4
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What in the name of the USSR does this mean?!

59 103 106 138 -1 106 138 -1 81 -1 135 81 141 103 94 135 -1 106 122 141 94 135 94 138 141 106 122 100 -1 87 125 91 94 -2 -1 91 125 122 -3 141 -1 157 125 144 -1 141 103 106 122 113 -4

It seems so easy, doesn't it? Just another Caesar Cipher, right?

Hints:

$-n$ denotes some sort of punctuation mark.
$+n$ denotes a letter.
It uses a mathematical constant for encrypting.
It also uses a non-injective and surjective function for encrypting.
All characters are necessary for the answer (meaning, there are no nulls in it).

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  • $\begingroup$ Why did you use \idots when you can simply use an ellipse as 1 character (…)? $\endgroup$ – warspyking Dec 17 '14 at 2:19
  • $\begingroup$ @warspyking ^^" I forgot about those... $\endgroup$ – Conor O'Brien Dec 17 '14 at 3:22
5
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The solution:

This is a rather interesting code, don't you think?

How I solved it:

It should be obvious that -1 is a space and that -3 is an apostrophe. I replaced the spaces between numbers with exclamation marks, to get this:

59!103!106!138! !106!138! !81! !135!81!141!103!94!135! !106!122!141!94!135!94!138!141!106!122!100! !87!125!91!94!-2! !91!125!122!'!141! !157!125!144! !141!103!106!122!113!-4
I then mapped all the numbers to a different letter, and replaced all the unknown punctuation with an asterisk, producing this:
abcd cd e fegbhf cighfhdgcij klmh* mli'g nlo gbcip*
Some substitution analysis later, we have our solution.

This was clearly not the intended solution: I suspect the actual method of encipherment revolved around the digits of pi, as the punctuation is mapped to -3, -1 and -4, the first 3 digits of pi, and some suspicious pi sequences appear in the ciphered text (141, for example). But I guess this just goes to show that no matter how clever you are in your method of encipherment, a substitution is just a substitution.

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  • $\begingroup$ Spot on, though your means, as you predicted, were different than my own. I used an encoding function that took every letter, multiplied its position in an array [A,B...Y,Z,a,b...y,z] by $\pi$, then floored the value. The punctuation came back as a negative. $\endgroup$ – Conor O'Brien Dec 17 '14 at 0:26
  • $\begingroup$ How on earth did you change block text through spoiler? $\endgroup$ – warspyking Dec 17 '14 at 2:21
  • 2
    $\begingroup$ @warspyking I used HTML pre tags. $\endgroup$ – Tryth Dec 17 '14 at 5:46

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