Comments for Duncan Jauncey's Blog http://www.duncanjauncey.com/blog Pushing electrons Sat, 09 Jul 2011 09:41:31 +0000 hourly 1 http://wordpress.org/?v=3.1.3 Comment on Zephyr #1 by john http://www.duncanjauncey.com/blog/archives/119/comment-page-1#comment-55046 john Sat, 09 Jul 2011 09:41:31 +0000 http://www.duncanjauncey.com/blog/?p=119#comment-55046 Except that the cypher is commutative, so basically you're just making a very long key. It's much the same as if you just applied the vigenere cypher on top of itself, but in this case it's shifted along. So for example to decrypt the first two parts you could use this key: <code>NEILARMSTRONGNEILAGMJBJDNXVWXLRZE</code> Which is attained by inverse decrypting <code>RMSTRONGNEILARM</code> using the 2nd key <code>PARIS</code> (adding rather than subtracting modulo 26). This give you: <code>THECAPITALOFFRANCETHELONGESTRIVERXRPTFEZHWYSHFAJDNZLBBFRWC</code> The same process could be applied to create a key that would decrypt the entire message in one go. So assuming the various keys line up right and don't overlap you could end up with a single key that's the length of the encrypted text (with no repetition). I believe if you were using a one time pad etc then it'd be impossible to decrypt a vigener cypher with a key as long as the message (that was suitably random), as there'd be no repetition. Not 100% sure about that though (never do cryptology alone). Though of course as the message itself contains clues to it's own decryption it's definitely more of a puzzle than a cypher... Except that the cypher is commutative, so basically you’re just making a very long key. It’s much the same as if you just applied the vigenere cypher on top of itself, but in this case it’s shifted along.

So for example to decrypt the first two parts you could use this key:

NEILARMSTRONGNEILAGMJBJDNXVWXLRZE

Which is attained by inverse decrypting RMSTRONGNEILARM using the 2nd key PARIS (adding rather than subtracting modulo 26). This give you:

THECAPITALOFFRANCETHELONGESTRIVERXRPTFEZHWYSHFAJDNZLBBFRWC

The same process could be applied to create a key that would decrypt the entire message in one go.

So assuming the various keys line up right and don’t overlap you could end up with a single key that’s the length of the encrypted text (with no repetition). I believe if you were using a one time pad etc then it’d be impossible to decrypt a vigener cypher with a key as long as the message (that was suitably random), as there’d be no repetition. Not 100% sure about that though (never do cryptology alone).

Though of course as the message itself contains clues to it’s own decryption it’s definitely more of a puzzle than a cypher…

]]> Comment on Zephyr #1 by duncanj http://www.duncanjauncey.com/blog/archives/119/comment-page-1#comment-55045 duncanj Mon, 04 Jul 2011 21:02:36 +0000 http://www.duncanjauncey.com/blog/?p=119#comment-55045 It seems to be that the key doesn't repeat very often over the plaintext question before it starts to repeat over the next block of ciphertext, hence frequency analysis doesn't have much to go on. It would be even stronger if the keys were generally longer. The brute force approach might produce good results though - unless the questions or keys were not normal english words, which they need not be. There must already be a name for this kind of cipher/puzzle - I can't have invented it first. :-) It seems to be that the key doesn’t repeat very often over the plaintext question before it starts to repeat over the next block of ciphertext, hence frequency analysis doesn’t have much to go on. It would be even stronger if the keys were generally longer. The brute force approach might produce good results though – unless the questions or keys were not normal english words, which they need not be.

There must already be a name for this kind of cipher/puzzle – I can’t have invented it first. :-)

]]> Comment on Zephyr #1 by john http://www.duncanjauncey.com/blog/archives/119/comment-page-1#comment-55044 john Mon, 04 Jul 2011 09:28:27 +0000 http://www.duncanjauncey.com/blog/?p=119#comment-55044 Without knowing the answers it'd be fairly tricky I guess, though the fact the answers tend to be words or names would make brute-forcing things a possibility. Plus it looks like the vignere cipher has weaknesses that are well know: http://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher#Cryptanalysis Basically the repeating key is the weakness. Without knowing the answers it’d be fairly tricky I guess, though the fact the answers tend to be words or names would make brute-forcing things a possibility. Plus it looks like the vignere cipher has weaknesses that are well know:

http://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher#Cryptanalysis

Basically the repeating key is the weakness.

]]> Comment on Zephyr #1 by duncanj http://www.duncanjauncey.com/blog/archives/119/comment-page-1#comment-55043 duncanj Sun, 03 Jul 2011 16:40:11 +0000 http://www.duncanjauncey.com/blog/?p=119#comment-55043 Well done. :-) I'll have to think up another one. How hard do you think these would be to crack without knowing the answers? Well done. :-) I’ll have to think up another one. How hard do you think these would be to crack without knowing the answers?

]]> Comment on Zephyr #1 by john http://www.duncanjauncey.com/blog/archives/119/comment-page-1#comment-55042 john Sat, 02 Jul 2011 10:19:24 +0000 http://www.duncanjauncey.com/blog/?p=119#comment-55042 Ah, of course. So in that case: <code> crypt='GLMNAGULTCCSLEEVNEZTNMXQTBNPOTMDVKVXEFVLZPPGULNNLYZRNKGAZP' from itertools import cycle def vigenere(crypt, key): return ''.join([ chr(ord('A') + (ord(l) - ord(k)) % 26) for (l,k) in zip(crypt, cycle(key)) ]) res1 = vigenere(crypt, 'NEILARMSTRONG') # THECAPITALOFFRANCEIHVTGCGVALGIMMJSCGQSPYVHEGDZVUUKMLAGYPZY res2 = vigenere(res1.replace('THECAPITALOFFRANCE', ''), 'PARIS') # THELONGESTRIVERDCPIAAYEZMRDINCFKVDIRYYRG res3 = vigenere(res2.replace('THELONGESTRIVER', ''), 'NILE') # QUEENSNAMEGZVFRBXNSEEQNNT res4 = vigenere(res3.replace('QUEENSNAME', ''), 'ELIZABETH') # CONGRATULATIONS </code> Ah, of course. So in that case:


crypt='GLMNAGULTCCSLEEVNEZTNMXQTBNPOTMDVKVXEFVLZPPGULNNLYZRNKGAZP'

from itertools import cycle

def vigenere(crypt, key):
return ''.join([ chr(ord('A') + (ord(l) - ord(k)) % 26) for (l,k) in zip(crypt, cycle(key)) ])

res1 = vigenere(crypt, 'NEILARMSTRONG') # THECAPITALOFFRANCEIHVTGCGVALGIMMJSCGQSPYVHEGDZVUUKMLAGYPZY

res2 = vigenere(res1.replace('THECAPITALOFFRANCE', ''), 'PARIS') # THELONGESTRIVERDCPIAAYEZMRDINCFKVDIRYYRG

res3 = vigenere(res2.replace('THELONGESTRIVER', ''), 'NILE') # QUEENSNAMEGZVFRBXNSEEQNNT

res4 = vigenere(res3.replace('QUEENSNAME', ''), 'ELIZABETH') # CONGRATULATIONS

]]> Comment on Zephyr #1 by duncanj http://www.duncanjauncey.com/blog/archives/119/comment-page-1#comment-55041 duncanj Fri, 01 Jul 2011 19:21:58 +0000 http://www.duncanjauncey.com/blog/?p=119#comment-55041 Very good, but you should chop off the question part before decrypting the remaining code. Very good, but you should chop off the question part before decrypting the remaining code.

]]> Comment on Zephyr #1 by john http://www.duncanjauncey.com/blog/archives/119/comment-page-1#comment-55040 john Fri, 01 Jul 2011 12:18:22 +0000 http://www.duncanjauncey.com/blog/?p=119#comment-55040 Using a "Vigenère cipher": <code> crypt='GLMNAGULTCCSLEEVNEZTNMXQTBNPOTMDVKVXEFVLZPPGULNNLYZRNKGAZP' from itertools import cycle ''.join([ chr(ord('a') + (ord(l) - ord(k)) % 26) for (l,k) in zip(crypt, cycle('NEILARMSTRONG')) ]) </code> Gives me: <code> 'thecapitaloffranceihvtgcgvalgimmjscgqspyvhegdzvuukmlagypzy' </code> But then trying to feed PARIS in as another key, or adding it to existing key doesn't seem to yield much. Using a “Vigenère cipher”:


crypt='GLMNAGULTCCSLEEVNEZTNMXQTBNPOTMDVKVXEFVLZPPGULNNLYZRNKGAZP'
from itertools import cycle
''.join([ chr(ord('a') + (ord(l) - ord(k)) % 26) for (l,k) in zip(crypt, cycle('NEILARMSTRONG')) ])

Gives me:


'thecapitaloffranceihvtgcgvalgimmjscgqspyvhegdzvuukmlagypzy'

But then trying to feed PARIS in as another key, or adding it to existing key doesn’t seem to yield much.

]]> Comment on Zephyr #1 by duncanj http://www.duncanjauncey.com/blog/archives/119/comment-page-1#comment-55039 duncanj Wed, 29 Jun 2011 09:30:40 +0000 http://www.duncanjauncey.com/blog/?p=119#comment-55039 Any takers? Any takers?

]]> Comment on Blog migration & a new ‘retro’ theme by john http://www.duncanjauncey.com/blog/archives/75/comment-page-1#comment-54908 john Thu, 31 Jul 2008 08:30:11 +0000 http://www.duncanjauncey.com/blog/?p=75#comment-54908 Good to see you've upgraded and we can leave comments again! Good to see you’ve upgraded and we can leave comments again!

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