I'm a bit confused. I thought that the first column Z's could be either 2 or 6, but the 3rd column Z's could be 2, 3, or 6. If he threw out the 3 possibility, then the puzzle would have two (putative) solutions, so by uniqueness, he would know that one of the 3rd column Z's was, in fact, a 3. Of course, if the solution is truly unique, this should become apparent at some later point, but I think that he's looking for a short-cut (is that correct, Rich?). I keep expecting to run across one that has more than one solution, so I try not to throw out anything based on uniqueness. Can we capture a similar mathematical statement? Mathematically, if I know that a problem has a unique solution, and I have a partial solution where one set of choices appears to lead to consistent, multiple solutions, can I exclude that set? It doesn't seem so, but perhaps my phrasing is inadequate, or perhaps I'm misinterpreting the question. --BIll Cordwell ----- Original Message ----- From: "Stan E. Isaacs" <stan@isaacs.com> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Friday, March 03, 2006 12:02 AM Subject: [math-fun] Re: sudoku
That's a perfectly good deduction, but you've stated it wrong. If the 2 Z's in the first row can only be 2 and 6, and similarly for the last row, than no other number in either row can be 2 or 6. The same holds for the two columns. It even has a name, but I'm not certain what it is. Perhaps "X-wing".
-- Stan Isaacs
Sudoku: I ran across a moral dilemma in a Sudoku last night: Can I use the uniqueness of the solution to exclude some lines of argument? My partial solution looked like this:
Z - Z | * * * | * * * x - x | * * * | * * * x - x | * * * | * * * --------------------- x - x | * * * | * * * x x ? | * * * | * * * x - x | * * * | * * * --------------------- x x x | * * * | * * * x x x | * * * | * * * Z x Z | * * * | * * *
x is a known value, - is an empty cell (so far), * could be either.
I knew that the left column Z,Z values had to be 2,6 in some order, and the third column Z,?,Z values had to be 2,3,6 in some order. I observed that, if I assigned 3 to the ?, that the Zs could be either 2,6 and 6,2, or vice versa. If one solution worked, the other would also, since all the puzzle constraints on uniqueness etc. would be satisfied in either case. Hence, if I used my meta-knowledge of a unique solution, I could exclude the value 3 in the ? cell. Dilemma: Is this fair? In a race, there's no way to exclude the inference; and we used to do similar things on multiple choice tests. But for self solving, should the conclusion be allowed?
Rich
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-- Stan Isaacs 210 East Meadow Drive Palo Alto, CA 94306 stan@isaacs.com
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