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Subject: "Fifteen puzzle and 9-puzzle variant"     Previous Topic | Next Topic
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Gary Bradshaw
Jun-27-05, 01:00 PM (EST)
"Fifteen puzzle and 9-puzzle variant"
   I'm having trouble determining whether the 3x3 version of Sam Loyd's fifteen puzzle has the same even/odd parity rule that the 4x4 version does, even though the page implies that it'should and experience says that it does.

The argument that puzzles have parity is based on the notion that a row shift does not change the number of inversions (and so does not change parity) and that a column-shift is also parity neutral. However, consider the case where the original board:

1 2 3
4 5 6
7 8

is replaced by board

1 2 3
4 5
7 8 6

By my understanding of the inversion rule, there are no inversions in the first board and two inversions in the second. The first board has the blank square in row 3, so it has a parity of one. But the second board has a blank square in row 2 and 2 inversions, for a total parity count of 4. 4%2 = 0, so this one move has apparently changed the parity
of the board.

Where am I going wrong?

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Charter Member
1564 posts
Jun-27-05, 01:54 PM (EST)
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1. "RE: Fifteen puzzle and 9-puzzle variant"
In response to message #0
   >Where am I going wrong?

The 3×3 puzzle is only mentioned in the form of a question: Will the proposition above apply to the 3×3 puzzle?

The answer is No, it will not.

For a 3×3 puzzle, one does not have to add the row number. The reason is that, when the rows are stringed up in a single file, a vertical move causes a counter to jump over an odd number of counters for the 4×4 puzzle and an even number of counters for the 3×3 puzzle. In the latter case, the parity of the number of inversions is preserved automatically. In the original puzle it does not.

I added a remark about that to the page. Thank you for asking.

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