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unsolvable game of free cell

Name: Lindsay Lohan 2007-08-27 9:06 ID:81T2hD1T

" It is believed (although not proven) that every game is winnable."

As Ad Ac Ah
Qs Qd Qc Qh 6s 6d 6c 6h
Ts Td Tc Th 8s 8d 8c 8h
2s 2d 2c 2h 4s 4d 4c 4h
5s 5d 5c 5h 3s 3d 3c 3h
9s 9d 9c 9h 7s 7d 7c 7h
Ks Kd Kc Kh Js Jd Jc Jh

tell me if i'm wrong about this. i'm far too lazy to write out a proof. the general idea is that, throwing one does nothing, throwing 2 does nothing, throwing 3 does nothing, if you take a good look at it. therefore you must throw 4. this can be done wrt columns as 4-0, 1-1-1-1, 3-1, or 2-2, or 2-1-1. 2-2 and 1-1-1-1 and 2-1-1 obviously do nothing. 4-0 i.e. 4 from the same column does nothing if you inspect it for a bit. 3-1 is the trickiest, it can be done in two different ways, by the symetry, 3-1 on the same side, or 3-1 on different sides. inspecting both leads to 2 dead ends.

Name: Anonymous 2007-08-27 13:44 ID:yUFCetOa

Welcome to 1996
http://people.cs.uu.nl/hansb/d.freecell/freecellhtml.html

But yes, if I am reading your (up-side down) notation right, FreeCell Pro confirms that game to be impossible. Well done.

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