Recently in class a computer science related deciphering riddle was shown which I could not solve. For the course itself it is not obligatory to solve it but I am interested in the topic and I wonder if it is actually solveable with the given details of the riddle.

It is about finding out encryption mapping of the following two strings, they both are encrypted according to the same mapping:

String 1:
C : \ 0 0 0 0 0 0 . c p p
Encrypted: b9b5564e04199314fee1c93587

String 2:
D : \ 0 0 0 0 0 0 . c p p
Encrypted: be6dca92656480aa051269aa27

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    $\begingroup$ If string really is "C : \ 0 0 0 0 0 0 . c p p " then encryption is in fact hash function… $\endgroup$ – Jan Ivan Mar 15 '17 at 7:53
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    $\begingroup$ I agree it looks like a hash function. Something like md5 perhaps? $\endgroup$ – dave Mar 16 '17 at 4:26
  • $\begingroup$ Your replies now make this more and more exciting! @JanIvan and dave ,what hints can be found in the strings and encrypted strings, that this ismd5? $\endgroup$ – Bruder Lustig Mar 16 '17 at 9:41
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    $\begingroup$ if string="C : \ 0 0 0 0 0 0 . c p p " then it has to be hash function, proly not md5, but some. Because there is no way to get back from hex numbers to ASCII with same amount of characters. but if string is "C:\000000.cpp" then it is solvable… hard, but solvable $\endgroup$ – Jan Ivan Mar 16 '17 at 10:07
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    $\begingroup$ @JanIvan you can say that it is not simple substitution of whole alphabet to ASCII codes, but IMHO it is not completely correct that it has to be some sort of hash function. There are only 9 characters presented (CD :\0.cp) so one hex number is enough to convert one character. It could also contain some sort of lossless compression. Since both strings has same length it is hard to tell more than the first character influence the rest of the result. $\endgroup$ – Artholl Mar 30 '17 at 12:03

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