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"UR" in May 16, 2006 "Nightmare"

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David Bryant

Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

Posted: Wed May 17, 2006 6:55 pm Post subject: "UR" in May 16, 2006 "Nightmare"

This is the "nightmare" puzzle for Tuesday, May 16.

Code:

2..8.3...
13...2..6
..9.5....
.2....7..
.4.7.1.3.
..6....5.
....2.8..
8..4...79
...3.9..1

This puzzle is fairly simple if one assumes the solution is unique. It also provides an excellent "DIC" exercise if one chooses not to make that assumption.

After making the "obvious" moves we reach this position.

Code:

2 567 457 8 1467 3 1459 149 57
1 3 457 9 47 2 45 8 6
467 8 9 16 5 46 1234 124 37
39 2 13 56 3469 4568 7 169 48
5 4 8 7 69 1 69 3 2
379 179 6 2 39 48 19 5 48
3469 1569 1345 156 2 7 8 46 35
8 156 23 4 16 56 23 7 9
467 567 2457 3 8 9 2456 246 1

Now the puzzle is easily solved if we apply the "non-unique rectangle" appearing in r46c69. Clearly neither "4" nor "8" can appear at r4c6, and when we eliminate those two values the {5, 6} pair is revealed -- the rest of the puzzle can be resolved by using "simple coloring" to find a few naked pairs.

What if we don't make the uniqueness assumption? Well, the solution is quite a bit more difficult, but a handful of "DIC"s will get us through it.

Let's warm up with r7c9 as the "alpha star".
A. r7c9 = 3 ==> r1c9 = 5 ==> r2c3 = 5
B. r7c9 = 5 ==> r7c3 <> 5
We conclude that r7c3 <> 5

A. r7c9 = 3 ==> r1c9 = 5 ==> r2c7 = 4 ==> r2c5 = 7
B. r7c9 = 5 ==> r1c9 = 7 ==> r2c5 = 7
We conclude that r2c5 = 7.

A. r7c9 = 3 ==> r8c3 = 3 ==> r4c3 = 1 ==> r6c7 = 1
B. r7c9 = 5 ==> r1c9 = 7 ==> r3c1 = 7 ==> {3, 9} pair in row 6 ==> r6c7 = 1
We conclude that r6c7 = 1, and that r4c3 = 1, as well. Now the grid looks like this.

Code:

2 567 457 8 146 3 459 149 57
1 3 45 9 7 2 45 8 6
467 8 9 16 5 46 234 124 37
39 2 1 56 3469 4568 7 69 48
5 4 8 7 69 1 69 3 2
379 79 6 2 39 48 1 5 48
3469 1569 34 156 2 7 8 46 35
8 156 23 4 16 56 23 7 9
467 567 2457 3 8 9 2456 246 1

We can make more deductions from the same "alpha star".
A. r7c9 = 3 ==> r7c3 = 4 ==> r2c3 = 5
B. r7c9 = 5 ==> r1c9 = 7 ==> {4, 5} pair in c3r12 ==> r9c3 <> 5
We conclude that r9c3 cannot contain a "5". So the "5" in column 3 must lie in row 1 or row 2, and we can also eliminate "5" at r1c2.

Now we find a "6-star constellation" rooted in r6c2.
A. r6c2 = 9 ==> r6c5 = 3
B. r6c2 = 7 ==> r7c2 = 9 ==> r8c2 = 1 ==> r8c5 = 6 ==> r5c5 = 9 ==> r6c5 = 3
We conclude that r6c5 = 3, and this forces r4c1 = 3. Now the grid looks like this.

Code:

2 67 457 8 146 3 459 149 57
1 3 45 9 7 2 45 8 6
467 8 9 16 5 46 234 124 37
3 2 1 56 469 4568 7 69 48
5 4 8 7 69 1 69 3 2
79 79 6 2 3 48 1 5 48
469 1569 34 156 2 7 8 46 35
8 156 23 4 16 56 23 7 9
467 567 247 3 8 9 2456 246 1

Next we can use an "8-star constellation" to show that r1c2 = 6.
A. r6c2 = 7 ==> r1c2 = 6
B. r6c2 = 9 ==> r6c1 = 7 ==> r3c9 = 7 ==> r3c7 = 3 ==> r8c3 = 3 ==> r9c3 = 2 ==> r1c3 = 7 ==> r1c2 = 6

Finally, a "6-star constellation" from r3c6 cracks it wide open.
A. r3c6 = 6 ==> r8c6 = 5 ==> r4c6 <> 5
B. r3c6 = 4 ==> the "4" in box 7 lies in column 1 ==> r7c3 = 3 ==> r7c9 = 5 ==> r4c4 = 5 ==> r4c6 <> 5

This leaves the "5" at r4c4 as unique in row 5, and the "5" at r8c6 as unique in column 6. And with these values entered, the rest is a piece of cake.

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