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Yet another logic puzzle

Posted By: Stein Kulseth
Date: Monday, 24 August 2009, at 10:06 a.m.

In Response To: Yet another logic puzzle (Steve Mellen)

Your version of stating the problem is equal to the original and a little less obfuscated.

And yes the game will always end, and the reasoning behind is recursive, a bit similar to Bob's blue-eyed and green-eyed persons puzzle.

In the given example where A chooses a=10, B chooses b=20, and C states that a+b=30 or a+b=32, A can deduce that:

  • b is either 20 or 22 -let's denote this by b[20,22]
  • B can deduce that a[10,12] or a[8,10]
  • B cdt A cdt b[22,24], b[20,22] or b[18,20]
  • etc. etc

    The recursion stops at 0 and 32 which are impossible candidates for either a and b. That is, an imagined number 30 is impossible to start with, as this would have allowed an imagined immediate deduction of the other's number. Thus after the first announcement, an imagined 2 is also impossible, as then the other number could have been deduced to be 28. Etc. etc. Each non-deduction will collapse the tree of imagined deductions one step, until at last 1 person will know the other persons number. It is possible that the other will be left in the dark though.

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