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decyphing vigener cypher without key

Scary Sheep answered on January 31, 2022 Popularity 1/10 Helpfulness 8/10

answer decyphing vigener cypher without key

decyphing vigener cypher without key

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from string import uppercase
from operator import itemgetter
 
def vigenere_decrypt(target_freqs, input):
    nchars = len(uppercase)
    ordA = ord('A')
    sorted_targets = sorted(target_freqs)
 
    def frequency(input):
        result = [[c, 0.0] for c in uppercase]
        for c in input:
            result[c - ordA][1] += 1
        return result
 
    def correlation(input):
        result = 0.0
        freq = frequency(input)
        freq.sort(key=itemgetter(1))
 
        for i, f in enumerate(freq):
            result += f[1] * sorted_targets[i]
        return result
 
    cleaned = [ord(c) for c in input.upper() if c.isupper()]
    best_len = 0
    best_corr = -100.0
 
    # Assume that if there are less than 20 characters
    # per column, the key's too long to guess
    for i in xrange(2, len(cleaned) // 20):
        pieces = [[] for _ in xrange(i)]
        for j, c in enumerate(cleaned):
            pieces[j % i].append(c)
 
        # The correlation seems to increase for smaller
        # pieces/longer keys, so weigh against them a little
        corr = -0.5 * i + sum(correlation(p) for p in pieces)
 
        if corr > best_corr:
            best_len = i
            best_corr = corr
 
    if best_len == 0:
        return ("Text is too short to analyze", "")
 
    pieces = [[] for _ in xrange(best_len)]
    for i, c in enumerate(cleaned):
        pieces[i % best_len].append(c)
 
    freqs = [frequency(p) for p in pieces]
 
    key = ""
    for fr in freqs:
        fr.sort(key=itemgetter(1), reverse=True)
 
        m = 0
        max_corr = 0.0
        for j in xrange(nchars):
            corr = 0.0
            c = ordA + j
            for frc in fr:
                d = (ord(frc[0]) - c + nchars) % nchars
                corr += frc[1] * target_freqs[d]
 
            if corr > max_corr:
                m = j
                max_corr = corr
 
        key += chr(m + ordA)
 
    r = (chr((c - ord(key[i % best_len]) + nchars) % nchars + ordA)
         for i, c in enumerate(cleaned))
    return (key, "".join(r))
 
 
def main():
    encoded = """
        MOMUD EKAPV TQEFM OEVHP AJMII CDCTI FGYAG JSPXY ALUYM NSMYH
        VUXJE LEPXJ FXGCM JHKDZ RYICU HYPUS PGIGM OIYHF WHTCQ KMLRD
        ITLXZ LJFVQ GHOLW CUHLO MDSOE KTALU VYLNZ RFGBX PHVGA LWQIS
        FGRPH JOOFW GUBYI LAPLA LCAFA AMKLG CETDW VOELJ IKGJB XPHVG
        ALWQC SNWBU BYHCU HKOCE XJEYK BQKVY KIIEH GRLGH XEOLW AWFOJ
        ILOVV RHPKD WIHKN ATUHN VRYAQ DIVHX FHRZV QWMWV LGSHN NLVZS
        JLAKI FHXUF XJLXM TBLQV RXXHR FZXGV LRAJI EXPRV OSMNP KEPDT
        LPRWM JAZPK LQUZA ALGZX GVLKL GJTUI ITDSU REZXJ ERXZS HMPST
        MTEOE PAPJH SMFNB YVQUZ AALGA YDNMP AQOWT UHDBV TSMUE UIMVH
        QGVRW AEFSP EMPVE PKXZY WLKJA GWALT VYYOB YIXOK IHPDS EVLEV
        RVSGB JOGYW FHKBL GLXYA MVKIS KIEHY IMAPX UOISK PVAGN MZHPW
        TTZPV XFCCD TUHJH WLAPF YULTB UXJLN SIJVV YOVDJ SOLXG TGRVO
        SFRII CTMKO JFCQF KTINQ BWVHG TENLH HOGCS PSFPV GJOKM SIFPR
        ZPAAS ATPTZ FTPPD PORRF TAXZP KALQA WMIUD BWNCT LEFKO ZQDLX
        BUXJL ASIMR PNMBF ZCYLV WAPVF QRHZV ZGZEF KBYIO OFXYE VOWGB
        BXVCB XBAWG LQKCM ICRRX MACUO IKHQU AJEGL OIJHH XPVZW JEWBA
        FWAML ZZRXJ EKAHV FASMU LVVUT TGK"""
 
    english_frequences = [
        0.08167, 0.01492, 0.02782, 0.04253, 0.12702, 0.02228, 0.02015,
        0.06094, 0.06966, 0.00153, 0.00772, 0.04025, 0.02406, 0.06749,
        0.07507, 0.01929, 0.00095, 0.05987, 0.06327, 0.09056, 0.02758,
        0.00978, 0.02360, 0.00150, 0.01974, 0.00074]
 
    (key, decoded) = vigenere_decrypt(english_frequences, encoded)
    print "Key:", key
    print "\nText:", decoded
 
main()

Popularity 1/10 Helpfulness 8/10 Language python

Source: rosettacode.org

Tags: cypher key python

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Contributed on Jan 31 2022

Scary Sheep

0 Answers Avg Quality 2/10

decyphing vigener cypher without key

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