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# Blind man's jigsaw puzzle

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### #1 mathmagician

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Posted 05 February 2013 - 07:15 PM

A blind old man loves jigsaw puzzles.  He starts them by picking a piece at random and putting it down. He then picks another piece at random and tries to attach it to each side of the piece he started with (trying all four rotations and flipping it over) and repeats until the piece fits.  Then he picks another piece at random and tries to attach it to the part of the puzzle he already has.  He continues doing this until the puzzle is solved.  What is the expected number of pieces he will try in order to solve an m by n puzzle?

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### #2 phaze

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Posted 06 February 2013 - 09:08 PM

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Perfecting Mafia suicide since August 2008

### #3 mathmagician

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Posted 08 February 2013 - 01:22 AM

Technically, you're right.  What i mean to ask though is how many times will he pick up a piece.

Note: When a piece does not fit, he puts it back into the pile and may pick it up again immediately.  If so, it still counts as picking up another piece.

The solution I found is recursive.  I have closed forms for m=1 and m=2.  I'm still working on generalizing it.

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### #4 phil1882

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Posted 09 February 2013 - 10:43 PM

Spoiler for so it seems to me

Edited by phil1882, 09 February 2013 - 10:44 PM.

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### #5 Prime

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Posted 10 February 2013 - 01:40 AM

The solution would be straightforward, but...

Spoiler for complications

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Past prime, actually.

### #6 Prime

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Posted 10 February 2013 - 03:31 AM

Actually, it's more complicated. Like Phil1882 noted.

Spoiler for more complicated

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Past prime, actually.

### #7 superprismatic

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Posted 11 February 2013 - 02:41 AM

Spoiler for my simulation results

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### #8 mathmagician

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Posted 22 February 2013 - 04:56 PM

Below are closed-form expressions for 1xn and 2xn cases

Spoiler for Some simple cases

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