Shoaib - 1 year ago 50

Python Question

`while 1 == 1 :`

print("")

option = input("Do you want to encrypt or decrypt? Write 'E' for encrypt or 'D' for decrypt! ").upper()

if option == "E":

file = open("message.txt", "r")

plaintext = file.read()

key = input("Key: ").upper()

encrypt = ''

for n in range(0, len(plaintext)):

if plaintext[n] == ' ':

new = ord(plaintext[n])

else:

new = ord(plaintext[n]) + ord(key[n%len(key)]) - 65

if new > 90:

new -= 26

encrypt += chr(new)

ET = (encrypt)

plaintext = ET

key = input("Key: ").upper()

encrypt = ''

for n in range(0, len(plaintext)):

if plaintext[n] == ' ':

new = ord(plaintext[n])

else:

new = ord(plaintext[n]) + ord(key[n%len(key)]) - 65

if new > 90:

new -= 26

encrypt += chr(new)

ET = (encrypt)

print(ET)

file = open("encrypt.txt", "w")

file.write(ET)

file.close()

elif option == "D":

file = open("encrypt.txt", 'r')

plaintext = file.read()

key = input("Key: ").upper()

decrypt = ''

for n in range(0, len(plaintext)):

if plaintext[n] == ' ':

new = ord(plaintext[n])

else:

new = ord(plaintext[n]) - ord(key[n%len(key)]) + 65

if new < 65:

new += 26

decrypt += chr(new)

plaintext = (decrypt)

key = input("Key: ").upper()

decrypt = ''

for n in range(0, len(plaintext)):

if plaintext[n] == ' ':

new = ord(plaintext[n])

else:

new = ord(plaintext[n]) - ord(key[n%len(key)]) + 65

if new < 65:

new += 26

decrypt += chr(new)

DT = decrypt

print (decrypt)

file = open("message.txt", "w")

file.write(DT)

file.close()

else:

print(option)

So i want to know how i can make sure that all punctuation isn't encrypted and only letters are encrypted? I know that this isn't a good way to encrypt but it is for a school project so it would be great if someone can help me. Also when i decrypt it doesn't decrypt properly and sometimes forgets full stops and stuff how can i fix this? Thanks.

Answer

Define all punctuation;

```
punctuation = " ',.;:.!?\r\n"
```

Replace *all* instances of

```
if plaintext[n] == ' ':
```

by

```
if plaintext[n] in punctuation:
```

**Addition**

While your code is functional, it doesn't use a lot of the powerful tools that Python puts at your disposal. Let me illustrate. The encryption/decryption (with punctuation stripped from the text) could be done like this, using list comprehensions;

```
In [42]: plaintext = 'THISISAPLAINTEXT' # Your algorithm only works for capitals.
In [43]: key = 'SPAMEGGS'
In [44]: count = int(len(plaintext)/len(key))+1
In [45]: stretchedkey = [ord(c) for c in key*count]
In [46]: # Encryption
In [47]: plainnum = [ord(c) for c in plaintext]
In [48]: ciphernum = [a+b-65 for a, b in zip(plainnum, stretchedkey)]
In [49]: ciphertext = ''.join([chr(c) if c <= 90 else chr(c-26) for c in ciphernum])
In [50]: ciphertext
Out[50]: 'LWIEMYGHDPIZXKDL'
In [51]: # Decryption
In [52]: ciphernum = [ord(c) for c in ciphertext]
In [53]: decryptnum = [a-b+65 for a, b in zip(ciphernum, stretchedkey)]
In [54]: decrypt = ''.join([chr(c) if c >= 65 else chr(c+26) for c in decryptnum])
In [55]: decrypt
Out[55]: 'THISISAPLAINTEXT'
```