Timeline for A combinatorial problem - counting the solutions

5 events

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Mar 10, 2014 at 18:11	comment	added	The Masked Avenger		Indeed, but tighter bounds arise from coloring interior edges and points.
Mar 10, 2014 at 17:57	comment	added	Roland Bacher		Another approach for an upper bound is by a transfer-matrix method: Dropping the condition that all squares are different, one gets an upper bound $2^{24}$ for the baby case and $2^{930}$ for the hard case.
Mar 8, 2014 at 10:15	comment	added	Alda		You can also have only two diagonals meeting at a corner - look at the example solution I added. My estimation went as follows: there are exactly 128 horizontal connections: pairs of tiles joined by a horizontal line. There are 240 places for such a connection, so $\binom {240} {128}$ possibilities. Same for verticals. For diagonals, again 128 for each of the two orientations, with 225 possible positions, $\binom {225} {128}$ possibilities. Multiply them all together and you get $3.29\times 10^{272}$.
Mar 7, 2014 at 17:51	comment	added	The Masked Avenger		The corner analysis assumes that a correct tiling induces a set of 128 vertices with four red corner lines emanating from each vertex. The rules do not say whether other diagonal line configurations are allowed.
Mar 7, 2014 at 16:41	history	answered	The Masked Avenger	CC BY-SA 3.0