On Wed, May 13, 2015 at 9:43 PM, Phoenix <[email protected]> wrote:
>
> I am not sure how to include "rep".
Can you type in a short example of what it could be?
> Its a list generated by a set of operations somewhere else.
> I am generating "Edge" by extracting the edges of the graph K_{n,n}
>
Can you type in a short example of what it could be?
>
>
>
> On Wednesday, May 13, 2015 at 8:28:57 PM UTC-5, David Joyner wrote:
>>
>> On Wed, May 13, 2015 at 9:13 PM, Phoenix <[email protected]> wrote:
>> >
>> >
>> > So I have this SAGE code which takes in two integers q and n and it
>> > generates ~ (q^3)^(n^2) 0/1 matrices and does a sum of their
>> > characteristic
>> > polynomials.
>> >
>> > I got answers for (q = 3, n = 2), (q =2 , n = 3) and (q=2, n=2)
>> >
>> > For any higher number the code runs for ~1hr and then it says,
>> >
>> > "
>> >
>> > Traceback (most recent call last):
>> > File "<stdin>", line 1, in <module>
>> > File "_sage_input_2.py", line 10, in <module>
>> > exec compile(u'open("___code___.py","w").write("# -*- coding: utf-8
>> > -*-\\n" +
>> >
>> > _support_.preparse_worksheet_cell(base64.b64decode("UA=="),globals())+"\\n");
>> > execfile(os.path.abspath("___code___.py"))
>> > File "", line 1, in <module>
>> >
>> > File "/tmp/tmpPiPIoq/___code___.py", line 2, in <module>
>> > exec compile(u'P
>> > File "", line 1, in <module>
>> >
>> > "
>> >
>> >
>> > Can someone help understand what is going on?
>> >
>> > Like any suggestion about how to go about it?
>> >
>> >
>> >
>> > EDIT:
>> >
>> > This is basically the main time-consuming step.
>> >
>> > At this point "rep" comes in as a ~q^3 size list of q^2 dimensional 0/1
>> > permutation matrices.
>> > And "Edge" is a list of integer tuples where each tuple is of the form
>> > (a,b)
>> > with a,b being from the set {0,1,2,3...,(2n-1)}
>> >
>> > 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 _ 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)
>> >
>>
>> Please include rep and Edge, if if it's in a simple case.
>>
>>
>> > The polynomial P is the final expected answer.
>> >
>> > (and for any larger value of q and n than the 3 cases mentioned the
>> > program
>> > basically stops with no output for P and with the above pasted message)
>> >
>> > --
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>
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