| Python script for search for factors of M1277 using random k-intervals [message #1861276] |
Tue, 03 October 2023 12:57  |
Eclipse User |
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import numpy as np
from random import randint
f = open("m1277.txt", "a")
def tf1277(k, r):
expos = np.zeros((4, r), int)
p = 1277
y = 0
z = 0
e1 = 2555
e2 = 3832
e3 = 7663
e4 = 8940
pbinary = []
q = p
while q > 0:
if q % 2 == 0:
pbinary.append(0)
q //= 2
if q % 2 == 1:
pbinary.append(1)
q //= 2
pbinary.reverse()
for x in range(4):
for prime in [3, 5, 7, 11, 13, 17, 19, 23, 29]:
for a in range(prime + 5):
if x == 0:
e = e1
elif x == 1:
e = e2
elif x == 2:
e = e3
else:
e = e4
if ((a + k) * 10216 + e) % prime == 0:
c = a
while c < r:
expos[x][c] = 1
c += prime
break
for a in range(len(expos[0])):
if expos[0][a] == 0:
c = 1
kp = (k + a) * 10216 + 2555
for d in pbinary:
c **= 2
if d == 1:
c *= 2
c %= kp
if c == 1:
print("-------------------------")
print("factor = ", kp)
print("k = ", (kp - 1) // 2 // 1277)
f.write("\n")
f.write("factor = " + str(kp))
f.write("\n")
f.write("k = " + str((kp - 1) // 2 // 1277))
f.write("\n")
quit()
if y == 1000:
y = 0
z += 1
print(z * 1000)
y += 1
for a in range(len(expos[1])):
if expos[1][a] == 0:
c = 1
kp = (k + a) * 10216 + 3832
for d in pbinary:
c **= 2
if d == 1:
c *= 2
c %= kp
if c == 1:
print("-------------------------")
print("factor = ", kp)
print("k = ", (kp - 1) // 2 // 1277)
f.write("\n")
f.write("factor = " + str(kp))
f.write("\n")
f.write("k = " + str((kp - 1) // 2 // 1277))
f.write("\n")
quit()
if y == 1000:
y = 0
z += 1
print(z * 1000)
y += 1
for a in range(len(expos[2])):
if expos[2][a] == 0:
c = 1
kp = (k + a) * 10216 + 7663
for d in pbinary:
c **= 2
if d == 1:
c *= 2
c %= kp
if c == 1:
print("-------------------------")
print("factor = ", kp)
print("k = ", (kp - 1) // 2 // 1277)
f.write("\n")
f.write("factor = " + str(kp))
f.write("\n")
f.write("k = " + str((kp - 1) // 2 // 1277))
f.write("\n")
quit()
if y == 1000:
y = 0
z += 1
print(z * 1000)
y += 1
for a in range(len(expos[3])):
if expos[3][a] == 0:
c = 1
kp = (k + a) * 10216 + 8940
for d in pbinary:
c **= 2
if d == 1:
c *= 2
c %= kp
if c == 1:
print("-------------------------")
print("factor = ", kp)
print("k = ", (kp - 1) // 2 // 1277)
f.write("\n")
f.write("factor = " + str(kp))
f.write("\n")
f.write("k = " + str((kp - 1) // 2 // 1277))
f.write("\n")
quit()
if y == 1000:
y = 0
z += 1
print(z * 1000)
y += 1
while True:
tf1277(randint(58000000000000000, 2 ** (1277 / 2) / 1277 / 2), 1000000)
Hello, I have made the above Python script, that is called with random starting k value, and then tests about 1263000 sived-out k's above that value. It uses a grid of zeroes to sieve k's and combines the facts that factors must be 2*k*p+1 and also 1 or 7 (mod 8) - so it only tests factors that are 2555, 3832, 7663, 8940 (mod 10216).
If there is a mistake (either logical or mathematical) in my script, I'll be grateful for pointing out. I would also be grateful if somebody could help me with making the code runnable on a Nvidia GPU with CUDA (or in OpenCL, if it's easier), without necessarily using FFT or Barrett multiplication. (But if somebody is willing to explain, I'll be happy to listen, or read...).
And for the last, if you want, you can try to find the factor yourself with that code. It takes my CPU about 12 seconds on average to go through random range of size 10216000000 (that is difference between largest and smallest k tested). So if I was to randomly check the equivalent number of ranges (untested range from 2^67 to 2^639 divided by range size 10216000000), it would take me... about 2*10^172 years . But if more people engage in search, probability increases (unfortunately it's only slight change).
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