# Latin Squares¶

A latin square of order $$n$$ is an $$n \times n$$ array such that each symbol $$s \in \{ 0, 1, \dots, n-1\}$$ appears precisely once in each row, and precisely once in each column. A partial latin square of order $$n$$ is an $$n \times n$$ array such that each symbol $$s \in \{ 0, 1, \dots, n-1\}$$ appears at most once in each row, and at most once in each column. Empty cells are denoted by $$-1$$. A latin square $$L$$ is a completion of a partial latin square $$P$$ if $$P \subseteq L$$. If $$P$$ completes to just $$L$$ then $$P$$ has unique completion.

A latin bitrade $$(T_1,\, T_2)$$ is a pair of partial latin squares such that:

1. $$\{ (i,\,j) \mid (i,\,j,\,k) \in T_1 \text{ for some symbol }k \} = \{ (i,\,j) \mid (i,\,j,\,k') \in T_2 \text{ for some symbol }k' \};$$

2. for each $$(i,\,j,\,k) \in T_1$$ and $$(i,\,j,\,k') \in T_2$$, $$k \neq k'$$;

3. the symbols appearing in row $$i$$ of $$T_1$$ are the same as those of row $$i$$ of $$T_2$$; the symbols appearing in column $$j$$ of $$T_1$$ are the same as those of column $$j$$ of $$T_2$$.

Intuitively speaking, a bitrade gives the difference between two latin squares, so if $$(T_1,\, T_2)$$ is a bitrade for the pair of latin squares $$(L_1,\, L_2)$$, then $$L1 = (L2 \setminus T_1) \cup T_2$$ and $$L2 = (L1 \setminus T_2) \cup T_1$$.

This file contains

1. LatinSquare class definition;

2. some named latin squares (back circulant, forward circulant, abelian $$2$$-group);

3. functions is_partial_latin_square and is_latin_square to test if a LatinSquare object satisfies the definition of a latin square or partial latin square, respectively;

4. tests for completion and unique completion (these use the C++ implementation of Knuth’s dancing links algorithm to solve the problem as a instance of $$0-1$$ matrix exact cover);

5. functions for calculating the $$\tau_i$$ representation of a bitrade and the genus of the associated hypermap embedding;

6. Markov chain of Jacobson and Matthews (1996) for generating latin squares uniformly at random (provides a generator interface);

7. a few examples of $$\tau_i$$ representations of bitrades constructed from the action of a group on itself by right multiplication, functions for converting to a pair of LatinSquare objects.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: B = back_circulant(5)
sage: B
[0 1 2 3 4]
[1 2 3 4 0]
[2 3 4 0 1]
[3 4 0 1 2]
[4 0 1 2 3]
sage: B.is_latin_square()
True
sage: B[0, 1] = 0
sage: B.is_latin_square()
False

sage: (a, b, c, G) = alternating_group_bitrade_generators(1)
sage: (T1, T2) = bitrade_from_group(a, b, c, G)
sage: T1
[ 0 -1  3  1]
[-1  1  0  2]
[ 1  3  2 -1]
[ 2  0 -1  3]
sage: T2
[ 1 -1  0  3]
[-1  0  2  1]
[ 2  1  3 -1]
[ 0  3 -1  2]
sage: T1.nr_filled_cells()
12
sage: genus(T1, T2)
1


Todo

1. Latin squares with symbols from a ring instead of the integers $$\{ 0, 1, \dots, n-1 \}$$.

2. Isotopism testing of latin squares and bitrades via graph isomorphism (nauty?).

AUTHORS:

• Carlo Hamalainen (2008-03-23): initial version

class sage.combinat.matrices.latin.LatinSquare(*args)

Bases: object

Latin squares.

This class implements a latin square of order n with rows and columns indexed by the set 0, 1, …, n-1 and symbols from the same set. The underlying latin square is a matrix(ZZ, n, n). If L is a latin square, then the cell at row r, column c is empty if and only if L[r, c] < 0. In this way we allow partial latin squares and can speak of completions to latin squares, etc.

There are two ways to declare a latin square:

Empty latin square of order n:

sage: n = 3
sage: L = LatinSquare(n)
sage: L
[-1 -1 -1]
[-1 -1 -1]
[-1 -1 -1]


Latin square from a matrix:

sage: M = matrix(ZZ, [[0, 1], [2, 3]])
sage: LatinSquare(M)
[0 1]
[2 3]

actual_row_col_sym_sizes()

Bitrades sometimes end up in partial latin squares with unused rows, columns, or symbols. This function works out the actual number of used rows, columns, and symbols.

Warning

We assume that the unused rows/columns occur in the lower right of self, and that the used symbols are in the range {0, 1, …, m} (no holes in that list).

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: B = back_circulant(3)
sage: B[0,2] = B[1,2] = B[2,2] = -1
sage: B[0,0] = B[2,1] = -1
sage: B
[-1  1 -1]
[ 1  2 -1]
[ 2 -1 -1]
sage: B.actual_row_col_sym_sizes()
(3, 2, 2)

apply_isotopism(row_perm, col_perm, sym_perm)

An isotopism is a permutation of the rows, columns, and symbols of a partial latin square self. Use isotopism() to convert a tuple (indexed from 0) to a Permutation object.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: B = back_circulant(5)
sage: B
[0 1 2 3 4]
[1 2 3 4 0]
[2 3 4 0 1]
[3 4 0 1 2]
[4 0 1 2 3]
sage: alpha = isotopism((0,1,2,3,4))
sage: beta  = isotopism((1,0,2,3,4))
sage: gamma = isotopism((2,1,0,3,4))
sage: B.apply_isotopism(alpha, beta, gamma)
[3 4 2 0 1]
[0 2 3 1 4]
[1 3 0 4 2]
[4 0 1 2 3]
[2 1 4 3 0]

clear_cells()

Mark every cell in self as being empty.

EXAMPLES:

sage: A = LatinSquare(matrix(ZZ, [[0, 1], [2, 3]]))
sage: A.clear_cells()
sage: A
[-1 -1]
[-1 -1]

column(x)

Return column x of the latin square.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: back_circulant(3).column(0)
(0, 1, 2)

contained_in(Q)

Return True if self is a subset of Q?

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: P = elementary_abelian_2group(2)
sage: P[0, 0] = -1
sage: P.contained_in(elementary_abelian_2group(2))
True
sage: back_circulant(4).contained_in(elementary_abelian_2group(2))
False


Internal function for find_disjoint_mates.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: B = back_circulant(4)
([[0, 16, 32],
[1, 17, 32],
[2, 18, 32],
[3, 19, 32],
[4, 16, 33],
[5, 17, 33],
[6, 18, 33],
[7, 19, 33],
[8, 16, 34],
[9, 17, 34],
[10, 18, 34],
[11, 19, 34],
[12, 16, 35],
[13, 17, 35],
[14, 18, 35],
[15, 19, 35],
[0, 20, 36],
[1, 21, 36],
[2, 22, 36],
[3, 23, 36],
[4, 20, 37],
[5, 21, 37],
[6, 22, 37],
[7, 23, 37],
[8, 20, 38],
[9, 21, 38],
[10, 22, 38],
[11, 23, 38],
[12, 20, 39],
[13, 21, 39],
[14, 22, 39],
[15, 23, 39],
[0, 24, 40],
[1, 25, 40],
[2, 26, 40],
[3, 27, 40],
[4, 24, 41],
[5, 25, 41],
[6, 26, 41],
[7, 27, 41],
[8, 24, 42],
[9, 25, 42],
[10, 26, 42],
[11, 27, 42],
[12, 24, 43],
[13, 25, 43],
[14, 26, 43],
[15, 27, 43],
[0, 28, 44],
[1, 29, 44],
[2, 30, 44],
[3, 31, 44],
[4, 28, 45],
[5, 29, 45],
[6, 30, 45],
[7, 31, 45],
[8, 28, 46],
[9, 29, 46],
[10, 30, 46],
[11, 31, 46],
[12, 28, 47],
[13, 29, 47],
[14, 30, 47],
[15, 31, 47]],
{(0, 16, 32): (0, 0, 0),
(0, 20, 36): (1, 0, 0),
(0, 24, 40): (2, 0, 0),
(0, 28, 44): (3, 0, 0),
(1, 17, 32): (0, 0, 1),
(1, 21, 36): (1, 0, 1),
(1, 25, 40): (2, 0, 1),
(1, 29, 44): (3, 0, 1),
(2, 18, 32): (0, 0, 2),
(2, 22, 36): (1, 0, 2),
(2, 26, 40): (2, 0, 2),
(2, 30, 44): (3, 0, 2),
(3, 19, 32): (0, 0, 3),
(3, 23, 36): (1, 0, 3),
(3, 27, 40): (2, 0, 3),
(3, 31, 44): (3, 0, 3),
(4, 16, 33): (0, 1, 0),
(4, 20, 37): (1, 1, 0),
(4, 24, 41): (2, 1, 0),
(4, 28, 45): (3, 1, 0),
(5, 17, 33): (0, 1, 1),
(5, 21, 37): (1, 1, 1),
(5, 25, 41): (2, 1, 1),
(5, 29, 45): (3, 1, 1),
(6, 18, 33): (0, 1, 2),
(6, 22, 37): (1, 1, 2),
(6, 26, 41): (2, 1, 2),
(6, 30, 45): (3, 1, 2),
(7, 19, 33): (0, 1, 3),
(7, 23, 37): (1, 1, 3),
(7, 27, 41): (2, 1, 3),
(7, 31, 45): (3, 1, 3),
(8, 16, 34): (0, 2, 0),
(8, 20, 38): (1, 2, 0),
(8, 24, 42): (2, 2, 0),
(8, 28, 46): (3, 2, 0),
(9, 17, 34): (0, 2, 1),
(9, 21, 38): (1, 2, 1),
(9, 25, 42): (2, 2, 1),
(9, 29, 46): (3, 2, 1),
(10, 18, 34): (0, 2, 2),
(10, 22, 38): (1, 2, 2),
(10, 26, 42): (2, 2, 2),
(10, 30, 46): (3, 2, 2),
(11, 19, 34): (0, 2, 3),
(11, 23, 38): (1, 2, 3),
(11, 27, 42): (2, 2, 3),
(11, 31, 46): (3, 2, 3),
(12, 16, 35): (0, 3, 0),
(12, 20, 39): (1, 3, 0),
(12, 24, 43): (2, 3, 0),
(12, 28, 47): (3, 3, 0),
(13, 17, 35): (0, 3, 1),
(13, 21, 39): (1, 3, 1),
(13, 25, 43): (2, 3, 1),
(13, 29, 47): (3, 3, 1),
(14, 18, 35): (0, 3, 2),
(14, 22, 39): (1, 3, 2),
(14, 26, 43): (2, 3, 2),
(14, 30, 47): (3, 3, 2),
(15, 19, 35): (0, 3, 3),
(15, 23, 39): (1, 3, 3),
(15, 27, 43): (2, 3, 3),
(15, 31, 47): (3, 3, 3)})

dlxcpp_has_unique_completion()

Check if the partial latin square self of order n can be embedded in precisely one latin square of order n.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: back_circulant(2).dlxcpp_has_unique_completion()
True
sage: P = LatinSquare(2)
sage: P.dlxcpp_has_unique_completion()
False
sage: P[0, 0] = 0
sage: P.dlxcpp_has_unique_completion()
True

dumps()

Since the latin square class does not hold any other private variables we just call dumps on self.square:

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
True

filled_cells_map()

Number the filled cells of self with integers from {1, 2, 3, …}

INPUT:

• self - Partial latin square self (empty cells have negative values)

OUTPUT: A dictionary cells_map where cells_map[(i,j)] = m means that (i,j) is the m-th filled cell in P, while cells_map[m] = (i,j).

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: (a, b, c, G) = alternating_group_bitrade_generators(1)
sage: (T1, T2) = bitrade_from_group(a, b, c, G)
sage: D = T1.filled_cells_map()
sage: {i: v for i,v in D.items() if i in ZZ}
{1: (0, 0),
2: (0, 2),
3: (0, 3),
4: (1, 1),
5: (1, 2),
6: (1, 3),
7: (2, 0),
8: (2, 1),
9: (2, 2),
10: (3, 0),
11: (3, 1),
12: (3, 3)}
sage: {i: v for i,v in D.items() if i not in ZZ}
{(0, 0): 1,
(0, 2): 2,
(0, 3): 3,
(1, 1): 4,
(1, 2): 5,
(1, 3): 6,
(2, 0): 7,
(2, 1): 8,
(2, 2): 9,
(3, 0): 10,
(3, 1): 11,
(3, 3): 12}


Warning

If allow_subtrade is True then we may return a partial latin square that is not disjoint to self. In that case, use bitrade(P, Q) to get an actual bitrade.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: B = back_circulant(4)
sage: g = B.find_disjoint_mates(allow_subtrade = True)
sage: B1 = next(g)
sage: B0, B1 = bitrade(B, B1)
sage: print(B0)
[-1  1  2 -1]
[-1  2 -1  0]
[-1 -1 -1 -1]
[-1  0  1  2]
sage: print(B1)
[-1  2  1 -1]
[-1  0 -1  2]
[-1 -1 -1 -1]
[-1  1  2  0]

gcs()

A greedy critical set of a latin square self is found by successively removing elements in a row-wise (bottom-up) manner, checking for unique completion at each step.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: A = elementary_abelian_2group(3)
sage: G = A.gcs()
sage: A
[0 1 2 3 4 5 6 7]
[1 0 3 2 5 4 7 6]
[2 3 0 1 6 7 4 5]
[3 2 1 0 7 6 5 4]
[4 5 6 7 0 1 2 3]
[5 4 7 6 1 0 3 2]
[6 7 4 5 2 3 0 1]
[7 6 5 4 3 2 1 0]
sage: G
[ 0  1  2  3  4  5  6 -1]
[ 1  0  3  2  5  4 -1 -1]
[ 2  3  0  1  6 -1  4 -1]
[ 3  2  1  0 -1 -1 -1 -1]
[ 4  5  6 -1  0  1  2 -1]
[ 5  4 -1 -1  1  0 -1 -1]
[ 6 -1  4 -1  2 -1  0 -1]
[-1 -1 -1 -1 -1 -1 -1 -1]

is_completable()

Return True if the partial latin square can be completed to a latin square.

EXAMPLES:

The following partial latin square has no completion because there is nowhere that we can place the symbol 0 in the third row:

sage: B = LatinSquare(3)

sage: B[0, 0] = 0
sage: B[1, 1] = 0
sage: B[2, 2] = 1

sage: B
[ 0 -1 -1]
[-1  0 -1]
[-1 -1  1]

sage: B.is_completable()
False

sage: B[2, 2] = 0
sage: B.is_completable()
True

is_empty_column(c)

Check if column c of the partial latin square self is empty.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: L = back_circulant(4)
sage: L.is_empty_column(0)
False
sage: L[0,0] = L[1,0] = L[2,0] = L[3,0] = -1
sage: L.is_empty_column(0)
True

is_empty_row(r)

Check if row r of the partial latin square self is empty.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: L = back_circulant(4)
sage: L.is_empty_row(0)
False
sage: L[0,0] = L[0,1] = L[0,2] = L[0,3] = -1
sage: L.is_empty_row(0)
True

is_latin_square()

self is a latin square if it is an n by n matrix, and each symbol in [0, 1, …, n-1] appears exactly once in each row, and exactly once in each column.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: elementary_abelian_2group(4).is_latin_square()
True

sage: forward_circulant(7).is_latin_square()
True

is_partial_latin_square()

self is a partial latin square if it is an n by n matrix, and each symbol in [0, 1, …, n-1] appears at most once in each row, and at most once in each column.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: LatinSquare(4).is_partial_latin_square()
True
sage: back_circulant(3).gcs().is_partial_latin_square()
True
sage: back_circulant(6).is_partial_latin_square()
True

is_uniquely_completable()

Return True if the partial latin square self has exactly one completion to a latin square. This is just a wrapper for the current best-known algorithm, Dancing Links by Knuth. See dancing_links.spyx

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: back_circulant(4).gcs().is_uniquely_completable()
True

sage: G = elementary_abelian_2group(3).gcs()
sage: G.is_uniquely_completable()
True

sage: G[0, 0] = -1
sage: G.is_uniquely_completable()
False

latex()

Return LaTeX code for the latin square.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: print(back_circulant(3).latex())
\begin{array}{|c|c|c|}\hline 0 & 1 & 2\\\hline 1 & 2 & 0\\\hline 2 & 0 & 1\\\hline\end{array}

list()

Convert the latin square into a list, in a row-wise manner.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: back_circulant(3).list()
[0, 1, 2, 1, 2, 0, 2, 0, 1]

ncols()

Number of columns in the latin square.

EXAMPLES:

sage: LatinSquare(3).ncols()
3

nr_distinct_symbols()

Return the number of distinct symbols in the partial latin square self.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: back_circulant(5).nr_distinct_symbols()
5
sage: L = LatinSquare(10)
sage: L.nr_distinct_symbols()
0
sage: L[0, 0] = 0
sage: L[0, 1] = 1
sage: L.nr_distinct_symbols()
2

nr_filled_cells()

Return the number of filled cells (i.e. cells with a positive value) in the partial latin square self.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: LatinSquare(matrix([[0, -1], [-1, 0]])).nr_filled_cells()
2

nrows()

Number of rows in the latin square.

EXAMPLES:

sage: LatinSquare(3).nrows()
3

permissable_values(r, c)

Find all values that do not appear in row r and column c of the latin square self. If self[r, c] is filled then we return the empty list.

INPUT:

• self - LatinSquare

• r - int; row of the latin square

• c - int; column of the latin square

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: L = back_circulant(5)
sage: L[0, 0] = -1
sage: L.permissable_values(0, 0)


random_empty_cell()

Find an empty cell of self, uniformly at random.

INPUT:

• self - LatinSquare

OUTPUT:

• [r, c] - cell such that self[r, c] is empty, or returns None if self is a (full) latin square.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: P = back_circulant(2)
sage: P[1,1] = -1
sage: P.random_empty_cell()
[1, 1]

row(x)

Return row x of the latin square.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: back_circulant(3).row(0)
(0, 1, 2)

set_immutable()

A latin square is immutable if the underlying matrix is immutable.

EXAMPLES:

sage: L = LatinSquare(matrix(ZZ, [[0, 1], [2, 3]]))
sage: L.set_immutable()
sage: {L : 0}   # this would fail without set_immutable()
{[0 1]
[2 3]: 0}

top_left_empty_cell()

Return the least [r, c] such that self[r, c] is an empty cell. If all cells are filled then we return None.

INPUT:

• self - LatinSquare

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: B = back_circulant(5)
sage: B[3, 4] = -1
sage: B.top_left_empty_cell()
[3, 4]

vals_in_col(c)

Return a dictionary with key e if and only if column c of self has the symbol e.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: B = back_circulant(3)
sage: B[0, 0] = -1
sage: back_circulant(3).vals_in_col(0)
{0: True, 1: True, 2: True}

vals_in_row(r)

Return a dictionary with key e if and only if row r of self has the symbol e.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: B = back_circulant(3)
sage: B[0, 0] = -1
sage: back_circulant(3).vals_in_row(0)
{0: True, 1: True, 2: True}

sage.combinat.matrices.latin.LatinSquare_generator(L_start, check_assertions=False)

Generator for a sequence of uniformly distributed latin squares, given L_start as the initial latin square.

This code implements the Markov chain algorithm of Jacobson and Matthews (1996), see below for the BibTex entry. This generator will never throw the StopIteration exception, so it provides an infinite sequence of latin squares.

EXAMPLES:

Use the back circulant latin square of order 4 as the initial square and print the next two latin squares given by the Markov chain:

sage: from sage.combinat.matrices.latin import *
sage: g = LatinSquare_generator(back_circulant(4))
sage: next(g).is_latin_square()
True


REFERENCES:

JacMat96

Mark T. Jacobson and Peter Matthews, “Generating uniformly distributed random Latin squares”, Journal of Combinatorial Designs, 4 (1996)

Construct generators a, b, c for the alternating group on 3m+1 points, such that a*b*c = 1.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: a, b, c, G = alternating_group_bitrade_generators(1)
sage: (a, b, c, G)
((1,2,3), (1,4,2), (2,4,3), Permutation Group with generators [(1,2,3), (1,4,2)])
sage: a*b*c
()

sage: (T1, T2) = bitrade_from_group(a, b, c, G)
sage: T1
[ 0 -1  3  1]
[-1  1  0  2]
[ 1  3  2 -1]
[ 2  0 -1  3]
sage: T2
[ 1 -1  0  3]
[-1  0  2  1]
[ 2  1  3 -1]
[ 0  3 -1  2]

sage.combinat.matrices.latin.back_circulant(n)

The back-circulant latin square of order n is the Cayley table for (Z_n, +), the integers under addition modulo n.

INPUT:

• n – int; order of the latin square.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: back_circulant(5)
[0 1 2 3 4]
[1 2 3 4 0]
[2 3 4 0 1]
[3 4 0 1 2]
[4 0 1 2 3]

sage.combinat.matrices.latin.beta1(rce, T1, T2)

Find the unique (x, c, e) in T2 such that (r, c, e) is in T1.

INPUT:

• rce - tuple (or list) (r, c, e) in T1

• T1, T2 - latin bitrade

OUTPUT: (x, c, e) in T2.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: T1 = back_circulant(5)
sage: x = isotopism( (0,1,2,3,4) )
sage: y = isotopism(5) # identity
sage: z = isotopism(5) # identity
sage: T2 = T1.apply_isotopism(x, y, z)
True
sage: beta1([0, 0, 0], T1, T2)
(1, 0, 0)

sage.combinat.matrices.latin.beta2(rce, T1, T2)

Find the unique (r, x, e) in T2 such that (r, c, e) is in T1.

INPUT:

• rce - tuple (or list) (r, c, e) in T1

• T1, T2 - latin bitrade

OUTPUT:

• (r, x, e) in T2.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: T1 = back_circulant(5)
sage: x = isotopism( (0,1,2,3,4) )
sage: y = isotopism(5) # identity
sage: z = isotopism(5) # identity
sage: T2 = T1.apply_isotopism(x, y, z)
True
sage: beta2([0, 0, 0], T1, T2)
(0, 1, 0)

sage.combinat.matrices.latin.beta3(rce, T1, T2)

Find the unique (r, c, x) in T2 such that (r, c, e) is in T1.

INPUT:

• rce - tuple (or list) (r, c, e) in T1

• T1, T2 - latin bitrade

OUTPUT:

• (r, c, x) in T2.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: T1 = back_circulant(5)
sage: x = isotopism( (0,1,2,3,4) )
sage: y = isotopism(5) # identity
sage: z = isotopism(5) # identity
sage: T2 = T1.apply_isotopism(x, y, z)
True
sage: beta3([0, 0, 0], T1, T2)
(0, 0, 4)


Form the bitrade (Q1, Q2) from (T1, T2) by setting empty the cells (r, c) such that T1[r, c] == T2[r, c].

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: B1 = back_circulant(5)
sage: alpha = isotopism((0,1,2,3,4))
sage: beta  = isotopism((1,0,2,3,4))
sage: gamma = isotopism((2,1,0,3,4))
sage: B2 = B1.apply_isotopism(alpha, beta, gamma)
sage: T1, T2 = bitrade(B1, B2)
sage: T1
[ 0  1 -1  3  4]
[ 1 -1 -1  4  0]
[ 2 -1  4  0  1]
[ 3  4  0  1  2]
[ 4  0  1  2  3]
sage: T2
[ 3  4 -1  0  1]
[ 0 -1 -1  1  4]
[ 1 -1  0  4  2]
[ 4  0  1  2  3]
[ 2  1  4  3  0]


Given group elements a, b, c in G such that abc = 1 and the subgroups a, b, c intersect (pairwise) only in the identity, construct a bitrade (T1, T2) where rows, columns, and symbols correspond to cosets of a, b, and c, respectively.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: a, b, c, G = alternating_group_bitrade_generators(1)
sage: (T1, T2) = bitrade_from_group(a, b, c, G)
sage: T1
[ 0 -1  3  1]
[-1  1  0  2]
[ 1  3  2 -1]
[ 2  0 -1  3]
sage: T2
[ 1 -1  0  3]
[-1  0  2  1]
[ 2  1  3 -1]
[ 0  3 -1  2]

sage.combinat.matrices.latin.cells_map_as_square(cells_map, n)

Return a LatinSquare with cells numbered from 1, 2, … to given the dictionary cells_map.

Note

The value n should be the maximum of the number of rows and columns of the original partial latin square

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: (a, b, c, G) = alternating_group_bitrade_generators(1)
sage: (T1, T2) = bitrade_from_group(a, b, c, G)
sage: T1
[ 0 -1  3  1]
[-1  1  0  2]
[ 1  3  2 -1]
[ 2  0 -1  3]


There are 12 filled cells in T:

sage: cells_map_as_square(T1.filled_cells_map(), max(T1.nrows(), T1.ncols()))
[ 1 -1  2  3]
[-1  4  5  6]
[ 7  8  9 -1]
[10 11 -1 12]


Three group elements a, b, c will generate a bitrade if a*b*c = 1 and the subgroups a, b, c intersect (pairwise) in just the identity.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: a, b, c, G = p3_group_bitrade_generators(3)
True
False

sage.combinat.matrices.latin.coin()

Simulate a fair coin (returns True or False) using ZZ.random_element(2).

EXAMPLES:

sage: from sage.combinat.matrices.latin import coin
sage: x = coin()
sage: x == 0 or x == 1
True

sage.combinat.matrices.latin.column_containing_sym(L, r, x)

Given an improper latin square L with L[r, c1] = L[r, c2] = x, return c1 or c2 with equal probability. This is an internal function and should only be used in LatinSquare_generator().

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: L = matrix([(1, 0, 2, 3), (0, 2, 3, 0), (2, 3, 0, 1), (3, 0, 1, 2)])
sage: L
[1 0 2 3]
[0 2 3 0]
[2 3 0 1]
[3 0 1 2]
sage: c = column_containing_sym(L, 1, 0)
sage: c == 0 or c == 3
True

sage.combinat.matrices.latin.direct_product(L1, L2, L3, L4)

The ‘direct product’ of four latin squares L1, L2, L3, L4 of order n is the latin square of order 2n consisting of

-----------
| L1 | L2 |
-----------
| L3 | L4 |
-----------


where the subsquares L2 and L3 have entries offset by n.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: direct_product(back_circulant(4), back_circulant(4), elementary_abelian_2group(2), elementary_abelian_2group(2))
[0 1 2 3 4 5 6 7]
[1 2 3 0 5 6 7 4]
[2 3 0 1 6 7 4 5]
[3 0 1 2 7 4 5 6]
[4 5 6 7 0 1 2 3]
[5 4 7 6 1 0 3 2]
[6 7 4 5 2 3 0 1]
[7 6 5 4 3 2 1 0]

sage.combinat.matrices.latin.dlxcpp_find_completions(P, nr_to_find=None)

Return a list of all latin squares L of the same order as P such that P is contained in L. The optional parameter nr_to_find limits the number of latin squares that are found.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: dlxcpp_find_completions(LatinSquare(2))
[[0 1]
[1 0], [1 0]
[0 1]]

sage: dlxcpp_find_completions(LatinSquare(2), 1)
[[0 1]
[1 0]]

sage.combinat.matrices.latin.dlxcpp_rows_and_map(P)

Internal function for dlxcpp_find_completions. Given a partial latin square P we construct a list of rows of a 0-1 matrix M such that an exact cover of M corresponds to a completion of P to a latin square.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: dlxcpp_rows_and_map(LatinSquare(2))
([[0, 4, 8],
[1, 5, 8],
[2, 4, 9],
[3, 5, 9],
[0, 6, 10],
[1, 7, 10],
[2, 6, 11],
[3, 7, 11]],
{(0, 4, 8): (0, 0, 0),
(0, 6, 10): (1, 0, 0),
(1, 5, 8): (0, 0, 1),
(1, 7, 10): (1, 0, 1),
(2, 4, 9): (0, 1, 0),
(2, 6, 11): (1, 1, 0),
(3, 5, 9): (0, 1, 1),
(3, 7, 11): (1, 1, 1)})

sage.combinat.matrices.latin.elementary_abelian_2group(s)

Return the latin square based on the Cayley table for the elementary abelian 2-group of order 2s.

INPUT:

• s – int; order of the latin square will be 2s.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: elementary_abelian_2group(3)
[0 1 2 3 4 5 6 7]
[1 0 3 2 5 4 7 6]
[2 3 0 1 6 7 4 5]
[3 2 1 0 7 6 5 4]
[4 5 6 7 0 1 2 3]
[5 4 7 6 1 0 3 2]
[6 7 4 5 2 3 0 1]
[7 6 5 4 3 2 1 0]

sage.combinat.matrices.latin.forward_circulant(n)

The forward-circulant latin square of order n is the Cayley table for the operation r + c = (n-c+r) mod n.

INPUT:

• n – int; order of the latin square.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: forward_circulant(5)
[0 4 3 2 1]
[1 0 4 3 2]
[2 1 0 4 3]
[3 2 1 0 4]
[4 3 2 1 0]

sage.combinat.matrices.latin.genus(T1, T2)

Return the genus of hypermap embedding associated with the bitrade (T1, T2).

Informally, we compute the [tau_1, tau_2, tau_3] permutation representation of the bitrade. Each cycle of tau_1, tau_2, and tau_3 gives a rotation scheme for a black, white, and star vertex (respectively). The genus then comes from Euler’s formula.

For more details see Carlo Hamalainen: Partitioning 3-homogeneous latin bitrades. To appear in Geometriae Dedicata, available at arXiv 0710.0938

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: (a, b, c, G) = alternating_group_bitrade_generators(1)
sage: (T1, T2) = bitrade_from_group(a, b, c, G)
sage: genus(T1, T2)
1
sage: (a, b, c, G) = pq_group_bitrade_generators(3, 7)
sage: (T1, T2) = bitrade_from_group(a, b, c, G)
sage: genus(T1, T2)
3

sage.combinat.matrices.latin.group_to_LatinSquare(G)

Construct a latin square on the symbols [0, 1, …, n-1] for a group with an n by n Cayley table.

EXAMPLES:

sage: from sage.combinat.matrices.latin import group_to_LatinSquare

sage: group_to_LatinSquare(DihedralGroup(2))
[0 1 2 3]
[1 0 3 2]
[2 3 0 1]
[3 2 1 0]

sage: G = gap.Group(PermutationGroupElement((1,2,3)))
sage: group_to_LatinSquare(G)
[0 1 2]
[1 2 0]
[2 0 1]


Combinatorially, a pair (T1, T2) of partial latin squares is a bitrade if they are disjoint, have the same shape, and have row and column balance. For definitions of each of these terms see the relevant function in this file.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: T1 = back_circulant(5)
sage: x = isotopism( (0,1,2,3,4) )
sage: y = isotopism(5) # identity
sage: z = isotopism(5) # identity
sage: T2 = T1.apply_isotopism(x, y, z)
True

sage.combinat.matrices.latin.is_disjoint(T1, T2)

The partial latin squares T1 and T2 are disjoint if T1[r, c] != T2[r, c] or T1[r, c] == T2[r, c] == -1 for each cell [r, c].

EXAMPLES:

sage: from sage.combinat.matrices.latin import is_disjoint, back_circulant, isotopism
sage: is_disjoint(back_circulant(2), back_circulant(2))
False

sage: T1 = back_circulant(5)
sage: x = isotopism( (0,1,2,3,4) )
sage: y = isotopism(5) # identity
sage: z = isotopism(5) # identity
sage: T2 = T1.apply_isotopism(x, y, z)
sage: is_disjoint(T1, T2)
True


A bitrade generated from elements a, b, c is primary if a, b, c = G.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: (a, b, c, G) = p3_group_bitrade_generators(5)
True

sage.combinat.matrices.latin.is_row_and_col_balanced(T1, T2)

Partial latin squares T1 and T2 are balanced if the symbols appearing in row r of T1 are the same as the symbols appearing in row r of T2, for each r, and if the same condition holds on columns.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: T1 = matrix([[0,1,-1,-1], [-1,-1,-1,-1], [-1,-1,-1,-1], [-1,-1,-1,-1]])
sage: T2 = matrix([[0,1,-1,-1], [-1,-1,-1,-1], [-1,-1,-1,-1], [-1,-1,-1,-1]])
sage: is_row_and_col_balanced(T1, T2)
True
sage: T2 = matrix([[0,3,-1,-1], [-1,-1,-1,-1], [-1,-1,-1,-1], [-1,-1,-1,-1]])
sage: is_row_and_col_balanced(T1, T2)
False

sage.combinat.matrices.latin.is_same_shape(T1, T2)

Two partial latin squares T1, T2 have the same shape if T1[r, c] = 0 if and only if T2[r, c] = 0.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: is_same_shape(elementary_abelian_2group(2), back_circulant(4))
True
sage: is_same_shape(LatinSquare(5), LatinSquare(5))
True
sage: is_same_shape(forward_circulant(5), LatinSquare(5))
False

sage.combinat.matrices.latin.isotopism(p)

Return a Permutation object that represents an isotopism (for rows, columns or symbols of a partial latin square).

Technically, all this function does is take as input a representation of a permutation of $$0,...,n-1$$ and return a Permutation object defined on $$1,...,n$$.

For a definition of isotopism, see the wikipedia section on isotopism.

INPUT:

According to the type of input (see examples below):

• an integer $$n$$ – the function returns the identity on $$1,...,n$$.

• a string representing a permutation in disjoint cycles notation, e.g. $$(0,1,2)(3,4,5)$$ – the corresponding permutation is returned, shifted by 1 to act on $$1,...,n$$.

• list/tuple of tuples – assumes disjoint cycle notation, see previous entry.

• a list of integers – the function adds $$1$$ to each member of the list, and returns the corresponding permutation.

• a PermutationGroupElement p – returns a permutation describing p without any shift.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: isotopism(5) # identity on 5 points
[1, 2, 3, 4, 5]

sage: G = PermutationGroup(['(1,2,3)(4,5)'])
sage: g = G.gen(0)
sage: isotopism(g)
[2, 3, 1, 5, 4]

sage: isotopism([0,3,2,1]) # 0 goes to 0, 1 goes to 3, etc.
[1, 4, 3, 2]

sage: isotopism( (0,1,2) ) # single cycle, presented as a tuple
[2, 3, 1]

sage: x = isotopism( ((0,1,2), (3,4)) ) # tuple of cycles
sage: x
[2, 3, 1, 5, 4]
sage: x.to_cycles()
[(1, 2, 3), (4, 5)]

sage.combinat.matrices.latin.next_conjugate(L)

Permute L[r, c] = e to the conjugate L[c, e] = r.

We assume that L is an n by n matrix and has values in the range 0, 1, …, n-1.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: L = back_circulant(6)
sage: L
[0 1 2 3 4 5]
[1 2 3 4 5 0]
[2 3 4 5 0 1]
[3 4 5 0 1 2]
[4 5 0 1 2 3]
[5 0 1 2 3 4]
sage: next_conjugate(L)
[0 1 2 3 4 5]
[5 0 1 2 3 4]
[4 5 0 1 2 3]
[3 4 5 0 1 2]
[2 3 4 5 0 1]
[1 2 3 4 5 0]
sage: L == next_conjugate(next_conjugate(next_conjugate(L)))
True


Generators for a group of order p3 where p is a prime.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
((2,6,7)(3,8,9), (1,2,3)(4,7,8)(5,6,9), (1,9,2)(3,7,4)(5,8,6), Permutation Group with generators [(2,6,7)(3,8,9), (1,2,3)(4,7,8)(5,6,9)])


Generators for a group of order pq where p and q are primes such that (q % p) == 1.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
((2,3,5)(4,7,6), (1,2,3,4,5,6,7), (1,4,2)(3,5,6), Permutation Group with generators [(2,3,5)(4,7,6), (1,2,3,4,5,6,7)])

sage.combinat.matrices.latin.row_containing_sym(L, c, x)

Given an improper latin square L with L[r1, c] = L[r2, c] = x, return r1 or r2 with equal probability. This is an internal function and should only be used in LatinSquare_generator().

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: L = matrix([(0, 1, 0, 3), (3, 0, 2, 1), (1, 0, 3, 2), (2, 3, 1, 0)])
sage: L
[0 1 0 3]
[3 0 2 1]
[1 0 3 2]
[2 3 1 0]
sage: c = row_containing_sym(L, 1, 0)
sage: c == 1 or c == 2
True

sage.combinat.matrices.latin.tau1(T1, T2, cells_map)

The definition of $$\tau_1$$ is

$\begin{split}\tau_1 : T1 \rightarrow T1 \\ \tau_1 = \beta_2^{-1} \beta_3\end{split}$

where the composition is left to right and $$\beta_i : T2 \rightarrow T1$$ changes just the $$i^{th}$$ coordinate of a triple.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: T1 = back_circulant(5)
sage: x = isotopism( (0,1,2,3,4) )
sage: y = isotopism(5) # identity
sage: z = isotopism(5) # identity
sage: T2 = T1.apply_isotopism(x, y, z)
True
sage: (cells_map, t1, t2, t3) = tau123(T1, T2)
sage: t1 = tau1(T1, T2, cells_map)
sage: t1
[2, 3, 4, 5, 1, 7, 8, 9, 10, 6, 12, 13, 14, 15, 11, 17, 18, 19, 20, 16, 22, 23, 24, 25, 21]
sage: t1.to_cycles()
[(1, 2, 3, 4, 5), (6, 7, 8, 9, 10), (11, 12, 13, 14, 15), (16, 17, 18, 19, 20), (21, 22, 23, 24, 25)]

sage.combinat.matrices.latin.tau123(T1, T2)

Compute the tau_i representation for a bitrade (T1, T2).

See the functions tau1, tau2, and tau3 for the mathematical definitions.

OUTPUT:

• (cells_map, t1, t2, t3)

where cells_map is a map to/from the filled cells of T1, and t1, t2, t3 are the tau1, tau2, tau3 permutations.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: (a, b, c, G) = pq_group_bitrade_generators(3, 7)
sage: (T1, T2) = bitrade_from_group(a, b, c, G)
sage: T1
[ 0  1  3 -1 -1 -1 -1]
[ 1  2  4 -1 -1 -1 -1]
[ 2  3  5 -1 -1 -1 -1]
[ 3  4  6 -1 -1 -1 -1]
[ 4  5  0 -1 -1 -1 -1]
[ 5  6  1 -1 -1 -1 -1]
[ 6  0  2 -1 -1 -1 -1]
sage: T2
[ 1  3  0 -1 -1 -1 -1]
[ 2  4  1 -1 -1 -1 -1]
[ 3  5  2 -1 -1 -1 -1]
[ 4  6  3 -1 -1 -1 -1]
[ 5  0  4 -1 -1 -1 -1]
[ 6  1  5 -1 -1 -1 -1]
[ 0  2  6 -1 -1 -1 -1]
sage: (cells_map, t1, t2, t3) = tau123(T1, T2)
sage: D = cells_map
sage: {i: v for i,v in D.items() if i in ZZ}
{1: (0, 0),
2: (0, 1),
3: (0, 2),
4: (1, 0),
5: (1, 1),
6: (1, 2),
7: (2, 0),
8: (2, 1),
9: (2, 2),
10: (3, 0),
11: (3, 1),
12: (3, 2),
13: (4, 0),
14: (4, 1),
15: (4, 2),
16: (5, 0),
17: (5, 1),
18: (5, 2),
19: (6, 0),
20: (6, 1),
21: (6, 2)}
sage: {i: v for i,v in D.items() if i not in ZZ}
{(0, 0): 1,
(0, 1): 2,
(0, 2): 3,
(1, 0): 4,
(1, 1): 5,
(1, 2): 6,
(2, 0): 7,
(2, 1): 8,
(2, 2): 9,
(3, 0): 10,
(3, 1): 11,
(3, 2): 12,
(4, 0): 13,
(4, 1): 14,
(4, 2): 15,
(5, 0): 16,
(5, 1): 17,
(5, 2): 18,
(6, 0): 19,
(6, 1): 20,
(6, 2): 21}
sage: cells_map_as_square(cells_map, max(T1.nrows(), T1.ncols()))
[ 1  2  3 -1 -1 -1 -1]
[ 4  5  6 -1 -1 -1 -1]
[ 7  8  9 -1 -1 -1 -1]
[10 11 12 -1 -1 -1 -1]
[13 14 15 -1 -1 -1 -1]
[16 17 18 -1 -1 -1 -1]
[19 20 21 -1 -1 -1 -1]
sage: t1
[3, 1, 2, 6, 4, 5, 9, 7, 8, 12, 10, 11, 15, 13, 14, 18, 16, 17, 21, 19, 20]
sage: t2
[4, 8, 15, 7, 11, 18, 10, 14, 21, 13, 17, 3, 16, 20, 6, 19, 2, 9, 1, 5, 12]
sage: t3
[20, 18, 10, 2, 21, 13, 5, 3, 16, 8, 6, 19, 11, 9, 1, 14, 12, 4, 17, 15, 7]

sage: t1.to_cycles()
[(1, 3, 2), (4, 6, 5), (7, 9, 8), (10, 12, 11), (13, 15, 14), (16, 18, 17), (19, 21, 20)]
sage: t2.to_cycles()
[(1, 4, 7, 10, 13, 16, 19), (2, 8, 14, 20, 5, 11, 17), (3, 15, 6, 18, 9, 21, 12)]
sage: t3.to_cycles()
[(1, 20, 15), (2, 18, 4), (3, 10, 8), (5, 21, 7), (6, 13, 11), (9, 16, 14), (12, 19, 17)]


The product t1*t2*t3 is the identity, i.e. it fixes every point:

sage: len((t1*t2*t3).fixed_points()) == T1.nr_filled_cells()
True

sage.combinat.matrices.latin.tau2(T1, T2, cells_map)

The definition of $$\tau_2$$ is

$\begin{split}\tau_2 : T1 \rightarrow T1 \\ \tau_2 = \beta_3^{-1} \beta_1\end{split}$

where the composition is left to right and $$\beta_i : T2 \rightarrow T1$$ changes just the $$i^{th}$$ coordinate of a triple.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: T1 = back_circulant(5)
sage: x = isotopism( (0,1,2,3,4) )
sage: y = isotopism(5) # identity
sage: z = isotopism(5) # identity
sage: T2 = T1.apply_isotopism(x, y, z)
True
sage: (cells_map, t1, t2, t3) = tau123(T1, T2)
sage: t2 = tau2(T1, T2, cells_map)
sage: t2
[21, 22, 23, 24, 25, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
sage: t2.to_cycles()
[(1, 21, 16, 11, 6), (2, 22, 17, 12, 7), (3, 23, 18, 13, 8), (4, 24, 19, 14, 9), (5, 25, 20, 15, 10)]

sage.combinat.matrices.latin.tau3(T1, T2, cells_map)

The definition of $$\tau_3$$ is

$\begin{split}\tau_3 : T1 \rightarrow T1 \\ \tau_3 = \beta_1^{-1} \beta_2\end{split}$

where the composition is left to right and $$\beta_i : T2 \rightarrow T1$$ changes just the $$i^{th}$$ coordinate of a triple.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: T1 = back_circulant(5)
sage: x = isotopism( (0,1,2,3,4) )
sage: y = isotopism(5) # identity
sage: z = isotopism(5) # identity
sage: T2 = T1.apply_isotopism(x, y, z)
True
sage: (cells_map, t1, t2, t3) = tau123(T1, T2)
sage: t3 = tau3(T1, T2, cells_map)
sage: t3
[10, 6, 7, 8, 9, 15, 11, 12, 13, 14, 20, 16, 17, 18, 19, 25, 21, 22, 23, 24, 5, 1, 2, 3, 4]
sage: t3.to_cycles()
[(1, 10, 14, 18, 22), (2, 6, 15, 19, 23), (3, 7, 11, 20, 24), (4, 8, 12, 16, 25), (5, 9, 13, 17, 21)]


Given permutations t1, t2, t3 that represent a latin bitrade, convert them to an explicit latin bitrade (T1, T2). The result is unique up to isotopism.

EXAMPLES:

sage: from sage.combinat.matrices.latin import *
sage: T1 = back_circulant(5)
sage: x = isotopism( (0,1,2,3,4) )
sage: y = isotopism(5) # identity
sage: z = isotopism(5) # identity
sage: T2 = T1.apply_isotopism(x, y, z)
sage: _, t1, t2, t3 = tau123(T1, T2)
sage: U1, U2 = tau_to_bitrade(t1, t2, t3)