Combinatorics

Proportion of 1s in a Hadamard matrix

Posted on by John

The first post in the recent series of posts on Hadamard matrices describes a way of constructing new Hadamard matrices from two other Hadamard matrices by taking their Kronecker product.

Starting with a Hadamard matrix H0 and a Hadamard matrix G, you can construct a sequence of Hadamard matrices by

Hn+1 = G ⊗ Hn

for  positive integers n. This is known as the generalized Sylvester method.

Let pn be the proportion of 1s in Hn and let q be the proportion of 1s in G. Then you can show that the recurrence holds

pn+1 = q pn + (1 − q)(1 − pn).

You can solve the recurrence to show that

limn → ∞ pn = ½

and so as the iterations proceed, the ratio of number of 1s to the number of −1s approaches 1.

This doesn’t say anything Hadamard matrices in general, but it does apply to all Hadamard matrices created by repeatedly applying the generalized Sylvester method.

If you set G and H equal to the matrix

\begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix}

then p0 = q = ¾. Then for n = 1, 2, 3, …, 8 the values of pn are

0.625
0.5625
0.53125
0.515625
0.5078125
0.50390625
0.501953125
0.5009765625.

 

Compressing a Hadamard matrix

Posted on by John

Hadamard matrices are in the news following the recent announcement of a newly discovered Hadamard matrix. I’ve written three posts on Hadamard matrices recently, one as a sort of introduction and two on applications: the error correcting code used in the Mariner 9 probe and constructing sphere packings.

A Hadamard matrix is an orthogonal matrix with all entries equal to ±1. Jacques Hadamard conjectured that there exist Hadamard matrices of order 4n for all positive integers n. It’s necessary that the order be divisible by 4, and Hadamard conjectured that this is sufficient [1].

How could you compactly represent a Hadamard matrix? Since the entries are all either 1 or − 1 each entry could be represented by a single bit, and n² bits could store an n × n Hadamard matrix. But we can do better.

Methodical matrices

If the matrix can be produced by an algorithm, you only need to store the name of the algorithm and the argument to the algorithm. So, for a 1024 × 1024 matrix applied by iterating Sylvester’s algorithm could be stored by saying “Apply Sylvester’s algorithm 10 times” rather than storing a megabyte of data.

Paley’s method can create a Hadamard matrix corresponding to every prime power. So you could determine a Paley type matrix by storing the prime and the exponent.

Next in complexity would be hybrid algorithms, such as start with the Paley method applied to 376 and then apply Sylvester’s method 3 times.

There are more methods of creating Hadamard matrices than Sylvester’s method and Paley’s method, though they’re harder to describe and parameterize.

Sporadic matrices

If a Hadamard matrix cannot be constructed using an algorithm, you can still store the matrix in fewer than n² bits. Since the rows are orthogonal, the last row of the matrix is determined by all the previous rows, up to sign. So you could store a Hadamard matrix using n(n − 1) + 1 bits.

Some Hadamard matrices are symmetric or skew. A symmetric matrix is determined by its diagonal and the elements above the diagonal. So a symmetric Hadamard matrix could be represented by n(n + 1)/2 bits.

A skew Hadamard matrix isn’t quite skew-symmetric. A matrix M is skew symmetric if

MT = −M.

This implies the diagonal elements are 0, and Hadamard matrices cannot contain 0s. A Hadamard matrix H is called skew if

H + HT = 2I.

This implies the diagonal elements are all 1s and the elements below the diagonal have the opposite sign of the elements above the diagonal. Since the elements on the diagonal are determined, a skew Hadamard matrix can be sorted using n(n − 1)/2 bits.

Incidentally, there is a conjecture that there exist skew Hadamard matrices of order 4n for all positive n.

 

[1] There are Hadamard matrices of order 1 and 2, but larger orders must be divisible by 4.

Constructing Hadamard matrices

Posted on by John

A Hadamard matrix is an orthogonal matrix whose entries are all either 1 or − 1. For example

\begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix}

is a Hadamard matrix of order 2. True to Stigler’s law of eponymy, James Joseph Sylvester investigated Hadamard matrices before Jacques Hadamard. Sylvester saw how to bootstrap the example above into more examples. If H is a Hadamard matrix, then the partitioned matrix

\begin{bmatrix} H & H\\ H & -H \end{bmatrix}

Sylvester’s construction can be generalized as follows. If Hm is a Hadamard matrix of order m and Hn is a Hadamard matrix of order n, the the Kronecker product Hm⊗Hn is a Hadamard matrix of order mn. That is, you can form a new Hadamard matrix by taking the matrix Hm and replacing ±1 with the matrix ±Hn.

Let S be the set of all possible Hadamard matrix orders. By the construction above, this set is closed under multiplication. Since 2 is in S, every power of 2 is in S. Hadamard proved that all n ≥ 4 in S are multiples of 4. That is, the condition 4 | n is necessary. He conjectured that it was also sufficient, though that has not been proven.

So the big question is what is the set S. Is there some multiple of 4 not in S? Until that question is answered, what is the smallest multiple of 4 not known to be in S? Hadamard matrices are useful in applications, so constructing Hadamard matrices of various orders is useful even while Hadamard’s conjecture remains open. For example, see the next post for how NASA used Hadamard matrices to transmit photographic images back from Mars.

Paley’s method

Raymond Paley came up with a way of constructing Hadamard matrices of size q + 1 if q is a prime power congruent to 3 mod 4, and of size 2(q + 1) if q is a prime power congruent to 1 mod 4. Let’s see what we can squeeze out of this.

If p is a prime congruent to 1 mod 4, every power of p is also congruent to 1 mod 4, and so there exist Hadamard matrices of order 2(pk + 1) for every k.

If p is a prime with p = 3 mod 4, then even powers of p are congruent to 1 mod 4 and odd powers of p are congruent to 3 mod 4. So there are Hadamard matrices of order 2(p2k + 1) and of order p2k+1 + 1.

Let’s run a script to see what we can learn from this.

from sympy import primerange

s = set()

for p in primerange(20):
    if p % 4 == 1:
        s.update([2*(p**k + 1) for k in range(1, 10)])
    if p % 4 == 3:
        s.update([2*(p**(2*k) + 1) for k in range(1, 6)])
        s.update([p**(2*k + 1) + 1 for k in range(1, 6)])
print(sorted(s)[:20])

This prints

12, 20, 28, 36, 52, 100, 164, 244, 252, 340]

We can add 16 to the list because it’s a power of 2, and we can add 24 because it’s 2 × 12, etc. But there doesn’t seem to be any way to get 44. There is a way to create a Hadamard matrix of order 44, but it doesn’t follow from anything we’ve seen so far.

New records

I have a book published in 1996 that says Hadamard’s conjecture had been verified for n up to 428. Until yesterday, the smallest multiple of 4 for which nobody had found a corresponding Hadamard matrix was 668. Then Levent Alpöge announced that he and his and collaborators found an example of size 668 and filled in all remaining gaps below 2000.

So now the set S is known to contain {1, 2, 4, 8, 12, 16, …, 2000}. It also contains all orders that can be obtained by Paley’s method and other methods. And it contains all products of its elements. But it is not yet known to contain 2004.

Related posts

Counting rooted trees

Posted on by John

Combinatorial problems can be interesting for their own sake, but they are more interesting when there is a connection to a problem outside combinatorics, and the more unexpected the connection the better.

Counting the number of unlabeled rooted trees [1] with n nodes is a pure mathematics problem. Designing numerical methods for solving differential equations is an applied mathematics problem. And yet the two are closely linked.

Let t(n) be the number of distinct unlabeled rooted trees with n nodes. The diagram below shows that the first few terms of this sequence are 1, 1, 2, and 4.

Recursive calculation

The values of t(n) can be computed recursively using

\begin{align*} g_k &= \sum_{d\mid k} d t_d \\ t_1 &= 1 \\ t_n &= \frac{1}{n-1} \sum_{k=1}^{n-1} g_k t_{n-k} \text{ for } n > 1<br /> \end{align*}<br />

You can implement this in Python as follows.

from sympy import divisors

def t(n):
    if n <= 1:
        return 1 if n == 1 else 0
    return sum(g(k) * t(n - k) for k in range(1, n)) // (n - 1)

def g(k):
    return sum(d * t(d) for d in divisors(k))

This code is correct, but it will run more efficiently if you cache function values to avoid calculating the same values over and over. You can do this by adding

from functools import lru_cache

and writing @lru_cache(maxsize=None) above both function definitions.

Connection to Runge-Kutta

In an earlier post I showed that designing a 4-stage explicit Runge-Kutta method required solving a system of 8 equations in 10 unknowns, leaving two degrees of freedom in the solutions.

The number of constraints c(s) needed to design an s-stage explicit RK method is equal to the number of rooted trees with up to s nodes:

c(s) = t(1) + t(2) + t(3) + … + t(s)

This is because there is a one-to-one correspondence between constraints on the nth derivative of an RK formula and rooted trees, and an s stage method has to satisfy the constraints of all stages up to s. In the example of the 4th order RK method, we have

c(4) = t(1) + t(2)  + t(3) + t(4) = 1 + 1 + 2 + 4 = 8.

The first few values [2] of t(n) are

1, 1, 2, 4, 9, 20, 48, 115, 286, 719, 1842, 4766, 12486, 32973, …

and so you can see that t(n) grows quickly. In fact, it grows exponentially [3].

However, the number of parameters in an s stage RK method is s(s + 1)/2. The number of equations grows exponentially and the number of variables grows only quadratically, so at some point you have more equations than variables. That’s already the case for s = 5 because you have 17 constraints on 15 variables. The system has a solution because symmetry considerations render some of the equations redundant.

A 10th order RK method requires 17 stages. (See the previous post for why the number of stages exceeds the order when the order is greater than 4.) Designing such a method would require solving over a million equations in 153 variables, and yet it can be done. [4]

Related posts

[1] This is a slightly contradictory term. Unlabeled means the we don’t distinguish the nodes. But we do distinguish one node, namely the root.

[2] See OEIS A000081.

[2] Richard Otter proved in 1948 that the number of unlabeled rooted trees with n nodes is asymptotically C αn / n−3/2 where C = 0.4399… and α = 2.9557…. The cumulative sum is at least this large since Otter’s estimate gives the size of the last term in the sum.

[3] E. Hairer. A Runge-Kutta Method of Order 10. J. Inst. Maths Applics (1978) 21, 47-59

Cryptographic Keys and Decks of Cards

Posted on by John

The previous post looked at the idea of storing a cryptographic key in the order of a deck of cards. A deck of 52 cards can store 225 bits of data because

⌊log2(52!)⌋ = 225.

Here ⌊x⌋ is x rounded down to the nearest integer.

If we want to store bigger keys, we’re going to need a bigger deck of cards.

Bitcoin

A Bitcoin key has 256 bits, which would require a deck of 58 cards. There is a card game called Zwicker that uses a deck of 58 cards, the usual 52 cards plus six jokers. So you could store a Bitcoin key in the permutation of a Zwicker deck.

You could also use a deck of 52 cards, plus 2 jokers, if you also consider orientation. 30 cards are rotationally symmetric, 22 are not, and neither are jokers. So, including two asymmetric jokers, you could add 24 additional bits. Permutations of a 54 card deck can encode 237 bits, and with 24 orientation bits, this is a total of 261 bits.

RSA

RSA key sizes vary, but 2048 and 3072 are common. A 2048-bit key would require a deck of 301 cards. Casinos often use a shoe of 312 cards, combining six decks of 52 cards, to deal Baccarat or Blackjack. However, casinos combine identical decks. If you were to combine six unique decks, you could store a 2048-bit key.

Storing a 3072-bit key would require a deck of 422 cards. You could make a deck of 432 cards by combining 8 distinguishable packs of 54 cards (52 + 2 jokers).

ML-KEM

ML-KEM is a proposed quantum-resistant replacement for RSA. As with RSA, key sizes for ML-KEM vary, the smallest being ML-KEM-512 with a key size of 1632 bytes, which equals 13056 bits. This would require a deck of 1442 cards. You could combine 28 distinct packs of 52 cards, but that’s unwieldy.

This illustrates one of the difficult trade-offs with post-quantum cryptography: key sizes are much bigger. If you wanted to create a deck of 1442 cards, you’d probably want to make your “cards” something other than standard playing cards. You’d want to use permutations of something else.

Verification

The following Python code verifies the calculations above.

from math import log2, factorial, floor

def capacity(cards):
    return floor(log2(factorial(cards)))

def verify(bits, cards):
    return capacity(cards) >= bits and capacity(cards-1) < bits

print(verify(237, 54))
print(verify(256, 58))
print(verify(2048, 301))
print(verify(3072, 422))
print(verify(1632*8, 1442))

For more on how I came up with the deck sizes, see the next post on computing the inverse factorial.

Hiding data in permutations

Posted on by John

The latest issue of Paged Out! has an article by Stephen Hewitt “An off-line backup of your cryptographic key using playing cards.” The idea is to use a deck of 52 to store a 128-bit cryptographic key. To erase the key, shuffle the deck. Hewitt gives his algorithm for embedding a key, one that can be carried out manually but isn’t maximally efficient.

You could store a 225-bit key as a permutation of 52 cards because

log2(52!) = 225.581.

But then how would you number permutations so you could go from a number to a particular permutation and later decode the permutation to a number? Is this even practical? For a small number n, you could encode a number k < n by enumerating the first k permutations of a set of n items, and you could decode by enumerating permutations until you find the one you have. But this is completely impractical for large n, such as n = 52.

The process of mapping permutation to an integer is called ranking, and the mapping from an integer to a permutation is called unranking. How efficiently can rankings and unrankings be calculated?

Let n be the number of symbols being permuted. Then there are simple algorithms for ranking and unranking with respect to lexicographical order that have complexity O(n²) and more sophisticated algorithms that have complexity O(n log n). There are also O(n) algorithms that do not preserve lexicographical order.

The Permutations class in SymPy has methods unrank_lex and rank to unrank and rank permutations according to lexicographical order.

The notation the Permutations class uses requires a little explanation. For example, suppose we unrank 2026.

>>> from sympy.combinatorics import Permutation
>>> Permutation.unrank_lex(52, 2026)
Permutation(45, 47, 51, 48, 46, 50)

The output is not a full list of 52 numbers in permuted order; it is only a cycle. The notation refers to the permutation that sends 45 to 47, 47 to 51, …, 50 to 45 and leaves everything else fixed.

If we rank the permutation given above, we get 2026 back.

>>> Permutation.rank(Permutation(45, 47, 51, 48, 46, 50))
2026

Note that we didn’t say how many elements (45, 47, 51, 48, 46, 50) is a permutation of. Because of lexicographical order, the rank would be the same whether we viewed this as a permutation of 52 objects or of more objects.

Now let’s do something larger. Let’s generate a 220-bit number and encode it as a permutation.

>>> n = random.getrandbits(225)
>>> a = Permutation.unrank_lex(52, n)
>>> n
40234719030664563684489051530416964877785781669439875437823431388841
>>> a
Permutation(0, 25, 32, 15, 8, 28)(1, 48, 34, 14, 10, 51, 38, 31, 21, 5, 42, 47, 29, 26, 46, 30, 50, 49, 37, 22, 18, 23)(2, 45, 17, 20, 36, 40, 11, 4, 7, 41, 33, 3, 43, 44, 19, 16, 35, 39, 12, 6, 9)
>>> Permutation.rank(a) == n
True

Now just for fun, let’s display the permutation above applied to a standard (French) deck of 52 cards. As explained here, symbols associated with these cards have a range of Unicode values. By printing these values, we can visualize the permuted deck.

Here’s the code that made the image above.

spades = list(range(0x1F0A1, 0x1F0AF))
spades.remove(0x1F0AC) # take out the knight
cards = [s + 16*i for s in spades for i in range(4)]

a = Permutation.unrank_lex(52, n)
p = a(cards)

for i in range(4):
    for j in range(13):
        print(chr(p[13*i + j]), end="")
    print()

The code above is plenty fast, but Permutation has methods rank_nonlex and unrank_nonlex that run in O(n) time, which could be useful for n much larger than 52.

Counting permutations with roots

Posted on by John

My post from yesterday on permutation roots ends with a Mathematica code for finding the probability that a permutation of n elements has a kth root. This is done by finding the coefficient of xn in the generating function

\prod_{m=1}^\infty \exp_{\text{gcd}(m, k)} \left\frac{x^m}{m}\right)

I wanted to say more about this, and look at implementing the same code in SymPy. I was curious how well SymPy would do because I’ve noticed that LLMs often generate SymPy code since it’s an open source CAS.

Wilf [1] describes the infinite product above as the exponential generating function (egf) of f(n, k), the number of permutations of n objects that have a kth root. Since egfs have a n! term in the denominator, this is also the ordinary generating function (ogf) of the probability that a randomly chosen permutation on n objects has a kth root.

My first attempt at using Mathematica to probe the generating function was

expq[x_, q_] := MittagLefflerE[q, x^q]	 
p[n_, k_] :=  SeriesCoefficient[	 
    Product[expq[x^m/m, GCD[m, k]], {m, 1, Infinity}], {x, 0, n}]

This hung forever when I tried to use it on a small example. I realized, but apparently Mathematica did not, that Infinity could be replaced by n since terms higher than n do not contribute to the coefficient of xn. With that change, the code ran quickly.

This morning I tried converting the Mathematica code to Sympy; Claude did this in one shot. I also reproduced the table of f(n, k) values on page 150 of [1] to test the code. Since Wilf tabulated f(n, k), not f(n, k)/n!, I multiplied the results by n!.

Here is the output:

k = 2 [1, 1, 3, 12, 60, 270, 1890, 14280, 128520, 1096200]
k = 3 [1, 2, 4, 16, 80, 400, 2800, 22400, 181440, 1814400]
k = 4 [1, 1, 3, 12, 60, 270, 1890, 13020, 117180, 1039500]
k = 5 [1, 2, 6, 24, 96, 576, 4032, 32256, 290304, 2612736]
k = 6 [1, 1, 1, 4, 40, 190, 1330, 8680, 52920, 340200]
k = 7 [1, 2, 6, 24, 120, 720, 4320, 34560, 311040, 3110400]

and here is the SymPy code. I edited the main but the rest is verbatim from Claude.

from sympy import symbols, gcd, factorial, Rational, S

x = symbols('x')

def expq_coeffs(m, q, n):
    """
    Truncated (degree <= n) series coefficients of
        expq(x**m/m, q) = MittagLefflerE(q, (x**m/m)**q)
    Since q is a positive integer:
        E_q(y^q) = sum_j y^(q*j) / (q*j)!
    with y = x**m/m, so the term of degree m*q*j has coefficient
        1 / ( m**(q*j) * (q*j)! ).
    Returns a list c[0..n] of coefficients.
    """
    c = [S.Zero] * (n + 1)
    j = 0
    while m * q * j <= n:
        deg = m * q * j
        c[deg] += Rational(1, m**(q * j) * factorial(q * j))
        j += 1
    return c

def poly_mult_trunc(a, b, n):
    """Multiply two series (lists of coeffs, index = degree) truncated to degree n."""
    c = [S.Zero] * (n + 1)
    for i, ai in enumerate(a):
        if ai == 0:
            continue
        max_j = n - i
        for j2 in range(max_j + 1):
            bj = b[j2]
            if bj != 0:
                c[i + j2] += ai * bj
    return c

def p(n, k):
    """
    SymPy equivalent of:
        expq[x_, q_] := MittagLefflerE[q, x^q]
        p[n_, k_] := SeriesCoefficient[
            Product[expq[x^m/m, GCD[m, k]], {m, 1, n}], {x, 0, n}]
    """
    result = [S.Zero] * (n + 1)
    result[0] = S.One
    for m in range(1, n + 1):
        q = gcd(m, k)
        factor = expq_coeffs(m, q, n)
        result = poly_mult_trunc(result, factor, n)
    return result[n]

# example
if __name__ == "__main__":
    for k in range(2, 8):
        print("k =", k, [factorial(n)*p(n, k) for n in range(1,11)])

[1] Herbert Wilf. Generatingfunctionology. Available online here.

DNA Sequence Alignment and Kings

Posted on by John

This morning I wrote a post that included the central Delannoy numbers. The nth central Delannoy number Dn counts the number of ways a king can move from one corner of a chessboard to the diagonally opposite corner without backtracking.

The more general Delannoy numbers Dm,n are the analogy for an m × n rectangular board, not necessarily square.

Dm,n is also the number of possible sequence alignments for a strand of DNA with m base pairs and a strand with n base pairs [1]. At each step in the alignment process, you can introduce a gap in the first strand, the second strand or neither, which is analogous to the king who can move N, E, or NE at each step.

The Delannoy numbers can be computed recursively:

def D(m, n):
    if m == 0 or n == 0:
        return 1
    return D(m - 1, n) + D(m, n - 1) + D(m - 1, n - 1)

The code above can be sped up tremendously by adding the decorator

@lru_cache(maxsize=None)

above the function definition to turn on memoization. I did an experiment computing D12,15 with and without memoization and the times were 77.1805 seconds and 0.000062 seconds respectively, i.e. memoization made the code over a million times faster.

Incidentally, D12,15 = 2653649025 and so there are a lot of ways to align even short sequences unless you place some restriction on the permissible alignments.

Update: Here’s a heatmap plotting log10(Dm,n). Obviously the function increases with m and n: bigger chessboards have more possible paths. Moreover, it’s larger along the diagonal (i.e. the central Delannoy numbers). If you look along northeast to southwest diagonals, the function is largest in the middle where m = n.

[1] Torres, A., Cabada, A., & Nieto, J. J. (2003). An exact formula for the number of alignments between two DNA sequences. DNA Sequence, 14(6), 427–430. https://doi.org/10.1080/10425170310001617894

Silver Rectangles and the Ways of Kings

Posted on by John

Golden rectangles

The defining property of golden rectangle is that if you stick a square on its longer side, you get another golden rectangle.

The smaller vertical rectangle is similar to the larger horizontal rectangle. This means

φ / 1 = (1 + φ) / φ

which tells us φ² = 1 + φ and so the golden ratio φ equals (1 + √5)/2.

Silver rectangles

A silver rectangle is one that if you stick two squares on its longer side you get another rectangle with the same aspect ratio.

This tells us

σ / 1 = (1 + 2σ) / σ

and so σ² = 1 + 2σ and the silver ratio is σ = 1 + √2.

Just as you can define a golden ratio and a silver ratio, there’s an analogous way to define a sequence of metallic ratios.

Kings and Delannoy numbers

The silver ratio has several connections to the ways of ways kings. By that I mean the number of ways a king can go from one corner of a chessboard to the diagonally opposite corner without backtracking.

A king can move one space in any direction. If we start with a king in the bottom left corner of the board, the no-backtracking requirement means the king can move up, right, or up and right.

The number of paths a king can take from one corner to the opposite corner of an n × n chessboard is the nth central Delannoy number Dn. more generally Delannoy numbers are defined for an m × n chessboard, but I’ll stick to the case m = n called the central Delannoy number, or just Delannoy numbers for short.

The first Delannoy number is 1 because there’s only one way for a king to get from one corner to the other: do nothing, because the opposite corner is the same corner. The second Delannoy number is 3 because the king can move up then right, or right then up, or move diagonally up and right.

For a 3 × 3 grid things are significantly more complicated, and D3 = 13. For an 8 × 8 grid the number of paths is 48,639.

Generating function

How would you estimate the number of paths on an n × n board for large values of n without calculating it exactly? You might start by finding a generating function for the Delannoy numbers, which works out to be

(x² − 6x + 1)−1/2

The radius of convergence r for the generating function series is the distance from 0 to the closest singularity of the generating function, which is the smaller root of

x² − 6x + 1

which is

3 − √8 = (3 + √8)−1 = (1 + √2)−2 = 1/σ²

i.e. the radius of convergence is the reciprocal of the silver ratio squared.

Asymptotic estimate

The radius of convergence gives us a first approximation to the asymptotic size of the series coefficients. Since we’re working with the generating function of the Delannoy numbers, these coefficients are the Delannoy numbers. That is,

Dn ~ r−n = (σ2)n = σ2n.

That’s as good as you can do just knowing the radius of convergence. A more careful analysis would refine this estimate by dividing by a factor proportional to √n.

Related posts

Queens on a prime order board

Posted on by John

The n queens problem is to place on an n × n chessboard n queens so that none attacks any other. This means there is only one queen on every horizontal, vertical, and diagonal line.

When n is a prime number ≥ 5, it is sufficient to place the queens on a line that has slope 2, 3, 4, …, n − 2. (The slope cannot be 1 because that’s a diagonal. And it cannot be n − 1 because n − 1 = −1 mod n is also a diagonal.) [1]

Here we imagine opposite edges of the board being joined together. Geometrically, this makes the chessboard a torus (donut). Algebraically, the points on a line of slope s have the coordinates

(a + k, b + ks)

where addition is carried out mod n.

All solutions to the n queens problem have this form when n = 5. Some solutions will have this form for larger prime values of n but not all.

For example, when n = 7, here is a solution where all the queens are on a line of slope 2.

But here is another solution where the queens do not all lie on a line of constant slope.

Related posts

[1] W. H. Bussey. A Note on the Problem of the Eight Queens. The American Mathematical Monthly, Vol. 29, No. 7 (August 1922), pp. 252–253