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r""" 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:
#. `\{ (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' \};`
#. for each `(i,\,j,\,k) \in T_1` and `(i,\,j,\,k') \in T_2`, `k \neq k'`;
#. 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
#. LatinSquare class definition;
#. some named latin squares (back circulant, forward circulant, abelian `2`-group);
#. 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;
#. 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);
#. functions for calculating the `\tau_i` representation of a bitrade and the genus of the associated hypermap embedding;
#. Markov chain of Jacobson and Matthews (1996) for generating latin squares uniformly at random (provides a generator interface);
#. 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
To do:
#. Latin squares with symbols from a ring instead of the integers `\{ 0, 1, \dots, n-1 \}`.
#. Isotopism testing of latin squares and bitrades via graph isomorphism (nauty?).
#. Combinatorial constructions for bitrades.
AUTHORS:
- Carlo Hamalainen (2008-03-23): initial version
TESTS::
sage: L = elementary_abelian_2group(3) sage: L == loads(dumps(L)) True
"""
#***************************************************************************** # Copyright (C) 2008 Carlo Hamalainen <carlo.hamalainen@gmail.com>, # # Distributed under the terms of the GNU General Public License (GPL) # # This code is distributed in the hope that it will be useful, # but WITHOUT ANY WARRANTY; without even the implied warranty of # MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU # General Public License for more details. # # The full text of the GPL is available at: # # http://www.gnu.org/licenses/ #***************************************************************************** from __future__ import print_function, absolute_import from six.moves import range
from sage.matrix.all import matrix from sage.rings.all import ZZ from sage.rings.all import Integer from sage.matrix.matrix_integer_dense import Matrix_integer_dense from sage.groups.perm_gps.permgroup_element import PermutationGroupElement from sage.interfaces.gap import GapElement from sage.combinat.permutation import Permutation from sage.interfaces.gap import gap from sage.groups.perm_gps.permgroup import PermutationGroup from sage.arith.all import is_prime from sage.rings.finite_rings.finite_field_constructor import FiniteField from sage.misc.misc import uniq from sage.misc.flatten import flatten
#load "dancing_links.spyx" #load "dancing_links.sage"
from .dlxcpp import DLXCPP from functools import reduce
class LatinSquare: def __init__(self, *args): """ 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] """
else: raise TypeError("bad input for latin square")
def dumps(self): """ Since the latin square class doesn't hold any other private variables we just call dumps on self.square:
EXAMPLES::
sage: from sage.combinat.matrices.latin import * sage: back_circulant(2) == loads(dumps(back_circulant(2))) True """
return dumps(self.square)
def __str__(self): """ The string representation of a latin square is the same as the underlying matrix.
EXAMPLES::
sage: print(LatinSquare(matrix(ZZ, [[0, 1], [2, 3]])).__str__()) [0 1] [2 3] """
def __repr__(self): """ The representation of a latin square is the same as the underlying matrix.
EXAMPLES::
sage: print(LatinSquare(matrix(ZZ, [[0, 1], [2, 3]])).__repr__()) [0 1] [2 3] """
def __getitem__(self, rc): """ If L is a LatinSquare then this method allows us to evaluate L[r, c].
EXAMPLES::
sage: from sage.combinat.matrices.latin import * sage: B = back_circulant(3) sage: B[1, 1] 2 """
def __setitem__(self, rc, val): """ If L is a LatinSquare then this method allows us to set L[r, c].
EXAMPLES::
sage: from sage.combinat.matrices.latin import * sage: B = back_circulant(3) sage: B[1, 1] = 10 sage: B[1, 1] 10 """
def set_immutable(self): """ 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} """
def __hash__(self): """ The hash of a latin square is precisely the hash of the underlying matrix.
EXAMPLES::
sage: L = LatinSquare(matrix(ZZ, [[0, 1], [2, 3]])) sage: L.set_immutable() sage: L.__hash__() 1677951251422179082 # 64-bit -479138038 # 32-bit """
def __eq__(self, Q): """ Two latin squares are equal if the underlying matrices are equal.
EXAMPLES::
sage: A = LatinSquare(matrix(ZZ, [[0, 1], [2, 3]])) sage: B = LatinSquare(matrix(ZZ, [[0, 4], [2, 3]])) sage: A == B False sage: B[0, 1] = 1 sage: A == B True """
def __copy__(self): """ To copy a latin square we must copy the underlying matrix.
EXAMPLES::
sage: A = LatinSquare(matrix(ZZ, [[0, 1], [2, 3]])) sage: B = copy(A) sage: B [0 1] [2 3] """
def clear_cells(self): """ 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] """
def nrows(self): """ Number of rows in the latin square.
EXAMPLES::
sage: LatinSquare(3).nrows() 3 """
def ncols(self): """ Number of columns in the latin square.
EXAMPLES::
sage: LatinSquare(3).ncols() 3 """
def row(self, x): """ Returns row x of the latin square.
EXAMPLES::
sage: from sage.combinat.matrices.latin import * sage: back_circulant(3).row(0) (0, 1, 2) """
def column(self, x): """ Returns column x of the latin square.
EXAMPLES::
sage: from sage.combinat.matrices.latin import * sage: back_circulant(3).column(0) (0, 1, 2) """
def list(self): """ 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] """
def nr_filled_cells(self): """ Returns 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 """
def actual_row_col_sym_sizes(self): """ 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) """
def is_empty_column(self, c): """ Checks 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 """
def is_empty_row(self, r): """ Checks 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 """
def nr_distinct_symbols(self): """ Returns 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 """
def apply_isotopism(self, 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] """
#Q = matrix(ZZ, self.nrows(), self.ncols())
except IndexError: s2 = self[r, c] # we must be leaving the symbol fixed?
def filled_cells_map(self): """ 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: T1.filled_cells_map() {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), (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} """
def top_left_empty_cell(self): """ Returns 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] """
return None
def is_partial_latin_square(self): """ 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 """
# Entry out of range 0, 1, ..., n-1:
# Entry has already appeared in this row:
# Entry out of range 0, 1, ..., n-1:
# Entry has already appeared in this column:
def is_latin_square(self): """ 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 """
# We don't allow latin rectangles: return False
# Every cell must be filled: return False
# By necessity self must be a partial latin square:
def permissable_values(self, 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) [0] """
return []
# We may have already removed a symbol # in the previous for-loop.
def random_empty_cell(self): """ 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] """
return None
def is_uniquely_completable(self): """ Returns 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 """
def is_completable(self): """ Returns 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 """
def gcs(self): """ 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] """
def dlxcpp_has_unique_completion(self): """ 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 """
def vals_in_row(self, r): """ Returns 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} """
def vals_in_col(self, c): """ Returns 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} """
def latex(self): r""" Returns 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} """
def disjoint_mate_dlxcpp_rows_and_map(self, allow_subtrade): """ Internal function for find_disjoint_mates.
EXAMPLES::
sage: from sage.combinat.matrices.latin import * sage: B = back_circulant(4) sage: B.disjoint_mate_dlxcpp_rows_and_map(allow_subtrade = True) ([[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)}) """
# We will need 3n^2 columns in total: # # n^2 for the xCy columns # n^2 for the xRy columns # n^2 for the xy columns
# If this is an empty cell of self then we do nothing.
# These should be constants
# The disjoint mate has to be disjoint.
# The permissible symbols must come from this row/column.
# We will have missed some columns. We # have to add 'dummy' rows so that the C++ DLX solver will find # a solution. dlx_rows.append([i])
def find_disjoint_mates(self, nr_to_find = None, allow_subtrade = False): r""" .. 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: assert is_bitrade(B0, 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] """
if nr_to_find is not None and nr_found >= nr_to_find: return
def contained_in(self, Q): r""" Returns 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 """
def genus(T1, T2): """ Returns 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 """
def 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: cells_map {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), (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 """
def 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 :class:`Permutation` object defined on `1,...,n`.
For a definition of isotopism, see the :wikipedia:`wikipedia section on isotopism <Latin_square#Equivalence_classes_of_Latin_squares>`.
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 :class:`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)] """
# Identity isotopism on p points:
# fixme Ask the Sage mailing list about the tuple/list issue!
# We expect a list like [0,3,2,1] which means # that 0 goes to 0, 1 goes to 3, etc.
# We have a single cycle:
# We have a tuple of cycles:
# Not sure what we got! raise TypeError("unable to convert {!r} to isotopism".format(p))
def cells_map_as_square(cells_map, n): """ Returns 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] """
# There is no cell (r,c) so skip it
def 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) sage: is_bitrade(T1, T2) True sage: beta1([0, 0, 0], T1, T2) (1, 0, 0) """
raise PairNotBitrade
def 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) sage: is_bitrade(T1, T2) True sage: beta2([0, 0, 0], T1, T2) (0, 1, 0) """
raise PairNotBitrade
def 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) sage: is_bitrade(T1, T2) True sage: beta3([0, 0, 0], T1, T2) (0, 0, 4) """
# fixme this range could be iffy if we # work with latin bitrade rectangles...
raise PairNotBitrade
def tau1(T1, T2, cells_map): r""" The definition of `\tau_1` is
.. MATH::
\tau_1 : T1 \rightarrow T1 \\ \tau_1 = \beta_2^{-1} \beta_3
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) sage: is_bitrade(T1, T2) 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)] """
# The cells_map has both directions, i.e. integer to # cell and cell to integer, so the size of T1 is # just half of len(cells_map).
# have permutations on 1..(something).
def tau2(T1, T2, cells_map): r""" The definition of `\tau_2` is
.. MATH::
\tau_2 : T1 \rightarrow T1 \\ \tau_2 = \beta_3^{-1} \beta_1
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) sage: is_bitrade(T1, T2) 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)] """
# The cells_map has both directions, i.e. integer to # cell and cell to integer, so the size of T1 is # just half of len(cells_map).
# have permutations on 1..(something).
def tau3(T1, T2, cells_map): r""" The definition of `\tau_3` is
.. MATH::
\tau_3 : T1 \rightarrow T1 \\ \tau_3 = \beta_1^{-1} \beta_2
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) sage: is_bitrade(T1, T2) 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)] """
# The cells_map has both directions, i.e. integer to # cell and cell to integer, so the size of T1 is # just half of len(cells_map).
# have permutations on 1..(something).
def 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] """
def 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] """
def 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] """
def elementary_abelian_2group(s): """ Returns 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] """
else:
def coin(): """ Simulates a fair coin (returns True or False) using ZZ.random_element(2).
EXAMPLES::
sage: from sage.combinat.matrices.latin import * sage: x = coin() sage: x == 0 or x == 1 True """
def next_conjugate(L): """ Permutes 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 """
def 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 """
def 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 """
def 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) """
################################# # Update the other two conjugates
#################################
r1 = ZZ.random_element(n) c1 = ZZ.random_element(n) x = L[r1, c1]
y = x while y == x: y = ZZ.random_element(n)
# Now find y in row r1 and column c1. if check_assertions: r2 = 0 c2 = 0 while L[r1, c2] != y: c2 += 1 while L[r2, c1] != y: r2 += 1
assert L_erc[y, r1] == c2 assert L_cer[c1, y] == r2
c2 = L_erc[y, r1] r2 = L_cer[c1, y]
if check_assertions: assert L[r1, c2] == y if check_assertions: assert L[r2, c1] == y
L[r1, c1] = y L[r1, c2] = x L[r2, c1] = x
# Now deal with the unknown point. # We want to form z + (y - x) z = L[r2, c2]
if z == x: L[r2, c2] = y else: # z and y have positive coefficients # x is the improper term with a negative coefficient proper = False else: # improper square, # L[r2, c2] = y + z - x # y and z are proper while x is the # improper symbol in the cell L[r2, c2].
r1 = row_containing_sym(L, c2, x) c1 = column_containing_sym(L, r2, x)
if check_assertions: assert L[r1, c2] == x if check_assertions: assert L[r2, c1] == x
# choose one of the proper symbols # uniformly at random (we will use whatever # lands in variable y). if coin(): y, z = z, y
# Add/subtract the symbolic difference (y - x) L[r2, c2] = z L[r1, c2] = y L[r2, c1] = y
if L[r1, c1] == y: L[r1, c1] = x proper = True else: # got another improper square z = L[r1, c1] x, y = y, x r2 = r1 c2 = c1
# Now we have L[r2, c2] = z+y-x as # usual proper = False # for emphasis
def 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 * 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] """
# Otherwise we must have some kind of Sage permutation group object, # such as sage.groups.perm_gps.permgroup.PermutationGroup_generic # or maybe sage.groups.perm_gps.permgroup_named.
def alternating_group_bitrade_generators(m): """ 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] """
def pq_group_bitrade_generators(p, q): """ 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 * sage: pq_group_bitrade_generators(3,7) ((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)]) """
# beta is a primitive root of the # congruence x^p = 1 mod q
def p3_group_bitrade_generators(p): """ Generators for a group of order p3 where p is a prime.
EXAMPLES::
sage: from sage.combinat.matrices.latin import * sage: p3_group_bitrade_generators(3) ((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)]) """
def check_bitrade_generators(a, b, c): """ 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) sage: check_bitrade_generators(a, b, c) True sage: check_bitrade_generators(a, b, gap('()')) False """
def is_bitrade(T1, T2): """ 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) sage: is_bitrade(T1, T2) True """
def is_primary_bitrade(a, b, c, G): """ 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) sage: is_primary_bitrade(a, b, c, G) True """
def tau_to_bitrade(t1, t2, t3): """ 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) sage: assert is_bitrade(U1, U2) sage: U1 [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: U2 [4 0 1 2 3] [0 1 2 3 4] [1 2 3 4 0] [2 3 4 0 1] [3 4 0 1 2] """
[set(c1[r]), set(c2[c]), set(c3[s])]))
def bitrade_from_group(a, b, c, G): """ 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] """
def 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 """
def 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 """
def 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 """
def 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)}) """
# We will need 3n^2 columns in total: # # n^2 for the xCy columns # n^2 for the xRy columns # n^2 for the xy columns
# These should be constants
#if P[r, c] >= 0: continue
# We only want the correct value to pop in here
def dlxcpp_find_completions(P, nr_to_find = None): """ Returns 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]] """
else:
def bitrade(T1, T2): r""" 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] """
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