beaker wrote:

So, if you are talking about a 4x4 grid, does the total number of possibilities change when you "add" in the 4 different procedures??........not a Mathematician, but Iwouldn't think so??

If you have 1 and 4 in a two-cell cage, the operation could be 5+, 3-, 4x, or 4/. The grid, or Latin square wouldn't change, but it would be a different puzzle.

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http://www.maths.qmul.ac.uk/~pjc/comb/ch6s.pdf

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Next is how many of these have a single solution..

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pnm wrote:

I've been thinking on and off about the question: how many possible 4x4 Calcudoku puzzles are there?

That is an interesting question.

On top of the possible Latin squares are the possible cages overlaying the cells. Theoretically you can have 1 to 16 cages. And then each cage has an operation.

16 cages would mean each cage is a single cell and the operator is '=' for each cage.

A single cage including all 16 cells is possible, except two problems arise. First, - and / wouldn't work as operations. Second, neither +40 nor x331776 provide sufficient information for a unique solution.

And then there are a _lot_ of permutations in between.

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pnm wrote:

paulv66 wrote:

While it's an interesting question, there's not a straightforward solution. If you look up Latin square on Wikipedia, you will find considerable detail on the mathematics. I'd post a link, but I don't know if that's permitted.

Fine to post a link of course

Here you go - https://en.m.wikipedia.org/wiki/Latin_square

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paulv66 wrote:

While it's an interesting question, there's not a straightforward solution. If you look up Latin square on Wikipedia, you will find considerable detail on the mathematics. I'd post a link, but I don't know if that's permitted.

Fine to post a link of course

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I'm quite confident I've found the answer for 4x4 grids, but it isn't very elegant and doesn't easily generalise to bigger grids. This leaves me to wonder if there are nice ways answer these questions. I haven't given it too much thought, but I certainly will.

So if anyone wants a puzzle, I challenge you to take a stab at this one

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It wasn't really that helpful as it was subscriber only last Monday and available to all members on the previous 2 Mondays. But the puzzle 3 weeks previous was subscriber only and had 82 solvers. So 54 does appear to be somewhat lower than average. Maybe Patrick can provide longer term data?

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I was looking for it on top of the puzzle not where it is listed.

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