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Topic: Xword Clues (Read 493 times) |
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Noke Lieu
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Xword Clues
« on: Jan 21st, 2010, 6:30pm » |
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So, a crossword puzzle (following the usual rules of layout), that's n x n squares...
what can you say about the number of clues?
What if the cross word was (n x m) ?
(sorry, a slow start to the year...  )
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Ronno
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Re: Xword Clues
« Reply #1 on: Jan 21st, 2010, 7:58pm » |
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The minimum number of clues is probably Floor[(n+1)/2]*Floor[(m+1)/2]
The maximum depends on certain constraints, such as:
1. How short words are allowed?
2. Can the crossword be divided into separate sub-crosswords?
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Noke Lieu
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Re: Xword Clues
« Reply #2 on: Jan 23rd, 2010, 7:28pm » |
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Nice, but I was thinking of something much more pedestrian... (and now filled with a sense of uncertainty thinking through 2x2 a bit more openly...)
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Noke Lieu
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Re: Xword Clues
« Reply #3 on: Jan 24th, 2010, 10:55pm » |
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I was originally thinking that the crossword requires 4 fold rotational symmetry.
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towr
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Re: Xword Clues
« Reply #4 on: Jan 25th, 2010, 1:43am » |
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I've seen enough crosswords that aren't in any way symmetric.
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Grimbal
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Re: Xword Clues
« Reply #5 on: Jan 25th, 2010, 8:00am » |
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Around here (Geneva, french-speaking) they are never symmetric.
It seems to me UK or US crosswords tend to be symmetric. I guess building crosswords is more difficult in some languages, making it too troublesome to make them symmetric.
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rmsgrey
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Re: Xword Clues
« Reply #6 on: Jan 26th, 2010, 4:30am » |
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As long as both n and m are greater than 1, there are at least two clues.
Assuming answers must be at least 3 letters, and that black squares separate answers, there's an upper bound of (m+1)n/4+m(n+1)/4 - if you add an extra column of black squares to the right of the grid, each across answer uses at least 4 squares - 3 letters and a black square - and similarly for down answers.
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