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The New Coffee Room

  1. TNCR
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  3. Puzzle time - binaries

Puzzle time - binaries

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  • KlausK Offline
    KlausK Offline
    Klaus
    wrote on last edited by Klaus
    #1

    Consider a string of binary numbers s, such as s = 1011.

    Now consider a machine that works on s. It performs the following three steps:

    1. Append 1 to the right end of s.
    2. Add the new s from step 1 to the old s with binary addition
    3. Remove all trailing 0s.

    The machine iterates these steps and stops only when s = 1.

    For instance, for s = 1011 we have s = 10111 after step 1, s = 10111+1011 = 100010 after step 2, s = 10001 after step 3.

    After the next iteration we get s = 01101, then 0101, then 1. The machine stops.

    Can you find an s such that the machine doesn't stop? Or conversely, can you make an argument why the machine will always stop for any s?

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    • jon-nycJ Offline
      jon-nycJ Offline
      jon-nyc
      wrote on last edited by jon-nyc
      #2

      I think it will always Collatz to 1 in the end. But that's just a conjecture.

      Only non-witches get due process.

      • Cotton Mather, Salem Massachusetts, 1692
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      • jon-nycJ Offline
        jon-nycJ Offline
        jon-nyc
        wrote on last edited by
        #3

        Kind of a dick move, by the way

        Only non-witches get due process.

        • Cotton Mather, Salem Massachusetts, 1692
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        • KlausK Offline
          KlausK Offline
          Klaus
          wrote on last edited by
          #4

          Damn, my superficial masquerade of the problem failed.

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