Python凱撒密碼解碼器

Question

在我的課程中，我的任務是創建一個凱撒密碼解碼器，它需要一串輸入並使用字母頻率找到最好的字符串。如果不知道多少意義，但讓問題發佈：

編寫一個程序，它執行以下操作。首先，它應該讀取一行輸入，這是編碼的消息，並將包含大寫字母和空格。您的程序必須嘗試使用​​所有26個可能的值S來解碼消息;在這26個可能的原始信息中，打印出最具善意的那一個。 爲方便起見，我們會預先爲您定義的變量letterGoodness，長度26的列表，它等於上述

頻率表中的值我到目前爲止這樣的代碼：

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) 

編輯：由Cristian Ciupitu建議更改 請忽略縮進錯誤，它們可能在我複製我的代碼時出現。

的程序是這樣的：

• 就拿輸入，並將其分割成一個列表
• 對於每一個列表值我將其提供給一個善良取景器。
• 它需要字符串的好處，並將其他所有內容進行比較，當有更高的善意時，它會使得更高的善良性比較。
• 它然後我偏移量文本的該字符串，看看優度較高或較低

我不太清楚是哪裏的問題，第一個測試：LQKP OG CV GKIJV DA VJG BQQ
打印正確的消息：JOIN ME AT在由ZOO

然而接下來的測試：UIJT JT乙TBNQMF MJOF PG UFYU GPS EFDSZQUJOH
給出的垃圾字符串：SGHR HRžRZLOKD KHMD NE SDWS ENQ CDBQXOSHMF
當它應該是：這是一個示例文本的DECRYPTIN摹

我知道我必須：
想盡一切移值
獲取字
返回字符串最高善良「善良」。

我希望我的解釋有意義，因爲我現在很困惑。

Answer 1

我的最終解決方案有效，這要感謝奇妙的克里斯蒂安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)) 

謝謝大家的幫助！

Answer 2

這是我的實施，它工作正常。

你應該打印每個可能的消息的好處，並看看你的程序輸出它的原因。

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，並使用了一個稍微不同的方法。這避免了創建和調用函數：

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，該代碼的很多地方可以（也應該）被壓縮，例如

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

他們離開擴大到「顯示我的工作」的教程練習。