You're absolutely correct - the cipher text must be long enough and in a recognizable langauge, because my code looks for repeated substrings at some interval, and compares the letter fequency distribution of every Nth letter to the standard letter frequency distribution for a given langauge (English in my case ;). In fact it looks for multiple repeating substrings to better try to zero in on the likely key length. Once repeated substrings are found, it computes the greatest common factor of the intervals for a given repeat, and then computes the greatest common factor for those greatest comman factors, and uses that as the key length.

That doesn't always work :)

I've had mixed results - if I rename to just "chargrill" and encode the same text, the above code can't find any common factors, and then fails. It would actually take a fair amount of refactoring to account for this.

Honestly, I think my algorithm for figuring out what key letter would make an encrypted cipher string of every Nth character matches the standard letter frequency distribution is the most ... interesting. I couldn't think of a better way to do it than with the human eye, and I think my algorithm gets as close to a visual inspection of peaks and valleys of letter frequency charts as I could think of...

$,=42;for(34,0,-3,9,-11,11,-17,7,-5){$*.=pack'c'=>$,+=$_}for(reverse s +plit//=>$* ){$%++?$ %%2?push@C,$_,$":push@c,$_,$":(push@C,$_,$")&&push@c,$"}$C[$# +C]=$/;($#C >$#c)?($ c=\@C)&&($ C=\@c):($ c=\@c)&&($C=\@C);$%=$|;for(@$c){print$_^ +$$C[$%++]}

In reply to Re^2: Breaking the indecipherable cipher, courtesy Charles Babbage. by chargrill
in thread Breaking the indecipherable cipher, courtesy Charles Babbage. by chargrill

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