代替解密方法交流

一：看密文文件，这是我的密文文件。

QurgrgwemousswcwkionwbridVurxiubvkixjxgmubmpurpybswektubfpyooierrwxgbikiftgmpvwwbwprmvkupgwsmpmbmruibmkpremrwftgmpsemqbwzrwbpucwmrrwbruibUrxmboeicuswbwqpwxyeurtrwxgbikifuwpdiedubmbxumkpwxyeurtvtcuerywidswxwbremkulmruibsmrmueewcwepuvukurtmbsoeifemnnmvukurtGiqwcwergwxinokwzmkfieurgnpidvkixjxgmubmewgufgktubrwbpucwmbsrgwewmewnmbtjubspidmrrmxjprgmrbwwsrivwswrwxrwsmbsoewcwbrwsqgukwdubmbxumkpwecuxwpmewrgwjwtrmefwrpidbwrqiejmrrmxjpRgwmookuxmruibidvkixjxgmubrwxgbikiftqukkubwcurmvktveubfbwqpwxyeurtrgewmrprirgwdubmbxumkpwxriembsoipwbwqxgmkkwbfwprirgwpwxyeurtiddubmbxumkbwrqiejpUbieswerifymembrwwrgwpwxyeurtiddubmbxumkptprwnpvmpwsibvkixjxgmubrwxgbikifturupyefwbrrixmeetiyrewpwmexgibrgwswrwxruibmbsoewcwbruibidmrrmxjvwgmcuiemfmubprrgwubrwfemruibiddubmbxumkptprwnpewoewpwbrwsvtrgwPQUDRptprwnmbsvkixjxgmubRgwnmubewpwmexgriouxpubxkysw:cykbwemvukurtmbmktpupmbsmrrmxjvwgmcuieswrwxruibidPQUDRptprwnmrrmxjvwgmcuieswrwxruibmbsoewcwbruibidvkixjxgmubptprwnpwxyeurtmbmktpupmbsmrrmxj-swdwbpwcweuduxmruibidPQUDRptprwnvmpwsibvkixjxgmubRgwoeihwxrgmpunoiermbrrgwiewruxmkpufbuduxmbxwmbsxkwmemookuxmruiboeipowxrpUrupwzowxrwsripikcwrgwpwxyeurtoeivkwnidrgwubrwfemruibidvkixjxgmubrwxgbikiftmbsPQUDRptprwnpgwkodykriprewbfrgwbxinokumbxwewayuewnwbrpdieubrwebmruibmkxeipp-viesweomtnwbrppyxgmpomtnwbrpwxyeurtmbru-nibwtkmybsweubfmbsmbru-rweeieupndubmbxubfmbswddwxrucwibpmdwfymesubfrgwpwxyeurtidrgwdubmbxumkptprwnqguxgmewidfewmrpufbuduxmbxwrinmubrmububfrgwpwxyeurtmbsprmvukurtidiyexiybretpbwrqiejp

拿到密文，首先是统计文件中字母出现的频率

1：统计字符出现频率的代码。如下：

#计算字母的出现频率
file = open("miwen","r",encoding="UTF-8")
sum = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,]  #创建list 第0 个存放总数  1-26   分别记录a-z 出现的次数
flag = 0
def calculate():
    global flag
    while flag < 30:
        data_char = file.read(1)
        # print(data_char)
        if len(data_char) == 0:
            flag +=1
            continue
        elif ord(data_char) in range(ord("a"),ord("z")):
            data_char = ord(data_char) - 96
            sum[data_char] += 1
            sum[0] += 1
        elif ord (data_char) in range (ord ("A"), ord ("Z")):
            data_char = ord(data_char) - 64
            sum[data_char] +=1
            sum[0] +=1
        else:
            pass
calculate()
for i in sum:
    print(i)

file.close()

我没有计算出百分比，sum[0]存放了总的字符数，用各个直接除以num【0】即可得到百分比。

统计结果为：

大数据测试的字母出现频率为：

e 16782 11.42% 11991 64.52%
a 12574 8.56% 10050 54.08%
i 11674 7.94% 9364 50.39%
r 11042 7.51% 9337 50.24%
t 10959 7.46% 8929 48.05%
o 10466 7.12% 8259 44.44%
n 9413 6.41% 7948 42.77%
s 8154 5.55% 6859 36.91%
l 8114 5.52% 6882 37.03%
c 6968 4.74% 6028 32.44%
u 5373 3.66% 4910 26.42%
p 4809 3.27% 4283 23.05%
m 4735 3.22% 4241 22.82%
d 4596 3.13% 4186 22.52%
h 4058 2.76% 3724 20.04%
g 3380 2.30% 3061 16.47%
b 3121 2.12% 2918 15.70%
y 2938 2.00% 2815 15.15%
f 2157 1.47% 1899 10.22%
v 1574 1.07% 1531 8.24%
w 1388 0.94% 1328 7.15%
k 1235 0.84% 1183 6.37%
x 507 0.35% 505 2.72%
z 356 0.24% 309 1.66%
q 343 0.23% 343 1.85%
j 220 0.15% 217 1.17%
作为单词首字母出现频率较高的字母依次是：
S, P, C, D, M, A...
#

常见单词表为：

从上面的三个频率表我们很快可以确定：

w 对应 e

g 对应 h

r 对应 t

之后几乎没有思路：即进行第二步：

二：找字符串，（借助word 工具）

如图是字符串ptprwn 出现的次数。

通过观察：我们还会找到，rwxgbikif, dubmbxumkp,等出现多次的字符串，通过结合字符的出现频率，我们最终得到如下对应关系;

"a":"q","b":"n","c":"v","d":"f","e":"r","f":"g","g":"h","h":"j","j":"k","k":"l","l":"z",
"i":"o","m":"a","n":"m","o":"p","p":"s",
"q":"w","r":"t","s":"d","t":"y","u":"i","v":"b","w":"e","x":"c","y":"u","z":"x",



第三步：写算法翻译原文

#codeing:UTF-8
#__author__:Duke
#date:2018/3/13/013

file1 = open("miwen","r",encoding="utf-8")
file2 = open("mingwen","a+",encoding="utf-8")

dict = {
    "a":"q","b":"n","c":"v","d":"f","e":"r","f":"g","g":"h","h":"j","j":"k","k":"l","l":"z",
    "i":"o","m":"a","n":"m","o":"p","p":"s",
    "q":"w","r":"t","s":"d","t":"y","u":"i","v":"b","w":"e","x":"c","y":"u","z":"x",

    "A": "Q", "B": "N", "C": "V", "D": "F", "E": "R", "F": "G", "G": "H", "H": "J", "J": "K", "K": "L", "L": "Z",
    "I": "O", "M": "A", "N": "M", "O": "P", "P": "S",
    "Q": "W", "R": "T", "S": "D", "T": "Y", "U": "I", "V": "B", "W": "E", "X": "C", "Y": "U", "Z": "X",
}
flag = 0
def tongji():
    global flag
    while flag <10:
        data_char = file1.read(1)
        if len(data_char) == 0:
            flag +=1
            continue
        if data_char in dict:
            print(dict[data_char],end="")
            file2.write(dict[data_char])

        else:
            pass
tongji()
file1.close()
file2.close()

最后得到明文：

WiththerapiddevelopmentofBitcoinblockchainasitsunderlyingsupporttechnologyhasbeenestablishedasanationalstrategyhasdrawnextensiveattentionItcanprovidenewsecuritytechnologiesforfinancialsecuritybyvirtueofdecentralizationdatairreversibilityandprogrammabilityHoweverthecomplexalgorithmsofblockchainarehighlyintensiveandtherearemanykindsofattacksthatneedtobedetectedandpreventedwhilefinancialservicesarethekeytargetsofnetworkattacksTheapplicationofblockchaintechnologywillinevitablybringnewsecuritythreatstothefinancialsectorandposenewchallengestothesecurityoffinancialnetworksInordertoguaranteethesecurityoffinancialsystemsbasedonblockchaintechnologyitisurgenttocarryoutresearchonthedetectionandpreventionofattackbehavioragainsttheintegrationoffinancialsystemsrepresentedbytheSWIFTsystemandblockchainThemainresearchtopicsincludevulnerabilityanalysisandattackbehaviordetectionofSWIFTsystemattackbehaviordetectionandpreventionofblockchainsystemsecurityanalysisandattackdefenseverificationofSWIFTsystembasedonblockchainTheprojecthasimportanttheoreticalsignificanceandclearapplicationprospectsItisexpectedtosolvethesecurityproblemoftheintegrationofblockchaintechnologyandSWIFTsystemshelpfultostrengthencompliancerequirementsforinternationalcrossborderpaymentssuchaspaymentsecurityantimoneylaunderingandantiterrorismfinancingandeffectiveonsafeguardingthesecurityofthefinancialsystemwhichareofgreatsignificancetomaintainingthesecurityandstabilityofourcountrysnetworks

posted @ 2018-03-13 22:24 Duke777 阅读(708) 评论(0) 编辑收藏举报

刷新页面返回顶部

Duke77

代替解密方法交流

公告