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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Fri Mar 03, 2006 12:04 am Post subject: Nightmare of Feb 28: A real treat! |
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Have you done the Sudo Cue Nightmare of 2/28? I had a great time with it. At least the way I solved it, it has coloring (a strongly linked chain) , two XYZ wings, and then coloring across two chains.
I am not sure the logic of my last step in linking the two chains is valid. I'll post a discussion this weekend, when I have more time.
Keith |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Fri Mar 03, 2006 12:57 am Post subject: It was a really good puzzle |
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I agree, Keith -- the 02/28/06 Nightmare was a real treat.
I didn't find any "XYZ-Wings" -- I've read about those, but I don't have much practice at finding them. I did notice two (prosaic) "XY-Wings." And the last tough step (for me) involved two binary chains on the digit "4", which solved seven or eight cells all at once.
I'll post a blow-by-blow description in a couple of days. dcb |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Mar 04, 2006 11:12 pm Post subject: XYZ wings, coloring - a great puzzle! |
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(I use the term "element" tor refer to any of a row, column or block.)
Here is the way I solved this one. I was biased to use coloring as I am learning the technique. From the starting point:
| Code: |
Puzzle: DSN022806
+-------+-------+-------+
| . . . | . . 3 | . . . |
| 5 . . | . 6 . | . . 1 |
| 9 8 3 | 2 . . | . 7 . |
+-------+-------+-------+
| . . . | 7 . . | . . 5 |
| . 2 . | . 4 . | . 8 . |
| 8 . . | . . 9 | . . . |
+-------+-------+-------+
| . 5 . | . . 1 | 9 6 8 |
| 1 . . | . 5 . | . . 7 |
| . . . | 3 . . | . . . |
+-------+-------+-------+
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After the available basic moves, we reach:
| Code: |
+-------------------+-------------------+-------------------+
| 2467 1467 12467 | 58 78 3 | 2458 245 9 |
| 5 47 247 | 9 6 478 | 2348 234 1 |
| 9 8 3 | 2 1 45 | 456 7 46 |
+-------------------+-------------------+-------------------+
| 46 1346 146 | 7 238 28 | 1246 9 5 |
| 67 2 9 | 156 4 56 | 167 8 3 |
| 8 13467 5 | 16 23 9 | 12467 124 246 |
+-------------------+-------------------+-------------------+
| 3 5 27 | 4 27 1 | 9 6 8 |
| 1 9 2468 | 68 5 268 | 234 234 7 |
| 2467 467 24678 | 3 9 2678 | 15 15 24 |
+-------------------+-------------------+-------------------+
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Now, there are two XYZ-wings here. An XYZ wing is a triple that spans two elements. One of the three possibilities must be present in all three squares. That possibility can then be excluded from squares that are neighbors of all of the three XYZ squares.
Squares R2C3, R2C2 and R7C3 are an XYZ-wing on <7>. R1C3 cannot have possibility <7>.
Squares R8C6, R8C4 and R4C6 are an XYZ-wing on <8>. R9C6 cannot have possibility <8>.
Noting that R9C3 is pinned to be <8>, the situation is now this:
| Code: |
+-------------------+-------------------+-------------------+
| 2467 1467 1246 | 58 78R 3 | 2458 245 9 |
| 5 47 247G | 9 6 478G | 2348 234 1 |
| 9 8 3 | 2 1 45 | 456 7 46 |
+-------------------+-------------------+-------------------+
| 46 1346 146 | 7 238 28 | 1246 9 5 |
| 67 2 9 | 156 4 56 | 167 8 3 |
| 8 13467 5 | 16 23 9 | 12467 124 246 |
+-------------------+-------------------+-------------------+
| 3 5 27R | 4 27G 1 | 9 6 8 |
| 1 9 246 | 68 5 268 | 234 234 7 |
| 2467 467 8 | 3 9 267R | 15 15 24 |
+-------------------+-------------------+-------------------+
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How does (simple) coloring work? For any value of a possibility, look in each element (row, column, block) of the puzzle to find where the possibility occurs twice. Draw a line connecting the two squares.
(Actually, I just use a separate piece of squared paper with a 9x9 empty grid to do this.)
When you are done, you may have a collection of "graphs", networks of lines that connect pairs of possibilities. Now, you can use two colors (say, red and green) to alternately label each square on the graph. Either the red squares have the possible value, or the green squares do.
In the grid above, I have labeled the squares that form one connected graph for the possibility <7> with R(ed) and G(reen).
Note that R2 has two Green squares, so the possibility <7> must be in the Red squares!
So, R1C5 = <7>. Following through with the resulting moves, we reach:
| Code: |
+--------------+--------------+---------------+
| 246 146A 16a | 58 7 3 | 28 245 9 |
| 5 7 24 | 9 6 48 | 238 234 1 |
| 9 8 3 | 2 1 45 | 456 7 46 |
+--------------+--------------+---------------+
| 46 3 16A | 7 8 2 | 146a 9 5 |
| 7 2 9 | 156c 4 56 | 16C 8 3 |
| 8 146a 5 | 16C 3 9 | 7 124B 246 |
+--------------+--------------+---------------+
| 3 5 7 | 4 2 1 | 9 6 8 |
| 1 9 24 | 68 5 68 | 23 234 7 |
| 246 46 8 | 3 9 7 | 15B 15b 24 |
+--------------+--------------+---------------+
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For the possibility <1>, there are three unconnected graphs, A, B, and C. I have labeled the alternate possibilities A,a, B,b, etc. Graph A has 5 squares, the other two have 3 squares each.
Now, take a look at R6 and C7. I concluded I could join graphs A and B. So, set B = A and b = a. Then, R6 and C7 have both possibilities: C cannot be <1>, c must be <1> (R5C4 = <1>), and the puzzle is solved!
I had great fun with this one. A good learning experience.
Keith |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Sun Mar 05, 2006 12:17 am Post subject: Here's how I worked it out |
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1. r3c5 = 1; r7c4 = 4 (sole candidate)
2. Naked pair {2, 7} in row 7; r7c1 = 3 (sole candidate)
3. Hidden pair {1, 5} in r9c8 & r9c9.
4. Naked triplet {1, 5, 6} in middle center 3x3 box.
5. The "3" in column 9 lies in middle right 3x3 box; r5c9 = 3 (unique horizontal)
6. r1c9 = 9; r4c8 = 9; r9c5 = 9 (unique vertical)
7. r2c4 = 9; r8c2 = 9; r5c3 = 9 (unique vertical)
8. r6c3 = 5 (unique vertical)
9. The "6" in row 3 lies in top right 3x3 box.
10. Coloring on "7" eliminates possible "7" at r2c3.
11. XY-Wing (r2c3, r2c2, r7c3) eliminates "7" at r9c2, r1c3.
12. The "7" in column 3 lies in bottom left 3x3 box.
13. Hidden pair {7, 8} in row 9.
14. The "6" in row 9 lies in bottom left 3x3 box.
15. Hidden pair {1, 6} in column 3.
16. XY-Wing (r9c6, r7c5, r4c6) eliminates "2" at r4c5 & r8c6.
17. Naked pair {6, 8} in row 8; naked pair {2, 4} in column 3.
18. r9c6 = 7; r8c5 = 2; r9c3 = 8; r7c3 = 7 (sole candidate)
19. r6c5 = 3; r4c5 = 8; r1c5 = 7; r4c6 = 2; r4c2 = 3 (sole candidate)
20. r2c2 = 7; r6c7 = 7; r5c1 = 7 (unique horizontal)
21. Coloring on "2" eliminates "2" at r2c8.
22. Hidden triplet {2, 3, 8} in column 7.
23. Coloring on "4" reveals that r3c9 = 6.
24. Re-coloring on "4" eliminates "4" at r9c1.
And now the puzzle is ready for the coup de grace:
| Code: | 24~6 146 16 58 7 3 28 245 9
5 7 24+ 9 6 4=8 238 34 1
9 8 3 2 1 4~5 4=5 7 6
4=6 3 16 7 8 2 14~6 9 5
7 2 9 156 4 56 16 8 3
8 14~6 5 16 3 9 7 124 24+
3 5 7 4 2 1 9 6 8
1 9 24- 68 5 68 23 234+ 7
26 4+6 8 3 9 7 15 15 24- |
There are two binary chains in the "4"s. Note the contradiction at r2c3: the "4+" is aligned with both an "=" and a "~". So we can eliminate "4" at r2c3, r9c2, r8c8, & r6c9 -- and we can place "4"s at r8c3 & r9c9. The rest is simple. dcb
PS Keith, I'm not sure (yet) that linking the two chains on "1" is logically valid. If you had an "A" opposite a "b" and an "a" opposite a "B" then I could see it. But in the grid you posted all I can find ia an "a" opposite a "B", meaning that "A or b" is always true. I don't see how to prove the other half, that "a or B" is always true.
I'll write more after I've had a chance to study your solution in more detail. Thanks for the XYZ Wing steps -- those are great! |
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TKiel
Joined: 22 Feb 2006
Posts: 292
Location: Kalamazoo, MI
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| Posted: Sun Mar 05, 2006 2:49 pm Post subject: |
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Keith,
I agree with David about your colouring solution using the 1's. Your A-a and B-b chains link in box 6 and what can be concluded is that both (a) and (B) can't be true, so either/both (A) and (b) are true but there is no cell that shares a group with cells with both those markers, so no exclusions can be made.
However, using the same candidate profile as contained in your last grid, there are three conjugate chains in 4's that can be linked in such a manner as to make an exclusion. They are labeled A-a, B-b and C-c.
| Code: |
*--------------------------------------------------*
| 246 146 16 | 58 7 3 | 28 245 9 |
| 5 7 24A | 9 6 48B | 238 234 1 |
| 9 8 3 | 2 1 45b | 456C 7 46 |
|----------------+----------------+----------------|
| 46C 3 16 | 7 8 2 | 146c 9 5 |
| 7 2 9 | 156 4 56 | 16 8 3 |
| 8 146c 5 | 16 3 9 | 7 124 246 |
|----------------+----------------+----------------|
| 3 5 7 | 4 2 1 | 9 6 8 |
| 1 9 24a | 68 5 68 | 23 234A 7 |
| 246 46 8 | 3 9 7 | 15 15 24a |
*--------------------------------------------------*
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If both A & C were true then both B & b would be false, so either/both of a & c must be true. That makes exclusions of 4 at r9c2 and r6c9. Now the A-a chain can be extended by one cell to include r3c9(A).
| Code: |
*--------------------------------------------------*
| 246 14 16 | 58 7 3 | 28 245 9 |
| 5 7 24A | 9 6 48B | 238 234 1 |
| 9 8 3 | 2 1 45b | 456 7 46A |
|----------------+----------------+----------------|
| 46 3 16 | 7 8 2 | 146 9 5 |
| 7 2 9 | 156 4 56 | 16 8 3 |
| 8 14 5 | 16 3 9 | 7 124 26 |
|----------------+----------------+----------------|
| 3 5 7 | 4 2 1 | 9 6 8 |
| 1 9 24a | 68 5 68 | 23 234A 7 |
| 24 6 8 | 3 9 7 | 15 15 24a |
*--------------------------------------------------*
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Now both ends of the B-b chain are linked to cells marked with (A), which we can conclude must be false and all the cells marked with (a) are true. Once the resulting assignments are made, there is an X-wing, in a different number, which solves the puzzle. |
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