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From: tycchow@math.mit.edu (Timothy Y. Chow)
Subject: Re: A proof of a solvability algorithm for a 15-puzzle, etc...
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In article <3mck0b$207@pooh.cs.ucsb.edu>,
Steve E. Chapel <schapel@pooh.cs.ucsb.edu> wrote:
>Igor Pechtchanski <ipechtch@its.brooklyn.cuny.edu> writes:
>
>>Does anyone know a good (easy to understand) proof......that there are two
>> equivalence classes of permutations for the 15-puzzle
>
>I thought there were TWELVE equivalence classes.

This latter comment sounds like a statement about Rubik's cube, not the
15-puzzle.

To prove that there are at least two equivalence classes, define the
"permutation of a configuration" to be the permutation of the numbers
1 to 15 obtained by reading off the digits of a configuration from
left to right along the top row, then from right to left along the
second row, from left to right along the third row, and from right
to left along the bottom row.  Ignore the space.  Thus the permutation
of the configuration

 4 12  3  9
   11  1 10
 5  2 14  7
13  8  6 15

is 4 12 3 9 10 1 11 5 2 14 7 15 6 8 13.  Then observe that any move
leaves the sign of this permutation unchanged.

Proving that there are no more than two equivalence classes is a little
more tedious; basically you just need to give an algorithm for solving
the puzzle.
-- 
Tim Chow     tycchow@math.mit.edu
Where a calculator on the ENIAC is equipped with 18,000 vacuum tubes and weighs
30 tons, computers in the future may have only 1,000 vacuum tubes and weigh
only 1 1/2 tons.                               ---Popular Mechanics, March 1949
