Susam's Combinatorics Pages

Langford Pairing

2011-09-17T00:00:00Z

Permutation Problem

A few days ago, I came across this problem:

There is a sequence of \( 2n \) numbers where each natural number from \( 1 \) to \( n \) is repeated twice, i.e. \[ (1, 1, 2, 2, \dots, n, n). \] Find a permutation of this sequence such that for each \( k \) where \( 1 \le k \le n, \) there are \( k \) numbers between two occurrences of \( k \) in the permutation.

Getting Started

In combinatorics, this problem has a name: Langford's problem. A permutation of \( (1, 1, 2, 2, \dots, n, n) \) that satisfies the condition given in the probem is known as a Langford pairing or Langford sequence.

For small \( n, \) say \( n = 4, \) Langford pairings can be obtained easily by trial and error: \( (4, 1, 3, 1, 2, 4, 3, 2). \) What if \( n \) is large? We need an algorithm to find a permutation that solves the problem in that case.

There is another question to consider: Is there a solution for every possible \( n? \) One can easily see that there are no Langford pairings for \( n = 1 \) and \( n = 2, \) i.e. the sequences \( (1, 1) \) and \( (1, 1, 2, 2) \) have no Langford pairings.

We need to understand two things:

For what values of \( n \) do Langford pairings exist?
For the values of \( n \) for which Langford pairings exist, how do we find the Langford pairings?

A simple Python 3 program I wrote to find Langford pairings for small values of \( n \) offers some clues. Here is the program:

def find_solutions(n, s=None):
    # If called from top-level (s=None), create a list of 2n zero
    # values.  Zeroes represent unoccupied cells.
    if s is None:
        s = [0] * (2 * n)

    # Next number to be placed.
    m = max(s) + 1

    # For each i, try to place m at s[i] and s[i + m + 1].
    for i in range(2 * n - m - 1):

        # If s[i] and s[i + m + 1] are unoccupied, ...
        if s[i] == s[i + m + 1] == 0:

            # first place m at s[i] and s[i + m + 1].
            s[i] = s[i + m + 1] = m

            # If m is the last number to be placed, ...
            if m == n:
                # then a solution has been found; yield it.
                yield s[:]
            else:
                # else try to place the next number.
                yield from find_solutions(n, s)

            # Undo placement of m.
            s[i] = s[i + m + 1] = 0


# Count solutions for 1 <= n <= 12.
for n in range(1, 13):
    count = sum(1 for s in find_solutions(n))
    print('n = {:2} => {:6} solutions'.format(n, count))

It takes a few minutes for this program to run. Here is the output of this program:

$ python3 langford.py
n =  1 =>      0 solutions
n =  2 =>      0 solutions
n =  3 =>      2 solutions
n =  4 =>      2 solutions
n =  5 =>      0 solutions
n =  6 =>      0 solutions
n =  7 =>     52 solutions
n =  8 =>    300 solutions
n =  9 =>      0 solutions
n = 10 =>      0 solutions
n = 11 =>  35584 solutions
n = 12 => 216288 solutions

Note that we always talk about Langford pairings in plural in this post. That's because either a sequence has no Langford pairings or it has two or more Langford pairings. There is never a sequence that has only one Langford pairing. That's because if we find at least one Langford pairing for a sequence, the reverse of that Langford pairing is also a Langford pairing. Therefore, when Langford pairings exist for a sequence, they must at least be two in number. Since they occur in pairs, they are always even in number. This is why we don't have to write "one or more Langford pairings" in this post. We can always write "Langford pairings" instead.

Conjecture

From the output above, we can form a conjecture:

  The sequence \( (1, 1, 2, 2, \dots, n, n) \) has Langford pairings
  if and only if \( n \equiv 0 \pmod{4} \) or
  \( n \equiv 3 \pmod{4}.  \)

For convenience, let us denote the sequence \( (1, 1, 2, 2, \dots, n, n) \) as \( S_n. \) We will now prove the above conjecture in two parts:

We will first show that the condition that either \( n \equiv 0 \pmod{4} \) or \( n \equiv 3 \pmod{4} \) must hold is a necessary condition for the sequence \( S_n \) to have Langford pairings.
We will then show that the condition that either \( n \equiv 0 \pmod{4} \) or \( n \equiv 3 \pmod{4} \) must hold is a sufficient condition for the sequence \( S_n \) to have Langford pairings.

Necessity

Let \( S_n = (1, 1, 2, 2, \dots, n, n) \) be a sequence such that it has Langford pairings. Let us pick an arbitrary Langford pairing \( s \) of \( S_n \) and split this Langford pairing \( s \) into two mutually exclusive subsequences \( s_1 \) and \( s_2 \) such that:

\( s_1 \) contains all numbers in odd numbered positions and
\( s_2 \) contains all numbers in even numbered positions.

For example, if we pick \( s = (1, 7, 1, 2, 5, 6, 2, 3, 4, 7, 5, 3, 6, 4) \) which is a Langford pairing of \( S_7, \) we split \( s \) into \begin{align*} s_1 & = (1, 1, 5, 2, 4, 5, 6), \\ s_2 & = (7, 2, 6, 3, 7, 3, 4). \end{align*}

We can make a few observations:

Both occurrences of an even number do not occur in the same subsequence.
There are \( \left\lfloor \frac{n}{2} \right\rfloor \) even numbers in each subsequence.
Both occurrences of an odd number occur in the same subsequence.
There are \( \left\lceil \frac{n}{2} \right\rceil \) odd numbers in each subsequence.

Do these observations hold good for every Langford pairing of any aribrary \( S_n \) for every positive integer value of \( n? \) Yes, they do. We will now prove them one by one:

Let us consider an even number \( k \) from a Langford pairing. If the first occurrence of \( k \) lies at the \( i \)th position in the pairing, then its second occurrence lies at the \( (i + k + 1) \)th position. Since \( k \) is even, \( i \) and \( i + k + 1 \) have different parities, i.e. if \( i \) is odd, then \( i + k + 1 \) is even and vice versa. Therefore, if the first occurrence of \( k \) lies at an odd numbered position, its second occurrence must lie at an even numbered position and vice versa. Thus one occurrence of \( k \) must belong to \( s_1 \) and the other must belong to \( s_2. \) This proves the first observation.
The number of even numbers between \( 1 \) and \( n, \) inclusive, is \( \left\lfloor \frac{n}{2} \right\rfloor. \) Each of these even numbers has been split equally between \( s_1 \) and \( s_2. \) This proves the second observation.
Now let us consider an odd number \( k \) from a Langford pairing. If the first occurrence of \( k \) lies at the \( i \)th position in the pairing, then its second occurrence lies at the \( (i + k + 1) \)th position. Since \( k \) is odd, \( i \) and \( i + k + 1 \) have the same parity. Therefore, either both occurrences of \( k \) belong to \( s_1 \) or both belong to \( s_2. \) This proves the third observation.
Each subsequence, \( s_1 \) or \( s_2 \) has \( n \) numbers because we split a Langford pairing \( s \) with \( 2n \) numbers equally between the two subsequences. We have shown that each subsequence has \( \left\lfloor \frac{n}{2} \right\rfloor \) even numbers. Therefore the number of odd numbers in each subsequence is \( n - \left\lfloor \frac{n}{2} \right\rfloor = \left\lceil \frac{n}{2} \right\rceil. \)

From the third observation, we know that the odd numbers always occur in pairs in each subsequence because both occurrences of an odd number occur together in the same subsequence. Therefore, the number of odd numbers in each subsequence must be even. Since the number of odd numbers in each subsequence is \( \left\lceil \frac{n}{2} \right\rceil \) as proven for the fourth observation, we conclude that \( \left\lceil \frac{n}{2} \right\rceil \) must be even.

Now let us see what must \( n \) be like so that \( \left\lceil \frac{n}{2} \right\rceil \) is even.

Let us express \( n \) as \( 4q + r \) where \( q \) is a nonnegative integer and \( r \in \{0, 1, 2, 3\}. \)

If \( n = 4q + 0, \) then\( \left\lceil \frac{n}{2} \right\rceil = \left\lceil \frac{4q}{2} \right\rceil = 2q. \)
If \( n = 4q + 1, \) then\( \left\lceil \frac{n}{2} \right\rceil = \left\lceil \frac{4q + 1}{2} \right\rceil = 2q + 1. \)
If \( n = 4q + 2, \) then\( \left\lceil \frac{n}{2} \right\rceil = \left\lceil \frac{4q + 2}{2} \right\rceil = 2q + 1. \)
If \( n = 4q + 3, \) then\( \left\lceil \frac{n}{2} \right\rceil = \left\lceil \frac{4q + 3}{2} \right\rceil = 2q + 2. \)

We see that \( \left\lceil \frac{n}{2} \right\rceil \) is even if and only if either \( n \equiv 0 \pmod{4} \) or \( n \equiv 3 \pmod{4} \) holds good.

We have shown that if a sequence \( S_n \) has Langford pairings, then either \( n \equiv 0 \pmod{4} \) or \( n \equiv 3 \pmod{4}. \) This proves the necessity of the condition.

Sufficiency

If we can show that we can construct a Langford pairing for \( (1, 1, 2, 2, \dots, n, n ) \) for both cases, i.e. \( n \equiv 0 \pmod{4} \) as well as \( n \equiv 3 \pmod{4}, \) then it would complete the proof.

Notation

Let us define some notation to make it easier to write sequences we will use in the construction of a Langford pairing:

\( (i \dots j)_{even} \) denotes a sequence of even positive integers from \( i \) to \( j, \) exclusive, arranged in ascending order.

For example, \( (1 \dots 8)_{even} = (2, 4, 6) \) and \( (1 \dots 2)_{even} = (). \)
\( (i \dots j)_{odd} \) denotes a sequence of odd positive integers from \( i \) to \( j, \) exclusive, arranged in ascending order.

For example, \( (1 \dots 8)_{odd} = (3, 5, 7) \) and \( (1 \dots 3)_{odd} = (). \)
\( s' \) denotes the reverse of the sequence \( s. \)

For example, for a sequence \( s = (2, 3, 4, 5), \) we have \( s' = (2, 3, 4, 5)' = (5, 4, 3, 2). \)
\( s \cdot t \) denotes the concatenation of sequences \( s \) and \( t. \)

For example, for sequences \( s = (1, 2, 3) \) and \( t = (4, 5), \) we have \( s \cdot t = (1, 2, 3) \cdot (4, 5) = (1, 2, 3, 4, 5). \)

Let \( x = \left\lceil \frac{n}{4} \right\rceil. \) Therefore, \[ x = \begin{cases} \frac{n}{4} & \text{if } n \equiv 0 \pmod{4}, \\ \frac{n + 1}{4} & \text{if } n \equiv 3 \pmod{4}. \end{cases} \] Let us now define the following eight sequences: \begin{align*} a & = (2x - 1), \\ b & = (4x - 2), \\ c & = (4x - 1), \\ d & = (4x), \\ p & = (0 \dots a)_{odd}, \\ q & = (0 \dots a)_{even}, \\ r & = (a \dots b)_{odd}, \\ s & = (a \dots b)_{even}. \end{align*} Now let us construct a Langford pairing for both cases: \( n \equiv 0 \pmod{4} \) and \( n \equiv 3 \pmod{4}. \) We will do this case by case.

Case \( n \equiv 0 \pmod{4} \)

If \( n \equiv 0 \pmod{4}, \) we construct a Langford pairing with the following concatenation: \[ s' \cdot p' \cdot b \cdot p \cdot c \cdot s \cdot d \cdot r' \cdot q' \cdot b \cdot a \cdot q \cdot c \cdot r \cdot a \cdot d. \] Let us do an example with \( n = 12. \)

For \( n = 12, \) we get \( x = \frac{n}{4} = 3. \) Therefore, \begin{alignat*}{2} a & = (2x - 1) && = (5), \\ b & = (4x - 2) && = (10), \\ c & = (4x - 1) && = (11), \\ d & = (4x) && = (12), \\ p & = (0 \dots a)_{odd} && = (1, 3), \\ q & = (0 \dots a)_{even} && = (2, 4), \\ r & = (a \dots b)_{odd} && = (7, 9), \\ s & = (a \dots b)_{even} && = (6, 8). \end{alignat*} After performing the specified concatenation, we get the following Langford pairing: \[ ( 8, 6, 3, 1, 10, 1, 3, 11, 6, 8, 12, 9, 7, 4, 2, 10, 5, 2, 4, 11, 7, 9, 5, 12 ). \] Let us now show that any construction of a sequence as per this specified concatenation always leads to a Langford pairing.

Each sequence \( a, \) \( b, \) \( c \) and \( d \) has one number each. Each sequence \( p, \) \( q, \) \( r \) and \( s \) has \( x - 1 \) numbers each.

The two occurrences of \( a \) have \( q, \) \( c \) and \( r \) in between, i.e. \[ (x - 1) + 1 + (x - 1) = 2x - 1 = a \] numbers in between. Similarly, we can check that the two occurrences of \( b \) have \( b \) numbers in between; likewise for \( c \) and \( d. \)

The two occurrences of \( 1 \) belong to \( p \) and \( p'. \) Between these two occurrences of \( 1, \) we have only one element of \( b. \)

We now show that for each \( k \) in \( p, \) there are \( k \) numbers in between. For any \( k \) in \( p, \) there is the sequence \( (0..k)'_{odd} \cdot b \cdot (0..k)_{odd} \) in between the two occurrences of \( k, \) i.e, there are \( \frac{k - 1}{2} + 1 + \frac{k - 1}{2} = k \) numbers in between. Similarly, we can check that for each \( k \) in \( q, \) there are \( k \) numbers in between.

Finally, we show that for each \( k \) in \( r, \) there are \( k \) numbers in between. For any \( k \) in \( r, \) there is the sequence \( (a..k)'_{odd} \cdot q' \cdot b \cdot a \cdot q \cdot c \cdot (a..k)_{odd} \) in between the two occurrences of \( k. \) Note that \( a \) is odd, so the number of integers in this sequence is \[ \frac{k - a - 2}{2} + (x - 1) + 1 + 1 + (x - 1) + 1 + \frac{k - a - 2}{2}. \] Simplifying the above expression and then substituting \( a = 2x - 1, \) we get \[ k - a - 2 + 2x + 1 = k. \] Similarly, we can check that for each \( k \) in \( s, \) there are \( k \) numbers in between.

Case \( n \equiv 3 \pmod{4} \)

If \( n \equiv 3 \pmod{4}, \) we construct a Langford pairing with the following concatenation: \[ s' \cdot p' \cdot b \cdot p \cdot c \cdot s \cdot a \cdot r' \cdot q' \cdot b \cdot a \cdot q \cdot c \cdot r. \] Note that this concatenation of sequences is almost the same as the concatenation in the previous section with the following two differences:

The sequence \( d \) is not used here.
The sequence \( a \) in the end has moved to replace the sequence \( d \) in the middle of the concatenation.

Let us do an example with \( n = 11. \) For \( n = 12, \) we get \( x = \frac{n + 1}{4} = 3. \) Therefore, the sequences \( a, \) \( b, \) \( c, \) \( p, \) \( q, \) \( r \) and \( s \) are same as those in the last example in the previous section. After performing the specified concatenation, we get the following Langford pairing: \[ ( 8, 6, 3, 1, 10, 1, 3, 11, 6, 8, 5, 9, 7, 4, 2, 10, 5, 2, 4, 11, 7, 9 ). \] We can verify that for every \( k \) in a Langford pairing constructed in this manner, there are \( k \) numbers in between. The verification steps are similar to what we did in the previous section.

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From Out Shuffles to Multiplicative Order

2011-03-23T00:00:00Z

Out Shuffles

A perfect riffle shuffle of a deck of cards involves splitting the deck into two halves (one in each hand) and then alternately dropping one card from each half till all cards have fallen. If the shuffling is done in such a manner that the bottom and the top cards remain preserved at their positions, then it is an out shuffle.

Problem

Here is a problem that a colleague asked me recently while discussing shuffling techniques:

  How many out shuffles do we need to perform on a deck of 52 cards to
  return the deck to its initial state?

Solution

The solution to this problem is rather long, so it has been split into three parts below.

Part 1: Congruence Relation

Let us assume that there are \( n \) cards where \( n \) is a positive even integer. Let us denote these cards as \( c_0, c_1, \dots, c_{n-1} \) where \( c_0 \) is the card at index \( 0 \) (top of the deck), \( c_{n - 1} \) is the card at index \( n - 1 \) (bottom of the deck) and \( c_i \) is the card at index \( i \) where \( 0 \le i \le n - 1. \)

For example, if we have \( 8 \) cards, then the cards are denoted as \[ c_0, c_1, c_2, c_3, c_4, c_5, c_6, c_7. \] After the first out shuffle, these cards are now in this order: \[ c_0, c_4, c_1, c_5, c_2, c_6, c_3, c_7. \]

From the problem description and the example above, we see that after a single out shuffle, a card at index \( i \) moves to index \( 2i \bmod (n - 1) \) for \( 0 \le i \lt n - 1 \) and the card at index \( n - 1 \) remains at the same place.

After \( m \) shuffles a card at index \( i \) moves to index \( (2^m i) \bmod (n - 1) \) for \( 0 \le i \lt n - 1. \) The card at index \( n - 1 \) always remains at the same place.

The solution to the problem then is the smallest positive integer \( m \) such that \[ 2^m i \equiv i \pmod{n - 1}. \] for all integers \( i \) that satisfy the inequality \( 0 \le i \lt n - 1. \)

In modular arithmetic, we know that \[ ac \equiv bc \pmod{n} \iff a \equiv b \pmod{n / \gcd(c, n)}. \] where \( a, \) \( b, \) \( c \) and \( n \) are integers. Therefore \[ 2^m i \equiv i \pmod{n - 1} \iff 2^m \equiv 1 \pmod{(n - 1) / \gcd(i, n - 1)}. \] Let \( F_{ni} = \frac{n - 1}{\gcd(i, n - 1)} \) where \( 0 \le i \lt n - 1. \) Now the above congruence relation can be written as \[ 2^m \equiv 1 \pmod{F_{ni}}. \]

Part 2: Multiplicative Order

The smallest positive integer \( m \) that satisfies the above congruence relation is known as the multiplicative order of \( 2 \) modulo \( F_{ni}. \) It is denoted as \( \ord_{F_{ni}}(2). \)

In general, given an integer \( a \) and a positive integer \( n \) with \( \gcd(a, n) = 1, \) the multiplicative order of \( a \) modulo \( n, \) denoted as \( \ord_{n}(a), \) is the smallest positive integer \( k \) such that \[ a^k \equiv 1 \pmod{n}. \] In this problem, \( n \) is even, so \( n - 1 \) is odd as a result of which \( F_{ni} \) is also odd. As a result, \( \gcd(2, F_{ni}) = 1. \) Therefore, the smallest positive integer \( m \) that satisfies the congruence relation \( 2^m \equiv 1 \pmod{F_{ni}} \) is denoted as \( \ord_{F_{ni}}(2). \)

If \( n = 2, \) \( F_{n0} = F_{n1} = 1, \) therefore \( 2^m \equiv 1 \pmod{F_{ni}} \) for \( 0 \le i \lt n - 1 \) and all positive integers \( m. \) This proves the trivial observation that when there are only two cards \( c_0 \) and \( c_1, \) they remain at index \( 0 \) and index \( 1 \) respectively, after any number of out shuffles, i.e. their positions do not change with out shuffles.

Case \( n \ge 4 \)

If \( n \ge 4, \) we know that there exists at least one integer between \( 1 \) and \( n - 1 \) that is coprime to \( n - 1 \) because \( \varphi(n) \ge 2 \) for \( n \ge 4 \) where \( \varphi(n) \) represents Euler's totient of \( n. \)

Subcase \( i \) is coprime to \( n - 1 \)

For any arbitrary \( n \ge 4, \) let \( i \) be an integer that is coprime to \( n - 1 \) such that \( 1 \lt i \lt n - 1. \) Now \( \gcd(i, n - 1 ) = 1, \) so \( F_{ni} = \frac{n - 1}{\gcd(i, n - 1)} = n - 1. \) As a result, \( \ord_{F_{ni}}(2) = \ord_{n - 1}({2}). \) This shows that any card at index \( i \) such that \( i \) is coprime to \( n - 1 \) requires a minimum of \( \ord_{n - 1}(2) \) out shuffles to return to its initial place.

Subcase \( i \) is not coprime to \( n - 1 \)

For any arbitrary \( n \ge 4, \) now let us consider the case of a card at index \( i \) such that \( i \) is not coprime to \( n - 1. \) Since \( n - 1 \) is odd, we have \( \gcd(2, n - 1) = 1. \) Therefore, by definition of multiplicative order, \[ 2^{\ord_{n - 1}(2)} \equiv 1 \pmod{n - 1}. \] Since \( F_{ni} \mid n - 1, \) we get, \[ 2^{\ord_{n - 1}(2)} \equiv 1 \pmod{F_{ni}}. \] This shows that a card at index \( i \) such that \( i \) is not coprime to \( n - 1 \) also return to its initial place after \( \ord_{n - 1}(2) \) out shuffles. Actually, it takes only \( \ord_{F_{ni}}(2) \) out shuffles to return the card to its initial place. But \( \ord_{F_{ni}}(2) \mid \ord_{n - 1}(2), \) so \( \ord_{n - 1}(2) = c \ord_{F_{ni}}(2) \) for some positive integer \( c. \) Therefore performing \( \ord_{n - 1}(2) \) out shuffles is same as repeating \( \ord_{F_{ni}}(2) \) out shuffles \( c \) times. Every \( \ord_{F_{ni}}(2) \) brings back the card to its initial position, so repeating it \( c \) times also brings it back to its initial position.

We have shown that a card at index \( i \) such that \( i \) is coprime to \( n - 1 \) takes a minimum of \( \ord_{n - 1}(2) \) out shuffles to return to its initial place. We have also shown that a card at other indices also return to its initial place after the same number of out shuffles. Therefore, it takes a minimum of \( \ord_{n - 1}{2} \) out shuffles to return the deck of cards to its initial state.

Case \( n = 2 \)

When there are only \( 2 \) cards in the deck, every out shuffle trivially returns the deck to its initial state. In other words, it takes only \( 1 \) out shuffle to return the deck to its initial state. Indeed \( \ord_{n - 1}(2) = \ord_{1}(2) = 1. \)

We have now shown that for any positive even integer \( n, \) a deck of \( n \) cards returns to its initial state after \( \ord_{n - 1}(2) \) out shuffles.

Part 3: Computing Multiplicative Order

For a deck of \( 52 \) cards, we have \( n = 52. \) A minimum of \( \ord_{51}(2) \) out shuffles are required to return the deck to its initial state. To compute \( \ord_{51}(2) \) we first enumerate the powers of \( 2 \) modulo \( 51 \): \begin{alignat*}{2} 2^0 & \equiv 1 && \pmod{51}, \\ 2^1 & \equiv 2 && \pmod{51}, \\ 2^2 & \equiv 4 && \pmod{51}, \\ 2^3 & \equiv 8 && \pmod{51}, \\ 2^4 & \equiv 16 && \pmod{51}, \\ 2^5 & \equiv 32 && \pmod{51}, \\ 2^6 & \equiv 13 && \pmod{51}, \\ 2^7 & \equiv 26 && \pmod{51}, \\ 2^8 & \equiv 1 && \pmod{51}. \end{alignat*} The smallest positive integer \( k \) such that \( 2^k \equiv 1 \pmod{51} \) is 8, so \( \ord_51(2) = 8. \) We need \( 8 \) out shuffles to return a deck of \( 52 \) cards to its initial state.

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From Diophantine Equation to Fermat's Last Theorem

2011-01-12T00:00:00Z

Here is a puzzle I created recently for my friends who love to indulge in recreational mathematics:

  Find all integer solutions to the equation

  \[
    y^2 + 3 = \frac{x^3}{18}.
  \]

If you want to think about this puzzle, this is a good time to pause and think about it. There are spoilers ahead.

It does not take long to realise that this is a Diophantine equation of the form \( a^n + b^n = c^n. \) Here is how the equation looks after rearranging the terms: \[ x^3 = 18y^2 + 54. \]

The right hand side is positive, so any \( x \) that satisfies this equation must also be positive, i.e. \( x \gt 0 \) must hold good for any solution \( x \) and \( y. \)

Also, if some \( y \) satisfies the equation, then \( -y \) also satisfies the equation because the right hand side value remains the same for both \( y \) and \( -y. \)

The right hand side is \( 2(9y^2 + 3^3). \) This is of the form \( 2(3a^2b + b^3) \) where \( a = y \) and \( b = 3. \) Now \( 2(3a^2b + b^3) = (a + b)^3 - (a - b)^3. \) Using these details, we get \[ x^3 = 18y^2 + 54 = 2(9y^2 + 3^3) = (y + 3)^3 - (y - 3)^3. \] Rearranging the terms, we get \[ x^3 + (y - 3)^3 = (y + 3)^3. \] From Fermat's Last Theorem, we know that an equation of the form \( a^n + b^n = c^n \) does not have any solution for positive integers \( a, \) \( b, \) \( c \) and positive integer \( n \gt 2. \) We know that \( x \gt 0. \) Therefore \( y \gt 3 \) contradicts Fermat's Last Theorem. Thus the inequality \( y \le 3 \) must hold good. Further since for every solution \( x \) and \( y, \) there is also a solution \( x \) and \( -y, \) the inequality \( -y \le 3 \) must also hold good. Therefore values of \( x \) and \( y \) that satisfy the above equation must satisfy the following inequalities: \begin{align*} x \gt 0, \\ -3 \le y \le 3. \end{align*} Since \( y \) must be one of the seven integers between \( -3 \) and \( 3, \) inclusive, we can try solving for \( x \) with each of these seven values of \( y. \) When we do so, we find that there are only two values of \( y \) for which we get integer solutions for \( x. \) They are \( y = 3 \) and \( y = -3. \) In both cases, we get \( x = 6. \) Therefore, the solutions to the given equation are: \[ x = 6, \qquad y = \pm 3. \]

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Combinatorics Coincidence

2008-09-14T00:00:00Z

Combinatorics for Fun

At my current workplace, there are several engineers who have an affinity for combinatorics. As a result, discussions about combinatorics problems often occur at the cafeteria. Probability theory is another popular topic of discussion. Of course, often combinatorics and probability theory go hand in hand.

Recurrence Relation

At the cafeteria one day, I joined in on a conversation about combinatorics problems. During the conversation, I happened to share the following problem:

For integers \( n \ge 1 \) and \( k \ge 1, \) \[ f_k(n) = \begin{cases} n & \text{if } k = 1, \\ \sum_{i=1}^n f_{k-1}(i) & \text{if } k \ge 2. \end{cases} \] Find a closed-form expression for \( f_k(n). \)

Nested Loops

Soon after I shared the above problem, a colleague of mine shared this problem with me:

Consider the following pseudocode with k nested loops:

x = 0
for c₁ in 0 to (n - 1):
    for c₂ in 0 to (c₁ - 1):
        ...
            for c_k in 0 to (c_k-1 - 1):
                x = x + 1

What is the final value of x after the outermost loop terminates?

Coincidence

With one problem each, we went back to our desks. As I began solving the nested loops problem shared by my colleague, I realised that the solution to his problem led me to the recurrence relation in the problem I shared with him.

In the nested loops problem, if \( k = 1, \) the final value of \( x \) after the loop terminates is \( x = n. \) This is also the value of \( f_1(n). \)

If \( k = 2, \) the inner loop with counter \( c_2 \) runs once when \( c_1 = 0, \) twice when \( c_1 = 1 \) and so on. When the loop terminates, \( x = 1 + 2 + \dots + n. \) Note that this series is same as \( f_2(n) = f_1(1) + f_1(2) + \dots + f_1(n). \)

Extending this argument, we now see that for any \( k \ge 1, \) the final value of \( x \) is \[ f_k(n) = f_{k-1}(1) + f_{k-1}(2) + \dots + f_{k-1}(n). \] In other words, the solution to his nested loops problem is the solution to my recurrence relation problem. It was an interesting coincidence that the problems we shared with each other had the same solution.

Closed-Form Expression

The closed form expression for the recurrence relation is \[ f_k(n) = \binom{n + k - 1}{k}. \] It is quite easy to prove this using the principle of mathematical induction. Since we know that this is also the result of the nested loops problem, we can also arrive at this result by another way.

In the nested loops problem, the following inequalities are always met due to the loop conditions: \[ n - 1 \ge c_1 \ge c_2 \ge \dots \ge c_k \ge 0. \] The variables \( c_1, c_2, \dots, c_k \) take all possible arrangements of integer values that satisfy the above inequalities. If we find out how many such arrangements are there, we will know how many times the variable \( x \) is incremented.

Let us consider \( n - 1 \) similar balls and \( k \) similar sticks. For every possible permutation of these balls and sticks, if we count the number of balls to the right of the \( i \)th stick where \( 1 \le i \le k, \) we get a number that the variable \( c_i \) holds in some iteration of the \( i \)th loop. Therefore the variable \( c_i \) is represented as the number of balls lying on the right side of the \( i \)th stick.

The above argument holds good because the number of balls on the right side of the first stick does not exceed \( n - 1, \) the number of balls on the right side of the second stick does not exceed the number of balls on the right side of the first stick and so on. Thus the inequalities mentioned earlier are always satisfied. Also, any set of valid values for \( c_1, c_2, \dots, c_k \) can be represented as an arrangement of these sticks and balls.

The number of permutations of \( n - 1 \) similar balls and \( k \) similar sticks is \[ \frac{(n + k - 1)!}{(n - 1)! \, k!} = \binom{n + k - 1}{k}. \] This closed-form expression is the solution to both the recurrence relation problem and the nested loops problem.

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