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105 changes: 105 additions & 0 deletions precomp.py
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"""
This generates bounds for products of coefficients of points in our fundamental domain.

Pypy to be used.
"""
import sys
sys.setrecursionlimit(2000)
import itertools
subsets = []
def findsubsets(S):
A = []
m = 0
while(m <= len(S)):
A.append(itertools.combinations(S,m))
m+=1
for k in A:
for l in k:
l = list(l)
subsets.append(l)
findsubsets([0,1,2,3,4,5,6,7,8,9,10,11])
gamma = [[0,1,1,1],[0,1,2,1],[0,1,3,1],[0,2,2,1],[0,2,3,1],[0,3,3,1],[0,1,1,-1],[0,1,2,-1],[0,1,3,-1],[0,2,2,-1],[0,2,3,-1],[0,3,3,-1]]
print len(subsets)
def check(list):
I = 0
J = 0
K = 0
AB = 0
i = 0
while(i < len(list)):
if(gamma[list[i]][1] == 1):
I+=1
if(gamma[list[i]][2] == 1):
I+=1
if(gamma[list[i]][1] == 2):
J+=1
if(gamma[list[i]][2] == 2):
J+=1
if(gamma[list[i]][1] == 3):
K+=1
if(gamma[list[i]][2] == 3):
K+=1
AB += gamma[list[i]][3]
if(I+J >=2*K and 2*I>=J+K and AB>=0):
return 0
change=0
while(I+J < 2*K or 2*I < J+K or AB < 0):
I+=2
AB+=1
change+=1
return change

abval = [[0,0,0,0,0,0,0,0,0,0,0,0],[0.5,0,0,0,0,0,0,0,0,0,0,0],[0.75,0.5,0,0,0,0,0,0,0,0,0,0],[0.25,1,0,0,0,0,0,0,0,0,0,0],[0.625,2.125,0.5,0.5,0,0,0,0,0,0,0,0],[23.0/16.0,3.625,1.5,0.25,1,0,0,0,0,0,0,0],[0.5,0,0,0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0.5,0,0,0,0,0],[0,0,0,0,0,0,0.75,0.5,0,0,0,0],[0,0,0,0,0,0,0.25,1,0,0,0,0],[0,0,0,0,0,0,0.625,2.125,0.5,0.5,0,0],[0,0,0,0,0,0,23.0/16.0,3.625,1.5,0.25,1,0]]

cdval = [[1,0,0,0,0,0,0,0,0,0,0,0],[0.5,1,0,0,0,0,0,0,0,0,0,0],[0.5,0.5,1,0,0,0,0,0,0,0,0,0],[0.25,1,0,1,0,0,0,0,0,0,0,0],[0.25,1,0.5,0.5,1,0,0,0,0,0,0,0],[0.25,0.25,1,0.25,1,1,0,0,0,0,0,0],[0.5,0,0,0,0,0,1,0,0,0,0,0],[0.25,0.5,0,0,0,0,0.5,1,0,0,0,0],[0.25,0.25,0.5,0,0,0,0.5,0.5,1,0,0,0],[0.125,0.5,0,0.5,0,0,0.25,1,0,1,0,0],[0.125,0.5,0.25,0.25,0.5,0,0.25,0.5,0.5,1,0,0],[0.125,0.125,0.5,0.125,0.5,0.5,0.25,0.25,1,0.25,1,1]]

R = range(0,12)
def loop_product(current, array, S,t1,t2,t3,t):
answer = 0
if(current == len(array) - 1):
for j in R:
S1=S
if(gamma[j][1] == 1):
t1+=1
if(gamma[j][2] == 1):
t1+=1
if(gamma[j][1] == 2):
t2+=1
if(gamma[j][2] == 2):
t2+=1
if(gamma[j][1] == 3):
t3+=1
if(gamma[j][2] == 3):
t3+=1
t+= gamma[j][3]
S1*=cdval[array[current]][j]
l = min(min(t1,t2),t3)
if(l!=0):
t1=t1%l
t2=t2%l
t3=t3%l
S1*= 1.07457**t
S1*= 1.54701**(t1-t3)
S1*= 0.895503**len(array)
if(S1!=0):
answer+=S1
if(current < len(array) - 1):
for j in R:
if(gamma[j][1] == 1):
t1+=1
if(gamma[j][2] == 1):
t1+=1
if(gamma[j][1] == 2):
t2+=1
if(gamma[j][2] == 2):
t2+=1
if(gamma[j][1] == 3):
t3+=1
if(gamma[j][2] == 3):
t3+=1
t+= gamma[j][3]
S*=cdval[array[current]][j]
answer += loop_product(current+1, array, S,t1,t2,t3,t)
return answer
for k in range(0, 4096):
print k, loop_product(1,subsets[k],1,0,0,0,0)
15 changes: 15 additions & 0 deletions todo.txt
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1. Send Kevin the code/GIT. [Today]

2. Check random pts outside the domain. (~10^6) [2nd]

3. Run races on bounds and see which ones run fastest. [For b_{ij}, where {ij} are large, especially] [4th]

4. See if decision tree is better than linear checks (flattening the decision tree) [3rd]

5. PyPy [Today]

6. Write reduction code in Cython. [Step 1, Section 4, jpv3.pdf] [1st]

7. Compute splitting types. [Given (A,B) and prime p, compute the splitting type of Q(A,B), precompute the sieve] [Soon]

8.