adding first pass substitution, affine brute force, vigenere brute force

crypto/affine_brute.py

@@ -1,3 +1,7 @@
+"""
+affine cipher brute force attempt
+"""
+
 from fractions import gcd
 import numpy as np
 from pprint import pprint 

crypto/ios.py

@@ -0,0 +1,64 @@
+"""
+Initial attempt to crack the ios puzzle.
+"""
+import numpy as np
+from pprint import pprint 
+
+p1 = "JK EJRN*U. I*T  G* NRFJNUC QJ QTLU NUEJNC UVUNY QNTI*TDTIQGE QUDULSJIU EJIVUN *TQGJI. FTYHU QSUY'C NUEJNCUC QSG* *U**GJI"
+p2 = "WO HJIBSJP HNIY MPO *TOIUBOQ. WO GBQSOI NGWNYQ. UNHO N VJBUO, JTS JM SCO ENEOG JM SJIATOQ, QKONFBIA SJ TQ. BS KGNYOR TQ N HBACSY RTE."
+p3 = "IUIMYFCI WTP PIEMIQP. DQ'P ASPQ T BTQQIM FR RDCGDCL F*Q VWTQ QWIY TMI."
+p4 = "SHB *GMKPSLJS SH*JO LIKUS QBTUP*SY. QYQSBGQ *QJ's HKW SHBY WKPD, *S'Q HKW SHBY RL *F."
+
+problems = [p1, p2, p3, p4]
+
+for problem in problems:
+    freq = {}
+    for c in problem:
+        if c in freq.keys():
+            freq[c] += 1
+        else:
+            freq[c] = 1
+    
+    pprint(freq)
+
+substitution = {}
+substitution['A'] = 'A'
+substitution['B'] = 'B'
+substitution['C'] = 'C'
+substitution['D'] = 'D'
+substitution['E'] = 'E'
+substitution['F'] = 'F'
+substitution['G'] = 'G'
+substitution['H'] = 'H'
+substitution['I'] = 'o'
+substitution['J'] = 'J'
+substitution['K'] = 'K'
+substitution['L'] = 'L'
+substitution['M'] = 'M'
+substitution['N'] = 'a'
+substitution['O'] = 'e'
+substitution['P'] = 'P'
+substitution['Q'] = 'Q'
+substitution['R'] = 'R'
+substitution['S'] = 'S'
+substitution['T'] = 'i'
+substitution['U'] = 'U'
+substitution['V'] = 'V'
+substitution['W'] = 'W'
+substitution['X'] = 'X'
+substitution['Y'] = 'Y'
+substitution['Z'] = 'Z'
+                      
+for problem in problems:
+    pprint( np.mean([len(x) for x in problem.split()]) )
+
+    result = ""
+    problemnew = problem
+    for c1, c2 in zip(substitution.keys(),substitution.values()):
+        problemnew = problemnew.replace(c1,c2)
+    print(problemnew)
+    print("\n")
+    
+
+
+

crypto/memo.txt

@@ -0,0 +1,25 @@
+cryptok.space/cpv
+
+
+how many prev characters taken into considerations
+
+search depth - limit iterations to top 100,000 results
+26^6 = 300M, so we limit this
+
+
+science
+
+
+
+
+
+
+featherduster:
+
+
+FeatherDuster > anlyze
+
+FeatherDuster > use vigenere
+
+
+

crypto/vigenere.py

@@ -0,0 +1,64 @@
+"""
+Attempt to crack the cipher 
+by treating it as a vignere cipher.
+
+This brute forces two- and three-character length vigenere keys
+
+http://simonsingh.net/The_Black_Chamber/vigenere_cracking_tool.html
+"""
+
+from fractions import gcd
+import numpy as np
+from pprint import pprint 
+
+p1 = "JK EJRNZU. IZT  GZ NRFJNUC QJ QTLU NUEJNC UVUNY QNTIZTDTIQGE QUDULSJIU EJIVUN ZTQGJI. FTYHU QSUY'C NUEJNCUC QSGZ ZUZZGJI"
+p2 = "WO HJIBSJP HNIY MPO ZTOIUBOQ. WO GBQSOI NGWNYQ. UNHO N VJBUO, JTS JM SCO ENEOG JM SJIATOQ, QKONFBIA SJ TQ. BS KGNYOR TQ N HBACSY RTE."
+p3 = "IUIMYFCI WTP PIEMIQP. DQ'P ASPQ T BTQQIM FR RDCGDCL FZQ VWTQ QWIY TMI."
+p4 = "SHB ZGMKPSLJS SHZJO LIKUS QBTUPZSY. QYQSBGQ ZQJ's HKW SHBY WKPD, ZS'Q HKW SHBY RL ZF."
+p5 = "MGV DGFN ZG XGS VCFQ QMTLT ETLLCNTL QG KTECIF LTYKTQ? ... I VCFQ QMTE QG KTECIF LTYKTQ OGK CL DGFN CL ETF CKT YCHCRDT GO TUID."
+p6 = "OH XDQ VNCP PD RGGJ N ZGELGP, XDQ BQZP NAZD FOTG OP HLDB XDQLZGAH."
+
+
+p1="JKEJRNZUIZTGZNRFJNUCQJQTLUNUEJNCUVUNYQNTIZTDTIQGEQUDULSJIUEJIVUNZTQGJIFTYHUQSUYCNUEJNCUCQSGZZUZZGJI"
+p2="WOHJIBSJPHNIYMPOZTOIUBOQWOGBQSOINGWNYQUNHONVJBUOJTSJMSCOENEOGJMSJIATOQQKONFBIASJTQBSKGNYORTQNHBACSYRTE"
+p3="IUIMYFCIWTPPIEMIQPDQPASPQTBTQQIMFRRDCGDCLFZQVWTQQWIYTMI"
+p4="SHBZGMKPSLJSSHZJOLIKUSQBTUPZSYQYQSBGQZQJsHKWSHBYWKPDZSQHKWSHBYRLZF"
+p5="MGVDGFNZGXGSVCFQQMTLTETLLCNTLQGKTECIFLTYKTQIVCFQQMTEQGKTECIFLTYKTQOGKCLDGFNCLETFCKTYCHCRDTGOTUID"
+p6="OHXDQVNCPPDRGGJNZGELGPXDQBQZPNAZDFOTGOPHLDBXDQLZGAH"
+
+
+'''
+Repeated sequences and likely key lengths:
+
+    problem 1: 2, 3, 6, 9, 18
+    problem 2: 2, 3, 6, 9, 18
+    problem 3: 19
+    problem 4: EXACTLY 2, 7, 14
+    problem 5: EXACTLY 2, 4, 8, 16
+    problem 6: 2, 3, 4, 5, 7, 10 (?)
+
+'''
+
+problems = [p5]
+
+for ip, problem in enumerate(problems):
+    original = problem
+    filename = "problem" + str(ip+1) + ".txt"
+
+    with open(filename,'w') as f:
+
+        for ii in range(26):
+            for jj in range(26):
+                key = ""
+                key += chr(ord('a')+ii)
+                key += chr(ord('a')+jj)
+        
+                ki = 0 # key index
+                newstr = ""
+                for c in original:
+                    newchr = chr( ord('a') + (ord(c) + ord(key[ki]) )%26 )
+                    ki = (ki + 1)%(len(key))
+                    newstr += newchr
+                
+                f.write(newstr + "\n")
+