Comments on: Magic Caesar autodecipher

By: Raskolnikov

Raskolnikov — Sat, 05 Jun 2010 21:25:22 +0000

todo este lío e incluso hasta poner una mini-entrevista realizada a Moot (ganador de la encuesta). La anotación titulada “Moot wins, Time Inc. loses” [en inglés] (bastante entretenida por cierto) hace énfasis en la traba que ponen en estos casos

By: narain

narain — Mon, 14 May 2007 08:19:51 +0000

I played around with your code a bit more, and I think your letter frequencies are off; not to mention, there are 27 of them.

Pseudonym: Good idea. With that you could probably decipher arbitrary substitution ciphers as well.

By: Andy

Andy — Mon, 14 May 2007 07:15:17 +0000

Whoops, here’s their code: http://www.cs.nott.ac.uk/~gmh/cipher.lhs

By: Andy

Andy — Mon, 14 May 2007 07:12:48 +0000

There’s a slightly more verbose version of this in Hutton’s, “Programming in Haskell”. They use the chi-squared method as narain suggested.

By: Pseudonym

Pseudonym — Mon, 14 May 2007 07:08:51 +0000

It’s even better if you use a more accurate model of English which takes into account inter-letter frequencies.

By: narain

narain — Mon, 14 May 2007 03:23:01 +0000

Forgot to mention, `sqr x = x*x`.

By: narain

narain — Mon, 14 May 2007 03:22:19 +0000

Using the chi-square seems to give better results. In particular, it gets “K DKVO DYVN LI KX SNSYD, PEVV YP CYEXN KXN PEBI, CSQXSPISXQ XYDRSXQ” right.

And all you have to do is replace (*) with (\o e -> sqr (o/(fromIntegral $ length w) – e) / e) in the definition of `score`.

By: Pseudonym

Pseudonym — Mon, 14 May 2007 00:04:51 +0000

echo “VS LBHGU, GUEBHTUBHG NYY UVFGBEL, UNQ UNQ N PUNZCVBA GB FGNAQ HC SBE VG; GB FUBJ N QBHOGVAT JBEYQ GUNG N PUVYQ PNA GUVAX; NAQ, CBFFVOYL, QB VG CENPGVPNYYL; LBH JBHYQA’G PBAFGNAGYL EHA NPEBFF SBYXF GBQNL JUB PYNVZ GUNG \”N PUVYQ QBA’G XABJ NALGUVAT.\”" | ./magic

On cases like this, you might as well do it by hand!