[wizzwizz4]

The five-colour theorem was first proven in the late 19th century. The known proof of the four-colour theorem is non-surveyable. There's probably no small mistake in your implementation, and the solution you should go with is unlikely to be a fully-general algorithm. (I'd suggest starting from the border on the outside, greedily filling in as many choices as are forced (up to isomorphism), then randomly choosing a candidate breadth-first or depth-first until you've found a four-colour solution.)

[frading]

actually, I realise now that the implementation I shared in my previous comment is a generic one. At some point, I decided to make a variation of it, specialised for this board. The reason wasn't the number of colors, though, it was to have better control of the colors of the bottom and top rows.

And I also realise that, contrary to what I said above, I do not set the colors on the left and right sides. So that will be an easy modification.

After that, I do go breadth first. But I do so by looking at all neighbours of the current tile, simply by taking them in the order given by my internal graph. Instead I should just take those that are connected to another tile that has been already visited.

So thank you for those advice, that's super helpful, I really appreciate. I had to look up "non-surveyable" as well, I'm learning a lot here. That said, I'm having a hard time understanding "as are forced (up to isomorphism)", is that something you could clarify?

[wizzwizz4]

Let's say your colours are black, white, green and purple. So far, you've managed to use only black and white, but you've reached a node with both black and white neighbours. You need (are forced) to choose a different colour! You could choose green, or purple ­– and whatever solutions you reach from that point, you could swap all the greens and all the purples to yield another solution. However, the next "free" choice you make could be the same colour as you chose last time, or a different one, so it is actually free (so isn't forced).

Since the "swap green and purple" operation is bijective (a one-to-one, reversible mapping), and preserves all the properties we care about, we can call it an isomorphism. If there's an isomorphism between two solutions, we say that those solutions are isomorphic. Swapping all the greens and purples of any isolated region of the graph (i.e., one surrounded by a strip of black and white) is an isomorphism, if the only property we care about is "is this a colouring of the original graph?".

I was saying it's pointless to search isomorphic solutions if you can help it. If you have isolated regions of the graph yet to be filled, you can halve the time it takes to explore each region's possible colourings.

However, if you consider isolated regions completely separately, you can solve them one after the other (and backtrack once you know any region's been made impossible by the choices of colours in its border), for a much larger speedup.

The whole thing about isomorphisms was a red herring, an artefact of my thought process rather than an actual insight. I doubt trying to exploit this will net you anything more than a 2× improvement, and that's only if you do it cleverly.

[frading]

ah I think I understand now. Thanks a lot for clarifying.

Although I'll have to re-read you when I rework the implementation. But in any case, the operation is already done very quickly, so I've never thought about monitoring it. So any performance improvement on it isn't a high priority. There are other operations in the full grid generation that are more time consuming, like the smoothing/squarification, which needs to be applied multiple times. But even then, since the grid is so small, the whole thing takes less than a second.

That said, I'm thinking of ways to improve the menu where players select the next board. Right now, when you want to generate a random one, I only display one. But it could be nice to display multiple ones. In which case 1 second generation becomes X seconds. So if I do that, any speed up could be welcome.