There are multiple solutions for each incomplete grid so how are you calculating the transitions for a grid w/ a non-unique solution?

Edit: I see you added questions for the ambiguities but modulo those choices your solution will almost work b/c it is not extensionally equivalent entirely. The transition graph and solver are almost extensionally equivalent but whereas the Prolog solver will backtrack there is no backtracking in the Markov chain and you have to re-run the chain multiple times to find all the solutions.

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Ukv 6 days ago | parent | next [-]

> but whereas the Prolog solver will backtrack there is no backtracking in the Markov chain and you have to re-run the chain multiple times to find all the solutions

If you want it to give all possible solutions at once, you can just expand the state space to the power-set of sudoku boards, such that the input board transitions to the state representing the set of valid solved boards.

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measurablefunc 6 days ago | parent | next [-]

That still won't work b/c there is no backtracking. The point is that there is no way to encode backtracking/choice points like in Prolog w/ a Markov chain. The argument you have presented is not extensionally equivalent to the Prolog solver. It is almost equivalent but it's missing choice points for starting at a valid solution & backtracking to an incomplete board to generate a new one. The typical argument for absorbing states doesn't work b/c sudoku is not a typical deterministic puzzle.

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Ukv 6 days ago | parent [-]

> That still won't work b/c there is no backtracking.

It's essentially just a lookup table mapping from input board to the set of valid output boards - there's no real way for it not to work (obviously not practical though). If board A has valid solutions B, C, D, then the transition matrix cell mapping {A} to {B, C, D} is 1.0, and all other entries in that row are 0.0.

> The point is that there is no way to encode backtracking/choice points

You can if you want, keeping the same variables as a regular sudoku solver as part of the Markov chain's state and transitioning instruction-by-instruction, rather than mapping directly to the solution - just that there's no particular need to when you've precomputed the solution.

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measurablefunc 6 days ago | parent [-]

My point is that your initial argument was missing several key pieces & if you specify the entire state space you will see that it's not as simple as you thought initially. I'm not saying it can't be done but that it's actually much more complicated than simply saying just take an incomplete board state s & uniform transitions between s, s' for valid solutions s' that are compatible with s. In fact, now that I spelled out the issues I still don't think this is a formal extensional equivalence. Prolog has interactive transitions between the states & it tracks choice points so compiling a sudoku solver to a Markov chain requires more than just tracking the board state in the context.

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Ukv 6 days ago | parent [-]

> My point is that your initial argument was missing several key pieces

My initial example was a response to "If you think it is possible then I'd like to see an implementation of a sudoku puzzle solver as Markov chain", describing how a Sudoku solver could be implemented as a Markov chain. I don't think there's anything missing from it - it solves all proper Sudokus, and I only left open the choice of how to handle improper Sudokus because that was unspecified (but trivial regardless of what's wanted).

> I'm not saying it can't be done but that it's actually much more complicated

If that's the case, then I did misinterpret your comments as saying it can't be done. But, I don't think it's really complicated regardless of whatever "ok but now it must encode choice points in its state" are thrown at it - it's just a state-to-state transition look-up table.

> so compiling a sudoku solver to a Markov chain requires more than just tracking the board state in the context.

As noted, you can keep all the same variables as a regular Sudoku solver as part of the Markov chain's state and transition instruction-by-instruction, if that's what you want.

If you mean inputs from a user, the same is true of LLMs which are typically ran interactively. Either model the whole universe including the user as part of state transition table (maybe impossible, depending on your beliefs about the universe), or have user interaction take the current state, modify it, and use it as initial state for a new run of the Markov chain.

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measurablefunc 6 days ago | parent [-]

> As noted, you can keep all the same variables as a regular Sudoku solver

What are those variables exactly?

	▲	Ukv 6 days ago \| parent [-]
		For a depth-first solution (backtracking), I'd assume mostly just the partial solutions and a few small counters/indices/masks - like for tracking the cell we're up to and which cells were prefilled. Specifics will depend on the solver, but can be made part of Markov chain's state regardless.

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Certhas 5 days ago | parent | prev [-]

People really don't appreciate what is possible in infinite (or more precisely: arbitrarily high) dimensional spaces.

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6 days ago | parent | prev [-]

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