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I decided to work my way through Simon Tatham's puzzle collection to broaden my puzzling skills, and have reached the game Dominosa

I have a subboard below, where I've eliminated the possibility of 9/3 and 9/5 dominoes being in this area.

2385/5990/5927/3565

May I infer that the 9/9 domino is the left half of the central 2x2 square, rather than the top half, and additionally that the right half is not a single 9/2 domino?

My thought is that if it were otherwise, the center square would be ambiguous in orientation between the horizontal and the vertical, but I'm not certain if such ambiguity is forbidden.

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    $\begingroup$ If ambiguity is indeed not possible (which seems the most logical to me) your conclusion seems correct to me. But I think only the creator of that website can answer that. He is the only one that knows if any ambiguity is possible. $\endgroup$ – Ivo Beckers Feb 27 '17 at 14:58
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In grid deduction puzzles, there is a unique solution to each puzzle. What you're using is known as "unique solution logic", using the meta-fact that the puzzle has a unique solution. There's nothing wrong with using it, but many people prefer not to do so in human-designed puzzles because it circumvents the logical path set by the designer, which could be more entertaining.

However, this is only valid if the puzzle is guaranteed to have a unique solution, and the algorithm used by Tatham to generate puzzles may not check for uniqueness. If so, this deduction would be invalid.

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  • $\begingroup$ The source code to this puzzle is available here git.tartarus.org/?p=simon/puzzles.git;a=tree . As I understand it, all these puzzles are set up by running a board through a solver with deterministic rules in the setup phase to make sure they are solvable deterministically. Some puzzles also have diagnostics which should make it possible (maybe with some code) to feed in a board to show you the solution route taken. I would be surprised if this "meta-rule" were included in this solver for the reasons you give, but it is verifiable. (I don't know this puzzle in the collection myself). $\endgroup$ – Dan Sheppard Nov 13 '20 at 9:56

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