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Binary Question

Posted by Karl Sharman on November 6, 2010 – 3:13 am

Heres a simple bit of binary code…. translate back into english, and answer the question…

Unsurprisingly, I’m giving no hints at this stage, and I’ve removed spaces and punctuation, except the question mark.

110011101001000100110011100110011110110
101100101100110010011101101001111010111
001010010000101011101011010110010110011
010010001111101011111000101001001101001
011111111010111001001000101110010000100
111010011100111011101100110011100111011
000011001011000110101101100111010010011
111111010111100011010010011001111111110
101100001100101011001111111110101?

Oh, alright, z=11010… is that enough help?

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8 Responds so far- Add one»

  1. 1. Spencer Said:

    There is no translation; not divisible by 8. Rethink your problem and use punctuation.

  2. 2. Random Guy Said:

    I saw this somewhere before… I can put my finger on it but I think its about a zero playing against a one…

  3. 3. Mohamed Said:

    I think that *z=bin(26)=11010* hint solves it in one way or another. For example, one can come up with a codebook where each character in the alphabet corresponds to the binary representation of its index. Whether the code words are padded or not, am not quite sure..both fits the given binary block (length is divisible by 5). Anyway, let’s go with the unpadded scenario as it’s more challenging to decode.

    So, here goes. We’ve a codebook of the alphabet with unpadded (variable length) non prefix-free code, as follows.

    1 = a
    10:11 = b:c
    100:111 = d:g
    1000:1111 = h:o
    10000:11010 = p:z

    Obviously, the extension of such code is not uniquely
    decodable, e.g., [101] can be seen as a sequence of the codewords [10-1] which corresponds to [b-a], or as the codeword [101] which corresponds to [e]. AFAIK, this can only be resolved by following some brute force techniques, or by knowing the statistics of the plain source.

    I tried a quick manual dictionary attack on the provided cipher, and even though the output seems garbaged, i think we can infer the question easily (provided that there is only one question in the statement ). Follows are my humble dictionary setup and the deciphered portions of the sequence.

    Dictionary:
    ————-
    [why]=[10111100011001]
    [would]=[101111111101011100100]
    [where]=[10111100010110010101]
    [what]=[101111000110100]
    [is]=[100110011]
    [it]=[100110100]
    [th]=[101001000]
    [so]=[100111111]
    [ne]=[1110101]

    Result:
    ——
    110011
    th
    is
    is
    110110101100101100110010011101101001
    ne
    1100
    th
    01010
    ne
    10101100101
    it
    100011
    ne
    111100010100
    it
    would
    1000101110010000100111010011100111011101
    is
    1001110110000110010110001101011011001110100
    so
    110
    what
    is
    1111
    ne
    10000110010101
    so
    ne?
    From the line marked by the and to the bottom of the listing, we can easily guess that the question is on the form [what is [1111]ne[10000110010101]s one?]. Following our codebook, [1111] maps to [o], and [10000-1100-10101] maps to [p-l-u]. So, our prey is [what is one plus one?], and the answer is 2. Phew!

    Nice one, Karl! Keep up!

  4. 4. Karys Said:

    @Spencer Actually you’re wrong, there is no reason why those should be octets. Just look at the hint which was given, z=11010, one letter is 5 digits.

    However the code
    1=A
    2=B

    26=Z

    doesn’t work.

  5. 5. What? Said:

    WHAT THE HECK? I HAVE NO IDEA WHAT THIS EVEN MEANS. NOT BY THE SLIGHTEST. :P

  6. 6. Karl Sharman Said:

    The full translation is:

    “As this is a very difficult puzzle, I thought it would help if I asked a simple question: What is one plus one?”

    And it would appear that Mohamed gets the question right. Being picky, I’m going to say the answer is 10…..

    @Karys – the code a-1-1, b-2-10, c-3-11… z-26-11010 is the base code used for this puzzle.

  7. 7. Karys Said:

    oh. My mistake :P

  8. 8. Spencer Said:

    Thanks Karys

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