Skip to content

Instantly share code, notes, and snippets.

Embed
What would you like to do?
#!/usr/bin/env python3
import re
import os
import sys
import random
from itertools import chain, product
testkey = 'q w e r'
message = 'ta kzxwgpdjtctpf prozif, d nop zs adpgidc agxjoin tn kdccom iokgintho, kzxwgpdjco zi moktmdjco ts proio tn da dcyzitprx ertkr poixtadpon dspoi d statpo dxzgap zs ptxo (ertkr xdf mowoam za pro ythoa agxjoi) dam kziiokpcf moktmon eroproi d ythoa agxjoi joczayn pz pro nop.'
sampledecrypt = 'ly dsourjwclqlje jxisbe, w hij sk ywjrbwq yrocibh lh dwqqig bidrbhlai, dsourjwcqi sb gidlgwcqi lk jxibi lh wy wqpsbljxo vxldx jibolywjih wkjib w klylji wosryj sk jloi (vxldx owe giuiyg sy jxi plaiy yrocib) wyg dsbbidjqe gidlgih vxijxib w plaiy yrocib ciqsyph js jxi hij.'
messagetarget = 'In computability theory, a set of natural numbers is called recursive, computable or decidable if there is an algorithm which terminates after a finite amount of time (which may depend on the given number) and correctly decides whether a given number belongs to the set.'.lower()
alpha = 'abcdefghijklmnopqrstuvwxyz'
'''
we first need to generate the master key (e50) from the initial key
seed the state (d) by concatenating the alphabet onto the key (all lowercase).
state = asdfabcdefghijklmnopqrstuvwxyz
loop through each letter of the alphabet.
call that letter's index (1-indexed) a
and call the letter itself b.
find that letter's position (1-indexed) in the state, this is c.
to update the state, we need to generate d_left and d_right.
d_left is the first c chars of the current state
d_right is the remainder of the chars (position c to the end)
with all occurrences of b deleted.
we update the state with the concatenation of d_left and d_right
to encrypt, we operate char by char.
we first get the index of the first occurrence of the plaintext char
in the master key, add 20 to that index, and call it findval.
the encrypted char is the findval'th character of e50
'''
def gen_e50(key):
col_a = [i for i in range(1, 271)] #
col_b = list(alpha) # =char(A51+96)
col_c = [] # =FIND(B51,D50)
col_d = [] # =CONCAT(LEFT(D50,C51), SUBSTITUTE(MID(D50,C51+1, LEN(D50)), B51, ""))
d50 = re.sub('[^a-z]', '', key.lower()[0:10]) + alpha
for a in col_a[0:len(col_b)]:
if a == 1:
d_prev = d50
else:
d_prev = col_d[a-2]
b = col_b[a-1]
c = d_prev.find(b) + 1
d_right = d_prev[c:len(d_prev)]
d_right = re.sub(b, '', d_right)
d_left = d_prev[0:c]
d = d_left + d_right
col_c.append(c)
col_d.append(d)
e50 = col_d[-1] + col_d[-1]
return e50
def encryptchar(char, e50):
# =IFERROR(MID($E$50, FIND(E51, $E$50)+20, 1), E51)
if not char in e50:
return char
findval = e50.find(char) + 20
return e50[findval]
def scoredecrypt(key):
testpt = ''.join([encryptchar(char, gen_e50(key)) for char in message])
score = 0
for c in range(len(testpt)):
if testpt[c] == messagetarget[c]:
score += 1
return score
assert gen_e50(testkey) == 'qwerabcdfghijklmnopstuvxyzqwerabcdfghijklmnopstuvxyz'
e50 = gen_e50(testkey)
testpt = ''.join([encryptchar(char, e50) for char in message])
assert testpt == sampledecrypt
key = list(' ' * 10)
scorehist = []
highscore = 0
while True:
print(''.join(key) + ' ' + str(highscore) + '\t[', end='')
progress = int((highscore / len(messagetarget)) * 100)
print('=' * progress, end='')
print(' ' * (100 - progress), end='')
print(']')
if highscore == len(messagetarget):
print("WOW!!!!!!!!!!!!!!!!!!!")
print(key)
break
if len(scorehist) > 2:
if scorehist[-1] == scorehist [-2]:
keyind = random.randint(0,9)
alphind = random.randint(0, len(alpha)-1)
# print("stuck - setting key index %s to alphabet index %s" % (keyind, alphind))
key[keyind] = alpha[alphind]
for k in range(len(key)):
keyscore = {}
for a in (alpha):
key[k] = a
score = scoredecrypt(''.join(key))
keyscore[a] = score
highscore = keyscore[max(keyscore, key=keyscore.get)]
key[k] = max(keyscore, key=keyscore.get)
scorehist.append((key, highscore))
Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment
You can’t perform that action at this time.