Break the Caesar!
Codewars Kata 36√
Description
https://www.codewars.com/kata/598e045b8c13926d8c0000e8
-简述：本题旨在破解凯撒密码的加密方法，将各字母在字母表中同时移动一定数量的位置使其加密，求破解该密码的方法。
-思路：分别测试将每个字母往回移动1-26次，句中单词的个数，求出最大个数。
-难点：1 生成凯撒密码的方法。
The Caesar cipher is a notorious (and notoriously simple) algorithm for encrypting a message: each letter is shifted a certain constant number of places in the alphabet. For example, applying a shift of 5 to the string “Hello, world!” yields “Mjqqt, btwqi!”, which is jibberish.
In this kata, your task is to decrypt Caesar-encrypted messages using nothing but your wits, your computer, and a set of the 1000 (plus a few) most common words in English in lowercase (made available to you as a preloaded variable named WORDS, which you may use in your code as if you had defined it yourself).
Given a message, your function must return the most likely shift value as an integer.
A few hints:
Be wary of punctuation
Shift values may not be higher than 25
My solution
abc = ‘abcdefghijklmnopqrstuvwxyz’
def caesar(s, k): trans = str.maketrans(abc, abc[k:] + abc[:k]) return s.translate(trans)
def break_caesar(message): message = ‘‘.join(c if c.isalpha() else ‘ ‘ for c in message).lower() hits = [] for k in range(26): dec = caesar(message, -k) cnt = 0 for word in dec.split(): if word in WORDS: cnt += 1 hits.append(cnt)
return hits.index(max(hits))
# 自己运行判断单词 可用enchant
# import enchant
# d = enchant.Dict("en_US")
# d.check("Hello")
# True/False
Given solution
abc = ‘abcdefghijklmnopqrstuvwxyz’
def break_caesar(message):
caesar = lambda s, key: s.translate(str.maketrans(abc, abc[key:] + abc[:key]))
message = ‘‘.join(c if c.isalpha() else ‘ ‘ for c in message).lower()
hits = [ sum(w in WORDS for w in caesar(message, -key).split()) for key in range(26) ]
return hits.index(max(hits))
Points
1 maketrans()方法语法：
str.maketrans(intab, outtab)
参数
intab – 字符串中要替代的字符组成的字符串。
outtab – 相应的映射字符的字符串。
返回值
返回字符串转换后生成的新字符串。
注：两个字符串的长度必须相同，为一一对应的关系。
2
3 自己运行判断单词 可用enchant
import enchant
d = enchant.Dict(“en_US”)
d.check(“Hello”)
True/False
