer edge contains \( n - 4 \) additional squares with attacking degree \( 4. \) Therefore, \[ D_4(n) = (4 + 4(n - 4))(4) = 16(n - 3). \] Consider a row or column that contains two inner corner squares of attacking degree \( 4. \) All \( n - 4 \) squares between the inner corner squares have attacking degree \( 6. \) There are two such rows and two such columns. Therefore, \[ D_6(n) = 4(n - 4)(6) = 24(n - 4). \] We have counted the attacking degrees of all squares in the first two columns and rows as well as the last two columns and rows. We are left with \( (n - 4)^2 \) squares in the middle and they all have attacking degree \( 8. \) Therefore, \[ D_8(n) = 8(n - 4)^2. \] Therefore, \begin{align*} g(n) & = D_2(n) + D_3(n) + D_4(n) + D_6(n) + D_8(n) \\ & = 8 + 24 + 16(n - 3) + 24(n - 4) + 8(n - 4)^2 \\ & = 8(n - 1)(n - 2). \end{align*} Even though we assumed \( n \ge 4 \) while obtaining the above formula, remarkably, it gives us the correct values for \( n = 1, 2 \) and \( 3. \) The number of mutually attacking knight placements for distinct knights on an \( n \times n \) board is \( 0 \) if \( n = 1 \) or \( 2. \) It is \( 16 \) if \( n = 3. \) Indeed the above formula gives us \[ g(1) = g(2) = 0, \quad g(3) = 16. \] Therefore, we can now generalise the above result as \[ g(n) = 8(n - 1)(n - 2) \] for all \( n \ge 1. \) Therefore, for all \( n \ge 1, \) \[ f(n) = \frac{g(n)}{2} = 4(n - 1)(n - 2). \]

Counting Placements From Minimal Attack Sections

Finally, in this section, we take a look at a simple and elegant solution that arrives at the closed-form solution in a more direct manner. The analysis begins by looking at the smallest section of the board where two knights can attack each other.

Minimal Attack Sections

Consider a \( 2 \times 3 \) section of a board of size \( 3 \times 3 \) or larger. Such a section has exactly two mutually attacking knight placements.

Similarly, a \( 3 \times 2 \) section of a board also has exactly two mutually attacking knight placements.

We call these \( 2 \times 3 \) and \( 3 \times 2 \) sections the minimal attack sections of a board, since no smaller section can contain a mutually attacking knight placement.

Two distinct \( 2 \times 3 \) sections can share at most a \( 1 \times 3 \) section, which is smaller than a minimal attack section. Consequently, no mutually attacking knight placement can be common to two distinct \( 2 \times 3 \) sections of a board.

Similarly, two distinct \( 3 \times 2 \) sections can share at most a \( 3 \times 1 \) section, again too small to contain a minimal attack section. Therefore, they share no mutually attacking knight placement.

A \( 2 \times 3 \) section and a \( 3 \times 2 \) section can share at most a \( 2 \times 2 \) section, which is still smaller than a minimal attack section, so they share no mutually attacking knight placement either.

To summarise, any two minimal attack sections of the board yield distinct pairs of mutually attacking knight placements. The total number of such placements is therefore exactly twice the number of minimal attack sections on the board.

Closed Form Expression

In an \( n \times n \) board where \( n \ge 3, \) the left edge of a \( 2 \times 3 \) section can be placed in any one of the first \( n - 2 \) columns of the board. Similarly, the top edge of such a section can be placed in any one of the first \( n - 1 \) rows of the board. Therefore, the total number of distinct \( 2 \times 3 \) sections on the board is \( (n - 2)(n - 1). \)

By similar reasoning, the number of distinct \( 3 \times 2 \) sections on an \( n \times n \) board, where \( n \ge 3, \) is also \( (n - 1)(n - 2). \)

Let \( h(n) \) be the total number of minimal attack sections we can find on an \( n \times n \) board where \( n \ge 1. \) From the discussion in the previous two paragraphs, we know that \( h(n) = 2(n - 1)(n - 2) \) for \( n \ge 3. \) Further, this formula for \( h(n) \) works for \( n = 1 \) and \( n = 2 \) as well since \( h(1) = h(2) = 0 \) and indeed a \( 1 \times 1 \) board or a \( 2 \times 2 \) board is too small to contain any minimal attack sections. Therefore, for all \( n \ge 1, \) we get \[ h(n) = 2(n - 1)(n - 2). \] Since each minimal attack section yields two mutually attacking knight placements, the total number of mutually attacking knight placements on an \( n \times n \) board is \[ f(n) = 2h(n) = 4(n - 1)(n - 2) \] for all \( n \ge 1. \)

References

• Two Knights from the CSES Problem Set
• Knight Graph by Eric W. Weisstein
• OEIS Entry A033996 by N. J. A. Sloane
• OEIS Entry A172132 by Vaclav Kotesovec

Read on website | #mathematics | #puzzle

Consider the following infinite grid of numbers, where the numbers are arranged in a spiral-like manner, but the spiral reverses direction each time it reaches the edge of the grid: \begin{array}{rcrcrcrcrl} 1 & \rt & 2 & \sp & 9 & \rt & 10 & \sp & 25 & \cd \\ \sp & \sp & \dn & \sp & \up & \sp & \dn & \sp & \up & \sp \\ 4 & \lf & 3 & \sp & 8 & \sp & 11 & \sp & 24 & \cd \\ \dn & \sp & \sp & \sp & \up & \sp & \dn & \sp & \up & \sp \\ 5 & \rt & 6 & \rt & 7 & \sp & 12 & \sp & 23 & \cd \\ \sp & \sp & \sp & \sp & \sp & \sp & \dn & \sp & \up & \sp \\ 16 & \lf & 15 & \lf & 14 & \lf & 13 & \sp & 22 & \cd \\ \dn & \sp & \sp & \sp & \sp & \sp & \sp & \sp & \up & \sp \\ 17 & \rt & 18 & \rt & 19 & \rt & 20 & \rt & 21 & \cd \\ \vd & \sp & \vd & \sp & \vd & \sp & \vd & \sp & \vd & \dd \end{array} Can we find a closed-form expression that tells us the number at the \( m \)th row and \( n \)th column?

Contents

• Introduction
• Patterns on the Edges
  • Computing Edge Numbers
  • Computing All Grid Numbers
  • Closed Form Expression
• Patterns on the Diagonal
  • Computing Diagonal Numbers
  • Computing All Grid Numbers
  • Closed Form Expression
• References

Introduction

Before we explore this problem further, let us rewrite the zigzag number spiral grid in a cleaner form, omitting the arrows: \begin{array}{rrrrrl} 1 & 2 & 9 & 10 & 25 & \cd \\ 4 & 3 & 8 & 11 & 24 & \cd \\ 5 & 6 & 7 & 12 & 23 & \cd \\ 16 & 15 & 14 & 13 & 22 & \cd \\ 17 & 18 & 19 & 20 & 21 & \cd \\ \vd & \vd & \vd & \vd & \vd & \dd \end{array} Let \( f(m, n) \) denote the number at the \( m \)th row and \( n \)th column. For example, \( f(1, 1) = 1 \) and \( f(2, 5) = 24. \) We want to find a closed-form expression for \( f(m, n). \)

Let us first clarify what we mean by a closed-form expression. There is no universal definition of a closed-form expression, but the term typically refers to a mathematical expression involving variables and constants, built using a finite combination of basic operations: addition, subtraction, multiplication, division, integer exponents, roots with integer index and functions such as exponentials, logarithms and trigonometric functions.

In this article, however, we need only addition, subtraction, division, squares and square roots. This may be a bit of a spoiler, but I must mention that the \( \max \) function appears in the closed-form expressions we are about to see. If you are concerned about whether functions like \( \max \) and \( \min \) are permitted in such expressions, note that \begin{align*} \max(m, n) & = \frac{m + n + \sqrt{(m - n)^2}}{2}, \\ \min(m, n) & = \frac{m + n - \sqrt{(m - n)^2}}{2}. \end{align*} So \( \max \) and \( \min \) are simply shorthand for expressions involving addition, subtraction, division, squares and square roots. In the discussion that follows, we will use only the \( \max \) function.

Patterns on the Edges

Let us begin by analysing the edge numbers. Number the rows as \( 1, 2, 3 \dots \) and the columns likewise. Observe where the spiral touches the left edge and changes direction. This happens only on even-numbered rows. Similarly, each time the spiral touches the top edge and changes direction, it does so on odd-numbered columns. In the following subsections, we take a closer look at this behaviour of the spiral.

I should mention that this section takes a rather long path to arrive at the closed-form solution. Personally, I enjoy such long tours. If you prefer a more direct approach, feel free to skip ahead to Patterns on the Diagonal for a shorter discussion that reaches the same result.

Computing Edge Numbers

Each time the spiral reaches the left edge of the grid, it does so at some \( m \)th row where \( m \) is even. The \( m \times m \) subgrid formed by the first \( m \) rows and the first \( m \) columns contains \( m^2 \) consecutive numbers. Since the numbers strictly increase as the spiral grows, the largest of these \( m^2 \) numbers must appear at the position where the spiral touches the left edge. This is illustrated in the figure below.

Whenever the spiral touches the left edge at the \( m \)th row (where \( m \) is even), the number in the first column of that row is \( m^2. \) Hence, we conclude that \( f(m, 1) = m^2 \) when \( m \) is even. Immediately after touching the left edge, the spiral turns downwards into the first column of the next row. Thus, in the next row, i.e. in the \( (m + 1) \)th row, we have \( f(m + 1, 1) = m^2 + 1, \) where \( m + 1 \) is odd. This can be restated as \( f(m, 1) = (m - 1)^2 + 1 \) when \( m \) is odd. Since \( f(1, 1) = 1, \) we can summarise the two formulas we have found here as: \[ f(m, 1) = \begin{cases} m^2 & \text{if } m \equiv 0 \pmod{2}, \\ (m - 1)^2 + 1 & \text{if } m \equiv 1 \pmod{2}. \end{cases} \]

We can perform a similar analysis for the numbers at the top edge and note that whenever the spiral touches the top edge at the \( n \)th column (where \( n \) is odd), the number in the first row of that column is \( n^2. \) This is illustrated below.

Immediately after touching the top edge, the spiral turns right into the next column. These observations give us the following formula for the numbers at the top edge: \[ f(1, n) = \begin{cases} n^2 & \text{if } n \equiv 1 \pmod{2}, \\ (n - 1)^2 + 1 & \text{if } n \equiv 0 \pmod{2}. \end{cases} \] Next we will find a formula for any arbitrary number anywhere in the grid.

Computing All Grid Numbers

Since the spiral touches the left edge on even-numbered rows, then turns downwards into the next (odd-numbered) row and then starts moving right until the diagonal (where it changes direction again), the following two rules hold:

• On every odd-numbered row, as we go from left to right, the numbers increase until we reach the diagonal.
• On every even-numbered row, as we go from left to right, the numbers decrease until we reach the diagonal.

Note that all the numbers we considered in the above two points lie on or below the diagonal (or equivalently, on or to the left of the diagonal). Therefore, on an odd-numbered row, we can find the numbers on or below the diagonal using the formula \( f(m, n) = f(m, 1) + (n - 1), \) where \( m \) is odd. Similarly, on even-numbered rows, we can find the numbers on or below the diagonal using the formula \( f(m, n) = f(m, 1) - (n - 1), \) where \( m \) is even.

By a similar analysis, the following rules hold when we consider the numbers in a column:

• On every even-numbered column, as we go from top to bottom, the numbers increase until we reach the diagonal.
• On every odd-numbered column, as we go from top to bottom, the numbers decrease until we reach the diagonal.

Now the numbers on or above the diagonal can be found using the formula \( f(m, n) = f(1, n) - (m - 1) \) when \( n \) is odd and \( f(m, n) = f(1, n) + (m - 1), \) when \( n \) is even.

Can we determine from the values of \( m \) and \( n \) if the number \( f(m, n) \) is above the diagonal or below it? Yes, if \( m \le n, \) then \( f(m, n) \) lies on or above the diagonal. However, if \( m \ge n, \) then \( f(m, n) \) lies on or below the diagonal.

We now have everything we need to write a general formula for finding the numbers anywhere in the grid. Using the four formulas and the two inequalities obtained in this section, we get \[ f(m, n) = \begin{cases} f(1, n) + (m - 1) & \text{if } m \le n \text{ and } n \equiv 0 \pmod{2}, \\ f(1, n) - (m - 1) & \text{if } m \le n \text{ and } n \equiv 1 \pmod{2}, \\ f(m, 1) - (n - 1) & \text{if } m \ge n \text{ and } m \equiv 0 \pmod{2}, \\ f(m, 1) + (n - 1) & \text{if } m \ge n \text{ and } m \equiv 1 \pmod{2}. \\ \end{cases} \] Using the equations for \( f(1, n) \) and \( f(m, 1) \) from the previous section, the above formulas can be rewritten as \[ f(m, n) = \begin{cases} (n - 1)^2 + 1 + (m - 1) & \text{if } m \le n \text{ and } n \equiv 0 \pmod{2}, \\ n^2 - (m - 1) & \text{if } m \le n \text{ and } n \equiv 1 \pmod{2}, \\ m^2 - (n - 1) & \text{if } m \ge n \text{ and } m \equiv 0 \pmod{2}, \\ (m - 1)^2 + 1 + (n - 1) & \text{if } m \ge n \text{ and } m \equiv 1 \pmod{2}. \\ \end{cases} \] Simplifying the expressions on the right-hand side, we get \[ f(m, n) = \begin{cases} (n - 1)^2 + m & \text{if } m \le n \text{ and } n \equiv 0 \pmod{2}, \\ n^2 - m + 1 & \text{if } m \le n \text{ and } n \equiv 1 \pmod{2}, \\ m^2 - n + 1 & \text{if } m \ge n \text{ and } m \equiv 0 \pmod{2}, \\ (m - 1)^2 + n & \text{if } m \ge n \text{ and } m \equiv 1 \pmod{2}. \\ \end{cases} \] This is pretty good. We now have a piecewise formula that works for any position in the grid. Let us now explore whether we can express it as a single closed-form expression.

Closed Form Expression

First, we will rewrite the piecewise formula from the previous section in the following form: \[ f(m, n) = \begin{cases} (n^2 - n + 1) + (m - n) & \text{if } m \le n \text{ and } n \equiv 0 \pmod{2}, \\ (n^2 - n + 1) - (m - n) & \text{if } m \le n \text{ and } n \equiv 1 \pmod{2}, \\ (m^2 - m + 1) + (m - n) & \text{if } m \ge n \text{ and } m \equiv 0 \pmod{2}, \\ (m^2 - m + 1) - (m - n) & \text{if } m \ge n \text{ and } m \equiv 1 \pmod{2}. \\ \end{cases} \] This is the same formula, rewritten to reveal common patterns between the four expressions on the right-hand side. In each expression, one variable plays the dominant role, occurring several times, while the other appears only once. For example, in the first two expressions, \( n \) plays the dominant role whereas \( m \) occurs only once. If we look closely, we realise that it is the variable that is greater than or equal to the other that plays the dominant role. Therefore the first and third expressions may be written as \[ \left( (\max(m, n))^2 - \max(m, n) + 1 \right) + (m - n). \] Similarly, the second and fourth expressions may be written as \[ \left( (\max(m, n))^2 - \max(m, n) + 1 \right) - (m - n). \] We have made some progress towards a closed-form expression. We have collapsed the four expressions in the piecewise formula to just two. The only difference between them lies in the sign of the second term: it is positive when the dominant variable is even and negative when it is odd. This observation allows us to unify both cases into a single expression: \[ f(m, n) = (\max(m, n))^2 - \max(m, n) + 1 + (-1)^{\max(m, n)} (m - n). \] Now we have a closed-form expression for \( f(m, n) \) that gives the number at any position in the grid.

Patterns on the Diagonal

As mentioned earlier, there is a shorter route to the same closed-form expression. This alternative approach is based on analysing the numbers along the diagonal of the grid. We still need to examine the edge numbers, but not all of them as we did in the previous section. Some of the reasoning about edge values will be repeated here to ensure this section is self-contained.

Computing Diagonal Numbers

A number on the diagonal has the same row number and column number. In other words, a diagonal number has the value \( f(n, n) \) for some positive integer \( n. \) Consider the case when \( n \) is even. In this case, the diagonal number is on a segment of the spiral that is moving to the left. The \( n \times n \) subgrid formed by the first \( n \) rows and the first \( n \) columns contains exactly \( n^2 \) consecutive numbers. Since the diagonal number is on the last row of this subgrid and the numbers in this row increase as we move from right to left, the largest number in the subgrid must be on the left edge of this row. Therefore the number at the left edge is \( f(n, 1) = n^2, \) where \( n \) is even. This is illustrated below.

From the diagonal to the edge of the subgrid, there are \( n \) consecutive numbers. In a sequence of \( n \) consecutive numbers, the difference between the maximum number and the minimum number is \( n - 1. \) Therefore, \( n^2 - f(n, n) = n - 1. \) This gives us \[ f(n, n) = n^2 - n + 1 \quad \text{if } n \equiv 0 \pmod{2}. \]

Now consider the case when \( n \) is odd.

By a similar reasoning, for odd \( n, \) the \( n \)th column has numbers that increase as we move up from the diagonal number towards the top edge. Therefore \( f(1, n) = n^2 \) and since \( n^2 - f(n, n) = n - 1, \) we again obtain \[ f(n, n) = n^2 - n + 1 \quad \text{if } n \equiv 1 \pmod{2}. \] Since \( f(n, n) \) takes the same form for both odd and even \( n, \) we can write \[ f(n, n) = n^2 - n + 1 \] for all positive integers \( n. \)

Computing All Grid Numbers

If \( m \le n, \) then the number \( f(m, n) \) lies on or above the diagonal number \( f(n, n). \) If \( n \) is even, then the numbers decrease as we go from the diagonal up to the top edge. Therefore \( f(m, n) \le f(n, n) \) and \( f(m, n) = f(n, n) - (n - m). \) If \( n \) is odd, then the numbers increase as we go from the diagonal up to the top edge and therefore \( f(m, n) \ge f(n, n) \) and \( f(m, n) = f(n, n) + (n - m). \)

If \( m \ge n, \) then the number \( f(m, n) \) lies on or below the diagonal number \( f(m, m). \) By a similar analysis, we find that \( f(m, n) = f(m, m) + (m - n) \) if \( n \) is even and \( f(m, n) = f(m, m) - (m - n) \) if \( n \) is odd. We summarise these results as follows: \[ f(m, n) = \begin{cases} f(n, n) - (n - m) & \text{if } m \le n \text{ and } n \equiv 0 \pmod{2}, \\ f(n, n) + (n - m) & \text{if } m \le n \text{ and } n \equiv 1 \pmod{2}, \\ f(m, m) + (m - n) & \text{if } m \ge n \text{ and } m \equiv 0 \pmod{2}, \\ f(m, m) - (m - n) & \text{if } m \ge n \text{ and } m \equiv 1 \pmod{2}. \\ \end{cases} \] Note that the above formula can be rewritten as \[ f(m, n) = \begin{cases} f(n, n) + (m - n) & \text{if } m \le n \text{ and } n \equiv 0 \pmod{2}, \\ f(n, n) - (m - n) & \text{if } m \le n \text{ and } n \equiv 1 \pmod{2}, \\ f(m, m) + (m - n) & \text{if } m \ge n \text{ and } m \equiv 0 \pmod{2}, \\ f(m, m) - (m - n) & \text{if } m \ge n \text{ and } m \equiv 1 \pmod{2}. \\ \end{cases} \]

Closed Form Expression

If we take a close look at the last formula in the previous section, we find that in each expression, one variable plays a dominant role, i.e. it occurs more frequently in the expression than the other. In the first two expressions \( n \) plays the dominant role whereas in the last two expressions \( m \) plays the dominant role. In fact, in each expression, the dominant variable is the one that is greater than or equal to the other. With this in mind, we can rewrite the above formula as \[ f(m, n) = \begin{cases} f(\max(m, n), \max(m, n)) + (m - n) & \text{if } \max(m, n) \equiv 0 \pmod{2}, \\ f(\max(m, n), \max(m, n)) - (m - n) & \text{if } \max(m, n) \equiv 1 \pmod{2}. \\ \end{cases} \] The only difference between the expressions is the sign of the second term: it is positive when \( \max(m, n) \) is even and negative when \( \max(m, n) \) is odd. As a result, we can rewrite the above formula as a single expression like this: \[ f(m, n) = f(\max(m, n), \max(m, n)) + (-1)^{\max(m, n)} (m - n). \] Using the formula \( f(n, n) = n^2 - n + 1 \) from the previous section, we get \[ f(m, n) = (\max(m, n))^2 - \max(m, n) + 1 + (-1)^{\max(m, n)} (m - n). \] We arrive again at the same closed-form expression, this time by focusing on the diagonal of the grid.

References

• Number Spiral from the CSES Problem Set
• Closed-Form Solution by Christopher Stover and Eric W. Weisstein
• Piecewise Function by Eric W. Weisstein

Read on website | #mathematics | #puzzle

Contents

• Illustration
• Ring Axioms
• Closure Properties
• Inverse of Inverse
• Multiplication by Zero
• Multiplication by Additive Inverse
• Product of Additive Inverses
• Alternate Proof
• Conclusion

Illustration

Let us begin with a quick illustration that shows why the product of two negative numbers must be positive for arithmetic to make sense. Consider \[ 7 \times 8 = 56. \] The above equation can also be written as \[ (10 - 3) \times (10 - 2) = 56. \] Using the distributive property of multiplication over subtraction, we get \[ (10 - 3) \times 10 + (10 - 3) \times (-2) = 56. \] Using the distributive property again, we have \[ 10 \times 10 + (-3) \times 10 + 10 \times (-2) + (-3) \times (-2) = 56. \] Now, we will take it for granted that a positive times a negative is negative. We will prove all of this rigorously later, but for now, we are just working through an illustration, so we will accept that rule and see where it leads. The equation becomes: \[ 100 + (-30) + (-20) + (-3) \times (-2) = 56. \] Adding the first three terms gives \[ 50 + (-3) \times (-2) = 56. \] Subtracting \( 50 \) from both sides, we get \[ (-3) \times (-2) = 6. \] What we have seen here is that if we accept \( 7 \times 8 = 56 \) and that positive times negative gives a negative result, then we must also accept that \( (-3) \times (-2) = 6. \)

Ring Axioms

From this section onwards, we take a rigorous approach. We want to show that the rule 'negative times negative equals positive' holds, in a general sense, for any set of elements that share certain properties with numbers. As it turns out, these elements do not need to possess all the properties of complex numbers, real numbers or even rational numbers. In fact, if they satisfy a small and specific set of properties held by the integers, then the rule still holds. These properties are known as the ring axioms.

A ring is an algebraic structure consisting of a set \( R \) with two binary operations \( + \) and \( \cdot, \) called addition and multiplication respectively, satisfying the following axioms:

1. Associativity of addition: For all \( a, b, c \in R, \) we have \( a + (b + c) = (a + b) + c. \)

2. Commutativity of addition: For all \( a, b \in R, \) we have \( a + b = b + a. \)

3. Additive identity: There exists an element \( 0 \in R \) such that for all \( a \in R, \) we have \( a + 0 = a = 0 + a. \)

4. Additive inverse: For each \( a \in R, \) there exists an element \( -a \in R \) such that \( a + (-a) = 0 = (-a) + a. \)

5. Associativity of multiplication: For all \( a, b, c \in R, \) we have \( a \cdot (b \cdot c) = (a \cdot b) \cdot c. \)

6. Left distributivity of multiplication over addition: For all \( a, b, c \in R, \) we have \( a \cdot (b + c) = (a \cdot b) + (a \cdot c). \)

7. Right distributivity of multiplication over addition: For all \( a, b, c \in R, \) we have \( (b + c) \cdot a = (b \cdot a) + (c \cdot a). \)

Note that we do not assume that the ring contains multiplicative identity, nor do we assume that multiplication is commutative. Many familiar types of numbers form rings. For example, the set of integers forms a ring with the usual addition and multiplication operations. The sets of rational numbers, real numbers and complex numbers satisfy the ring axioms too.

Rings need not consist of numbers; they may contain elements of any type. As long as a set of elements, together with suitable addition and multiplication operations, satisfies the seven axioms above, it forms a ring. For example, the set of all polynomials in the indeterminate \( t \) with coefficients in some ring \( R \) forms a ring under the usual addition and multiplication of polynomials. Such a ring is called a polynomial ring and it is denoted \( R[t]. \)

Closure Properties

Some texts include the following additional axioms for the closure properties of a ring:

1. Closure under addition: For all \( a, b \in R, \) we have \( a + b \in R. \)

2. Closure under multiplication: For all \( a, b \in R, \) we have \( a \cdot b \in R. \)

However, stating these axioms explicitly is usually considered redundant because a binary operation is closed by definition. A binary operation \( \circ \) on a set \( M \) is defined to be a function \[ \circ : M \times M \to M; \quad (a, b) \mapsto a \circ b. \] This definition automatically implies the closure property, since the domain and codomain are the same. The addition and multiplication operations on a ring \( R \) may be defined as \begin{align*} + &: R \times R \to R; \quad (a, b) \mapsto a + b, \\ \cdot &: R \times R \to R; \quad (a, b) \mapsto a \cdot b. \end{align*} These definitions imply that a ring is closed under addition and multiplication. In practice, while deciding if some set \( R \) forms a ring, we should always verify that the addition and multiplication operations indeed have \( R \) as the codomain to confirm that the closure property holds.

Inverse of Inverse

Theorem 1. Let \( R \) be a ring with \( + \) and \( \cdot \) operations. Then for all \( a \in R, \) we have \[ -(-a) = a. \]

Proof. This result follows directly from the additive inverse axiom. First, observe that \[ a + (-a) = 0. \] Therefore \( a \) is an additive inverse of \( -a, \) i.e. \[ -(-a) = a. \] This completes the proof.

Notice that this proof does not involve the multiplication operation of a ring at all. In fact, it holds true in a more general algebraic structure known as a group, which requires only a binary operation with associativity, an identity element and inverses. A ring, under addition, is also a group. Since the proof relies solely on these additive group properties, this theorem holds for all groups. However, for brevity and to avoid introducing group axioms separately, I have stated and proved this theorem in the context of rings.

It is also worth noting that the additive identity is unique in a ring (as well as in any group), but since this fact is not needed for later results, its proof has been omitted. Even if, hypothetically, there were two distinct additive identities, \( 0 \) and \( 0', \) in a ring (there are not, of course), the arguments below would still hold if we simply focus on \( 0. \)

Multiplication by Zero

Theorem 2. Let \( R \) be a ring with \( + \) and \( \cdot \) operations. Then for all \( a \in R, \) we have \[ a \cdot 0 = 0 \cdot a = 0. \]

Proof. Using the additive identity axiom, we get \[ 0 + 0 = 0. \] Multiplying both sides on the left by \( a, \) we get \[ a \cdot (0 + 0) = a \cdot 0. \] Using the left distributivity axiom, we get \[ a \cdot 0 + a \cdot 0 = a \cdot 0. \] Let \( b = a \cdot 0. \) Then \[ b + b = b. \] Since a ring is closed under multiplication, \( b \in R. \) By the additive inverse axiom, there exists \( -b \in R \) such that \( b + (-b) = 0. \) Adding \( -b \) to both sides of the above equation, we get \[ (b + b) + (-b) = b + (-b). \] By associativity of addition in a ring, we get \[ b + (b + (-b)) = b + (-b). \] Since \( b + (-b) = 0, \) the above equation becomes \[ b + 0 = 0. \] By the additive identity axiom, we get \[ b = 0. \] Since \( b = a \cdot 0, \) the above equation may be written as \[ a \cdot 0 = 0. \] A similar argument shows that \[ 0 \cdot a = 0. \] This completes the proof.

Multiplication by Additive Inverse

Theorem 3. Let \( R \) be a ring with \( + \) and \( \cdot \) operations. Then for all \( a, b \in R, \) we have \begin{align*} a \cdot (-b) &= -(a \cdot b), \\ (-a) \cdot b &= -(a \cdot b). \end{align*}

Proof. Using the left distributivity and additive inverse properties of a ring along with Theorem 2, we get \[ a \cdot b + a \cdot (-b) = a \cdot (b + (-b)) = a \cdot 0 = 0. \] Therefore \( a \cdot (-b) \) is an additive inverse of \( a \cdot b , \) i.e. \[ -(a \cdot b) = a \cdot (-b). \] Similarly \[ a \cdot b + (-a) \cdot b = (a + (-a)) \cdot b = 0 \cdot b = 0 \] and thus \[ -(a \cdot b) = (-a) \cdot b. \] This completes the proof.

Product of Additive Inverses

Theorem 4. Let \( R \) be a ring with \( + \) and \( \cdot \) operations. Then for all \( a, b \in R, \) we have \[ (-a) \cdot (-b) = a \cdot b. \]

Proof. From Theorem 3, we know that \[ a \cdot (-b) = -(a \cdot b). \] Substituting \( a \) with \( -a, \) we get \[ (-a) \cdot (-b) = -((-a) \cdot b). \] Again by Theorem 3, we have \( (-a) \cdot b = -(a \cdot b). \) Substituting this in the above equation, we obtain \[ (-a) \cdot (-b) = -(-(a \cdot b)). \] Now using Theorem 1, the right-hand side becomes \( a \cdot b, \) so we get \[ (-a) \cdot (-b) = a \cdot b. \] This completes the proof.

Alternate Proof

The above sequence of theorems is not the only way to arrive at Theorem 4. There are other ways to reach this result as well. Let us briefly discuss another such proof. From Theorem 3 we know that \( a \cdot b \) is the additive inverse of \( a \cdot (-b). \) In a very similar way, we can show that \( (-a) \cdot (-b) \) is also the additive inverse of \( a \cdot (-b). \) The proof goes as follows: \[ (-a) \cdot (-b) + a \cdot (-b) = (-a + a) \cdot (-b) = 0 \cdot (-b) = 0. \] Note that we used Theorem 2 again for the last equality. So now we know that both \( (-a) \cdot (-b) \) and \( a \cdot b \) are additive inverses of \( a \cdot (-b). \) Does this mean that \( (-a) \cdot (-b) = a \cdot b? \) Yes, since the additive inverse of an element is unique in a ring. Let us prove this now. Let \( b \) and \( c \) be additive inverses of \( a. \) Then \( a + b = b + a = 0 \) and \( a + c = c + a = 0. \) Using this, we get \[ b = b + 0 = b + (a + c) = (b + a) + c = 0 + c = c. \] Since \( (-a) \cdot (-b) \) and \( a \cdot b \) are additive inverses of the same element \(a \cdot (-b) \) and since the additive inverse of an element is unique in a ring, it follows that \[ (-a) \cdot (-b) = a \cdot b. \] Note that we do not need Theorem 1 in this alternate proof but we introduced a new theorem about the uniqueness of the additive inverse to complete this proof.

Conclusion

Theorems 1 to 4 establish certain algebraic properties that hold in any ring. Although these results were proven abstractly for rings, they reflect properties we are already familiar with from our experience with numbers. For example, in the ring of integers, we observe \( -(-2) = 2 \) which is a specific case of Theorem 1.

Similarly, Theorem 2 confirms the well-known fact that multiplying any integer by \( 0 \) yields \( 0. \) For example, \( 2 \cdot 0 = 0. \)

Then Theorem 3 implies the rule that multiplying a positive number by a negative number yields a negative result. For example, \( 2 \cdot (-3) = -(2 \cdot 3) = -6. \)

Finally, Theorem 4 implies that the product of two negative numbers is positive. For example, \( (-2) \cdot (-3) = 2 \cdot 3 = 6. \)

These familiar results are not limited to the ring of integers. The results hold in any ring, including polynomial rings, rings of integers modulo a fixed positive integer and many other algebraic systems. These results demonstrate how the ring axioms formalise familiar arithmetic rules within a more general algebraic framework.

Read on website | #mathematics

Contents

• Definition of Ideals
• Examples of Ideals
• Known Results
• Ideals of Fields
• Rings With Trivial Ideals
• Conclusion

Definition of Ideals

A left ideal of a ring \( R \) is a subset \( I \subseteq R \) such that \( I \) is an additive subgroup of \( R \) and for all \( a \in I \) and \( r \in R, \) we have \( r \cdot a \in I. \) We say that a left ideal absorbs multiplication from the left by any ring element; or equivalently, that it is closed under left multiplication by any ring element.

Similarly, a right ideal of a ring \( R \) is a subset \( I \subseteq R \) such that \( I \) is an additive subgroup of \( R \) and for all \( a \in I \) and \( r \in R, \) we have \( a \cdot r \in I. \) We say that a right ideal absorbs multiplication from the right by any ring element; or equivalently, that it is closed under right multiplication by any ring element.

In a commutative ring \( R, \) every left ideal is also a right ideal and vice versa. This is because for all \( a \in I \) and \( r \in R, \) we have \( r \cdot a = a \cdot r. \) Therefore, when working with commutative rings, we do not need to distinguish between left and right ideals and we simply refer to them as ideals. In this case, the ideal is said to absorb multiplication by any ring element; or equivalently, it is said to be closed under multiplication by any ring element.

Examples of Ideals

Consider the set of even integers \[ \langle 2 \rangle = \{ 2n : n \in \mathbb{Z} \}. \] This is an ideal of \( \mathbb{Z}. \) Indeed, if we multiply any even integer by any integer, the result is an even integer. In other words, the set of even integers absorbs multiplication by any integer. Equivalently, the set of even integers is closed under multiplication by any integer.

Let us see another example. Consider the ring of polynomials in the indeterminate \( t \) with integer coefficients, denoted \( \mathbb{Z}[t]. \) The set \[ \langle 2, t \rangle = \{ 2f + tg : f, g \in \mathbb{Z}[t] \} \] is an ideal of \( \mathbb{Z}[t]. \) Every element of this ideal is a linear combination of \( 2 \) and \( t \) with polynomial coefficients. If we take any element \( 2f + tg \in \langle 2, t \rangle \) where \( f, g \in \mathbb{Z}[t] \) and multiply it by any polynomial \( h \in \mathbb{Z}[t], \) we obtain \( 2fh + tgh, \) which is again an element of \( \langle 2, t \rangle. \) Hence \( \langle 2, t \rangle \) absorbs multiplication by any element of \( \mathbb{Z}[t], \) i.e. it is closed under multiplication by elements of \( \mathbb{Z}[t]. \)

Known Results

For the sake of brevity, we assume the following standard results.

Proposition 1. Let \( R \) be a ring. Then, for all \( a \in R, \) we have \[ a \cdot 0 = 0 \cdot a = 0. \]

Proposition 2. Let \( R \) be a ring and let \( a \in R. \) Then \begin{align*} I_L &= \{ r \cdot a : r \in R \}, \\ I_R &= \{ a \cdot r : r \in R \} \end{align*} are respectively a left ideal and a right ideal of \( R. \) If \( R \) is commutative, then \( I_L = I_R \) and we write \[ \langle a \rangle = \{ a \cdot r : r \in R \} \] and say that \( \langle a \rangle \) is an ideal of \( R \) generated by \( a. \)

Ideals of Fields

In this section, we show that a field \( K \) has only two ideals: \( \{ 0 \} \) and \( K \) itself.

Clearly \( \{ 0 \} \) is an ideal of \( K, \) as it satisfies the definition of an ideal. It is the trivial additive subgroup of \( K \) and by Proposition 1, for all \( r \in K, \) we have \( r \cdot 0 = 0 \in \{ 0 \}. \)

Now \( K \) is also an ideal of itself. Since \( K \) is an additive group by the definition of a field, it is an additive subgroup of itself. Moreover, as a field, \( K \) is closed under multiplication, so for all \( a, r \in K \) we have \( a \cdot r \in K. \)

We will now show that \( \{ 0 \} \) and \( K \) are the only ideals of \( K. \) Let \( I \) be an ideal of \( K. \) There are two cases to consider: \( I = \{ 0 \} \) and \( I \ne \{ 0 \}. \) Suppose \( I \ne \{ 0 \}. \) Then there exists a non-zero element \( b \in I. \) Since \( b \ne 0 \) and \( K \) is a field, \( b \) has a multiplicative inverse \( b^{-1} \in K. \) Since \( b \in I, \) \( b^{-1} \in K \) and \( I \) is closed under multiplication by any element of \( K, \) we have \[ 1 = b \cdot b^{-1} \in I. \] Now, let \( c \in K. \) Since \( 1 \in I, \) \( c \in K \) and \( I \) is an ideal of \( K, \) we get \[ c = 1 \cdot c \in I. \] Thus \( K \subseteq I \) and since \( I \subseteq K \) by definition, we conclude \( I = K. \) Therefore the only ideals of \( K \) are \( \{ 0 \} \) and \( K \) itself.

Rings With Trivial Ideals

We now show that if \( R \) is a commutative ring with \( 1 \ne 0 \) and the only ideals of \( R \) are \( \{ 0 \} \) and \( R \) itself, then \( R \) must be a field. To do this, we first show that every non-zero element of \( R \) has a multiplicative inverse in \( R. \) Let \( a \in R \) with \( a \ne 0. \) We now show that there exists a multiplicative inverse \( a^{-1} \in R. \) By Proposition 2, the set \[ \langle a \rangle = \{ a \cdot r : r \in R \}. \] is an ideal of \( R. \) Since \( a = a \cdot 1 \in \langle a \rangle, \) we have \( \langle a \rangle \ne \{ 0 \}. \) By assumption, the only ideals of \( R \) are \( \{ 0 \} \) and \( R, \) so it must be that \( \langle a \rangle = R. \) Therefore \( 1 \in \langle a \rangle \) and \[ 1 = a \cdot s \] for some \( s \in R. \) Thus \( a \) has a multiplicative inverse \( s \in R \) and this holds for every non-zero \( a \in R. \)

The remaining properties of fields, namely, associativity and commutativity of addition and multiplication, the existence of distinct additive and multiplicative identities, the existence of additive inverses and the distributivity of multiplication over addition, are inherited from the ring \( R. \) Therefore \( R \) is a field.

Conclusion

To summarise, any commutative ring with distinct additive and multiplicative identities that has only trivial ideals is a field and every field has only trivial ideals.

Note that every field is also a commutative ring with distinct additive and multiplicative identities. Therefore, we can say that every field is a commutative ring with distinct additive and multiplicative identities and only trivial ideals and vice versa. It is neat how the two facts align so nicely.

Read on website | #mathematics

Contents

• Definition of Integral Domain
• Examples of Integral Domains
• Known Results
• On Distinct Identities
• Every Field Is an Integral Domain
• Infinite Integral Domains
• Every Finite Integral Domain Is a Field
  • Alternate Proof
• Conclusion

Definition of Integral Domain

An integral domain is a commutative ring, with distinct additive and multiplicative identities, in which the product of any two non-zero elements is also non-zero.

Equivalently, an integral domain is a commutative ring, with distinct additive and multiplicative identities, such that if the product of two elements is zero, then one of the elements must be zero.

Using standard notation, we can write that a commutative ring \( R \) is an integral domain if \( 0 \ne 1 \) and for \( a, b \in R, \) \[ a \ne 0 \text{ and } b \ne 0 \implies a \cdot b \ne 0 \] or equivalently, \[ a \cdot b = 0 \implies a = 0 \text{ or } b = 0. \] There are many other alternative ways to define an integral domain which are all equivalent. In a ring \( R, \) a zero divisor is a non-zero element \( a \in R \) such that there exists a non-zero element \( b \in R \) with \( ab = 0. \) With this definition of a zero divisor, we can define an integral domain to be a unital commutative ring, with \( 0 \ne 1, \) that has no zero divisors.

Examples of Integral Domains

The ring of integers \( \mathbb{Z} \) is an integral domain since the product of two non-zero integers is non-zero. The field of rational numbers \( \mathbb{Q} \) is also an integral domain. The ring of polynomials in the indeterminate \( t \) with coefficients in an integral domain \( R, \) denoted \( R[t], \) is an integral domain as well.

The ring of integers modulo 5, denoted \( \mathbb{Z}_5 \) is an integral domain. However, the ring of integers modulo 6, denoted \( \mathbb{Z}_6, \) is not an integral domain since \( 2 \cdot 3 = 0 \) in \( \mathbb{Z}_6. \) In other words, \( \mathbb{Z}_6 \) has zero divisors, namely \( 2 \) and \( 3, \) so it is not an integral domain. In fact, the ring of integers modulo \( n, \) denoted \( \mathbb{Z}_n \) is an integral domain if and only if \( n \) is prime.

Known Results

For the sake of brevity, we assume the following known results.

Proposition 1. Let \( R \) be a ring. Then, for all \( a \in R, \) we have \[ a \cdot 0 = 0 \cdot a = 0. \]

Proposition 2. Let \( D \) be an integral domain. Then, for all \( a, b, c \in D \) such that \( a \ne 0, \) we have \[ a \cdot b = a \cdot c \implies b = c. \]

The second result is also known as the cancellation property of integral domains.

On Distinct Identities

We have been mentioning the distinctiveness of the additive and multiplicative identities as a property of an integral domain. Some texts express this more concisely by saying that an integral domain is a non-zero unital commutative ring without zero divisors, i.e. the zero ring \( \{ 0 \} \) is excluded from the definition.

This follows directly from Proposition 1. If \( 0 = 1 \in R, \) then for all \( r \in R, \) we get \[ r = r \cdot 1 = r \cdot 0 = 0 \] which means that every element of \( R \) is zero, i.e. \( R = \{ 0 \}. \) To summarise \[ 0 = 1 \in R \implies R = \{ 0 \} \] or equivalently, for a ring \( R \) with unity, \[ R \ne \{ 0 \} \implies 0 \ne 1. \] Further, if \( 0 \) and \( 1 \) are two distinct elements of \( R, \) then \( R \) has at least two elements, so for a ring \( R \) with unity, \[ 0 \ne 1 \implies R \ne \{ 0 \} \] Therefore, a ring with unity has distinct additive and multiplicative identities if and only if it is a non-zero ring. This is why an integral domain can also be defined as a non-zero unital commutative ring without zero divisors.

Every Field Is an Integral Domain

We now show that every field is indeed an integral domain. Let \(F \) be a field and let \( a, b \in F \) such that \( ab = 0. \) There are two cases to consider: \( a = 0 \) and \( a \ne 0. \) If \( a = 0, \) we are done.

Now suppose \( a \ne 0. \) Then by the properties of fields, there exists a multiplicative inverse \( a^{-1} \in F \) such that \( a \cdot a^{-1} = 1. \) Then using the properties of fields, we get \[ b = b \cdot 1 = b \cdot (a \cdot a^{-1}) = (a \cdot b) \cdot a^{-1} = 0 \cdot a^{-1} = 0. \] The last equality follows from Proposition 1. We have shown that if \( ab = 0, \) then either \( a = 0 \) or \( b = 0. \) Therefore, if both \( a \ne 0 \) and \( b \ne 0, \) then it must be that \( ab \ne 0. \) Therefore \( F \) is an integral domain.

Infinite Integral Domains

Now we arrive at the next natural question. Is every integral domain a field?

The ring of integers \( \mathbb{Z} \) is an integral domain but it is not a field since \( 2 \in \mathbb{Z}, \) but \( 2^{-1} \notin \mathbb{Z}. \) Therefore, \( \mathbb{Z} \) is an example of an infinite integral domain that is not a field.

Next we ask ourselves: Is every infinite integral domain not a field? Not quite! Some infinite integral domains are, in fact, fields. This follows directly from the result in the previous section. Every field is an integral domain and there are plenty of infinite fields, so they must all be integral domains too. Consider the field of rational numbers \( \mathbb{Q} \) or the field of complex numbers \( \mathbb{C}. \) Since these are fields, they are also integral domains. So, clearly, there are infinite integral domains that are also fields.

Every Finite Integral Domain Is a Field

We will now turn our attention to finite integral domains. Is every finite integral domain a field? Yes! This can be shown as follows.

Let \( D \) be a finite integral domain. Let \( a \in D \) with \( a \ne 0. \) Consider the set \[ A = \{ a, a^2, a^3, \dots \}. \] Since a ring is closed under multiplication, every element of \( A \) belongs to \( D, \) so \( A \subseteq D. \) Since \( D \) is finite, \( A \) is finite too. Therefore, by the pigeonhole principle, there exist integers \( m \gt n \ge 0 \) such that \[ a^m = a^n \] This equation can be rewritten as \[ a \cdot a^{m - n - 1} \cdot a^n = 1 \cdot a^n. \] Since \( a \) is a non-zero element of an integral domain, it follows that \( a^n \ne 0. \) Therefore we can use Proposition 2 (the cancellation property of integral domains) to get \[ a \cdot a^{m - n - 1} = 1. \] Since a ring is closed under multiplication and since \( m - n - 1 \ge 0, \) it follows that \( a^{m - n - 1} \in D. \) Thus every non-zero element \( a \in D \) has a multiplicative inverse in \( D. \) This establishes the multiplicative inverse property of a field.

Since an integral domain has distinct additive and multiplicative identities, it satisfies two additional field properties: the existence of an additive identity and a distinct multiplicative identity.

Finally, the remaining field properties are inherited from the ring structure, i.e. associativity and commutativity of addition and multiplication, the existence of additive inverses and the distributivity of multiplication over addition all hold in \( D, \) since they hold in any ring. Thus, \( D \) satisfies all the field properties. Therefore \( D \) is a field.

Alternate Proof

The proof in the previous section presents what I initially came up with while working through these concepts and proving these results for myself. However, I later found that there is another proof that is quite popular in the literature. This alternate proof differs in one key aspect: it does not invoke the cancellation property of integral domains stated in Proposition 2. Let us examine this alternate proof.

As before, we consider the set \( A = \{ a, a^2, a^3, \dots \} \subseteq D, \) where \( a \in D \) and \( a \ne 0 \) and we obtain the equation \[ a^m = a^n \] for some integers \( m \gt n \ge 0. \) As before, we use the fact that \( a \) is an element of an integral domain to conclude that \( a^n \ne 0. \) Now adding the additive inverse of \( a^n \) to both sides we get \[ a^m - a^n = 0. \] Using the distributivity property of rings, we get \[ a^n (a^{m - n} - 1) = 0 \] Since a ring is closed under addition and multiplication, both \( a^n \) and \( a^{m - n} - 1 \) belong to \( D. \) As \( D \) is an integral domain and \( a^n \ne 0, \) we conclude that \( a^{m - n} - 1 = 0. \) Therefore \( a^{m - n} = 1. \) Since \( m - n \ge 1, \) we can write: \[ a \cdot a^{m - n - 1} = 1. \] Therefore every non-zero element \( a \in D \) has a multiplicative inverse in \( D. \) The remaining properties of a field are established in the same manner as in the previous section. Hence, if \( D \) is a finite integral domain, then it is also a field.

Conclusion

We now summarise all the results here before concluding the article:

• Every field is an integral domain.
• Every finite integral domain is a field.
• Some infinite integral domains are not fields. A convenient example is the set of integers \( \mathbb{Z}. \)
• Some infinite integral domains are fields. Every infinite field, such as \( \mathbb{Q}, \) \( \mathbb{R} \) or \( \mathbb{C} \) is an example.

It is worth reiterating here that the fourth result in the summary above follows from the fact that every field is an integral domain. These results reveal how structure and size interact in algebraic systems. It is interesting how simply being finite guarantees that an integral domain is a field.

Read on website | #mathematics

This post illustrates a key lemma that is used in proving the fundamental theorem of Galois theory (FTGT). Note that FTGT is not covered in this post. The focus of this post is on understanding and proving this lemma only. Here is the lemma from the book Galois Theory, 5th ed. by Stewart (2023):

The notation \( M^* \) denotes the group of all \( M \)-automorphisms of \( L \) with composition as the group operation. Note that Stewart writes \( \tau(M)^{*} = \tau M^* \tau^{-1} \) while stating the lemma but I have reversed the LHS and RHS to maintain consistency with the equations that appear in the discussion below.

To build intuition for this lemma, I'll first present an illustration, followed by a proof. The discussion below assumes familiarity with field extensions and field automorphisms, as several notations and results from these areas will be used implicitly without detailed justification. This post is meant to serve as a set of notes on the lemma, not a comprehensive tutorial.

Contents

• Introduction
• Illustration
  • Concrete Example
  • LHS ⊆ RHS
  • LHS ⊇ RHS
  • LHS = RHS
• Proof

Illustration

Concrete Example

Let \( L = \mathbb{Q}(\sqrt{2}, \sqrt{3}), \) \( K = \mathbb{Q} \) and \( M = \mathbb{Q}(\sqrt{2}). \) Note that \begin{align*} L &= \{ a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} : a, b, c, d \in \mathbb{Q} \}, \\ M &= \{ k + l \sqrt{2} : k, l \in \mathbb{Q} \}. \end{align*} Now the group of \( K \)-automorphisms of \( L \) is \[ K^* = \{\phi_1, \phi_2, \phi_3, \phi_4 \} \] where each \( \phi_i \) is given by \begin{align*} \phi_1 &: a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} \mapsto a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}, \\ \phi_2 &: a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} \mapsto a - b \sqrt{2} + c \sqrt{3} - d \sqrt{6}, \\ \phi_3 &: a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} \mapsto a + b \sqrt{2} - c \sqrt{3} - d \sqrt{6}, \\ \phi_4 &: a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} \mapsto a - b \sqrt{2} - c \sqrt{3} + d \sqrt{6}. \end{align*} Then \( M^* = \{ \phi_1, \phi_3 \}. \) Let \( \tau = \phi_2. \) Then \begin{align*} \tau(M) &= \{ \tau(x) : x \in \mathbb{M} \} \\ &= \{ \tau(k + l \sqrt{2}) : k, l \in \mathbb{Q} \} \\ &= \{ k - l \sqrt{2} : k, l \in \mathbb{Q} \}. \end{align*} Note that in this case we ended up with \( \tau(M) = M \) but we will be careful not to utilise this fact. We will ensure that the steps below work without assuming \( \tau(M) = M. \) Next we find \begin{equation} \tau(M)^* = \{ \phi_1, \phi_3 \}. \label{eq-tau-m-ast} \end{equation} Now \begin{align*} \tau M^* \tau^{-1} &= \{ \tau \gamma \tau^{-1} : \gamma \in {M^*} \} \\ &= \{ \tau \phi_1 \tau^{-1}, \tau \phi_3 \tau^{-1} \}. \end{align*} Let us now find out how each element of \( \tau M^* \tau^{-1} \) transforms the elements of \( L. \) For all \( a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} \in L, \) we get \begin{align*} (\tau \phi_1 \tau^{-1})(a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}) &= (\tau \phi_1)(a - b\sqrt{2} + c\sqrt{3} - d\sqrt{6}) \\ &= \tau (a - b\sqrt{2} + c\sqrt{3} - d\sqrt{6}) \\ &= a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}). \end{align*} Therefore \[ \tau \phi_1 \tau^{-1} = \phi_1. \] Similarly, \begin{align*} (\tau \phi_3 \tau^{-1})(a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}) &= (\tau \phi_3)(a - b\sqrt{2} + c\sqrt{3} - d\sqrt{6}) \\ &= \tau (a - b\sqrt{2} - c\sqrt{3} + d\sqrt{6}) \\ &= a + b\sqrt{2} - c\sqrt{3} - d\sqrt{6}. \end{align*} Therefore \[ \tau \phi_3 \tau^{-1} = \phi_3. \] We have shown that \begin{equation} \tau M^* \tau^{-1} = \{ \phi_1, \phi_3 \}. \label{eq-tau-coset} \end{equation} From \( \eqnref{eq-tau-m-ast}{1} \) and \( \eqnref{eq-tau-coset}{2} \) we see that \[ \tau M^* \tau^{-1} = \tau(M)^*. \] Since we are working with a concrete example of \( \tau \) here, we know exactly how it behaves, so we succeeded in demonstrating the above equality. However, in a general proof, \( \tau \) is going to be an arbitrary \( K \)-automorphism of \( L, \) so we cannot know exactly how it behaves and as a result, we cannot obtain the above equation directly. Therefore, in a general proof, we we will first show that \( \tau M^* \tau^{-1} \subseteq \tau(M)^* \) and then we will show that \( \tau M^* \tau^{-1} \supseteq \tau(M)^* \) in order to prove the above equation.

LHS ⊆ RHS

Once again, let us see how each element of \( \tau M^* \tau^{-1} \) transforms the elements of \( \tau(M). \) Note that this time we are not going to examine how they transform arbitrary elements of \( L. \) We are only going to see how they transform the elements of \( \tau(M). \) For all \( k - l \sqrt{2} \in \tau(M), \) we get \begin{align*} (\tau \phi_1 \tau^{-1})(k - l \sqrt{2}) &= (\tau \phi_1)(k + l \sqrt{2}) \\ &= \tau(k + l \sqrt{2}) \\ &= k - l \sqrt{2}. \end{align*} Similarly, for all \( k - l \sqrt{2} \in \tau(M), \) we get \begin{align*} (\tau \phi_3 \tau^{-1})(k - l \sqrt{2}) &= (\tau \phi_3)(k + l \sqrt{2}) \\ &= \tau(k + l \sqrt{2}) \\ &= k - l \sqrt{2}. \end{align*} Note above that both \( \phi_1 \) and \( \phi_3 \) fix \( k + l \sqrt{2} \in M \) because \( \phi_1, \phi_2 \in M^*, \) the set of \( M \)-automorphisms of \( L. \) This detail will be used in the general proof.

Since both \( \tau \phi_1 \tau^{-1} \) and \( \tau \phi_3 \tau^{-1} \) fix the elements of \( \tau(M), \) they are both \( \tau(M) \)-automorphisms of \( L. \) Therefore \( \tau M^* \tau^{-1} \subseteq \tau(M)^{*}. \)

LHS ⊇ RHS

Consider the set \( \tau^{-1} \tau(M)^* \tau \) and examine how its elements transform the elements of \( M. \) For all \( k + l \sqrt{2} \in M, \) we get \begin{align*} (\tau^{-1} \phi_1 \tau)(k + l \sqrt{2}) &= (\tau^{-1} \phi_1)(k - \sqrt{2}) \\ &= \tau^{-1}(k - \sqrt{2}) \\ &= k + l \sqrt{2}. \end{align*} Similarly, for all \( k + l \sqrt{2} \in M, \) we get \begin{align*} (\tau^{-1} \phi_3 \tau)(k + l \sqrt{2}) &= (\tau^{-1} \phi_3)(k - \sqrt{2}) \\ &= \tau^{-1}(k - \sqrt{2}) \\ &= k + l \sqrt{2}. \end{align*} Here both \( \phi_1 \) and \( \phi_3 \) fix \( k - l \sqrt{2} \in \tau(M) \) because \( \phi_1, \phi_2 \in \tau(M)^*, \) the set of \( \tau(M) \)-automorphisms of \( L. \)

Since both \( \tau^{-1} \phi_1 \tau \) and \( \tau^{-1} \phi_3 \tau \) fix the elements of \( M, \) they are both \( M \)-automorphisms of \( L. \) Therefore \( \tau^{-1} \tau(M)^* \tau \subseteq M^* \) which implies \( \tau M^* \tau^{-1} \supseteq \tau(M)^*. \)

LHS = RHS

The previous two sections complete the illustration of the lemma with the chosen example. We have shown that \( \tau M^* \tau^{-1} \subseteq \tau(M)^{*} \) and \( \tau M^* \tau^{-1} \supseteq \tau(M)^*. \) Therefore \( \tau M^* \tau^{-1} = \tau(M)^*. \)

Proof

The ideas presented in the previous sections will now be extended to formulate a general proof. For clarity, the lemma is stated once again below before proceeding with the proof.

Lemma 12.1. Suppose that \( L/K \) is a field extension, \( M \) is an intermediate field, and \( \tau \) is a \( K \)-automorphism of \( L. \) Then \( \tau M^* \tau^{-1} = \tau(M)^{*}. \)

Proof. For all \( \gamma \in M^*, \) \( x' \in \tau(M), \) we use the notation \( x = \tau^{-1}(x') \in M \) and get \[ (\tau \gamma \tau^{-1})(x') = (\tau \gamma)(x) = \tau(x) = x'. \] In the second equality above, we have used the fact that \( \gamma \in M^* \) which implies that \( \gamma \) is an \( M \)-automorphism of \( L \) which allows us to conclude that \( \gamma(x) = x \) for \( x \in M. \) Since every \( \tau \gamma \tau^{-1} \in \tau M^* \tau^{-1} \) fixes all elements \( x' \in \tau(M), \) each \( \tau \gamma \tau^{-1} \) must be a \( \tau(M) \)-automorphism of \( L. \) Thus \( \tau M^* \tau^{-1} \subseteq \tau(M)^*. \)

Similarly, for all \( \gamma' \in \tau(M)^*, \) \( x \in M, \) we use the notation \( x' = \tau(x) \in \tau(M) \) and get \[ (\tau^{-1} \gamma' \tau)(x) = (\tau^{-1} \gamma')(x') = \tau^{-1}(x') = x. \] In the second equality above, we have used the fact that \( \gamma' \in \tau(M)^* \) which implies that \( \gamma' \) is an \( \tau(M) \)-automorphism of \( L \) which allows us to conclude that \( \gamma'(x') = x' \) for \( x' \in \tau(M). \) Since every \( \tau^{-1} \gamma' \tau \in \tau^{-1} \tau(M)^* \tau \) fixes all elements \( x \in M, \) each \( \tau^{-1} \gamma' \tau \) must be an \( M \)-automorphism of \( L. \) Thus \( \tau^{-1} \tau(M)^* \tau \subseteq M^*. \) This implies \( \tau M^* \tau^{-1} \supseteq \tau(M)^*. \)

We have shown that \( \tau M^* \tau^{-1} \subseteq \tau(M)^* \) and \( \tau M^* \tau^{-1} \supseteq \tau(M)^*. \) Therefore \( \tau M^* \tau^{-1} = \tau(M)^*. \)

Read on website | #mathematics

Contents

• Definition
• Notation
• Domain, Codomain and Image
• Injection, Surjection and Bijection

Definition

A function \( f \) from a set \( X \) to a set \( Y \) is a binary relation \( R \) that satisfies the following conditions:

• \( R \subseteq \{ (x, y) \mid x \in X, y \in Y \} = X \times Y. \)
• For every \( x \in X, \) there exists \( y \in Y \) such that \( (x, y) \in R. \)
• If \( (x, y) \in R \) and \( (x, z) \in R, \) then \( y = z. \)

The set \( X \) is called the domain of \( f \) and the set \( Y \) is called the codomain of \( f. \) The relation \( R \) is also known as the graph of \( f. \)

Notation

Let \( f \) be a function from a set \( X \) to a set \( Y. \) Then the name \( f \) represents the function and the notation \( f(x) \) represents the application of the function to the argument \( x, \) i.e. \( f(x) \) represents the value of \( f \) for the element \( x \in X. \) In other words, for all \( x \in X, \) we have \( (x, f(x)) \in R. \)

A function \( f \) with domain \( X \) and codomain \( Y \) is also written as \( f : X \to Y. \) The function \( f \) may also be written as \( x \mapsto f(x). \) This notation specifies a function that maps \( x \) to \( f(x). \)

Formally, \( f(x) \) denotes the application of the function \( f \) to the argument \( x. \) However, in practice, it is common to use the expression \( f(x) \) to refer to both the function itself and its output for a given \( x, \) which is a slight deviation from strict notation. Similarly, the function \( x \mapsto g(x) \) is often written as \( f(x) = g(x). \) For example, the function \( x \mapsto x^2 - 1 \) may also be written as \( f(x) = x^2 - 1. \)

Consider a function \( f \) that returns the square of a real number. The following are common notations used to define this function, roughly ordered from the most formal form to the least formal one:

• \( f : \mathbb{R} \to \mathbb{R} ; \; x \mapsto x^2, \)
• \( f : \mathbb{R} \to \mathbb{R} : x \mapsto x^2, \)
• \( f : x \mapsto x^2, \)
• \( f: \mathbb{R} \to \mathbb{R} \) where \( f(x) = x^2, \)
• \( f(x) = x^2. \)

Domain, Codomain and Image

The domain of a function is the set of all values for which the function is defined.

A codomain of a function \( f \) is a set within which the values \( f(x) \) for all \( x \in X \) must lie, where \( X \) is the domain of \( f. \)

The image of a function \( f \) is the set \( \{ f(x) \mid x \in X \} \) where \( X \) is the domain of \( f. \)

The term range is often used as a synonym of image. However the use of this term is inconsistent across literature. Some old books use the term range to mean codomain while other books use this term to mean the image. Therefore it is best to use the term image because it is free from such ambiguity.

Injection, Surjection and Bijection

A function \( f : X \to Y \) is injective if \( \forall a, b \in X, a \neq b \implies f(a) \neq f(b). \) A function is injective, if each element of the codomain is mapped to by at most one element of the domain. An injective function is also known as a one-to-one function or an injection.

A function \( f : X \to Y \) is surjective if \( \forall y \in Y, \exists x \in X \) such that \( y = f(x). \) A function is surjective, if each element of the codomain is mapped to by at least one element of the domain. A surjective function is also known as an onto function or a surjection.

A function \( f : X \to Y \) is bijective if \( \forall y \in Y, \) there exists exactly one \( x \in X, \) such that \( y = f(x). \) A function is bijective, if each element of the codomain is mapped to by exactly one element of the domain. A bijective function is also known as a one-to-one correspondence or bijection. A bijection is both injective and surjective. In other words, a bijection is both one-to-one and onto.

The function \( f : \mathbb{R} \to \mathbb{R}; \; x \mapsto e^x \) is injective but not surjective. It is injective because distinct values of \( x \) produce distinct values of \( e^x. \) However, it is not surjective as no value in the domain maps to negative numbers in the codomain, leaving some elements in the codomain unmapped.

The function \( f : \mathbb{R} \to \mathbb{R}; \; x \mapsto x^3 - x \) is surjective but not injective. It is surjective because every value in the codomain is mapped to by at least one value in the domain. However, it is not injective, as distinct values in the domain can map to the same value in the codomain. For example, \( f(-1) = f(0) = f(1) = 0. \)

The function \( f : \mathbb{R} \to \mathbb{R}; \; x \mapsto x + 1 \) is bijective. It is both injective and surjective. This function is invertible with the inverse given by the function \( f^{-1} : \mathbb{R} \to \mathbb{R}; \; x \mapsto x - 1. \)

The function \( f : \mathbb{R} \to \mathbb{R}; \; x \mapsto x^2 \) is neither injective nor surjective. First, the function is not injective because distinct values in the domain can map to the same value in the codomain. For example, \( f(-2) = f(2) = 4. \) Additionally, the function is not surjective because no value in the domain maps to the negative numbers in the codomain.

Read on website | #mathematics | #definition

We get this absurd result because of the incorrect assumption that there exists an integer that is the largest of all the integers.

Read on website | #mathematics

Let us focus on \( \ln x \) again. Is it not peculiar? What does \( \ln \) stand for really? Logarithm natural? Sounds very unnatural.

Well, as a kid I learnt that \( \ln \) here stands for the Latin phrase "logarithmus naturalis". It is only recently that I bothered to verify if this expansion of \( \ln x \) that I learnt as a kid is really true. The most credible discussion of this that I could find online is this thread on Mathematics Stack Exchange: math.stackexchange.com/q/1694. The answer by Dan Velleman points us to page 277 of an 1875 book Lehrbuch der Mathematik by Anton Steinhauser. Quoting the relevant portion from the page:

Man pflegt nun, um Verwechslungen dieser beiden Systeme vorzubeugen, mit log.nat. a (gesprochen: logarithmus naturalis a) oder ln. a, oder am einfachsten mit la den natürlichen, mit log.brigg. a (gesprochen: Logarithmus briggus a) oder log.a, oder am einfachsten mit lg. a den gemeinen Logarithmus (von a) zu bezeichnen.

Translated to English, it says:

One is accustomed now, in order to prevent confusion between these two systems, to use log.nat. a (pronounced: logarithmus naturalis a) or ln. a, or most simply la for the natural, and log.brigg. a (pronounced: logarithmus briggus a) or log. a, or most simply lg. a to denote the common logarithm (of a).

So it does look like what I learnt as a kid is correct and the earliest possible reference of this the Internet is able to find for us is the 1875 book quoted above.

Read on website | #mathematics

William Paul Thurston (October 30, 1946 – August 21, 2012) was an American mathematician. He was a pioneer in the field of low-dimensional topology and was awarded the Fields Medal in 1982 for his contributions to the study of 3-manifolds.

Thurston was a professor of mathematics at Princeton University, University of California at Davis and Cornell University. He was also a director of the Mathematical Sciences Research Institute.

MathOverflow makes all answers posted to the website available under a Creative Commons licence. In particular, all answers posted before 08 Apr 2011 (UTC) are available under the terms of the Creative Commons Attribution-ShareAlike 2.5 Generic (CC BY-SA 2.5) licence. Thurston wrote the answer I am about to share on 30 Oct 2010. Due to the licence terms, this post too is available under the terms of the same licence.

Thurston posted his answer while replying to a MathOverflow question: What's a mathematician to do?. The question enquires about how an ordinary mathematician can contribute to mathematics. Thurston's answer from mathoverflow.net/a/44213 is reproduced below:

It's not mathematics that you need to contribute to. It's deeper than that: how might you contribute to humanity, and even deeper, to the well-being of the world, by pursuing mathematics? Such a question is not possible to answer in a purely intellectual way, because the effects of our actions go far beyond our understanding. We are deeply social and deeply instinctual animals, so much that our well-being depends on many things we do that are hard to explain in an intellectual way. That is why you do well to follow your heart and your passion. Bare reason is likely to lead you astray. None of us are smart and wise enough to figure it out intellectually.

The product of mathematics is clarity and understanding. Not theorems, by themselves. Is there, for example any real reason that even such famous results as Fermat's Last Theorem, or the Poincaré conjecture, really matter? Their real importance is not in their specific statements, but their role in challenging our understanding, presenting challenges that led to mathematical developments that increased our understanding.

The world does not suffer from an oversupply of clarity and understanding (to put it mildly). How and whether specific mathematics might lead to improving the world (whatever that means) is usually impossible to tease out, but mathematics collectively is extremely important.

I think of mathematics as having a large component of psychology, because of its strong dependence on human minds. Dehumanized mathematics would be more like computer code, which is very different. Mathematical ideas, even simple ideas, are often hard to transplant from mind to mind. There are many ideas in mathematics that may be hard to get, but are easy once you get them. Because of this, mathematical understanding does not expand in a monotone direction. Our understanding frequently deteriorates as well. There are several obvious mechanisms of decay. The experts in a subject retire and die, or simply move on to other subjects and forget. Mathematics is commonly explained and recorded in symbolic and concrete forms that are easy to communicate, rather than in conceptual forms that are easy to understand once communicated. Translation in the direction conceptual -> concrete and symbolic is much easier than translation in the reverse direction, and symbolic forms often replaces the conceptual forms of understanding. And mathematical conventions and taken-for-granted knowledge change, so older texts may become hard to understand.

In short, mathematics only exists in a living community of mathematicians that spreads understanding and breaths life into ideas both old and new. The real satisfaction from mathematics is in learning from others and sharing with others. All of us have clear understanding of a few things and murky concepts of many more. There is no way to run out of ideas in need of clarification. The question of who is the first person to ever set foot on some square meter of land is really secondary. Revolutionary change does matter, but revolutions are few, and they are not self-sustaining --- they depend very heavily on the community of mathematicians.

In the comments to the answer, one of the commenters was Suresh Venkatasubramanian who was a professor in the School of Computing at the University of Utah back then. He is now a professor of Computer Science and Data Science at Brown University. In his comment, Suresh proposed that this answer be called Thurston's Paean. Here is his complete comment:

This seems like an ideal counterpoint to Hardy's Lament. I'm calling it Thurston's Paean :). Seems poignant now that he has passed.

Thurston's answer does appear to be a perfect complement to Hardy's lament in the 1940 essay A Mathematician's Apology. While Hardy's lament is remarkably beautiful and introspective, it may also feel a little depressing at places. Thurston's post on the other hand is full of hope and purpose that goes beyond the actual work of doing mathematics. Indeed Thurston's Paean is a befitting title for his answer.

Read on website | #mathematics | #miscellaneous | #quote

One of the many techniques for solving ordinary differential equations involves using an integrating factor. An integrating factor is a function that a differential equation is multiplied by to simplify it and make it integrable. It almost appears to work like magic!

The Method

Let us first see how the integrating factor method works. In this post, we will work with linear first-order ordinary differential equations of type \[ \frac{dy}{dx} + y P(x) = Q(x) \] to discuss, reason about and illustrate this method. We will also often use the Leibniz's notation \( dy/dx \) and the Lagrange's notation \( y'(x) \) or simply \( y' \) interchangeably as is typical in calculus. They all mean the same thing: the derivative of the function \( y \) with respect to \( x. \) Thus the above differential equation may also be written as \[ y' + y P(x) = Q(x). \] Given a differential equation of this form, we first find an integrating factor \( M(x) \) using the formula \[ M(x) = e^{\int P(x) \, dx}. \] Then we multiply both sides of the differential equation by this integrating factor. Now remarkably, the left-hand side (LHS) reduces to a single term consisting only of a derivative. As a result, we can get rid of that derivative by integrating both sides of the equation and we then proceed to obtain a solution.

An Example

Here is an example that demonstrates the method of using an integrating factor. Let us say we want to solve the differential equation \[ y' + y \left( \frac{x + 1}{x} \right) = \frac{1}{x}. \] Indeed this is in the form \( y' + y P(x) = Q(x) \) with \( P(x) = (x + 1)/x \) and \( Q(x) = 1/x. \) We first obtain the integrating factor \[ M(x) = e^{\int P(x) \, dx} = e^{\int (x + 1)/x \, dx} = e^{\int (1 + 1/x) \, dx} = e^{x + \ln x} = x e^x. \] Now we multiply both sides of the differential equation by this integrating factor and get \[ y' x e^x + y (x + 1) e^x = e^x. \] The LHS can now be simplified to \( \frac{d}{dx} (y x e^x). \) This can be verified using the product rule for derivatives. This simplification of the LHS is the remarkable feature of this method. Therefore the above equation can be written as \[ \frac{d}{dx} (y x e^x) = e^x. \] Note that the expression on the LHS is a product of the function \( y \) and the integrating factor \( x e^x. \) We will discuss this observation in more detail a little later. Let us first complete solving this differential equation. Since the LHS is now a single term that consists of a derivative, obtaining a solution now simply involves integrating both sides with respect to \( x. \) Integrating both sides we get \[ y x e^x = e^x + C \] where \( C \) is the constant of integration. Finally, we divide both sides by the integrating factor \( x e^x \) to get \[ y = \frac{1}{x} + \frac{C}{x e^x}. \] We have now obtained a solution for the differential equation. If we review the steps above, we will find that after multiplying both sides of the given differential equation by the integrating factor, the differential equation becomes significantly simpler and integrable. In fact, after multiplying both sides of the given differential equation by the integrating factor, the LHS always becomes the derivative of the product of the function \( y \) and the integrating factor. We will now see why this is so.

An Interesting Relationship

Consider once again the linear first-order differential equation \begin{equation} \label{eq-if-diff} y' + yP(x) = Q(x). \end{equation} We first find the integrating factor \begin{equation} \label{eq-if-integrating-factor} M(x) = e^{\int P(x)\, dx}. \end{equation} The integrating factor obtained like this satisfies an interesting relationship: \begin{equation} \label{eq-if-property} M'(x) = M(x) P(x). \end{equation} We can prove this relationship easily by differentiating both sides of \( \eqnref{eq-if-integrating-factor}{2} \) as follows: \[ M'(x) = \frac{d}{dx} \left( e^{\int P(x)\, dx} \right) = e^{\int P(x)\, dx} \frac{d}{dx} \left( \int P(x)\, dx \right) = M(x) P(x). \] Note that we use the chain rule to work out the derivative above. This beautiful result is due to how the derivative of the exponential function works. When we apply the chain rule to obtain the derivative of \( e^{f(x)} \) we get \[ \frac{d}{dx} e^{f(x)} = e^{f(x)} f'(x). \] This nice property of the exponential function leads to the interesting relationship in \( \eqnref{eq-if-property}{3}. \)

Simplification of LHS

Now let us multiply both sides of the differential equation \( \eqnref{eq-if-diff}{1} \) by the integrating factor \( M(x) \) By doing so, we get \[ y' M(x) + y P(x) M(x) = Q(x) M(x). \] But from \( \eqnref{eq-if-property}{3} \) we know that \( P(x) M(x) = M'(x), \) so the above equation can be written as \[ y' M(x) + y M'(x) = Q(x) M(x). \] Look what we have got on the LHS! We have the expansion of \( \frac{d}{dx}(yM(x)) \) on the LHS. By product rule of differentiation, we have \( \frac{d}{dx}(yM(x)) = y' M(x) + y M'(x). \) Therefore the above equation can be written as \[ \frac{d}{dx}(yM(x)) = Q(x) M(x). \] The "magic" has occurred here! Multiplying both sides of the differential equation by the integrating factor has led us to an equation that has got a single derivative only on the LHS. As a result, finding the solution is now a simple matter of integrating both sides, i.e. \[ y M(x) = \int Q(x) M(x) \, dx. \] Thus \[ y = \frac{1}{M(x)} \int Q(x) M(x) \, dx. \] Note that the result of indefinite integral on the RHS will contain the constant of integration, which we will denote as \( C, \) so the final solution looks like \begin{equation} \label{eq-if-general-solution} y = \frac{1}{M(x)} \int Q(x) M(x) \, dx + \frac{C}{M(x)}. \end{equation}

Illustration

Let us illustrate the method and its magic with a very simple differential equation: \[ y' + \frac{y}{x} = x. \] First we note that this equation is in the form \( y' + yP(x) = Q(x) \) with \( P(x) = 1/x \) and \( Q(x) = x. \) We then find the integrating factor \[ M(x) = e^{\int P(x) \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x. \] Then we multiply both sides of the differential equation by the integrating factor to get \[ y'x + y = x^2. \] Now indeed the LHS can be written down as a single derivative as shown below: \[ \frac{d}{dx} yx = x^2. \] Note that the LHS is the derivative of the product of \( y \) and the integrating factor \( x. \) This is exactly what we discussed in the previous section. We integrate both sides of the above equation to get \[ yx = \frac{x^3}{3} + C. \] Finally we divide both sides by the integrating factor \( x \) to get \[ y = \frac{x^2}{3} + \frac{C}{x}. \] We have arrived at the solution \( y(x) \) for the differential equation.

Conclusion

In this post, we used very simple and convenient differential equations that led to nice closed-form solutions. In practice, differential equations can be quite complicated and may not always lead to closed-form solutions. In such cases, we leave the result in the form of an expression that contains an unsolved integral. Such solutions may resemble the form shown in \( \eqnref{eq-if-general-solution}{4}. \)

The method of using integrating factors to solve differential equations can also be extended to linear higher-order differential equations. That is something we did not discuss in this post. However, I hope that the intuition gained from understanding how and why this method works for linear first-order differential equations will be useful while studying such extensions of this method.

Read on website | #mathematics