Homework 1: Password Cracking¶

#list out all the possible scenarios
def encry(t):
    if t>ord('Z'):
        t = t - 26
    return chr(t)

s = input()

for i in range(1, 26):
    print(f"向右偏移{i:2}位: ", end='')
    for char in s:
        print(encry(ord(char)+i), end='')
    print()

# 对于vigenere密码，当key长度为1时，相当于caesar加密
# 直接copy过来

def encry(t):
    if t>ord('z'):
        t = t - 26
    return chr(t)

s = input()

for i in range(0, 26):
    print("key:"+chr(ord('a')+i), end='')
    print()
    for char in s:
        print(encry(ord(char)+i), end='')
    print()

ktbueluegvitnthuexmonveggmrcgxptlyhhjaogchoemqchpdnetxupbqntietiabpsmaoncnwvoutiugtagmmqsxtvxaoniiogtagmbpsmtuvvihpstpdvcrxhokvhxotawswquunewcgxptlcrxtevtubvewcnwwsxfsnptswtagakvoyyak 
key:a
ktbueluegvitnthuexmonveggmrcgxptlyhhjaogchoemqchpdnetxupbqntietiabpsmaoncnwvoutiugtagmmqsxtvxaoniiogtagmbpsmtuvvihpstpdvcrxhokvhxotawswquunewcgxptlcrxtevtubvewcnwwsxfsnptswtagakvoyyak 
key:b
lucvfmvfhwjuouivfynpowfhhnsdhyqumziikbphdipfnrdiqeofuyvqcroujfujbcqtnbpodoxwpvujvhubhnnrtyuwybpojjphubhncqtnuvwwjiqtuqewdsyiplwiypubxtxrvvofxdhyqumdsyufwuvcwfxdoxxtygtoqutxubhblwpzzbl!
key:c
mvdwgnwgixkvpvjwgzoqpxgiioteizrvnajjlcqiejqgosejrfpgvzwrdspvkgvkcdruocqpepyxqwvkwivcioosuzvxzcqpkkqivciodruovwxxkjruvrfxetzjqmxjzqvcyuyswwpgyeizrvnetzvgxvwdxgyepyyuzhuprvuyvcicmxqaacm"
key:d
nwexhoxhjylwqwkxhaprqyhjjpufjaswobkkmdrjfkrhptfksgqhwaxsetqwlhwldesvpdrqfqzyrxwlxjwdjpptvawyadrqllrjwdjpesvpwxyylksvwsgyfuakrnykarwdzvztxxqhzfjaswofuawhywxeyhzfqzzvaivqswvzwdjdnyrbbdn#
key:e
oxfyipyikzmxrxlyibqsrzikkqvgkbtxpcllneskglsiquglthrixbytfurxmixmeftwqesrgrazsyxmykxekqquwbxzbesrmmskxekqftwqxyzzmltwxthzgvblsozlbsxeawauyyriagkbtxpgvbxizxyfziagraawbjwrtxwaxekeozscceo$
key:f
pygzjqzjlanysymzjcrtsajllrwhlcuyqdmmoftlhmtjrvhmuisjyczugvsynjynfguxrftshsbatzynzlyflrrvxcyacftsnntlyflrguxryzaanmuxyuiahwcmtpamctyfbxbvzzsjbhlcuyqhwcyjayzgajbhsbbxckxsuyxbyflfpatddfp%
key:g
qzhakrakmboztznakdsutbkmmsximdvzrennpguminukswinvjtkzdavhwtzokzoghvysgutitcbuazoamzgmsswydzbdgutooumzgmshvyszabbonvyzvjbixdnuqbnduzgcycwaatkcimdvzrixdzkbzahbkcitccydlytvzyczgmgqbueegq&
key:h
raiblsblncpauaobletvuclnntyjnewasfooqhvnjovltxjowkulaebwixuaplaphiwzthvujudcvbapbnahnttxzeacehvuppvnahntiwztabccpowzawkcjyeovrcoevahdzdxbbuldjnewasjyealcabicldjuddzemzuwazdahnhrcvffhr'
key:i
sbjcmtcmodqbvbpcmfuwvdmoouzkofxbtgppriwokpwmuykpxlvmbfcxjyvbqmbqijxauiwvkvedwcbqcobiouuyafbdfiwvqqwobioujxaubcddqpxabxldkzfpwsdpfwbieaeyccvmekofxbtkzfbmdbcjdmekveeafnavxbaebioisdwggis(
key:j
tckdnudnpercwcqdngvxwenppvalpgycuhqqsjxplqxnvzlqymwncgdykzwcrncrjkybvjxwlwfexdcrdpcjpvvzbgcegjxwrrxpcjpvkybvcdeerqybcymelagqxteqgxcjfbfzddwnflpgyculagcnecdkenflwffbgobwycbfcjpjtexhhjt)
key:k
udleoveoqfsdxdreohwyxfoqqwbmqhzdvirrtkyqmryowamrznxodhezlaxdsodsklzcwkyxmxgfyedseqdkqwwachdfhkyxssyqdkqwlzcwdeffsrzcdznfmbhryufrhydkgcgaeexogmqhzdvmbhdofdelfogmxggchpcxzdcgdkqkufyiiku*
key:l
vemfpwfprgteyesfpixzygprrxcnriaewjssulzrnszpxbnsaoypeifambyetpetlmadxlzynyhgzfetfrelrxxbdiegilzyttzrelrxmadxefggtsadeaognciszvgsizelhdhbffyphnriaewnciepgefmgphnyhhdiqdyaedhelrlvgzjjlv+
key:m
wfngqxgqshufzftgqjyazhqssydosjbfxkttvmasotaqycotbpzqfjgbnczfuqfumnbeymazozihagfugsfmsyycejfhjmazuuasfmsynbeyfghhutbefbphodjtawhtjafmieicggzqiosjbfxodjfqhfgnhqioziiejrezbfeifmsmwhakkmw,
key:n
xgohryhrtivgaguhrkzbairttzeptkcgyluuwnbtpubrzdpucqargkhcodagvrgvnocfznbapajibhgvhtgntzzdfkgiknbavvbtgntzocfzghiivucfgcqipekubxiukbgnjfjdhharjptkcgypekgrighoirjpajjfksfacgfjgntnxibllnx-
key:o
yhpiszisujwhbhvislacbjsuuafquldhzmvvxocuqvcsaeqvdrbshlidpebhwshwopdgaocbqbkjcihwiuhouaaeglhjlocbwwcuhouapdgahijjwvdghdrjqflvcyjvlchokgkeiibskquldhzqflhsjhipjskqbkkgltgbdhgkhouoyjcmmoy.
key:p
ziqjtajtvkxiciwjtmbdcktvvbgrvmeianwwypdvrwdtbfrwesctimjeqfcixtixpqehbpdcrclkdjixjvipvbbfhmikmpdcxxdvipvbqehbijkkxwehieskrgmwdzkwmdiplhlfjjctlrvmeiargmitkijqktlrcllhmuhceihlipvpzkdnnpz/
key:q
ajrkubkuwlyjdjxkuncedluwwchswnfjboxxzqewsxeucgsxftdujnkfrgdjyujyqrficqedsdmlekjykwjqwccginjlnqedyyewjqwcrficjkllyxfijftlshnxealxnejqmimgkkdumswnfjbshnjuljkrlumsdmminvidfjimjqwqaleooqa0
key:r
bkslvclvxmzkekylvodfemvxxditxogkcpyyarfxtyfvdhtyguevkolgshekzvkzrsgjdrfetenmflkzlxkrxddhjokmorfezzfxkrxdsgjdklmmzygjkgumtioyfbmyofkrnjnhllevntxogkctiokvmklsmvntennjowjegkjnkrxrbmfpprb1
key:s
cltmwdmwynalflzmwpegfnwyyejuyphldqzzbsgyuzgweiuzhvfwlpmhtiflawlasthkesgfufongmlamylsyeeikplnpsgfaagylsyethkelmnnazhklhvnujpzgcnzpglsokoimmfwouyphldujplwnlmtnwoufookpxkfhlkolsyscngqqsc2
key:t
dmunxenxzobmgmanxqfhgoxzzfkvzqimeraacthzvahxfjvaiwgxmqniujgmbxmbtuilfthgvgpohnmbnzmtzffjlqmoqthgbbhzmtzfuilfmnoobailmiwovkqahdoaqhmtplpjnngxpvzqimevkqmxomnuoxpvgpplqylgimlpmtztdohrrtd3
key:u
envoyfoyapcnhnboyrgihpyaaglwarjnfsbbduiawbiygkwbjxhynrojvkhncyncuvjmguihwhqpioncoanuaggkmrnpruihccianuagvjmgnoppcbjmnjxpwlrbiepbrinuqmqkoohyqwarjnfwlrnypnovpyqwhqqmrzmhjnmqnuauepissue4
key:v
fowpzgpzbqdoiocpzshjiqzbbhmxbskogtccevjbxcjzhlxckyizospkwliodzodvwknhvjixirqjpodpbovbhhlnsoqsvjiddjbovbhwknhopqqdcknokyqxmscjfqcsjovrnrlppizrxbskogxmsozqopwqzrxirrnsanikonrovbvfqjttvf5
key:w
gpxqahqacrepjpdqatikjraccinyctlphuddfwkcydkaimydlzjaptqlxmjpeapewxloiwkjyjsrkqpeqcpwciimotprtwkjeekcpwcixloipqrredloplzryntdkgrdtkpwsosmqqjasyctlphyntparpqxrasyjssotbojlpospwcwgrkuuwg6
key:x
hqyrbirbdsfqkqerbujlksbddjozdumqiveegxldzelbjnzemakbqurmynkqfbqfxympjxlkzktslrqfrdqxdjjnpuqsuxlkffldqxdjympjqrssfempqmaszouelhseulqxtptnrrkbtzdumqizouqbsqrysbtzkttpucpkmqptqxdxhslvvxh7
key:y
irzscjscetgrlrfscvkmltceekpaevnrjwffhymeafmckoafnblcrvsnzolrgcrgyznqkymlalutmsrgseryekkoqvrtvymlggmeryekznqkrsttgfnqrnbtapvfmitfvmryuquosslcuaevnrjapvrctrsztcualuuqvdqlnrquryeyitmwwyi8
key:z
jsatdktdfuhsmsgtdwlnmudfflqbfwoskxggiznfbgndlpbgocmdswtoapmshdshzaorlznmbmvuntshtfszfllprwsuwznmhhnfszflaorlstuuhgorsocubqwgnjugwnszvrvpttmdvbfwoskbqwsdustaudvbmvvrwermosrvszfzjunxxzj9

# 对于key长度大于1的情况进行分析

print("please enter the string to be encry")
s = input()

print("please enter the width of the key")
key_len = input()
key_len = int(key_len)

# 对于key = len的情况，将其分为len组，每组进行频率分析，得到每组的频率最高的字母，认为其对应的明文字母为e，计算出每组的key字母
for i in range(key_len):
    # 统计每组的频率
    freq = {}
    for j in range(i, len(s), key_len):
        if s[j] in freq:
            freq[s[j]] += 1
        else:
            freq[s[j]] = 1
    # 找到频率前三的字母，打印出来(格式：字母+频率)，并存在新数组中
    freq = sorted(freq.items(), key=lambda x: x[1], reverse=True)
    print("the frequency of the group " + str(i) + " is:")
    for j in range(min(3, len(freq))):
        print(freq[j][0] + ": " + str(freq[j][1]))

# 输入字符串，不断尝试各种key，直到找到想要的结果，输入quit退出
print("please enter the string to be encry")
s = input()

while True:
# 这里输入的不是key，而是几组中对应的e的字母组成的字符串
    print("please enter the key to be processed")
    pre_key = input()
    if pre_key == "quit":
        break
    key = ""
    for c in pre_key:
        idx = (ord(c)-ord('e'))%26
        key += chr(idx+ord('a')) #此时key就是我们想要验证的key
# 对于key长度大于等于2的情况，开始解密，输出解密得到的字符串
    res = ""
    for i in range(len(s)):
        idx = (ord(s[i])-ord(key[i%len(key)]))%26
        res += chr(idx+ord('a'))

    print("the result is:")
    print(res)
    print("the key is:")
    print(key)