Encryption

Project Euler 79: By analysing a user’s login attempts, can you determine the secret numeric passcode?

Project Euler 79: By analysing a user’s login attempts, can you determine the secret numeric passcode?

Before even starting this problem I would like to say that I have solved the major part of Problem 79 of Project Euler by hand. The problem reads

A common security method used for online banking is to ask the user for three random characters from a passcode. For example, if the passcode was 531278, they may ask for the 2nd, 3rd, and 5th characters; the expected reply would be: 317.

The text file, keylog.txt, contains fifty successful login attempts.

Given that the three characters are always asked for in order, analyse the file so as to determine the shortest possible secret passcode of unknown length.

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Posted by Kristian in Project Euler, 20 comments
Project Euler 59: Using a brute force attack, can you decrypt the cipher using XOR encryption?

Project Euler 59: Using a brute force attack, can you decrypt the cipher using XOR encryption?

People who know me would also know how much I have been looking forward to this exercise. Problem 59 of Project Euler is about encryption, and modern days encryption relies heavily on mathematics. I had much fun trying to crack the code of this problem. The problems reads:

Each character on a computer is assigned a unique code and the preferred standard is ASCII (American Standard Code for Information Interchange). For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.

A modern encryption method is to take a text file, convert the bytes to ASCII, then XOR each byte with a given value, taken from a secret key. The advantage with the XOR function is that using the same encryption key on the cipher text, restores the plain text; for example, 65 XOR 42 = 107, then 107 XOR 42 = 65.

For unbreakable encryption, the key is the same length as the plain text message, and the key is made up of random bytes. The user would keep the encrypted message and the encryption key in different locations, and without both “halves”, it is impossible to decrypt the message.

Unfortunately, this method is impractical for most users, so the modified method is to use a password as a key. If the password is shorter than the message, which is likely, the key is repeated cyclically throughout the message. The balance for this method is using a sufficiently long password key for security, but short enough to be memorable.

Your task has been made easy, as the encryption key consists of three lower case characters. Using cipher1.txt (right click and ‘Save Link/Target As…’), a file containing the encrypted ASCII codes, and the knowledge that the plain text must contain common English words, decrypt the message and find the sum of the ASCII values in the original text.

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Posted by Kristian in Project Euler, 16 comments