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When using mod (%), you don't need to check if e.g.

You don't need to manually increment the counter in a for loop (and if you wanted to, i + 1 without assignment effectively does nothing, you need i += 1).

Your while loop will never start if the key is longer than the plaintext, and never end otherwise, as you never shorten text1 (and doing so would break your indexing).

#Memperkuat sinyal modem telkomsel flash code#

You code currently has the following problems (I am assuming that var = 0, encr =, encr1 =, key1 = list(key) and text1 = list(text) happen before the code you have posted): This is the function that is making the encryption: a = len(key) b = len(text) while (len(key1) 25): encr.append(num3%25) else: encr.append(num3) i + 1 for i in range(0,a): encr1.append(chr(encr+97)) i + 1 for i in range(0,a): key1.append(encr1) i + 1 var = var + a.

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Here is an example when the keyword is ”Carbondale” with k = 10: Plaintext: SIU_CS-Department_is_the_best Key: CarbondaleUIHAQBBDPTUZ,MUOUCX Ciphertext: UIHAQBBDPTUZ,MUOUCXHTODQTPYUM So, I want to know how to tackle the part of the extra characters ',' '.' If the key length is shorter than the plaintext, the ciphertext of block size k of the previous block is concatenated to the key. Basically, the plaintext and ciphertext will have blocks of size k which is same as the key size. Instead, we will follow a block cipher-based idea. The key is not allowed to be repeated as in the standard Vigenere cipher. It should prompt the user for the encryption key of size k. Your script should read from standard input and write to standard output. In addition to letters, there will be four other characters in the plain text: comma (26), dot (27), dash (28), underscore (29) changing the encryption function to be under mod 30.

Your script should distinguish lowercase and uppercase letters (i.e., the encryption key and the plaintext are allowed to be composed of lowercase and uppercase letters but the ciphertext should be uppercase). This is what I have to do: Write a script in Python that is an implementation of a version of the Vigenere cipher for English text. KeyIndices) cipher = encrypt(plaintext, key, alphabet) print(plaintext). You can use translation to restore the original exception-raising behaviour.įor an approach without using dict s in encryption/decryption, see below. For encrypt(), translation = dict(zip(key, string.ascii_lowercase)) ciphertext = '.join(translation.get(c, c) for c in original) Here, translation.get(c, c) causes unrecognized characters to be passed through. Note that any character that is not ASCII lowercase will cause a ValueError.

Making a dictionary for the translation table would be better.

Whenever you have this pattern: some_list = for dummy in some_iterable: some_list.append(some_function_of(dummy)) that's a candidate for replacement with a list comprehension.Īlpha.index(letter) and its inverse could be a performance issue when processing longer messages. 'abcdefghijklmnopqrstuvwxyz' is just the constant. Import random alpha = 'abcdefghijklmnopqrstuvwxyz' def encrypt(original, key=None): if key is None: l = list(alpha) random.shuffle(l) key = '.join(l) new = for letter in original: new.append(key) return def decrypt(cipher, key=None): if key is not None: new = for letter in cipher: new.append(alpha) return '.join(new) # we'll have an else here for the actual Community Challenge! I'm nowhere near an expert with Python, so if I'm not following coding standards, please guide me there too.Ī simple test might be: e = encrypt('helloworld', None) print(e) print(decrypt(e, e)) and we'd get something along the lines of: helloworld. I'd like to know if what I have is the most efficient, the most well written, and the simplest form of these functions. What we have here is your simple encryption and decryption (with a given key) tool. Keep in mind, this isn't a submission for the challenge, just a utility to help in the later process of completing the challenge. In light of, I've taken a stab at building something related.