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I was staring at my crossword this morning when I had this question: How many possible arrangement of grid squares are there in an $N\times N$ crossword, provided we impose certain constraints? I take the following standard constraints present in leading dailies.

A) Only four-letter words or more are permissible. B) Every alternate letter of every word must be check-able, i.e., must have an intersecting word. For example, for a five-letter word, letters $1$, $3$ and $5$ must intersect with other words, or letters $2$ and $4$ must do so.

I took up the example of $N=4$. I obtained 4 possible grid-fills. If $A=[a_{ij}]$ is the grid matrix, I shaded the following sets of elements in the $4$ solutions:

1) $a_{22},\ a_{24},\ a_{42},\ a_{44}$
2) $a_{11},\ a_{13},\ a_{31},\ a_{33}$
3) $a_{21},\ a_{23},\ a_{41},\ a_{43}$
4) $a_{12},\ a_{14},\ a_{32},\ a_{34}$

Is it possible to apply combinatoric theory to derive a general solution for arbitrary $N$?

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    The British crosswords I've seen have the across words in odd rows, the down words in odd columns, and each row and column has at least one word. This seems more restrictive, but probably easier to solve.2015-01-23

0 Answers 0