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 \mbox{ for some symbol $k$} \} = \{ (i,\,j) \mid (i,\,j,\,k') \in T_2 \mbox{ 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 2 -1 1]
[-1 0 1 3]
[ 3 -1 0 2]
[ 1 3 2 -1]
sage: T2
[ 1 0 -1 2]
[-1 3 0 1]
[ 0 -1 2 3]
[ 3 2 1 -1]
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
"""
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.rings.arith import is_prime
from sage.rings.finite_rings.constructor import FiniteField
from sage.misc.misc import uniq
from sage.misc.flatten import flatten
from dlxcpp import DLXCPP
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]
"""
if len(args) == 1 and (type(args[0]) == Integer or type(args[0]) == int):
self.square = matrix(ZZ, args[0], args[0])
self.clear_cells()
elif len(args) == 2 and (type(args[0]) == Integer or type(args[0]) == int) and (type(args[1]) == Integer or type(args[1]) == int):
self.square = matrix(ZZ, args[0], args[1])
self.clear_cells()
elif len(args) == 1 and type(args[0]) == Matrix_integer_dense:
self.square = args[0]
else:
raise NotImplemented
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]
"""
return self.square.__str__()
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]
"""
return self.square.__str__()
return self.square.__repr__()
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
"""
r = rc[0]
c = rc[1]
return self.square[r, c]
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
"""
r = rc[0]
c = rc[1]
self.square[r, c] = val
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}
"""
self.square.set_immutable()
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__()
12
"""
return self.square.__hash__()
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
"""
return self.square == Q.square
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]
"""
C = LatinSquare(self.square.nrows(), self.square.ncols())
from copy import copy
C.square = copy(self.square)
return C
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]
"""
for r in range(self.square.nrows()):
for c in range(self.square.ncols()):
self.square[r, c] = -1;
def nrows(self):
"""
Number of rows in the latin square.
EXAMPLES::
sage: LatinSquare(3).nrows()
3
"""
return self.square.nrows()
def ncols(self):
"""
Number of columns in the latin square.
EXAMPLES::
sage: LatinSquare(3).ncols()
3
"""
return self.square.ncols()
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)
"""
return self.square.row(x)
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)
"""
return self.square.column(x)
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]
"""
return self.square.list()
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
"""
s = 0
for r in range(self.nrows()):
for c in range(self.ncols()):
if self[r, c] >= 0: s += 1
return s
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).
EXAMPLE::
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)
"""
row_max = self.nrows()
col_max = self.ncols()
sym_max = self.nr_distinct_symbols()
while self.is_empty_row(row_max-1): row_max -= 1
while self.is_empty_column(col_max-1): col_max -= 1
return row_max, col_max, sym_max
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
"""
return uniq(self.column(c)) == [-1]
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
"""
return uniq(self.row(r)) == [-1]
def nr_distinct_symbols(self):
"""
Returns the number of distinct symbols in the partial latin square
self.
EXAMPLE::
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
"""
symbols = uniq(flatten(map(lambda x: list(x), list(self.square))))
symbols = filter(lambda x: x >= 0, symbols)
return len(symbols)
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 = LatinSquare(self.nrows(), self.ncols())
for r in range(self.nrows()):
for c in range(self.ncols()):
try:
if self[r, c] < 0: s2 = -1
else: s2 = sym_perm[self[r, c]] - 1
except IndexError:
s2 = self[r, c]
Q[row_perm[r]-1, col_perm[c]-1] = s2
return Q
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, 1), 3: (0, 3), 4: (1, 1), 5: (1, 2), 6: (1, 3), 7: (2, 0), 8: (2, 2), 9: (2, 3), 10: (3, 0), 11: (3, 1), 12: (3, 2), (1, 3): 6, (0, 3): 3, (1, 2): 5, (3, 0): 10, (2, 2): 8, (1, 1): 4, (3, 2): 12, (0, 0): 1, (2, 3): 9, (0, 1): 2, (3, 1): 11, (2, 0): 7}
"""
cells_map = {}
k = 1
for r in range(self.nrows()):
for c in range(self.ncols()):
e = self[r, c]
if e < 0: continue
cells_map[ (r,c) ] = k
cells_map[k] = (r,c)
k += 1
return cells_map
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]
"""
for r in range(self.nrows()):
for c in range(self.ncols()):
if self[r, c] < 0:
return [r, c]
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
"""
assert self.nrows() == self.ncols()
n = self.nrows()
for r in range(n):
vals_in_row = {}
for c in range(n):
e = self[r, c]
if e < 0: continue
if e >= n: return False
if vals_in_row.has_key(e): return False
vals_in_row[e] = True
for c in range(n):
vals_in_col = {}
for r in range(n):
e = self[r, c]
if e < 0: continue
if e >= n: return False
if vals_in_col.has_key(e): return False
vals_in_col[e] = True
return True
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
"""
if self.nrows() != self.ncols():
return False
if len(filter(lambda x: x >= 0, self.list())) != self.nrows()*self.ncols():
return False
if not self.is_partial_latin_square():
return False
return True
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]
"""
if self[r, c] >= 0:
return []
assert self.nrows() == self.ncols()
n = self.nrows()
vals = {}
for e in range(n):
vals[e] = True
for i in range(n):
if self[i, c] >= 0:
del vals[ self[i, c] ]
for j in range(n):
if self[r, j] >= 0:
try:
del vals[ self[r, j] ]
except KeyError:
pass
return vals.keys()
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]
"""
cells = {}
for r in range(self.nrows()):
for c in range(self.ncols()):
if self[r, c] < 0:
cells[ (r,c) ] = True
cells = cells.keys()
if len(cells) == 0: return None
rc = cells[ ZZ.random_element(len(cells)) ]
return [rc[0], rc[1]]
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
"""
return self.dlxcpp_has_unique_completion()
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
"""
return len(dlxcpp_find_completions(self, nr_to_find = 1)) > 0
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]
"""
n = self.nrows()
from copy import copy
G = copy(self)
for r in range(n-1, -1, -1):
for c in range(n-1, -1, -1):
e = G[r, c]
G[r, c] = -1
if not G.dlxcpp_has_unique_completion():
G[r, c] = e
return G
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
"""
return len(dlxcpp_find_completions(self, nr_to_find = 2)) == 1
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}
"""
n = self.ncols()
vals_in_row = {}
for c in range(n):
e = self[r, c]
if e >= 0: vals_in_row[e] = True
return vals_in_row
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}
"""
n = self.nrows()
vals_in_col = {}
for r in range(n):
e = self[r, c]
if e >= 0: vals_in_col[e] = True
return vals_in_col
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}
"""
a = ""
a += r"\begin{array}{" + self.ncols()*"|c" + "|}"
for r in range(self.nrows()):
a += r"\hline "
for c in range(self.ncols()):
s = self[r, c]
if s < 0: a += "~"
else: a += str(s)
if c < self.ncols()-1: a += " & "
else: a += "\\\\"
a += r"\hline"
a += r"\end{array}"
return a
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]], {(9, 29, 46): (3, 2, 1), (13, 17, 35): (0, 3, 1), (7, 19, 33): (0, 1, 3), (14, 26, 43): (2, 3, 2), (0, 28, 44): (3, 0, 0), (5, 25, 41): (2, 1, 1), (11, 31, 46): (3, 2, 3), (14, 18, 35): (0, 3, 2), (11, 23, 38): (1, 2, 3), (5, 29, 45): (3, 1, 1), (13, 21, 39): (1, 3, 1), (1, 29, 44): (3, 0, 1), (0, 20, 36): (1, 0, 0), (12, 24, 43): (2, 3, 0), (8, 28, 46): (3, 2, 0), (12, 20, 39): (1, 3, 0), (11, 27, 42): (2, 2, 3), (6, 22, 37): (1, 1, 2), (1, 17, 32): (0, 0, 1), (10, 18, 34): (0, 2, 2), (12, 28, 47): (3, 3, 0), (1, 25, 40): (2, 0, 1), (10, 22, 38): (1, 2, 2), (5, 17, 33): (0, 1, 1), (3, 23, 36): (1, 0, 3), (6, 26, 41): (2, 1, 2), (9, 25, 42): (2, 2, 1), (7, 31, 45): (3, 1, 3), (15, 27, 43): (2, 3, 3), (3, 31, 44): (3, 0, 3), (8, 20, 38): (1, 2, 0), (2, 22, 36): (1, 0, 2), (3, 19, 32): (0, 0, 3), (9, 17, 34): (0, 2, 1), (15, 31, 47): (3, 3, 3), (8, 16, 34): (0, 2, 0), (14, 22, 39): (1, 3, 2), (4, 16, 33): (0, 1, 0), (14, 30, 47): (3, 3, 2), (2, 30, 44): (3, 0, 2), (4, 20, 37): (1, 1, 0), (6, 30, 45): (3, 1, 2), (12, 16, 35): (0, 3, 0), (15, 19, 35): (0, 3, 3), (5, 21, 37): (1, 1, 1), (4, 24, 41): (2, 1, 0), (13, 25, 43): (2, 3, 1), (0, 16, 32): (0, 0, 0), (15, 23, 39): (1, 3, 3), (7, 23, 37): (1, 1, 3), (6, 18, 33): (0, 1, 2), (10, 30, 46): (3, 2, 2), (13, 29, 47): (3, 3, 1), (11, 19, 34): (0, 2, 3), (1, 21, 36): (1, 0, 1), (7, 27, 41): (2, 1, 3), (0, 24, 40): (2, 0, 0), (10, 26, 42): (2, 2, 2), (3, 27, 40): (2, 0, 3), (2, 26, 40): (2, 0, 2), (9, 21, 38): (1, 2, 1), (8, 24, 42): (2, 2, 0), (4, 28, 45): (3, 1, 0), (2, 18, 32): (0, 0, 2)})
"""
assert self.nrows() == self.ncols()
n = self.nrows()
dlx_rows = []
cmap = {}
max_column_nr = -1
for r in range(n):
valsrow = self.vals_in_row(r)
for c in range(n):
valscol = self.vals_in_col(c)
if self[r, c] < 0: continue
for e in uniq(valsrow.keys() + valscol.keys()):
c_OFFSET = e + c*n
r_OFFSET = e + r*n + n*n
xy_OFFSET = 2*n*n + r*n + c
cmap[(c_OFFSET, r_OFFSET, xy_OFFSET)] = (r,c,e)
if (not allow_subtrade) and self[r, c] == e: continue
if not(valsrow.has_key(e)): continue
if not(valscol.has_key(e)): continue
dlx_rows.append([c_OFFSET, r_OFFSET, xy_OFFSET])
if max_column_nr < max(c_OFFSET, r_OFFSET, xy_OFFSET):
max_column_nr = max(c_OFFSET, r_OFFSET, xy_OFFSET)
used_columns = flatten(dlx_rows)
for i in range(0, max_column_nr+1):
if not i in used_columns:
dlx_rows.append([i])
return dlx_rows, cmap
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 = g.next()
sage: B0, B1 = bitrade(B, B1)
sage: assert is_bitrade(B0, B1)
sage: print B0, "\n,\n", B1
[-1 1 2 -1]
[-1 2 -1 0]
[-1 -1 -1 -1]
[-1 0 1 2]
,
[-1 2 1 -1]
[-1 0 -1 2]
[-1 -1 -1 -1]
[-1 1 2 0]
"""
assert self.nrows() == self.ncols()
n = self.nrows()
dlx_rows, cmap = self.disjoint_mate_dlxcpp_rows_and_map(allow_subtrade)
nr_found = 0
for x in DLXCPP(dlx_rows):
nr_found += 1
from copy import deepcopy
Q = deepcopy(self)
for y in x:
if len(dlx_rows[y]) == 1: continue
(r, c, e) = cmap[tuple(dlx_rows[y])]
Q[r, c] = e
yield Q
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
"""
for r in range(self.nrows()):
for c in range(self.ncols()):
if self[r, c] >= 0 and Q[r, c] < 0: return False
if self[r, c] >= 0 and (self[r, c] != Q[r, c]): return False
return True
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 http://arxiv.org/abs/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
"""
cells_map, t1, t2, t3 = tau123(T1, T2)
return (len(t1.to_cycles()) + len(t2.to_cycles()) + len(t3.to_cycles()) - T1.nr_filled_cells() - 2)/(-2)
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 2 -1 -1 -1 -1]
[ 2 4 3 -1 -1 -1 -1]
[ 4 0 5 -1 -1 -1 -1]
[ 3 5 1 -1 -1 -1 -1]
[ 6 3 0 -1 -1 -1 -1]
[ 1 6 4 -1 -1 -1 -1]
[ 5 2 6 -1 -1 -1 -1]
sage: T2
[ 2 0 1 -1 -1 -1 -1]
[ 3 2 4 -1 -1 -1 -1]
[ 5 4 0 -1 -1 -1 -1]
[ 1 3 5 -1 -1 -1 -1]
[ 0 6 3 -1 -1 -1 -1]
[ 4 1 6 -1 -1 -1 -1]
[ 6 5 2 -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), (2, 1): 8, 15: (4, 2), 16: (5, 0), 17: (5, 1), 18: (5, 2), 19: (6, 0), 20: (6, 1), 21: (6, 2), (5, 1): 17, (4, 0): 13, (1, 2): 6, (3, 0): 10, (5, 0): 16, (2, 2): 9, (4, 1): 14, (1, 1): 5, (3, 2): 12, (0, 0): 1, (6, 0): 19, 14: (4, 1), (4, 2): 15, (1, 0): 4, (0, 1): 2, (6, 1): 20, (3, 1): 11, (2, 0): 7, (6, 2): 21, (5, 2): 18, (0, 2): 3}
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
[2, 3, 1, 5, 6, 4, 8, 9, 7, 11, 12, 10, 14, 15, 13, 17, 18, 16, 20, 21, 19]
sage: t2
[4, 8, 12, 10, 20, 18, 19, 5, 15, 16, 14, 9, 1, 17, 6, 7, 2, 21, 13, 11, 3]
sage: t3
[15, 16, 20, 3, 7, 14, 18, 1, 11, 6, 19, 2, 21, 10, 8, 12, 13, 5, 9, 4, 17]
::
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)]
sage: t2.to_cycles()
[(1, 4, 10, 16, 7, 19, 13), (2, 8, 5, 20, 11, 14, 17), (3, 12, 9, 15, 6, 18, 21)]
sage: t3.to_cycles()
[(1, 15, 8), (2, 16, 12), (3, 20, 4), (5, 7, 18), (6, 14, 10), (9, 11, 19), (13, 21, 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
"""
assert is_bitrade(T1, T2)
cells_map = T1.filled_cells_map()
t1 = tau1(T1, T2, cells_map)
t2 = tau2(T1, T2, cells_map)
t3 = tau3(T1, T2, cells_map)
return (cells_map, t1, t2, t3)
def isotopism(p):
"""
Returns a Permutation object that represents an isotopism (for
rows, columns or symbols of a partial latin square). Since matrices
in Sage are indexed from 0, this function translates +1 to agree
with the Permutation class. We also handle
PermutationGroupElements.
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)]
"""
if type(p) == Integer or type(p) == int:
return Permutation(range(1, p+1))
if type(p) == PermutationGroupElement:
return Permutation(list(p.tuple()))
if type(p) == list:
return Permutation(map(lambda x: x+1, p))
if type(p) == tuple:
if type(p[0]) == Integer:
return Permutation(tuple(map(lambda x: x+1, p)))
if type(p[0]) == tuple:
x = isotopism(p[0])
for i in range(1, len(p)):
x = x * isotopism(p[i])
return x
raise NotImplemented
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 2 -1 1]
[-1 0 1 3]
[ 3 -1 0 2]
[ 1 3 2 -1]
There are 12 filled cells in T::
sage: cells_map_as_square(T1.filled_cells_map(), max(T1.nrows(), T1.ncols()))
[ 1 2 -1 3]
[-1 4 5 6]
[ 7 -1 8 9]
[10 11 12 -1]
"""
assert n > 1
L = LatinSquare(n, n)
for r in range(n):
for c in range(n):
try:
L[r, c] = cells_map[ (r,c) ]
except KeyError:
L[r, c] = -1
return L
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)
"""
r = rce[0]
c = rce[1]
e = rce[2]
assert T1[r, c] == e
assert e >= 0
for x in range(T1.nrows()):
if T2[x, c] == e: return (x, c, e)
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)
"""
r = rce[0]
c = rce[1]
e = rce[2]
assert T1[r, c] == e
assert e >= 0
for x in range(T1.ncols()):
if T2[r, x] == e: return (r, x, e)
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)
"""
r = rce[0]
c = rce[1]
e = rce[2]
assert T1[r, c] == e
assert e >= 0
for x in range(T1.nrows()):
if T2[r, c] == x: return (r, c, x)
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)]
"""
x = (int(len(cells_map)/2) + 1) * [-1]
for r in range(T1.nrows()):
for c in range(T1.ncols()):
e = T1[r, c]
if e < 0: continue
(r2, c2, e2) = beta2( (r,c,e), T1, T2)
(r3, c3, e3) = beta3( (r2,c2,e2), T2, T1)
x[ cells_map[(r,c)] ] = cells_map[ (r3,c3) ]
x.pop(0)
return Permutation(x)
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)]
"""
x = (int(len(cells_map)/2) + 1) * [-1]
for r in range(T1.nrows()):
for c in range(T1.ncols()):
e = T1[r, c]
if e < 0: continue
(r2, c2, e2) = beta3( (r,c,e), T1, T2)
(r3, c3, e3) = beta1( (r2,c2,e2), T2, T1)
x[ cells_map[(r,c)] ] = cells_map[ (r3,c3) ]
x.pop(0)
return Permutation(x)
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)]
"""
x = (int(len(cells_map)/2) + 1) * [-1]
for r in range(T1.nrows()):
for c in range(T1.ncols()):
e = T1[r, c]
if e < 0: continue
(r2, c2, e2) = beta1( (r,c,e), T1, T2)
(r3, c3, e3) = beta2( (r2,c2,e2), T2, T1)
x[ cells_map[(r,c)] ] = cells_map[ (r3,c3) ]
x.pop(0)
return Permutation(x)
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]
"""
assert n >= 1
L = LatinSquare(n, n)
for r in range(n):
for c in range(n):
L[r, c] = (r + c) % n
return L
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]
"""
assert n >= 1
L = LatinSquare(n, n)
for r in range(n):
for c in range(n):
L[r, c] = (n-c+r) % n
return L
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]
"""
assert L1.nrows() == L2.nrows() == L3.nrows() == L4.nrows()
assert L1.ncols() == L2.ncols() == L3.ncols() == L4.ncols()
assert L1.nrows() == L1.ncols()
n = L1.nrows()
D = LatinSquare(2*n, 2*n)
for r in range(n):
for c in range(n):
D[r, c] = L1[r, c]
D[r, c+n] = L2[r, c] + n
D[r+n, c] = L3[r, c] + n
D[r+n, c+n] = L4[r, c]
return D
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]
"""
assert s > 0
if s == 1:
L = LatinSquare(2, 2)
L[0, 0] = 0
L[0, 1] = 1
L[1, 0] = 1
L[1, 1] = 0
return L
else:
L_prev = elementary_abelian_2group(s-1)
L = LatinSquare(2**s, 2**s)
offset = L.nrows()/2
for r in range(L_prev.nrows()):
for c in range(L_prev.ncols()):
L[r, c] = L_prev[r, c]
L[r+offset, c] = L_prev[r, c] + offset
L[r, c+offset] = L_prev[r, c] + offset
L[r+offset, c+offset] = L_prev[r, c]
return L
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
"""
return ZZ.random_element(2) == 0
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
"""
assert L.nrows() == L.ncols()
n = L.nrows()
C = LatinSquare(n, n)
for r in range(n):
for c in range(n):
e = L[r, c]
assert e >= 0 and e < n
C[c, e] = r
return C
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
"""
r1 = -1
r2 = -1
for r in range(L.nrows()):
if r1 >= 0 and r2 >= 0: break
if L[r, c] == x and r1 < 0:
r1 = r
continue
if L[r, c] == x and r2 < 0:
r2 = r
break
assert r1 >= 0 and r2 >= 0
if coin(): return r1
else: return r2
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
"""
c1 = -1
c2 = -1
for c in range(L.ncols()):
if c1 >= 0 and c2 >= 0: break
if L[r, c] == x and c1 < 0:
c1 = c
continue
if L[r, c] == x and c2 < 0:
c2 = c
break
assert c1 >= 0 and c2 >= 0
if coin(): return c1
else: return c2
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: g.next().is_latin_square()
True
REFERENCE::
@article{MR1410617,
AUTHOR = {Jacobson, Mark T. and Matthews, Peter},
TITLE = {Generating uniformly distributed random {L}atin squares},
JOURNAL = {J. Combin. Des.},
FJOURNAL = {Journal of Combinatorial Designs},
VOLUME = {4},
YEAR = {1996},
NUMBER = {6},
PAGES = {405--437},
ISSN = {1063-8539},
MRCLASS = {05B15 (60J10)},
MRNUMBER = {MR1410617 (98b:05021)},
MRREVIEWER = {Lars D{\o}vling Andersen},
}
"""
if check_assertions: assert L_start.is_latin_square()
n = L_start.nrows()
r1 = r2 = c1 = c2 = x = y = z = -1
proper = True
from copy import copy
L = copy(L_start)
L_rce = L
L_cer = LatinSquare(n, n)
L_erc = LatinSquare(n, n)
while True:
if proper:
if check_assertions: assert L.is_latin_square()
for r in range(n):
for c in range(n):
e = L[r, c]
L_cer[c, e] = r
L_erc[e, r] = c
yield L
r1 = ZZ.random_element(n)
c1 = ZZ.random_element(n)
x = L[r1, c1]
y = x
while y == x:
y = ZZ.random_element(n)
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
z = L[r2, c2]
if z == x:
L[r2, c2] = y
else:
proper = False
else:
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
if coin():
y, z = z, y
L[r2, c2] = z
L[r1, c2] = y
L[r2, c1] = y
if L[r1, c1] == y:
L[r1, c1] = x
proper = True
else:
z = L[r1, c1]
x, y = y, x
r2 = r1
c2 = c1
proper = False
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]
"""
if isinstance(G, GapElement):
rows = map(lambda x: list(x), list(gap.MultiplicationTable(G)))
new_rows = []
for x in rows:
new_rows.append(map(lambda x: int(x)-1, x))
return matrix(new_rows)
T = G.cayley_table()
return matrix(ZZ, T.table())
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 2 -1 1]
[-1 0 1 3]
[ 3 -1 0 2]
[ 1 3 2 -1]
sage: T2
[ 1 0 -1 2]
[-1 3 0 1]
[ 0 -1 2 3]
[ 3 2 1 -1]
"""
assert m >= 1
a = tuple(range(1, 2*m+1 + 1))
b = tuple(range(m+1, 0, -1) + range(2*m+2, 3*m+1 + 1))
a = PermutationGroupElement(a)
b = PermutationGroupElement(b)
c = PermutationGroupElement((a*b)**(-1))
G = PermutationGroup([a, b])
return (a, b, c, G)
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)])
"""
assert is_prime(p)
assert is_prime(q)
assert (q % p) == 1
F = FiniteField(q)
fgen = F.multiplicative_generator()
beta = fgen**((q-1)/p)
assert beta != 1
assert (beta**p % q) == 1
Q = tuple(range(1, q+1))
P = []
seenValues = {}
for i in range(2, q):
if seenValues.has_key(i):
continue
cycle = []
for k in range(p):
x = (1 + (i-1)*beta**k) % q
if x == 0: x = q
seenValues[x] = True
cycle.append(x)
P.append(tuple(map(Integer,cycle)))
G = PermutationGroup([P, Q])
assert G.order() == p*q
assert not G.is_abelian()
a = PermutationGroupElement(P)
b = PermutationGroupElement(Q)
c = PermutationGroupElement((a*b)**(-1))
return (a, b, c, PermutationGroup([P, 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)])
"""
assert is_prime(p)
F = gap.new("FreeGroup(3)")
a = F.gen(1)
b = F.gen(2)
c = F.gen(3)
rels = []
rels.append( a**p )
rels.append( b**p )
rels.append( c**p )
rels.append( a*b*((b*a*c)**(-1)) )
rels.append( c*a*((a*c)**(-1)) )
rels.append( c*b*((b*c)**(-1)) )
G = F.FactorGroupFpGroupByRels(rels)
iso = gap.IsomorphismPermGroup(G)
x = PermutationGroupElement(gap.Image(iso, G.gen(1)))
y = PermutationGroupElement(gap.Image(iso, G.gen(2)))
return (x, y, (x*y)**(-1), PermutationGroup([x, y]))
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
"""
A = PermutationGroup([a])
B = PermutationGroup([b])
C = PermutationGroup([c])
if a*b != c**(-1): return False
X = gap.Intersection(gap.Intersection(A, B), C)
return X.Size() == 1
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
"""
if not is_disjoint(T1, T2): return False
if not is_same_shape(T1, T2): return False
if not is_row_and_col_balanced(T1, T2): return False
return 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
"""
H = PermutationGroup([a, b, c])
return G == H
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.
EXAMPLE::
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]
"""
c1 = t1.to_cycles()
c2 = t2.to_cycles()
c3 = t3.to_cycles()
pt_to_cycle1 = {}
pt_to_cycle2 = {}
pt_to_cycle3 = {}
for i in range(len(c1)):
for j in range(len(c1[i])):
pt_to_cycle1[c1[i][j]] = i
for i in range(len(c2)):
for j in range(len(c2[i])):
pt_to_cycle2[c2[i][j]] = i
for i in range(len(c3)):
for j in range(len(c3[i])):
pt_to_cycle3[c3[i][j]] = i
n = max(len(c1), len(c2), len(c3))
T1 = LatinSquare(n)
T2 = LatinSquare(n)
for r in range(len(c1)):
for c in range(len(c2)):
for s in range(len(c3)):
nr_common = len(reduce(set.intersection, \
[set(c1[r]), set(c2[c]), set(c3[s])]))
assert nr_common in [0, 1]
if nr_common == 1: T1[r, c] = s
for cycle in c1:
for pt1 in cycle:
pt2 = t1[pt1 - 1]
pt3 = t2[pt2 - 1]
assert t3[pt3 - 1] == pt1
r = pt_to_cycle1[pt1]
c = pt_to_cycle2[pt2]
s = pt_to_cycle3[pt3]
T2[r, c] = s
return T1, T2
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 2 -1 1]
[-1 0 1 3]
[ 3 -1 0 2]
[ 1 3 2 -1]
sage: T2
[ 1 0 -1 2]
[-1 3 0 1]
[ 0 -1 2 3]
[ 3 2 1 -1]
"""
hom = gap.ActionHomomorphism(G, gap.RightCosets(G, gap.TrivialSubgroup(G)), gap.OnRight)
t1 = gap.Image(hom, a)
t2 = gap.Image(hom, b)
t3 = gap.Image(hom, c)
t1 = Permutation(str(t1).replace('\n', ''))
t2 = Permutation(str(t2).replace('\n', ''))
t3 = Permutation(str(t3).replace('\n', ''))
return tau_to_bitrade(t1, t2, t3)
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
"""
for i in range(T1.nrows()):
for j in range(T1.ncols()):
if T1[i, j] < 0 and T2[i, j] < 0: continue
if T1[i, j] == T2[i, j]:
return False
return 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
"""
for i in range(T1.nrows()):
for j in range(T1.ncols()):
if T1[i, j] < 0 and T2[i, j] < 0: continue
if T1[i, j] >= 0 and T2[i, j] >= 0: continue
return False
return True
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
"""
for r in range(T1.nrows()):
val1 = set(filter(lambda x: x >= 0, T1.row(r)))
val2 = set(filter(lambda x: x >= 0, T2.row(r)))
if val1 != val2: return False
for c in range(T1.ncols()):
val1 = set(filter(lambda x: x >= 0, T1.column(c)))
val2 = set(filter(lambda x: x >= 0, T2.column(c)))
if val1 != val2: return False
return True
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]], {(2, 4, 9): (0, 1, 0), (1, 5, 8): (0, 0, 1), (1, 7, 10): (1, 0, 1), (0, 6, 10): (1, 0, 0), (3, 7, 11): (1, 1, 1), (2, 6, 11): (1, 1, 0), (0, 4, 8): (0, 0, 0), (3, 5, 9): (0, 1, 1)})
"""
assert P.nrows() == P.ncols()
n = P.nrows()
dlx_rows = []
cmap = {}
for r in range(n):
valsrow = P.vals_in_row(r)
for c in range(n):
valscol = P.vals_in_col(c)
for e in range(n):
c_OFFSET = e + c*n
r_OFFSET = e + r*n + n*n
xy_OFFSET = 2*n*n + r*n + c
cmap[(c_OFFSET, r_OFFSET, xy_OFFSET)] = (r,c,e)
if P[r, c] >= 0 and P[r, c] != e: continue
if P[r, c] < 0 and valsrow.has_key(e): continue
if P[r, c] < 0 and valscol.has_key(e): continue
dlx_rows.append([c_OFFSET, r_OFFSET, xy_OFFSET])
return dlx_rows, cmap
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]]
"""
assert P.nrows() == P.ncols()
n = P.nrows()
dlx_rows, cmap = dlxcpp_rows_and_map(P)
SOLUTIONS = {}
for x in DLXCPP(dlx_rows):
x.sort()
SOLUTIONS[tuple(x)] = True
if nr_to_find is not None and len(SOLUTIONS) >= nr_to_find: break
comps = []
for i in SOLUTIONS.keys():
soln = list(i)
from copy import deepcopy
Q = deepcopy(P)
for x in soln:
(r, c, e) = cmap[tuple(dlx_rows[x])]
if Q[r, c] >= 0:
assert Q[r, c] == e
else:
Q[r, c] = e
comps.append(Q)
return comps
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]
"""
assert T1.nrows() == T1.ncols()
assert T2.nrows() == T2.ncols()
assert T1.nrows() == T2.nrows()
n = T1.nrows()
from copy import copy
Q1 = copy(T1)
Q2 = copy(T2)
for r in range(n):
for c in range(n):
if T1[r, c] == T2[r, c]:
Q1[r, c] = -1
Q2[r, c] = -1
return Q1, Q2
