Develop Reference

Finding null space of binary matrix in python

In factoring methods based on the quadratic sieve, finding the left null space of a binary matrix (values computed mod 2) is a crucial step. (This is also the null space of the transpose.) Does numpy or scipy have tools to do this quickly?
For reference, here is my current code:
# Row-reduce binary matrix
def binary_rr(m):
rows, cols = m.shape
l = 0
for k in range(min(rows, cols)):
print(k)
if l >= cols: break
# Swap with pivot if m[k,l] is 0
if m[k,l] == 0:
found_pivot = False
while not found_pivot:
if l >= cols: break
for i in range(k+1, rows):
if m[i,l]:
m[[i,k]] = m[[k,i]] # Swap rows
found_pivot = True
break
if not found_pivot: l += 1
if l >= cols: break # No more rows
# For rows below pivot, subtract row
for i in range(k+1, rows):
if m[i,l]: m[i] ^= m[k]
l += 1
return m
It is pretty much a straightforward implementation of Gaussian elimination, but since it's written in python it is very slow.

qwr, I found a very fast gaussian elimination routine that finishes so qiuckly that the slow point is the Quadratic Sieving or SIQS Sieving step. The gaussian elimination functions were taken from skollmans factorise.py at https://raw.githubusercontent.com/skollmann/PyFactorise/master/factorise.py
I'll soon be working on a SIQS/GNFS implementation from scratch, and hope to write something super quick for python with multithreading and possiblly cython. In the meantime, if you want something that compiles C (Alpertons ECM Engine) but uses python, you can use: https://github.com/oppressionslayer/primalitytest/ which requires you to cd into calculators directory and run make before importing p2ecm with from sfactorint import p2ecm. With that you can factorise 60 digit numbers in a few seconds.
# Requires sympy and numpy to be installed
# Adjust B and I accordingly. Set for 32 length number
# Usage:
# N=1009732533765251*1896182711927299
# factorise(N, 5000, 25000000) # Takes about 45-60 seconds on a newer computer
# N=1009732533765251*581120948477
# Linear Algebra Step finishes in 1 second, if that
# N=factorise(N, 5000, 2500000) # Takes about 5 seconds on a newer computer
# #Out[1]: 581120948477
import math
import numpy as np
from sympy import isprime
#
# siqs_ functions are the Gaussian Elimination routines right from
# skollmans factorise.py. It is the fastest Gaussian Elimination that i have
# found in python
#
def siqs_factor_from_square(n, square_indices, smooth_relations):
"""Given one of the solutions returned by siqs_solve_matrix_opt,
return the factor f determined by f = gcd(a - b, n), where
a, b are calculated from the solution such that a*a = b*b (mod n).
Return f, a factor of n (possibly a trivial one).
"""
sqrt1, sqrt2 = siqs_calc_sqrts(square_indices, smooth_relations)
assert (sqrt1 * sqrt1) % n == (sqrt2 * sqrt2) % n
return math.gcd(abs(sqrt1 - sqrt2), n)
def siqs_find_factors(n, perfect_squares, smooth_relations):
"""Perform the last step of the Self-Initialising Quadratic Field.
Given the solutions returned by siqs_solve_matrix_opt, attempt to
identify a number of (not necessarily prime) factors of n, and
return them.
"""
factors = []
rem = n
non_prime_factors = set()
prime_factors = set()
for square_indices in perfect_squares:
fact = siqs_factor_from_square(n, square_indices, smooth_relations)
if fact != 1 and fact != rem:
if isprime(fact):
if fact not in prime_factors:
print ("SIQS: Prime factor found: %d" % fact)
prime_factors.add(fact)
while rem % fact == 0:
factors.append(fact)
rem //= fact
if rem == 1:
break
if isprime(rem):
factors.append(rem)
rem = 1
break
else:
if fact not in non_prime_factors:
print ("SIQS: Non-prime factor found: %d" % fact)
non_prime_factors.add(fact)
if rem != 1 and non_prime_factors:
non_prime_factors.add(rem)
for fact in sorted(siqs_find_more_factors_gcd(non_prime_factors)):
while fact != 1 and rem % fact == 0:
print ("SIQS: Prime factor found: %d" % fact)
factors.append(fact)
rem //= fact
if rem == 1 or sfactorint_isprime(rem):
break
if rem != 1:
factors.append(rem)
return factors
def add_column_opt(M_opt, tgt, src):
"""For a matrix produced by siqs_build_matrix_opt, add the column
src to the column target (mod 2).
"""
M_opt[tgt] ^= M_opt[src]
def find_pivot_column_opt(M_opt, j):
"""For a matrix produced by siqs_build_matrix_opt, return the row of
the first non-zero entry in column j, or None if no such row exists.
"""
if M_opt[j] == 0:
return None
return lars_last_powers_of_two_trailing(M_opt[j] + 1)
def siqs_build_matrix_opt(M):
"""Convert the given matrix M of 0s and 1s into a list of numbers m
that correspond to the columns of the matrix.
The j-th number encodes the j-th column of matrix M in binary:
The i-th bit of m[i] is equal to M[i][j].
"""
m = len(M[0])
cols_binary = [""] * m
for mi in M:
for j, mij in enumerate(mi):
cols_binary[j] += "1" if mij else "0"
return [int(cols_bin[::-1], 2) for cols_bin in cols_binary], len(M), m
def siqs_solve_matrix_opt(M_opt, n, m):
"""
Perform the linear algebra step of the SIQS. Perform fast
Gaussian elimination to determine pairs of perfect squares mod n.
Use the optimisations described in [1].
[1] Koç, Çetin K., and Sarath N. Arachchige. 'A Fast Algorithm for
Gaussian Elimination over GF (2) and its Implementation on the
GAPP.' Journal of Parallel and Distributed Computing 13.1
(1991): 118-122.
"""
row_is_marked = [False] * n
pivots = [-1] * m
for j in range(m):
i = find_pivot_column_opt(M_opt, j)
if i is not None:
pivots[j] = i
row_is_marked[i] = True
for k in range(m):
if k != j and (M_opt[k] >> i) & 1: # test M[i][k] == 1
add_column_opt(M_opt, k, j)
perf_squares = []
for i in range(n):
if not row_is_marked[i]:
perfect_sq_indices = [i]
for j in range(m):
if (M_opt[j] >> i) & 1: # test M[i][j] == 1
perfect_sq_indices.append(pivots[j])
perf_squares.append(perfect_sq_indices)
return perf_squares
def sqrt_int(N):
Nsqrt = math.isqrt(N)
assert Nsqrt * Nsqrt == N
return Nsqrt
def siqs_calc_sqrts(square_indices, smooth_relations):
"""Given on of the solutions returned by siqs_solve_matrix_opt and
the corresponding smooth relations, calculate the pair [a, b], such
that a^2 = b^2 (mod n).
"""
res = [1, 1]
for idx in square_indices:
res[0] *= smooth_relations[idx][0]
res[1] *= smooth_relations[idx][1]
res[1] = sqrt_int(res[1])
return res
def quad_residue(a,n):
l=1
q=(n-1)//2
x = q**l
if x==0:
return 1
a =a%n
z=1
while x!= 0:
if x%2==0:
a=(a **2) % n
x//= 2
else:
x-=1
z=(z*a) % n
return z
def STonelli(n, p):
assert quad_residue(n, p) == 1, "not a square (mod p)"
q = p - 1
s = 0
while q % 2 == 0:
q //= 2
s += 1
if s == 1:
r = pow(n, (p + 1) // 4, p)
return r,p-r
for z in range(2, p):
#print(quad_residue(z, p))
if p - 1 == quad_residue(z, p):
break
c = pow(z, q, p)
r = pow(n, (q + 1) // 2, p)
t = pow(n, q, p)
m = s
t2 = 0
while (t - 1) % p != 0:
t2 = (t * t) % p
for i in range(1, m):
if (t2 - 1) % p == 0:
break
t2 = (t2 * t2) % p
b = pow(c, 1 << (m - i - 1), p)
r = (r * b) % p
c = (b * b) % p
t = (t * c) % p
m = i
return (r,p-r)
def build_smooth_relations(smooth_base, root_base):
smooth_relations = []
for xx in range(len(smooth_base)):
smooth_relations.append((root_base[xx], smooth_base[xx], xx))
return smooth_relations
def strailing(N):
return N>>lars_last_powers_of_two_trailing(N)
def lars_last_powers_of_two_trailing(N):
p,y=1,2
orign = N
#if orign < 17: N = N%16
N = N&15
if N == 1:
if ((orign -1) & (orign -2)) == 0: return orign.bit_length()-1
while orign&y == 0:
p+=1
y<<=1
return p
if N in [3, 7, 11, 15]: return 1
if N in [5, 13]: return 2
if N == 9: return 3
return 0
def build_matrix(factor_base, smooth_base):
factor_base = factor_base.copy()
factor_base.insert(0, 2)
sparse_matrix = []
col = 0
for xx in smooth_base:
sparse_matrix.append([])
for fx in factor_base:
count = 0
factor_found = False
while xx % fx == 0:
factor_found = True
xx=xx//fx
count+=1
if count % 2 == 0:
sparse_matrix[col].append(0)
continue
else:
if factor_found == True:
sparse_matrix[col].append(1)
else:
sparse_matrix[col].append(0)
col+=1
return np.transpose(sparse_matrix)
def get_mod_congruence(root, N, withstats=False):
r = root - N
if withstats==True:
print(f"{root} ≡ {r} mod {N}")
return r
def primes_sieve2(limit):
a = np.ones(limit, dtype=bool)
a[0] = a[1] = False
for (i, isprime) in enumerate(a):
if isprime:
yield i
for n in range(i*i, limit, i):
a[n] = False
def remove_singletons(XX):
no_singletons = []
for xx in XX:
if len(xx) != 1:
no_singletons.append(xx)
return no_singletons
def fb_sm(N, B, I):
factor_base, sieve_base, sieve_list, smooth_base, root_base = [], [], [], [], []
primes = list(primes_sieve2(B))
i,root=-1,math.isqrt(N)
for x in primes[1:]:
if quad_residue(N, x) == 1:
factor_base.append(x)
for x in range(I):
xx = get_mod_congruence((root+x)**2, N)
sieve_list.append(xx)
if xx % 2 == 0:
xx = strailing(xx+1) # using lars_last_modulus_powers_of_two(xx) bit trick
sieve_base.append(xx)
for p in factor_base:
residues = STonelli(N, p)
for r in residues:
for i in range((r-root) % p, len(sieve_list), p):
while sieve_base[i] % p == 0:
sieve_base[i] //= p
for o in range(len(sieve_list)):
# This is set to 350, which is only good for numbers
# of len < 32. Modify
# to be more dynamic for larger numbers.
if len(smooth_base) >= 350:
break
if sieve_base[o] == 1:
smooth_base.append(sieve_list[o])
root_base.append(root+o)
return factor_base, smooth_base, root_base
def isSquare(hm):
cr=math.isqrt(hm)
if cr*cr == hm:
return True
return False
def find_square(smooth_base):
for x in smooth_base:
if isSquare(x):
return (True, smooth_base.index(x))
else:
return (False, -1)
t_matrix=[]
primes=list(primes_sieve2(1000000))
def factorise(N, B=10000, I=10000000):
global primes, t_matrix
if isprime(N):
return N
for xx in primes:
if N%xx == 0:
return xx
factor_base, smooth_base, root_base = fb_sm(N,B,I)
issquare, t_matrix = find_square(smooth_base)
if issquare == True:
return math.gcd(math.isqrt(smooth_base[t_matrix])+get_mod_congruence(root_base[t_matrix], N), N)
t_matrix = build_matrix(factor_base, smooth_base)
smooth_relations = build_smooth_relations(smooth_base, root_base)
M_opt, M_n, M_m = siqs_build_matrix_opt(np.transpose(t_matrix))
perfect_squares = remove_singletons(siqs_solve_matrix_opt(M_opt, M_n, M_m))
factors = siqs_find_factors(N, perfect_squares, smooth_relations)
return factors

Solving a function that measures distance between points in a list of points (x, y)

I am writing a function route. This function has a mandatory parameter points that takes a list of points. The function must return the total distance traveled if each of the points in the given list is visited in turn. Apart from the mandatory parameter, the function also has two optional parameters:
cycle: takes a Boolean value that indicates whether the end of the route is equal to its starting point (True) or not (False); the default value of this parameter is False
distance: takes a distance function that is used for the calculation of the total distance between two consecutive points in the given route; if no explicit value is passed to this parameter, the Euclidean distance must be used
Problem: Anybody knows with the last definition route() how to solve it for the case:
route([(41.79, 13.59), (41.68, 14.65), (21.16, -4.79)], distance=lambda p1, p2: abs(p1[0] + p2[0]))
correct answer : 146.31
Part of my code I refer to:
if cycle == False and distance is λ(p1, p2): abs(p1[0] + p2[0]):
l = list()
count = 0
for items in range(len(points)-1):
a = points[items]
b = points[items+1]
d = euclidean(a[0], b[0])
l.append(d)
count += 1
return sum(l)
In this part I got stuck at the first rule and further.
Complete code which works fine (except for the part above):
def euclidean(a, b):
'''
>>> euclidean((42.36, 56.78), (125.65, 236.47))
198.05484139500354
'''
from math import sqrt
return sqrt(sum((a - b)**2 for a, b in zip(a, b)))
def manhattan(c, d):
'''
>>> manhattan((42.36, 56.78), (125.65, 236.47))
262.98
'''
return sum(abs(c - d) for c, d in zip(c, d))
def chessboard(e, f):
'''
>>> chessboard((42.36, 56.78), (125.65, 236.47))
179.69
'''
return max(abs(e - f) for e, f in zip(e, f))
def route(points, cycle=False, distance=None):
'''
>>> route([(6.59, 6.73), (4.59, 5.54), (5.33, -13.98)])
21.861273201261746
>>> route(cycle=True, points=[(6.59, 6.73), (4.59, 5.54), (5.33, -13.98)])
42.60956710702662
>>> route([(6.59, 6.73), (4.59, 5.54), (5.33, -13.98)], distance=manhattan)
23.45
>>> route([(6.59, 6.73), (4.59, 5.54), (5.33, -13.98)], cycle=True, distance=manhattan)
45.42
'''
if cycle == False and distance is None:
l = list()
count = 0
for items in range(len(points)-1):
a = points[items]
b = points[items+1]
d = euclidean(a, b)
l.append(d)
count += 1
return sum(l)
if cycle == False and distance is euclidean:
l = list()
count = 0
for items in range(len(points)-1):
a = points[items]
b = points[items+1]
d = euclidean(a, b)
l.append(d)
count += 1
return sum(l)
if cycle == False and distance is λ(p1, p2): abs(p1[0] + p2[0]):
l = list()
count = 0
for items in range(len(points)-1):
a = points[items]
b = points[items+1]
d = euclidean(a[0], b[0])
l.append(d)
count += 1
return sum(l)
if cycle == True and distance is None:
l = list()
count = 0
for items in range(len(points)-1):
a = points[items]
b = points[items+1]
d = euclidean(a, b)
l.append(d)
count += 1
f = points[0]
g = points[-1]
r = euclidean(g, f)
k = sum(l) + r
return k
if cycle == True and distance is euclidean:
l = list()
count = 0
for items in range(len(points)-1):
a = points[items]
b = points[items+1]
d = euclidean(a, b)
l.append(d)
count += 1
f = points[0]
g = points[-1]
r = euclidean(g, f)
k = sum(l) + r
return k
if cycle is False and distance is manhattan:
l = list()
count = 0
for items in range(len(points)-1):
a = points[items]
b = points[items+1]
d = manhattan(a, b)
l.append(d)
count += 1
return sum(l)
if cycle is True and distance is manhattan:
l = list()
count = 0
for items in range(len(points)-1):
a = points[items]
b = points[items+1]
d = manhattan(a, b)
l.append(d)
count += 1
f = points[0]
g = points[-1]
r = manhattan(g, f)
k = sum(l) + r
return k

I Agree with Duncan. You have way too much duplication.
Here a more direct approach:
euclidean = lambda p1, p2: sqrt(sum((p1_i - p2_i)**2 for p1_i, p2_i in zip(p1, p2)))
manhattan = lambda p1, p2: sum(abs(p1_i - p2_i) for p1_i, p2_i in zip(p1, p2))
chessboard = lambda p1, p2: max(abs(p1_i - p2_i) for p1_i, p2_i in zip(p1, p2))
def route(points, cycle=False, metric=euclidean):
l = 0.0
for i in range(len(points) - 1):
l += metric(points[i], points[i + 1])
if cycle:
l += metric(points[-1], points[0])
return l
Any metric funtion can be passed and is then used instead of the euclidean metric.

Function which measures time M times

I am to measure how much time does it take for the function below to represent: C in range [0, 10] with the numbers in list N. (M measurements for each C).
import itertools
def amount(C):
N = [1, 2, 5]
#N = list(N)
N = sorted(N)
while C < max(N):
N.remove(max(N))
res = []
for i in range(1, C):
for j in list(itertools.combinations_with_replacement(N, i)):
res.append(sum(list(j)))
m = 0
for z in range (0, len(res)):
if res[z] == C:
m += 1
if N[0] == 1:
return m + 1
else:
return m
EDITED:
import itertools
def amount(C):
N = [1, 2, 5]
res = []
for i in range(1, C):
for j in list(itertools.combinations_with_replacement(N, i)):
res.append(sum(list(j)))
m = 0
for z in range (0, len(res)):
if res[z] == C:
m += 1
if N[0] == 1:
return m + 1
else:
return m
I would like to make 10 measurements and then take for example median of all those measurements.
There is my code but something unfortunately doesn't work properly and I have no idea what is wrong:
import time
def time_counter(amount, n=11, M=11):
res = list(range(n))
def count_once():
start = time.perf_counter()
amount(res)
return time.perf_counter() - start
return [count_once() for m in range(M)]

You are again passing a list and trying to do range(1,C) where C is a list
Here is how your program should be
import itertools
import time
def amount(C):
N = [1, 2, 5]
res = []
for i in range(1, C):
for j in list(itertools.combinations_with_replacement(N, i)):
res.append(sum(list(j)))
m = 0
for z in range (0, len(res)):
if res[z] == C:
m += 1
if N[0] == 1:
return m + 1
else:
return m
def time_counter(amount, n=11, M=11):
res = list(range(n))
def count_once(c):
start = time.perf_counter()
amount(c)
return time.perf_counter() - start
return [count_once(m) for m in range(M)]
#testing
print(time_counter(amount))

How it works in python? ruby code

I don't understand what it code doing, please help.
How it will be working in python or in another simple language?
a = []
a << [1]
for i in 2..10001
f = 0
a.each{ |group|
m = 1
group.each { |c|
m *= i % c
}
f += m
if m > 0
group << i
break
end
}
a << [i] if f == 0
end
p a
p a.size

Literally translated to python this is:
a = []
a.append([1])
for i in range(2,10001 + 1):
f = 0
for group in a:
m = 1
for c in group:
m *= i % c
f += m
if m > 0:
group.append(i)
break
if f == 0:
a.append([i])
print a
print len(a)

In python:
a = []
a.append(1)
for i in range(2, 10000):
f = 0
for group in a:
m = 1
for c in group:
m *= i%c
f += m
if m>0:
group.append(i)
break
if not f:
a.append(i)
print(a)
print(len(a))

Tail recursion to loop

I as working on a program that given a and b, computes the last 5 digits of a^b. I currently have it working as long as b is sufficiently low, but if b is large (>1000) this will crush the stack. Is there a way I can make this an iterative function? I have tried converting to iterative, but I can't figure it out.
def pow_mod(a,b):
if b == 0:
return 1
elif b == 2:
return a*a % 10000
return ((b%2*(a-1) + 1) * pow_mod(a,b//2)**2) % 10000

In order to do this computation, iteratively, you start with an answer=a, and then square it repeatedly (and multiply by a if b is odd). To exit the loop, divide b by 2 each time, and check for when b>1.
def pow_mod(a,b):
if b == 0:
return 1
c = 1;
while b > 1:
if b % 2:
c *= a
a *= a
b //= 2
a %= 10000
c %= 10000
return a * c % 10000

Use a while loop instead of recursion, to adjust b the same way that you do in the recursive call.
def pow_mod(a, b):
result = 1
while b > 0:
result = (b%2*(a-1) + 1) * result**2) % 10000
b = b//2
return result

The trick is to store the result in a temporary variable and then pass it to each function call.
lst = []
while b:
lst.insert(0, b)
b = b//2
def _pow_mod(a, b, answer):
if b == 2:
return a*a % 10000;
else:
return ((b%2*(a-1) + 1) * answer **2) % 10000
answer = 1 # b = 0 case
for b in lst:
answer = _pow_mod(a, b, answer)
Complete Code:
# Original recursive solution
def pow_mod_orig(a, b):
if b == 0:
return 1
elif b == 2:
return a*a % 10000
return ((b%2*(a-1) + 1) * pow_mod_orig(a,b//2)**2) % 10000
# Iterative solution
def pow_mod(a, b):
lst = []
while b:
lst.insert(0, b)
b = b//2
def _pow_mod(a, b, answer):
if b == 2:
return a*a % 10000;
else:
return ((b%2*(a-1) + 1) * answer **2) % 10000
answer = 1 # b = 0 case
for b in lst:
answer = _pow_mod(a, b, answer)
return answer
for i in range(1000):
assert(pow_mod_orig(3, i) == pow_mod(3, i))