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Thread: Sliding jigsaw configurations

  1. #1
    May 2009

    Sliding jigsaw configurations

    Consider an $\displaystyle N\times N$ sliding jigsaw (SJ) with an empty block and $\displaystyle {N}^{2}-1$ numbered blocks from $\displaystyle 1$ to $\displaystyle {N}^{2}-1$ (see the attachment for a configuration of a $\displaystyle 4 \times 4$ SJ, where blocks $\displaystyle 8,5,2$ or $\displaystyle 9$ can move to the empty block, leaving their blocks empty, which any of their adjacent neighbors can in turn move to, ect.)

    Definition 1: A configuration is standard, if the empty block is at the northwest corner of the board.

    Definition 2: A configuration is ordered, if every block is greater than the block to its left (the first block of a row is greater than the last block of the previous row), ignoring the empty block.

    Question 1: Starting from a standard configuration of an $\displaystyle N\times N$ SJ, how many different standard configurations can we reach?

    Question 2: Starting from an arbitary configuration, can we reach an ordered SJ?
    Attached Thumbnails Attached Thumbnails Sliding jigsaw configurations-jigsaw.jpg  
    Last edited by godelproof; Jul 26th 2011 at 04:38 AM. Reason: attachment added
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK

    Re: Sliding jigsaw configurations

    There is a very clear analysis of the 4x4 case here, with enough information to give you some good hints on tackling the NxN case.
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  3. #3
    May 2009

    Re: Sliding jigsaw configurations

    Thanks. So answer is:

    Q1: $\displaystyle ({N}^{2}-1)!/2$

    Q2: Yes.

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