2004-9-4
crypto night
"So I was sitting there on prom night, in my goddamn rented tuxedo, and the whole night played out in my mind, and I knew right then, for the first time, that I really wanted to kill somebody." -Martin Blank
I think I may devote myself to contributing to fortune. The unix program. I would still like to make a way to put a daily thought or fortune in a little place on my blog. I get so much joy out of these small subversive jokes...
So anyways, I get Dr. Dobbs' (Dobb's?) Journal, and this month there is a stiff cardboard insert that looks just like the old '70s middle school standardized tests, with the green print on white paper, the bubbles that you are supposed to fill in, etcetera. It's the GLAT- Google Labs Aptitude Test. They are very aggressively recruiting these days in the developer magazines and on pages like SourceForge. Anyways, it's a mind-bending but short questionnaire, that folds into an envelope to mail back to Google. I answered some of the more off-the-cuff type personality questions, and pondered over the technical ones. Of course one of them grabbed me. It's an alphabetic cipher or cryptosystem question. At least that's how it looks to me... Anyhow, it's:
WWWDOT - GOOGLE = DOTCOM
The only rules are T-E=M, which gives you the pattern they are looking at, and "no leading zeros". I don't get the second part, it makes me thing I'm barking completely up the wrong tree with my solution. I'm looking at a kind of decoder wheel with six wheels, the alphabet being on each. Well, I haven't made one, I've just been scribbling. I stayed up til the wee hours on freenode #perl bouncing ideas around. So far my solutions haven't been panning out. Sometimes they almost do. But the basic idea is:
It's a six column equation:
WWWDOT
GOOGLE
DOTCOM
In each column, each letter of the alphabet is attached to a number in numeric and alphabetic order. The problem is, you don't know which letter is assigned #1. So if in the 5th column for example, M is assigned #1, then N=2, O=3, etcetera. I use this example because O-L=O. That means that the difference from O all the way back around to O is L. That means that L=26, since there are 26 characters in the alphabet. But only in the 5th column apparently, because in the 2nd and 3rd columns, W-O=O and W-O=T, respectively. That means that the system skips so many numbers per column, such as 5. Why five? The difference between O and T is five, so in the 2nd column W-O=O, and in the third column the letter with the same value as W-O in the 2nd column is five characters farther down. So could we say that whatever letter = 1 in column 1 will = 21 in the next column? Checking that answer doesn't play out. So maybe each column skips ahead a different number? So between the 2nd and 3rd columns it jumps ahead 5, maybe between 3 and 4 it jumps ahead 6, or 7, or 20...
I found a theory that supports jumping ahead 6 from column 3 to 4. But it doesn't play out across all six columns.
I'm still sticking with the difference idea, and assuming for now that the "leading zeros" thing is for people coming up with purely mathematical answers.
I am enjoying my first foray into alphabetic ciphers :) The only thing I know about them I learned from Neil Stephenson's "The Cryptonomicon", and of that I mostly remember the system where Turing's bicycle was used for a cryptosystem concept, maybe it evolved into Enigma, I don't remember. But basically he had a bent wheel, and a busted chain link. So every so many revolutions there was a such-and-such chance that the tire would hit the bad link. That was the measure of how secure the cryptosystem would be...
Anyways, as I play out these hypotheses with six separate alphabets revolving in my mind, I might try to figure out a way to represent it in Perl or Python to make it easier to keep track. Or harder...
I think I may devote myself to contributing to fortune. The unix program. I would still like to make a way to put a daily thought or fortune in a little place on my blog. I get so much joy out of these small subversive jokes...
So anyways, I get Dr. Dobbs' (Dobb's?) Journal, and this month there is a stiff cardboard insert that looks just like the old '70s middle school standardized tests, with the green print on white paper, the bubbles that you are supposed to fill in, etcetera. It's the GLAT- Google Labs Aptitude Test. They are very aggressively recruiting these days in the developer magazines and on pages like SourceForge. Anyways, it's a mind-bending but short questionnaire, that folds into an envelope to mail back to Google. I answered some of the more off-the-cuff type personality questions, and pondered over the technical ones. Of course one of them grabbed me. It's an alphabetic cipher or cryptosystem question. At least that's how it looks to me... Anyhow, it's:
WWWDOT - GOOGLE = DOTCOM
The only rules are T-E=M, which gives you the pattern they are looking at, and "no leading zeros". I don't get the second part, it makes me thing I'm barking completely up the wrong tree with my solution. I'm looking at a kind of decoder wheel with six wheels, the alphabet being on each. Well, I haven't made one, I've just been scribbling. I stayed up til the wee hours on freenode #perl bouncing ideas around. So far my solutions haven't been panning out. Sometimes they almost do. But the basic idea is:
It's a six column equation:
WWWDOT
DOTCOM
In each column, each letter of the alphabet is attached to a number in numeric and alphabetic order. The problem is, you don't know which letter is assigned #1. So if in the 5th column for example, M is assigned #1, then N=2, O=3, etcetera. I use this example because O-L=O. That means that the difference from O all the way back around to O is L. That means that L=26, since there are 26 characters in the alphabet. But only in the 5th column apparently, because in the 2nd and 3rd columns, W-O=O and W-O=T, respectively. That means that the system skips so many numbers per column, such as 5. Why five? The difference between O and T is five, so in the 2nd column W-O=O, and in the third column the letter with the same value as W-O in the 2nd column is five characters farther down. So could we say that whatever letter = 1 in column 1 will = 21 in the next column? Checking that answer doesn't play out. So maybe each column skips ahead a different number? So between the 2nd and 3rd columns it jumps ahead 5, maybe between 3 and 4 it jumps ahead 6, or 7, or 20...
I found a theory that supports jumping ahead 6 from column 3 to 4. But it doesn't play out across all six columns.
I'm still sticking with the difference idea, and assuming for now that the "leading zeros" thing is for people coming up with purely mathematical answers.
I am enjoying my first foray into alphabetic ciphers :) The only thing I know about them I learned from Neil Stephenson's "The Cryptonomicon", and of that I mostly remember the system where Turing's bicycle was used for a cryptosystem concept, maybe it evolved into Enigma, I don't remember. But basically he had a bent wheel, and a busted chain link. So every so many revolutions there was a such-and-such chance that the tire would hit the bad link. That was the measure of how secure the cryptosystem would be...
Anyways, as I play out these hypotheses with six separate alphabets revolving in my mind, I might try to figure out a way to represent it in Perl or Python to make it easier to keep track. Or harder...
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