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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

PostPosted: Wed May 17, 2006 6:55 pm    Post subject: "UR" in May 16, 2006 "Nightmare" Reply with quote

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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