Excellent, Mike, your proposal satisfies the Dan's condition. Your smallest example is: 1 2 ? 2 1 ? ? ? ? However, you are using the particularity that Dan allows to not analyze all the lines, if they are empty (why, Dan?). If we extend the Dan's condition to all the lines, empty or not, we can directly see that the last row (and the last column) cannot be filled. Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Michael Reid Envoyé : vendredi 9 juin 2006 08:47 À : math-fun@mailman.xmission.com Objet : RE: [math-fun] Latin square question
I'm afraid I may not have expressed myself clearly. What I meant was, suppose a matrix is partially filled. These entries will never change in this question.
CONDITION (*): Suppose that any *individual* row that already contains at least one filled square satisfies the condition that its unfilled squares can be filled without repetition so that there is also no repetition in any column. *And* the same thing after switching the roles of rows & columns.
This is clearly a prerequisite to extending the originally partially-filled matrix to a Latin square.
Question: Is this sufficient to know the original partially-filled matrix can be extended to a Latin square?
how about putting an (n-1) x (n-1) latin square in the upper left corner? if i parsed it correctly, that appears to satisfy your condition (*), but obviously (well, for n > 2 ) cannot be extended to an n x n latin square. mike _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun