A deciphering riddle












3












$begingroup$


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






cipher cryptography computer-science







share|improve this question











$endgroup$








  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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:00000.cpp" then it is solvable… hard, but solvable
    $endgroup$
    – Jan Ivan
    Mar 16 '17 at 10:07








  • 1




    $begingroup$
    @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 :.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
















3












$begingroup$


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






cipher cryptography computer-science







share|improve this question











$endgroup$








  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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:00000.cpp" then it is solvable… hard, but solvable
    $endgroup$
    – Jan Ivan
    Mar 16 '17 at 10:07








  • 1




    $begingroup$
    @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 :.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














3












3








3


1



$begingroup$


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






cipher cryptography computer-science







share|improve this question











$endgroup$




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






cipher cryptography computer-science




cipher cryptography computer-science






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 14 '17 at 16:19







Bruder Lustig

















asked Mar 14 '17 at 16:14









Bruder LustigBruder Lustig

1194




1194








  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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:00000.cpp" then it is solvable… hard, but solvable
    $endgroup$
    – Jan Ivan
    Mar 16 '17 at 10:07








  • 1




    $begingroup$
    @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 :.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














  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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






  • 1




    $begingroup$
    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:00000.cpp" then it is solvable… hard, but solvable
    $endgroup$
    – Jan Ivan
    Mar 16 '17 at 10:07








  • 1




    $begingroup$
    @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 :.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








1




1




$begingroup$
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




$begingroup$
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




1




1




$begingroup$
I agree it looks like a hash function. Something like md5 perhaps?
$endgroup$
– dave
Mar 16 '17 at 4:26




$begingroup$
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




$begingroup$
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




1




1




$begingroup$
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:00000.cpp" then it is solvable… hard, but solvable
$endgroup$
– Jan Ivan
Mar 16 '17 at 10:07






$begingroup$
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:00000.cpp" then it is solvable… hard, but solvable
$endgroup$
– Jan Ivan
Mar 16 '17 at 10:07






1




1




$begingroup$
@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 :.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




$begingroup$
@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 :.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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$begingroup$

Covered in a liquid
Thicker than water
Covered in people
That no longer move
Riddled with metal
That once sang a song of death
Horrifying to most
Glorious to some
What am I?






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$begingroup$

Covered in a liquid
Thicker than water
Covered in people
That no longer move
Riddled with metal
That once sang a song of death
Horrifying to most
Glorious to some
What am I?






share|improve this answer








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$begingroup$

Covered in a liquid
Thicker than water
Covered in people
That no longer move
Riddled with metal
That once sang a song of death
Horrifying to most
Glorious to some
What am I?






share|improve this answer








New contributor




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Covered in a liquid
Thicker than water
Covered in people
That no longer move
Riddled with metal
That once sang a song of death
Horrifying to most
Glorious to some
What am I?







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