培根密码，培根所用的密码是一种本质上用二进制数设计的，没有用通常的0和1来表示，而是采用a和b
一、培根密码加密方式
第一种方式： A aaaaa B aaaab C aaaba D aaabb E aabaa F aabab G aabba H aabbb I abaaa J abaab K ababa L ababb M abbaa N abbab O abbba P abbbb Q baaaa R baaab S baaba T baabb U babaa V babab W babba X babbb Y bbaaa Z bbaab 第二种方式 a AAAAA g AABBA n ABBAA t BAABA b AAAAB h AABBB o ABBAB u-v BAABB c AAABA i-j ABAAA p ABBBA w BABAA d AAABB k ABAAB q ABBBB x BABAB e AABAA l ABABA r BAAAA y BABBA f AABAB m ABABB s BAAAB z BABBB
二、举例
例1、 baabaaabbbabaaabbaaaaaaaaabbabaaaabaaaaaabaaabaabaaaabaabbbaabbbaababb
baaba aabbb abaaa bbaaa aaaaa abbab aaaab aaaaa abaaa baaba aaaba abbba abbba ababb
s h i y a n b a i s c o o l
例2、明文是LOVE，用“随意选取句子和文”加密，得到密文就是“SuLyi XuanQ uJuZi HEwEN”（这里用大写字母代表a,小写字母代表b）
SuLyi XuanQ uJuZi HEwEN
ababbabbbabababaabaa
这里，加密者使用两种不同字体，准备好一篇包含相同ab字数的假信息后，即两种字体分别代表a型和b型。然后假信息中的每个字母按字体来决定其代表“a”还是“b”。
例3、明文:now is a good t.........密码术 密码学 密码论坛 密码术论坛, `, Q" J/ K2 F9 S
n o w i s a g o o d t abbaa abbab babaa abaaa baaab aaaaa aabba abbab abbab aaabb baaba 密文:BOWED ASTER PINED JOKED THEIR BLACK HASTE ARRAY INSET CHEST SLING加密规则：（1）a-m -> a ，也就是说 a ——m 的所有字母都能替换成 a。
n-z -> b ，也就是说 n ——z 的所有字母都能替换成 b。
（2）a-m <- a ，也就是说 a 可以替换 a——m 中的任何一个字母。
n-z <- a ，也就是说 b 可以替换 n——z 中的任何一个字母。
Ps:密文中的单词都是根据加密规则随便拼凑出来的，为了使加密更隐蔽，一般拼凑出来的单词可连成通顺的一句话
三、附加解密脚本如下：
#!/usr/bin/python# -*- coding: utf-8 -*-import realphabet = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']first_cipher = ["aaaaa","aaaab","aaaba","aaabb","aabaa","aabab","aabba","aabbb","abaaa","abaab","ababa","ababb","abbaa","abbab","abbba","abbbb","baaaa","baaab","baaba","baabb","babaa","babab","babba","babbb","bbaaa","bbaab"]second_cipher = ["aaaaa","aaaab","aaaba","aaabb","aabaa","aabab","aabba","aabbb","abaaa","abaaa","abaab","ababa","ababb","abbaa","abbab","abbba","abbbb","baaaa","baaab","baaba","baabb","baabb","babaa","babab","babba","babbb"]def encode(): string = raw_input("please input string to encode:\n") e_string1 = "" e_string2 = "" for index in string: for i in range(0,26): if index == alphabet[i]: e_string1 += first_cipher[i] e_string2 += second_cipher[i] break print "first encode method result is:\n"+e_string1 print "second encode method result is:\n"+e_string2 returndef decode(): e_string = raw_input("please input string to decode:\n") e_array = re.findall(".{5}",e_string) d_string1 = "" d_string2 = "" for index in e_array: for i in range(0,26): if index == first_cipher[i]: d_string1 += alphabet[i] if index == second_cipher[i]: d_string2 += alphabet[i] print "first decode method result is:\n"+d_string1 print "second decode method result is:\n"+d_string2 returnif __name__ == '__main__': while True: print "\t*******Bacon Encode_Decode System*******" print "input should be lowercase,cipher just include a b" print "1.encode\n2.decode\n3.exit" s_number = raw_input("please input number to choose\n") if s_number == "1": encode() raw_input() elif s_number == "2": decode() raw_input() elif s_number == "3": break else: continue
注：本笔记参考文献、、
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ps:萌新一枚，本文纯为学习笔记，以防某日犯傻忘记--。如有不妥处，望大佬们赐教~