Anybody got a way to solve it that doesnt involve uniqueness?
Laura
Feb 1 Nightmare
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Here's one way ...
Hi, Laura!
I hadn't planned to do this, but since you asked I went ahead and gave it a try.
After doing the obvious stuff the grid looks like this.
It's apparent that there aren't many ways to place the "9"s in this puzzle. And
there are only two ways to put a "9" in the top center 3x3 box. So let's see what
happens if we set r1c5 = 9. This immediately forces the rest of the "9"s, plus a "7"
at r2c6 and an "8" at r6c3.
After these hypothetical moves the grid looks like this.
Notice that setting r1c5 = 9 creates a naked pair {3, 6} in the top right 3x3 box,
and that allows us to identify another pair {2, 4} in column 8. The same move
(r1c5 = 9) also creates the {2, 7} pair in r7c5 & r9c5, and uncovers {4, 6} in
r7c4 & r9c6.
Now the path to a contradiction is fairly simple.
-- We can eliminate the "2" and the "6" at r1c4 because of the pairs in row 1. So r1c4 = 1.
-- r1c4 = 1 ==> r5c5 = 4 ==> r7c4 = 6 ==> r7c1 = 4.
And now it's clear that we can't complete the top left 3x3 box, because there are
only two values -- "2" and "8" -- that can possibly fit in the three squares r1c1,
r2c2, & r3c1. Therefore r1c5 <> 9 and we must have r2c6 = 9. dcb
I hadn't planned to do this, but since you asked I went ahead and gave it a try.
After doing the obvious stuff the grid looks like this.
Code: Select all
24 5 147 126 79 8 369 23469 36
3 278 6 25 4 79 58 29 1
248 9 148 1256 3 16 7 246 58
1 4678 478 9 568 2 3568 367 3568
5689 678 3 14 568 14 2 679 568
25689 268 89 7 568 3 15689 169 4
469 1346 5 246 279 4679 136 8 27
7 36 2 8 1 5 4 36 9
4689 1468 1489 3 279 4679 16 5 27
there are only two ways to put a "9" in the top center 3x3 box. So let's see what
happens if we set r1c5 = 9. This immediately forces the rest of the "9"s, plus a "7"
at r2c6 and an "8" at r6c3.
After these hypothetical moves the grid looks like this.
Code: Select all
24 5 147 126 9 8 36 24 36
3 28 6 25 4 7 58 9 1
248 9 14 1256 3 16 7 24 58
1 467 47 9 568 2 3568 367 3568
9 67 3 14 568 14 2 67 568
256 26 8 7 568 3 9 16 4
46 1346 5 46 27 9 136 8 27
7 36 2 8 1 5 4 36 9
468 1468 9 3 27 46 16 5 27
and that allows us to identify another pair {2, 4} in column 8. The same move
(r1c5 = 9) also creates the {2, 7} pair in r7c5 & r9c5, and uncovers {4, 6} in
r7c4 & r9c6.
Now the path to a contradiction is fairly simple.
-- We can eliminate the "2" and the "6" at r1c4 because of the pairs in row 1. So r1c4 = 1.
-- r1c4 = 1 ==> r5c5 = 4 ==> r7c4 = 6 ==> r7c1 = 4.
And now it's clear that we can't complete the top left 3x3 box, because there are
only two values -- "2" and "8" -- that can possibly fit in the three squares r1c1,
r2c2, & r3c1. Therefore r1c5 <> 9 and we must have r2c6 = 9. dcb
Cool, I did not see that.
Fantastic. I missed that altogether. thank you.We can eliminate the "2" and the "6" at r1c4 because of the pairs in row 1. So r1c4 = 1.
Laura