If there's an open square that is the only open square in its row or column, goto 3. Else, goto 4.
Pick an open square that is the only open square in its row or column. Calculate what the value should be. If the value is invalid (used already, greater than 20, less than 1, or different for the row and columns), return (from recursion, or overall -- in the latter case, it's unsolvable). If it's valid, fill in. Goto 1.
Find a row or a column with the least number of open squares (but one that still has open squares). Pick one of its open squares. For each unused value 1 .. 20 try filling it in in the open square. Recurse.
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horizontal scrolling (and possible janitor
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