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Old 2021-07-21, 14:18   #13
Viliam Furik
 
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"Viliam Furík"
Jul 2018
Martin, Slovakia

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Quote:
Originally Posted by LaurV View Post
Anyhow, as a way to go forward for your team, you could try to transform your algorithm to solve hamiltonian cycle or 3-sat, and I believe there are algorithms on the web to transform generalized sudoku into one of these or viceversa. Telling people about sudoku, they are reticent. Tell them about graphs and boolean circuits, you got them by the ears.
I tried that. It didn't pan out so far. There is almost a trivial relation to the Latin squares, but the rest is not as easy to do.

Quote:
Originally Posted by LaurV View Post
If you don't believe me, try this grid, without guessing. Courtesy of Mark and Simon, from Cracking the Cryptic youtube channel (google them, it is fun!). This grid is trivial to solve by guessing, but almost impossible to solve by logic, as they proved in two videos. Then think about a 169x169 grid (with 13x13 biscuits) and tell my how would you solve THAT one in O(n^6) You can have few hundreds of chains, every one of them with a hundred links, and you will have to make few hundred guesses, fill each chain, and if the guess was wrong and you got a contradiction, you will need to backtrack and refill with the correct chain. I don't see other way. But again, I may be too stupid to be a Mozart, haha
I know them, great channel indeed!

I'll send you more details on inner working of the algorithm in PM (as well as the results of the experiment), but this I will say "publicly":

I should be able to get a solution in O(n^12) - unless I found a counterexample to the algorithm. In O(n^6), I should be able to tell you whether it's uniquely solvable, non-uniquely solvable, or non-solvable. The part about being able to distinguish between unique solutions and non-unique ones is not well-studied yet. It's only simple observation. But the same observation tells me, that if it is uniquely solvable, I should be able to get a solution in O(n^6).

EDIT:
After running the experiment, I found that at least for the present version of the algorithm, the striken is not true.

Last fiddled with by Viliam Furik on 2021-07-21 at 14:32
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