Robin Houston writes Although I m not a complexity theorist, I very much enjoy reading your weblog. I also enjoy solving the Sudoku puzzles published in the

Message 1 of 1
, Aug 11, 2005

Robin Houston writes

Although I'm not a complexity theorist, I very much enjoy reading
your weblog. I also enjoy solving the Sudoku puzzles published in
the British press, so it was doubly nice to see your May 25 post
about the complexity of Sudoku!

As far as I can tell, it follows from Yato's work that the problem:

Given a partially completed grid, find a valid completion if
there is one; otherwise report that there isn't one.

is solvable in polynomial time iff P=NP.

That's interesting of course, and it's a problem that faces those
who set the puzzles; but the problem that we solvers are faced with
is not quite (1). It's:

Given a partially completed grid that has a unique valid
completion, find that completion.

Can anything be said about problem (2)? If there were a polynomial-
time algorithm for (2), would it follow that P=NP? If not, would
there be any other significant consequences for complexity theory?

Good question. Since Yato's reductions preserve solutions the problem
is equivalent to finding a satisfying assignment of a Boolean formula
that has exactly one satisfying assignment (Unique SAT).

We don't know if Unique SAT is NP-complete in the traditional sense.
However Valiant and Vazirani have a nice paper
that shows how to
randomly reduce SAT to Unique SAT. Putting it together we get the
following equivalence:

Given a partially completed grid that has a unique valid
completion, probabilistically find that completion in polynomial time.

NP=RP (i.e. all NP problems have efficient probabilistic
solutions).

Since we don't believe that NP has fast probabilistic algorithms, we
expect that there are no efficient procedures to completing a
generalized Sudoku grid, even if there is only one such completion.