zoltag wrote:Wow, thanks for the links.
You're welcome. I'm glad to help.
This example may be redundant, but I said I'd post something, so here it is. This is drawn from Ruud's "nightmare" puzzle for Wednesday, 3 May, 2006.
Code: Select all
9 3 246 7 25 1 58 24 268
1 5 26 3 8 4 7 9 26
24 8 7 9 25 6 135 124 23
3 7 28 6 1 289 89 5 4
25 9 1258 4 7 28 138 6 238
6 4 128 5 3 289 189 12 7
7 1 9 2 6 3 4 8 5
45 2 45 8 9 7 6 3 1
8 6 3 1 4 5 2 7 9
This is the position I reached after working on the puzzle for a while. A "non-unique rectangle" is beginning to take shape in r1&2, c3&9. We can eliminate the "4" at r1c3 with a simple forcing chain.
A. r1c3 = 4 ==> r1c8 = 2 ==> r2c9 = 6 ==> r1c9 = 8
B. r1c3 = 4 ==> r1c8 = 2 ==> r1c5 = 5 ==> r1c7 = 8
But we can't have two "8"s in row 1, so r1c3 <> 4. Now the grid looks like this.
Code: Select all
9 3 26 7 25 1 58 24 268
1 5 26 3 8 4 7 9 26
24 8 7 9 25 6 135 124 23
3 7 28 6 1 289 89 5 4
25 9 1258 4 7 28 138 6 238
6 4 128 5 3 289 189 12 7
7 1 9 2 6 3 4 8 5
45 2 45 8 9 7 6 3 1
8 6 3 1 4 5 2 7 9
To avoid the "deadly pattern" we must set r1c9 = 8, and the rest of the puzzle falls out pretty quickly. But since the solution to the puzzle is unique, there must be another way to solve it. In this case we can make the eliminations resulting from the {2, 6} pair in column 3, and the puzzle is a piece of cake. But I'd like to illustrate one other principle here.
We can usually avoid making the "uniqueness" assumption by picking one of the cells that's "conjugate" to the "unique" corner of the rectangle and reasoning our way to a contradiction. In this case the "unique" corner is r1c9, and the two "conjugate" cells (which might also contain the digit 8) are r1c7 and r5c9. Notice how assuming that either one of these is "8" leads to a contradiction.
A. r1c7 = 8 ==> r4c7 = 9 ==> r6c7 = 1
B. {2, 6} in c3r1&2 ==> r4c3 = 8 ==> r6c3 = 1
There can't be two "1"s in row 6, so r1c7 <> 8.
C. r5c9 = 8 ==> r5c7 = 3 ==> r5c3 = 1 (only spots left for "3", then "1", in row 5).
D. {2, 6} in c3r1&2 ==> r4c3 = 8 ==> r6c3 = 1
There can't be two "1"s in column 3, so r5c9 <> 8.
The logic for contradictions connected with "non-unique rectangles" isn't always as clean as in this example. But I've had a lot of fun with Ruud's puzzles (where these rectangles pop up pretty frequently) by consciously trying to avoid making the "uniqueness" assumption, and using the conjugate cells to reach the same conclusion (e.g., r1c9 = 8) via the back door. dcb