On Thu, May 14, 2015 at 4:43 PM, Phoenix <[email protected]> wrote:
> The time trouble starts at the point in the code where it doesn't matter
> where the permutation matrices comes from.
>
> What the main part does is basically this.
>
> Given any set of matrices of dimensional k it produces by the "product" and
> "zip" command all possible ways of distributing those matrices over the set
> of edges of the graph of size n .
> One fixes some arbitrary orientation for each of the edges. It doesn't
> matter what.
> Now it creates a matrix M of size kn as a n x n array of size k matrices as
> follows : for the (a,b) edge is in the oriented edge list it (1) places into
> the (a,b) array position the matrix that was assigned to that edge and (2)
> it places into the (b,a) position the inverse of that matrix. Now it
> calculates the characteristic polynomial of this M and keeps adding it up
> over all possible ways of distributing the matrices over the edges.
>
>
I ran the block below and it ran without error until I killed it (note
the additional print statement at the end) To me this suggests there
is no problem with your code. There might be a problem managing a
large list. If that is the case, you'll need to think more about other
ways to do the computation.
To the OP: if you had simply supplied the block below initially, as
William Stein suggested, you would have saved both your time and ours.
q=2; n=4
g = graphs.CompleteBipartiteGraph(n, n)
Edge = []
for (a,b,c) in g.edges ():
Edge.append ( (a,b) )
#This is how "rep" is created,
Fq2 = []
for i in range (q):
for j in range (q):
Fq2.append (matrix([[i],[j]]))
SL2Fq = []
for i1 in range (q):
for i2 in range (q):
for i3 in range(q):
for i4 in range (q):
if (i4*i1-i2*i3)% q == 1 :
SL2Fq.append ( matrix( [ [i1, i2],[i3,i4] ] ) )
rep = []
for Y in SL2Fq:
M = []
for A in Fq2:
B = (Y*A)%q
row = []
for C in Fq2:
if B == C:
row.append ([1])
else:
row.append ([0])
M.append(flatten(row))
rep.append (matrix(M).transpose())
P = 0
from itertools import product
from itertools import izip
for X in product(rep,repeat = len (Edge)):
k = izip(Edge,X)
M = [[matrix(q^2, q^2, 0)]*(2*n) for ell in range(2*n)]
for ((a,b),Y) in k:
M [a][b] = Y
M [b][a] = Y.inverse()
Z = block_matrix(M)
P = P + Z.charpoly(x)
print P
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