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The goal of Hidato is to fill the grid with a series of consecutive numbers adjacent to each other orthogonally or diagonally, and all tiles are required to be filled in.


So, as you may or may not know, I like creating Hidato puzzles sometimes. I usually try to make my Hidato puzzles big, but still rigid enough where there is still only one unique solution. However, every time I post one, it seems that there are always multiple solutions because of those interchangeable numbers that I somehow always miss, even though I test my puzzles to make sure that there aren't multiple solutions. I can make a good puzzle on its own, but then the puzzle usually ends up being a bit too easy, or I end up making it too small to ensure no contradictions/multiple solutions.


For example, this 5x5 that I made in ~1 minute:


enter image description here


This puzzle, of course has the unique solution


enter image description here


which in testing, took me ~1.5 minutes to solve at first, although I was able to do it in ~40 to 50 seconds once I realized how everything fell into place. (e.g. it should be automatically implied that 14 is placed at R4C3 since there's no other way to loop around to 19 without causing contradictions, which forces the 20 to be placed at R5C3, and so on. 6 is also forced at R2C3 in order to satisfy the connection between 5 and 7, which forces 2 in R1C2 and 4 in R2C1.)


The only problem with this now is: How do I expand the size of a Hidato puzzle while still keeping it rigid enough?

Trying this with a 7x7 that I created in ~5 minutes:


enter image description here


Which should have a unique solution that I arrived at in ~10 minutes, although I am unsure if this actually has only one solution or not:


enter image description here


So my question is: How does one go about creating a big, but good Hidato puzzle?


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    $\begingroup$ I don't have much experience with this type of puzzle but one thing I can see is that you shouldn't put numbers differing by 3 next to each other if the two numbers in between will be in the two adjacent squares. In your 7x7 the 33 and 36 are next to each other and the 34 and 35 are interchangeable. The 21 and 24 work because they are offset to one side. BTW the 5x5 was nice and my favorite puzzle that you've posted. $\endgroup$
    – caPNCApn
    Commented Nov 13, 2023 at 20:02

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