4
$\begingroup$

I got this puzzle from my grandpa who was developing a cipher system; this is his first creation and his first example. He doesn't remember the plaintext, so I want to crack it, but I suck so I leave it for you guys/girls...

Cipher text: 11001000000101010110010111110

The key: 111000111011001001101010001010101111011000000110010011100000100001010101111100111001111010111111001001011111101

This is a classic cipher text with XOR, my grandpa said to me 5 seconds ago... I am pretty sure that the plaintext was something important because they later used this (improved) version of the cipher and encrypted the location for money but he already found THIS one is harder...

Thanks for trying.

$\endgroup$
  • $\begingroup$ I don't think you can XOR two strings of unequal length...? $\endgroup$ – Joe Z. Dec 31 '14 at 15:30
  • $\begingroup$ It goes but some of the data will not me XOR:ed or you just fill it out with 0000000000000... until it matches the other one $\endgroup$ – tor Dec 31 '14 at 15:47
  • 1
    $\begingroup$ So 1950 preceded the use of ASCII encoding. Binary encoding would likely use 5 bit ITA2 code (also generically called Baudot code). The other possibility is Morse code but it cannot be XOR'd. The above strings are 29 bits and 111 bits which suggests some homemade encoding. If this is true, decoding this cipher is not possible (unless more clues are provided). $\endgroup$ – Len Jan 1 '15 at 0:26
  • $\begingroup$ Oh... Then this question is impossible to solve. $\endgroup$ – tor Jan 1 '15 at 9:54
  • 1
    $\begingroup$ 1. Are leading zeroes truncated on the key or the cipher text? 2. Is it possible that this uses Fieldata encoding? $\endgroup$ – Marty Apr 30 '15 at 15:20
1
$\begingroup$

The advantage with the XOR cipher is that using the same encryption key on the cipher text, restores the plain text.

Converting your cipher text to hexadecimal gives 1902ACBE.

Converting your key to hexadecimal gives 71D935157B032704.

XOR operation of cipher and key gives plain text as 71d9351562018bba.So, in binary, the plain text will be 0111000111011001001101010001010101100010000000011000101110111010.

Now, if you want to verify the cipher text, then you may apply XOR operation between your plain text and key.

XOR operation of plain text and key gives cipher text as 71d935157b032704.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.