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ArchiveGuest
Apr 18, 2006, 06:45 PM
(This post was originally by juggle7)
Hi, I'm new and very inerested in this forum. I'm having a debate with my brother about a puzzle, and the definition of what I call a guesser. Is it true that if you get to a point in a puzzle that 2 or more numbers can work in every available empty square that you must now start guessing? Please look at the following grid and tell me if it can be proven without a guess that the number 1 goes in the box address of 1-1. Thanks

090 536 072
037 000 650
562 700 903

009 287 536
306 050 007
705 163 009

901 370 265
653 000 700
270 605 390

Howard
Apr 18, 2006, 07:07 PM
Yes, its possible to make progress to prevent it from having multiple solutions. (That's what you call a guesser, I think :) )

The boxes are numbered
123
456
789

If you look in box 4, you'll see that the only two cells with an 8 as a possible candidate lie in a (vertical) line. That means that the 8 *has* to be in one of those places, so that entire line can't have any 8s in any other boxes (1 or 7). That means you can rule out 8 as a candidate in r7c2, leaving that with just 4 as its only candidate.

http://www.sudokuoftheday.com/image/image.php?size=230&sg=(148)W9(48)536(148)72(148)W37(489)(1249)(12489) 65(148)5W627(14)(148)9(148)3W(14)W(14)W9287536W3W( 1248)W6(49)5(49)(148)(1248)7W7W(248)W5163(48)(248) 99X(48)137(48)2656W53(489)(1249)(12489)7(148)(148) 2W7(48)6(14)539(148)

The rest from there is easy :)

Howard
Apr 18, 2006, 07:08 PM
Oh, and it isn't a 1 that ends up at the top left ;)