def solve(card_pk, door_pk):
base = 7
m = 20201227
card_loop_size = discrete_log(m, card_pk, base)
door_loop_size = discrete_log(m, door_pk, base)
return (
pow(door_pk, card_loop_size, m),
pow(card_pk, door_loop_size, m)
)
def part1(keys):
pubKDoor = int(keys[0])
pubKCard = int(keys[1])
# Polard Rho
key = discrete_log(p, pubKDoor, subjectNumber)
# Brute Force
# key = 0
# while(transformSubjectNumber(subjectNumber, key) != pubKDoor):
# key += 1
return transformSubjectNumber(pubKCard, key)
def part_1(card_public_key: int, door_public_key: int) -> int:
card_loop_size = discrete_log(20201227, card_public_key, 7)
card_encryption_key = pow(door_public_key, card_loop_size, 20201227)
return card_encryption_key
def handshake(data, g, p):
"""SymPy discrete_log on raw data."""
a, b = data
x = discrete_log(p, a, g)
return pow(b, x, p)
def crack2(pubkey):
return discrete_log(n, pubkey, 7)
def get_loop_sz(pub_key):
return discrete_log(20201227, pub_key, 7)
print()
''' # BruteForce
# A = 2^a mod p
# c = b^e mod m
for i in range(0,p): # not in the first 1 billion - took a few minutes to calculate
x = pow(2,i,p)
if(x==A):
print("a:",i)
elif(x==B):
print("b:",i)
'''
# 2. Find an integer x in [1,p-1] s.t. g^{x} = A mod p and g^{x} = B mod p.
# Note that the complexity of this computation is exponential as it is the discrete logarithm problem.
x = discrete_log(p, A, 2)
y = discrete_log(p, B, 2)
print("x:",x)
print("y:",y)
if (pow(2,x,p) == A):
print("a:",x)
a = x
if (pow(2,y,p) == B):
print("b:",y)
b = y
print()
# 3. Compute the secret common key. (to verify correct keys)
#!/usr/bin/python3
from sympy.ntheory.residue_ntheory import discrete_log
M = 491988559103692092263984889813697016406
P = 232042342203461569340683568996607232345
B = 5
A = discrete_log(M, P, B)
message = 12259991521844666821961395299843462461536060465691388049371797540470
bobkey = 76405255723702450233149901853450417505
text = bytes.fromhex(hex(pow(bobkey, A, M) ^ message)[2:]).decode('ASCII')
print(text)
first = 4500*3256+1829
quo, rest = fill_corpus(first,resp_first)
do_number(first)
iteration = 1
while not check_corpus(a_quo_sfactors,a_rest_sfactors,b_quo_sfactors,b_rest_sfactors) and iteration < NB_REQUESTS:
i = 1
prime = next(primes_N)
print("test",prime)
while prime in Number():
prime = next(primes_N)
while (prime * (i)) < p:
try:
power = discrete_log(p,prime*i,first)
break
except Exception as e:
i += 1
continue
if power:
iteration += 1
print("iteration :",iteration,"for",prime,end=" | ")
do_number(first,power)
print("nb corpus elts :",len(Number.corpus_i),"nb iteration :",iteration)
print("P :",p," Challenge :",A,"dans",B,"?")
if check_corpus(a_quo_sfactors,a_rest_sfactors,b_quo_sfactors,b_rest_sfactors):
A = Number.get_number(a_quo_sfactors,a_rest_sfactors)
B = Number.get_number(b_quo_sfactors,b_rest_sfactors)
def part_one(code: Tuple[int, int]) -> None:
subject_number = 7
cycle = 20201227
result = pow(code[1], discrete_log(cycle, code[0], subject_number), cycle)
print(f"Result for part one is {result}")