def get_random_rsa(cls, bits=512, num_prime_checks=121):
assert bits % 8 == 0 and bits >= 8
a = b = 0
while a==b:
a, b = [get_random_prime(bits/2+1, final_num_checks=num_prime_checks) for _ in xrange(2)]
modulus = a * b
period = (a-1)*(b-1)*gcd(a-1, b-1) # anything^(1+period) == anything (mod modulus)
exponent = period
while gcd(exponent, period) != 1:
exponent = get_random_number(16)+2**16
d, _ = solve_mod_problem(exponent, period, 1)
return cls(exponent, modulus, d)
def compat_check(p1, p2):
if (p1, p2) in compatCache:
return compatCache[(p1, p2)]
gcd1 = gcd(p1 - 1, p2)
if gcd1 != 1:
if (p1 - 4) % gcd1 != 0:
compatCache[(p1, p2)] = False
return False
gcd2 = gcd(p2 - 1, p1)
if gcd2 != 1:
if (p2 - 4) % gcd2 != 0:
compatCache[(p1, p2)] = False
return False
compatCache[(p1, p2)] = True
return True
def make_keys(n_primes = 100, pq_bound = 100):
'''
takes: int n
returns: tuples_int pub, prv
'''
n, p, q = 1, 1, 1
prms = primes.prime(n_primes)
while math.log(n, 2) <= 1:
while n < pq_bound:
p, q = np.random.choice(prms, size = 2, replace = False)
n = p * q
print "Key bit length: {}".format(math.log(n, 2))
print "Generated by p: {} and q: {}".format(p, q)
t = (p - 1) * (q - 1) # =: |{k | k < n && (k, n) == 0}|
print "Euler's totient function of {}x{}={}: {}".format(p, q, n, t)
e = t
while primes.gcd(e, t) != 1 or e == 1:
e = int(np.random.uniform(low = 1, high = t))
print "Public key: {}".format(e)
d = primes.inv(e, t)
print "Private key: {}".format(d)
return (n, e), (n, d)
def compat_check(p1, p2):
if (p1,p2) in compatCache:
return compatCache[(p1,p2)]
gcd1 = gcd(p1-1, p2)
if gcd1 != 1:
if (p1-4) % gcd1 != 0:
compatCache[(p1,p2)] = False
return False
gcd2 = gcd(p2-1, p1)
if gcd2 != 1:
if (p2-4) % gcd2 != 0:
compatCache[(p1,p2)] = False
return False
compatCache[(p1,p2)] = True
return True
def generate_keys(p, q):
# given two primes, calculate modulus M
M = p * q
# calculate phi(M) [(p - 1)(q - 1)]
phi = M - p - q + 1
# obtain exponent k with gcd(k, phi(M)) == 1
k = 1 + (M - 1) // 2
k = k // primes.gcd(k, phi)
# obtain multiplicative inverse of k, mod phi(M)
l = mult_inv_mod_m(k, phi)
return ((k, M), (l, M))
def computeBetter(a):
valDict = { 0 : 2*a }
for b in xrange(a, 0, -2*twoPowerList[a]):
addToDict(valDict, gcd(a,b), 2)
addToDict(valDict, 0, -2)
addToDict(valDict, a, -1)
sumz = (f(1) + f(0)) * valDict[0] * SIZE /2
del valDict[0]
for c in xrange(1,SIZE+1):
sumz += sum([f(pow(c,key, MOD-1)) * valDict[key] for key in valDict.keys()])
return sumz
def computeBetter(a):
valDict = {0: 2 * a}
for b in xrange(a, 0, -2 * twoPowerList[a]):
addToDict(valDict, gcd(a, b), 2)
addToDict(valDict, 0, -2)
addToDict(valDict, a, -1)
sumz = (f(1) + f(0)) * valDict[0] * SIZE / 2
del valDict[0]
for c in xrange(1, SIZE + 1):
sumz += sum(
[f(pow(c, key, MOD - 1)) * valDict[key] for key in valDict.keys()])
return sumz
def answer():
numerator = denominator = 1
for i in range(10, 100):
if not i % 10: continue
for j in range(i+1, 100):
if not j % 10: continue
k, l = str(i), str(j)
if(k.find(l[0]) >= 0):
k, l = float(k[(k.find(l[0])+1)%2]), float(l[1])
elif(k.find(l[1]) >= 0):
k, l = float(k[(k.find(l[1])+1)%2]), float(l[0])
else: continue
if(i / j == k / l):
numerator *= k
denominator *= l
return denominator / gcd(numerator, denominator)
totientVals = get_totient(int(math.sqrt(LIM) + 10))
# naive method. Not bad, .04737 seconds for 10^6
count = 0
for m in xrange(2,int(math.sqrt(LIM))+1):
if m**2 > LIM:
break
# \/ this part is the slowest.
if m**2 + (m-1)**2 > LIM:
nRange = xrange(1,m,2) if m%2 == 0 else xrange(2,m,2)
for n in nRange:
if m**2 + n**2 > LIM:
break
if gcd(m,n) != 1:
continue
count += 1
elif m % 2 == 1:
count += totientVals[m] / 2
else:
count += totientVals[m]
print "Answer:", count
print "Time taken:", time.time() - START
"""
Time taken: 0.120429039001 # time for 10^7
def compute(a, b, c):
if (a - b) % (2 * twoPowerList[b]) == 0:
return pow(c, gcd(a, b), MOD - 1)
return c % 2
def compute(a,b,c):
if (a-b) % (2 * twoPowerList[b]) == 0:
return pow(c, gcd(a,b), MOD-1)
return c % 2
def gcd3(a,b,c): return gcd(gcd(a,b),c)
import time
from fractions import Fraction
from primes import gcd
START = time.time()
LIMIT = 10**4
count = 0
for denominator in xrange(1, LIMIT):
numer_range = xrange(1, denominator /
100) if denominator % 2 == 1 else xrange(
1, denominator / 100, 2)
for numerator in numer_range:
if gcd(numerator, denominator) != 1:
continue
if gcd(numerator - 1, denominator) != 1 and gcd(
numerator + 1, denominator) != 1:
count += 1
#print Fraction(numerator, denominator)
print count
print "Time Taken:", time.time() - START
"""
i want to get the set of all numer, denom where:
numer < denom / 100
gcd(numer, denom) = 1
gcd(numer-1, denom) != 1
gcd(numer+1, denom) != 1
how the hell do I get the overlap of the two groups of:
count)
insertIntoDict(countDict, (iteration + 1, orderSide3 + 1,
orderTriangle + 1, orderSide3, orderSide3),
count)
gcdDict = dict()
count = -6 # ignore the case where size = 2
for order in xrange(2, SIZE + 1):
for key in filter((lambda key: key[0] == order - 1), countDict.keys()):
appendEntry(key)
val1 = sum(countDict[key] for key in filter((
lambda key: key[0] == order and key[1] == 1), countDict.keys()))
val3 = sum(countDict[key] for key in filter((
lambda key: key[0] == order and key[1] == 3), countDict.keys()))
count += gcd(val1, val3)
if gcd(val1, val3) > 30:
print order, gcd(val1, val3), time.time() - START
print count
print "Time Taken:", time.time() - START
for value in sorted(set(gcdDict.values())):
if not m_r(value / 6):
continue
prime = value / 6
minKey = min(filter(lambda x: gcdDict[x] == value, gcdDict.keys()))
print prime, '\t', \
minKey, '\t', \
modNum[prime], '\t', \
(minKey - modNum[prime]) / (prime -1.)
import time
from primes import get_totient, gcd, crt
START = time.time()
lower_bound = 10**6
upper_bound = lower_bound + 5000
answer = 0
totients = get_totient(upper_bound)
for n in xrange(lower_bound, upper_bound):
for m in xrange(lower_bound, n):
gcdVal = gcd(m,n)
if gcdVal == 1:
number = crt([n,m], [totients[n], totients[m]]) % (n * m)
answer += number
else:
# if the two totients aren't equal mod the gcd, continue
if totients[n] % gcdVal != totients[m] % gcdVal:
continue
values = [ (totients[n] / gcdVal), totients[m] / gcdVal ]
number = crt([n/ gcdVal, m / gcdVal], values) * gcdVal + totients[n] % gcdVal
answer += number
print answer
print "Time Taken:", time.time() - START
"""
Congratulations, the answer you gave to problem 531 is correct.
import time
from primes import gcd
START = time.time()
SIZE = 1000
sumz = 0
for i in xrange(1,int(SIZE**.5)+1):
sumz += i
for j in xrange(i+1, SIZE/i+1):
sumz += gcd(i,j) * 2
print "Answer:", sumz
print "Time Taken:", time.time() - START
"""
def f(n) = sum( gcd(n, d) for d in xrange(1,n+1))
f(p) = 2p - 1 if p is a prime
f(pq) = (2p-1)(2q-1) if p,q are primes
f(p^2) = p(3p-2) if p is a prime
f(p^3) = p^2(4p-3) if p is a prime
f(p^n) = p^(n-1) * ((n+1)p-n)
f(nm) = f(n) * f(m) if gcd(n,m) = 1
def simplify(n, d):
gcd = primes.gcd(n, d)
return (n // gcd, d // gcd)
# RSA demo
import primes
import sys
p = 3 # one prime number
q = 19 # another prime
n = p * q # the product of the primes
phi = (p - 1) * (q - 1) # the totient
# Try a sensible value for the encryption exponent
e = 5
# Check to see that this is OK
if primes.gcd(phi, e) != 1:
print 'e = %i was not a good choice' % e
exit(0)
# Find the value of the decryption exponent
d = primes.invmod(phi, e)
# Check that this is OK - should not have to do this
if d * e % phi != 1:
print '(d*e) mod (phi) != 1 - something really wrong here!!!'
sys.exit(0)
# Print out the prime numbers, n, the totient and the encryption and decryption exponents
print 'p = %i, q = %i, n = %i, phi = %i, e = %i, d = %i\n' % (p, q, n, phi, e,
d)
# Encryption and decryption using the pow function to carry out modulo exponentiation
ptext = 53 # NOTE: 53 decimal is the decimal ASCII value for the CHARACTER 5
ctext = pow(ptext, e, n) # encrypt the integer 53
print 'The ciphertext of %i is %i' % (ptext, ctext)
import time
from primes import gcd
START = time.time()
SIZE = 1000
sumz = 0
for i in xrange(1, int(SIZE**.5) + 1):
sumz += i
for j in xrange(i + 1, SIZE / i + 1):
sumz += gcd(i, j) * 2
print "Answer:", sumz
print "Time Taken:", time.time() - START
"""
def f(n) = sum( gcd(n, d) for d in xrange(1,n+1))
f(p) = 2p - 1 if p is a prime
f(pq) = (2p-1)(2q-1) if p,q are primes
f(p^2) = p(3p-2) if p is a prime
f(p^3) = p^2(4p-3) if p is a prime
f(p^n) = p^(n-1) * ((n+1)p-n)
f(nm) = f(n) * f(m) if gcd(n,m) = 1
"""
def cheat_crt(prime_list): prime_prod = reduce(lambda x,y: x*y, prime_list) prime_tot = lcm_list([x-1 for x in prime_list]) prime_prod /= gcd(prime_prod, prime_tot) return crt([prime_prod, prime_tot], [0, prime_tot - 3])
totientVals = get_totient(int(math.sqrt(LIM) + 10))
# naive method. Not bad, .04737 seconds for 10^6
count = 0
for m in xrange(2, int(math.sqrt(LIM)) + 1):
if m**2 > LIM:
break
# \/ this part is the slowest.
if m**2 + (m - 1)**2 > LIM:
nRange = xrange(1, m, 2) if m % 2 == 0 else xrange(2, m, 2)
for n in nRange:
if m**2 + n**2 > LIM:
break
if gcd(m, n) != 1:
continue
count += 1
elif m % 2 == 1:
count += totientVals[m] / 2
else:
count += totientVals[m]
print "Answer:", count
print "Time taken:", time.time() - START
"""
Time taken: 0.120429039001 # time for 10^7
Time taken: 0.00853085517883 # time for 10^7 excluding the slowest part... 15x faster to exclude it... gg
computing the numbers that don't let us go up to the limit for n are the slowest ones, by a huge factor.
def cheat_crt(prime_list):
prime_prod = reduce(lambda x, y: x * y, prime_list)
prime_tot = lcm_list([x - 1 for x in prime_list])
prime_prod /= gcd(prime_prod, prime_tot)
return crt([prime_prod, prime_tot], [0, prime_tot - 3])
insertIntoDict(countDict, (iteration + 1, orderTriangle, orderSide1, orderSide2, orderTriangle + 1), count)
insertIntoDict(countDict, (iteration + 1, orderTriangle, orderSide3, orderSide2, orderTriangle + 1), count)
insertIntoDict(countDict, (iteration + 1, orderTriangle, orderSide1, orderSide3, orderTriangle + 1), count)
insertIntoDict(countDict, (iteration + 1, orderSide1+1, orderTriangle+1, orderSide1, orderSide1), count)
insertIntoDict(countDict, (iteration + 1, orderSide2+1, orderTriangle+1, orderSide2, orderSide2), count)
insertIntoDict(countDict, (iteration + 1, orderSide3+1, orderTriangle+1, orderSide3, orderSide3), count)
gcdDict = dict()
count = -6 # ignore the case where size = 2
for order in xrange(2,SIZE+1):
for key in filter((lambda key: key[0] == order-1), countDict.keys()):
appendEntry(key)
val1 = sum(countDict[key] for key in filter( (lambda key: key[0] == order and key[1] == 1), countDict.keys()))
val3 = sum(countDict[key] for key in filter( (lambda key: key[0] == order and key[1] == 3), countDict.keys()))
count += gcd(val1, val3)
if gcd(val1, val3) > 30:
print order, gcd(val1, val3), time.time() - START
print count
print "Time Taken:", time.time() - START
for value in sorted(set(gcdDict.values())):
if not m_r(value/6):
continue
prime = value / 6
minKey = min(filter(lambda x: gcdDict[x] == value, gcdDict.keys()))
print prime, '\t', \
minKey, '\t', \
modNum[prime], '\t', \
import time
from fractions import Fraction
from primes import gcd
START = time.time()
LIMIT = 10**4
count = 0
for denominator in xrange(1, LIMIT):
numer_range = xrange(1, denominator / 100) if denominator % 2 == 1 else xrange(1, denominator / 100, 2)
for numerator in numer_range:
if gcd(numerator, denominator) != 1:
continue
if gcd(numerator - 1, denominator) != 1 and gcd(numerator +1, denominator) != 1:
count += 1
#print Fraction(numerator, denominator)
print count
print "Time Taken:", time.time() - START
"""
i want to get the set of all numer, denom where:
numer < denom / 100
gcd(numer, denom) = 1
gcd(numer-1, denom) != 1
gcd(numer+1, denom) != 1
how the hell do I get the overlap of the two groups of:
# pick two primes, calculate modulus M
# and phi(M)
# IN PRACTICE, find primes between 10^N and 10^M
P1 = 23
P2 = 91
M = P1*P2
x = P1-1
y = P2-1
print(P1, P2, x, y)
PHI = M-P1-P2+1 # not (p1-1)*(p2-1)
# obtain exponent k with gcd(k,phi(M))=1
k = 1 + (M-1)//2
k = k//primes.gcd(k, PHI)
# PUBLIC KEY is (k,M)
print("P1=", P1, "P2=", P2, "M=", M, "phi(M)=", PHI, "K=", k)
plaintext = "Test message"
digitext = [ord(x) for x in plaintext]
ciphertext = [modular_power(x, k, M) for x in digitext]
print(plaintext)
print(digitext)
print(ciphertext)
# calculate decode key
ans = primes.extended_gcd(k, PHI)
