Python凯撒密码解码器

Question

在我的课程中，我的任务是创建一个凯撒密码解码器，它需要一串输入并使用字母频率找到最好的字符串。如果不知道多少意义，但让问题发布：

编写一个程序，它执行以下操作。首先，它应该读取一行输入，这是编码的消息，并将包含大写字母和空格。您的程序必须尝试使用​​所有26个可能的值S来解码消息;在这26个可能的原始信息中，打印出最具善意的那一个。 为方便起见，我们会预先为您定义的变量letterGoodness，长度26的列表，它等于上述

频率表中的值我到目前为止这样的代码：

x = input() 
NUM_LETTERS = 26 #Can't import modules I'm using a web based grader/compiler 
def SpyCoder(S, N): 
    y = "" 
    for i in S: 
     x = ord(i) 
     x += N 
     if x > ord('Z'): 
      x -= NUM_LETTERS 
     elif x < ord('A'): 
      x += NUM_LETTERS 
     y += chr(x) 
    return y 

def GoodnessFinder(S): 
    y = 0 
    for i in S: 
     if x != 32: 
      x = ord(i) 
      x -= ord('A') 
      y += letterGoodness[x] 
    return y 

def GoodnessComparer(S): 
    goodnesstocompare = GoodnessFinder(S) 
    goodness = 0 
    v = '' 
    for i in range(0, 26): 
     v = SpyCoder(S, i) 
     goodness = GoodnessFinder(v) 
     if goodness > goodnesstocompare: 
      goodnesstocompare = goodness 
    return v 

y = x.split() 
z = '' 
for i in range(0, len(y)): 
    if i == len(y) - 1: 
     z += GoodnessComparer(y[i]) 
print(z) 

编辑：由Cristian Ciupitu建议更改 请忽略缩进错误，它们可能在我复制我的代码时出现。

的程序是这样的：

• 就拿输入，并将其分割成一个列表
• 对于每一个列表值我将其提供给一个善良取景器。
• 它需要字符串的好处，并将其他所有内容进行比较，当有更高的善意时，它会使得更高的善良性比较。
• 它然后我偏移量文本的该字符串，看看优度较高或较低

我不太清楚是哪里的问题，第一个测试：LQKP OG CV GKIJV DA VJG BQQ
打印正确的消息：JOIN ME AT在由ZOO

然而接下来的测试：UIJT JT乙TBNQMF MJOF PG UFYU GPS EFDSZQUJOH
给出的垃圾字符串：SGHR HRžRZLOKD KHMD NE SDWS ENQ CDBQXOSHMF
当它应该是：这是一个示例文本的DECRYPTIN摹

我知道我必须：
想尽一切移值
获取字
返回字符串最高善良“善良”。

我希望我的解释有意义，因为我现在很困惑。

Answer 1

我的最终解决方案有效，这要感谢奇妙的克里斯蒂安Ciupitu。

x = input() 
NUM_LETTERS = 26 #Can't import modules I'm using a web based grader/compiler 
def SpyCoder(S, N): 
    y = "" 
    for i in S: 
     if(i.isupper()): 
     x = ord(i) 
     x += N 
     if x > ord('Z'): 
      x -= NUM_LETTERS 
     elif x < ord('A'): 
      x += NUM_LETTERS 
     y += chr(x) 
     else: 
     y += " " 
    return y 

def GoodnessFinder(S): 
    y = 0 
    for i in S: 
     if i.isupper(): 
     x = ord(i) 
     x -= ord('A') 
     y += letterGoodness[x] 
     else: 
     y += 1 
    return y 

def GoodnessComparer(S): 
    goodnesstocompare = GoodnessFinder(S) 
    goodness = 0 
    v = '' 
    best_v = S 
    for i in range(0, 26): 
    v = SpyCoder(S, i) 
    goodness = GoodnessFinder(v) 
    if goodness > goodnesstocompare: 
     best_v = v 
     goodnesstocompare = goodness 
    return best_v 


print(GoodnessComparer(x)) 

谢谢大家的帮助！

Answer 2

这是我的实施，它工作正常。

你应该打印每个可能的消息的好处，并看看你的程序输出它的原因。

letterGoodness = dict(zip(string.ascii_uppercase, 
         [.0817,.0149,.0278,.0425,.1270,.0223,.0202, 
         .0609,.0697,.0015,.0077,.0402,.0241,.0675, 
         .0751,.0193,.0009,.0599,.0633,.0906,.0276, 
         .0098,.0236,.0015,.0197,.0007])) 

trans_tables = [ str.maketrans(string.ascii_uppercase, 
       string.ascii_uppercase[i:]+string.ascii_uppercase[:i]) 
       for i in range(26)] 

def goodness(msg): 
    return sum(letterGoodness.get(char, 0) for char in msg) 

def all_shifts(msg): 
    msg = msg.upper() 
    for trans_table in trans_tables: 
     txt = msg.translate(trans_table) 
     yield goodness(txt), txt 

print(max(all_shifts(input()))) 

Answer 3

我正在研究相同的教程，并使用了一个稍微不同的方法。这避免了创建和调用函数：

inp = input()  #to hold code text 
code = list(inp) #store code as a list 
soln = []   #store the 'Goodness' for each of 26 possible answers 
y=0    #variable to hold total goodness during calculations 
clear = []  #will hold decoded text 
pos=0    #position marker for a list 

#for every possible value of shift 
#note range as 0 to 25 are valid shifts and shift 26 = shift 0 

for shift in range(0,26): 
    for i in code:     #loop through each letter in code 
     if i == " ":     #spaces have no score so omit them 
     continue 
     else:      #if it's a letter 
     x = ord(i)-shift   #apply the test shift 
     if x < 65:    #prevent shifting outside A-Z range 
      x = x + 26    
     x = x - 64    #turn ord into character position in A-Z with A=1 
     x = letterGoodness[x-1] #turn this into the Goodness score 
     y = y + x     #add this to a running total 
    soln.insert(shift-1,y)   #AFTER decoding all letters in code, add total(y) to list of scores 
    y = 0       #reset y before next test value 

bestSoln=max(soln)     #find highest possible score 

for i in range(0,26):    #check the list of solutions for this score 
    if soln[i]==bestSoln:   #the position in this list is the shift we need 
     bestShift = i+1    #+1 as the first solution is 0 

for i in code:      #now decode the original text using our best solution 
    if i == " ":     #spaces are not encoded so just add these to the string 
     clear.insert(pos," ")  #pos used to track next position for final string 
     pos = pos + 1 
     continue 
    else: 
     x = ord(i)-bestShift   #same operation as before 
     if x < 65: 
     x = x + 26 
    z = chr(x) 
    clear.insert(pos,z)    #add the decoded letter to the clear text 
    pos = pos + 1 
print("".join(clear))    #join the list of clear text into one string and print it 

注意，该代码的很多地方可以（也应该）被压缩，例如

x = x - 64 
x = letterGoodness[x-1] 
y = y + x 

他们离开扩大到“显示我的工作”的教程练习。