by chipe » Sat Apr 30, 2005 8:25 pm
Thanks IJ. Your March 19, 2005, 1:53 PM, post did it for me. Thanks too to Pappocom's March 19, 2005, 3:24 PM, post confirming it and noting the col. 7/8 mistype. Wayne's remarks are appropriate as to using the correct terminology. Also, posters should register so they can be emailed, or at least so they can edit their errors.
For those still confused, let me say it is solved as follows;
(1) I started with Jane's grid, given at the start of this thread.
(2) I put in all of the pencil marks (candidates) in each cell.
(3) one particularly important set of pencil marks is the 6's at the bottom (r9) of Box 8. They lead us to eliminate the 6s pencil marks at the bottom (r9) of Box 9. (This is basic, but I'll say it anyway: the possible 6s in Box 8 can only go in row 9 of Box 8 [and 6s go all over Box 9], so the 6s of Box 8 (all in row 9) knock out the 6's in row 9 of Box 9.)
(4) Now, the only 6s possibilities (pencil marks) left in column 7 are in the cells in row 4 and 5 . That is, either r4,c7 or r5,c7 must be a 6. Both of these cells are in Box 6. Therefore, the other pencil mark 6s in Box 6 (in column 8 of Box 6) are eliminated.
(5) Now we switch our attention to column 8. All of the remarks in this paragraph are to cells in column 8. 3678 pencil marks are in row 3, and 1679 pencil marks are in row 8. These are the only 6s and 7s left in the column! So we can shorten the pencil marks to 67 pencil marks in row 3 and 67 pencil marks in row 8. Thus, we eliminated 3 and 8 pencil marks in row 3, and 1 and 9 pencil marks in row 8. As we will see in a moment, eliminating that 1 pencil mark from r8,c8 is crucial.
(6) In column 8, there were (once) only two 1 pencil marks -- at r7,c8 and r8,c8. So either cell could be a 1. But in (5), above, we eliminated the 1 from r8,c8, so we have 1 as the sole number in r7,c8. I assume that from here on the puzzle can be easily solved (I hope).