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Ancient puzzle, solved by abstract algebra?

  1. Jan 22, 2006 #1
    i reackon youv'e seen it already, the problem is to rearrange the next numbers in the fixed order:
    1 2 3 4
    5 6 7 8
    9 11 10
    when you have at the last entry a vacant place you need to put it in order.
    this is from the text of edwin h. connell, and i think it's impossible (after a lot of trial & error on my behalf), now my question is how do you prove/disprove my assertion?
     
  2. jcsd
  3. Jan 23, 2006 #2

    0rthodontist

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    It's impossible. I've seen it before but I couldn't remember how to prove it until I looked here
    http://members.tripod.com/~dogschool/permutation.html

    If on each move you could make a swap between _any_ two numbers (or between a number and the space) then to swap the 11 and the 10 while leaving the other pieces the same requires an odd number of swaps. But the way the game is set up, the empty space can only be at the bottom right if you have made an even number of swaps.
     
  4. Jan 24, 2006 #3
    oh, this old dogschool site is indeed great, this was my first insight into group theory.
     
  5. Jan 29, 2006 #4
    i did not understand the problem until i quoted it. it is poorly written...the 9 11 and 10 appear to be as long as the first columns. Okay...everyone says it is impossible but...

    5 6 7 8
    9 11 10

    9 6 11 10

    5 7 11 8
    9 6 10

    5 7 10 11
    9 6 8

    1 2 3 4
    5 6 7 10
    9 8 11

    almoust...almoust...

    5 6 10 11
    9 8 7

    5 6 11
    9 8 10 7

    1 2 3 4
    5 6 11
    9 8 10 7

    1 2 3 4
    5 6 7 10
    9 8 11

    once again...close

    5 8 6 10
    9 7 11

    5 7 8 10
    9 6 11

    that 10 just wont go down!!!!!!!!

    i give up. I'll let you with this lengthy post maybe someone uses a Rhiemann Space to teleport it into place.
     
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