OFFSET
0,3
COMMENTS
An alternating sign hypermatrix (ASHM) of order n is an n X n X n hypermatrix with entries from {0, 1, -1} such that the nonzero entries of each row, column, and vertical line alternate in sign, beginning and ending with +1.
ASHMs are the three-dimensional analog of alternating sign matrices and generalize Latin squares, as the set of n X n X n ASHMs containing no negative entries is in bijection with the set of n X n Latin squares.
LINKS
R. Brualdi and G. Dahl, Alternating sign matrices and hypermatrices, and a generalization of Latin squares, Advances in Applied Mathematics, 95 (2018), 116 - 151.
Cian O'Brien, Alternating sign hypermatrix decompositions of Latin-like squares, Advances in Applied Mathematics, 121 (2020), 102097.
FORMULA
Verified using 2 computer searches. The search given counts all sequences of n alternating sign matrices of order n that form an ASHM. The other search uses corner-sum matrices, which are known to be in bijection with alternating sign matrices, by counting all 3-dimensional analogs of corner-sum matrices.
EXAMPLE
For n = 1, the only n X n X n ASHM is [[[1]]].
For n = 2, the two n X n X n ASHMs are
[[[1,0],
[0,1]],
[[0,1],
[1,0]]]
and
[[[0,1],
[1,0]],
[[1,0],
[0,1]]].
PROG
(Sage)
# Program written in Sage
# Returns True if a given list of n n X n ASMs form an ASHM, returns False otherwise
def ASHM(L):
n = len(L)
# Searches through the vertical line in position (i, j) of the hypermatrix for each i and j
for i in range(n):
for j in range(n):
# Since the first nonzero entry in each line of an ASHM is +1, the alternating condition is checked
# as if the previous nonzero entry was -1
last = -1
for k in range(n):
# In each position of the current vertical line, if the sign of the current entry is the opposite
# of the previous, then the previous sign is updated
if L[k][i, j]*last == -1:
last *= -1
# Otherwise False is returned unless the current entry is 0
elif L[k][i, j] != 0:
return False
# If the most recent nonzero entry is not +1 by the time all entries have been checked, False is returned
if last != 1:
return False
# If False has not been returned, return True
return True
# Generates all combinations of one element from each list in L
def combos(L, current = [[]]):
# If there are no elements left which have not been combined, then return the combinations already made
if len(L) == 0:
return current
# Otherwise, each element of the next list in L is appended to the current list of combinations made
output = []
for K in current:
for a in L[0]:
output.append(K + [a])
return combos(L[1:], output)
# Counts all ASHMs of order n
def count_ASHMs(n):
# All ASMs of order n are imported as matrices
asms = []
for A in AlternatingSignMatrices(n):
asms.append(A.to_matrix())
# Initially, zero ASHMs have been counted
count = 0
# Every possible combination of n n X n ASMs is checked
for i in combos([[k for k in range(len(asms))] for m in range(n)]):
# If the current list of n n X n ASMs forms an ASHM, then it is counted
count += int(ASHM([asms[i[k]] for k in range(n)]))
# The final count is returned
return count
# Note: I ran a more efficient version of this program in Python to obtain the answer for n=5, and even then it took 6 hours.
print(count_ASHMs(0))
print(count_ASHMs(1))
print(count_ASHMs(2))
print(count_ASHMs(3))
print(count_ASHMs(4))
print(count_ASHMs(5))
# Cian O'Brien, May 31 2023
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Samuel Zbarsky (sa_zbarsky(AT)yahoo.com), Sep 07 2008
EXTENSIONS
a(4) corrected and a(5) added, and definition updated by Cian O'Brien, May 31 2023
STATUS
approved