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AZ Matt
Joined: 03 Nov 2005
Posts: 63
Location: Hiding under my desk in Phoenix AZ USA
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 Posted: Thu Jul 06, 2006 11:30 pm Post subject: Brainbashers Super Hard July 5: Colouring? |
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I got to here and then solved using a UR type 4 and then an "I don't know what to call it" solving technique.
| Code: | 37 6 9 5 4 8 137 137 2
347 2 34 1 6 9 5 8 37
8 1 5 7 2 3 6 9 4
1 359 2 369 79 67 4 356 8
349 389 34 3689 5 2 137 1367 137
6 358 7 38 1 4 2 35 9
359 379 1 2 79 567 17 2 167
59 4 8 69 3 567 17 2 167
2 37 6 4 8 1 9 37 5 |
It took awhile, but I found the UR Type 4 in the r25c13 and eliminated the 3s from r2c1 and r5c1.
Then I saw some promise for a swordfish or colouring on the 3 candidates, so I laid them out on a grid:
| Code: | 3xx xxx 33x
xx3 xxx xx3
xxx xxx xxx
x3x 3xx x3x
x33 3xx 333
x3x 3xx x3x
33x xxx xx3
xxx xxx xxx
x3x xxx x3x |
It is easy to see fairly quickly (and in a fascinating pattern) that r1c1 can't be 3 and r1c7 must be 3, and the puzzle falls from there.
Is that just colouring with a twist, a lazy man's swordfish, or am I engaging in "non-logical" solving techniques again?
BTW: Kevin Stone's solve used a swordfish on the 3s to elminate 3 from r7c2, revealing the 79 locked pairs in r7, removing the 9 from r7c1, and then a dandy xyz-wing with the r7c6, r8c1, ans r8c4 to remove the 5s from r7c1 and r8c6, leaving a 3 in r7c1 and resulting in a similar crumbling of the puzzle. |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Fri Jul 07, 2006 11:57 am Post subject: This technique is called "Nishio" |
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| AZ Matt wrote: | | Is that just colouring with a twist, a lazy man's swordfish, or am I engaging in "non-logical" solving techniques again? |
This technique is called "Nishio." The basic idea is that if a "3" is placed at r1c1, it is not possible to place all 9 "3"s on the grid. So the "3" in column 1 must appear at r7c1.
Ruud's "SudoCue" program calls this a "template" elimination. The name "Nishio" was coined in Japan, during the period when Sudoku was not very popular in Europe, Australia, and North America. dcb
PS The "Nishio" move also works before applying the "UR" elimination, and makes the "UR" much easier to analyze -- clearly r5c1 = 9. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Jul 07, 2006 1:37 pm Post subject: Valid puzzle? |
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The posted puzzle has two <2>'s in R7 and C8. Should R7C8 be 4?
Keith |
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Fri Jul 07, 2006 2:20 pm Post subject: Correcting the grid |
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Oh -- I noticed that, Keith, but didn't post the corrected grid. Here it is.
| Code: | 37 6 9 5 4 8 137 137 2
347 2 34 1 6 9 5 8 37
8 1 5 7 2 3 6 9 4
1 359 2 369 79 67 4 356 8
349 389 34 3689 5 2 137 1367 137
6 358 7 38 1 4 2 35 9
359 379 1 2 79 567 8 4 367
59 4 8 69 3 567 17 2 167
2 37 6 4 8 1 9 37 5 |
The puzzle can be solved without reference to the "non-unique rectangle" as follows.
1. The Nishio move. We have the following chain of inference.
r1c1 = 3 ==> r2c9 = 3 ==> r9c8 = 3 ==> r5c7 = 3
(r1c1 = 3 & r9c8 = 3) ==> r7c2 = 3
And now there's no possible way to fit a "3" in the middle left 3x3 box. So we can set r1c1 = 7 and r2c9 = 7, leaving the grid in this state.
| Code: | 7 6 9 5 4 8 13 13 2
34 2 34 1 6 9 5 8 7
8 1 5 7 2 3 6 9 4
1 359 2 369 79 67 4 356 8
349 389 34 3689 5 2 137 1367 13
6 358 7 38 1 4 2 35 9
359 379 1 2 79 567 8 4 36
59 4 8 69 3 567 17 2 16
2 37 6 4 8 1 9 37 5 |
Now there's a very nice double-implication chain sprouting from r8c1:
A. r8c1 = 5 ==> {3, 7, 9} triplet in row 7 ==> r7c9 = 6 ==> r8c9 = 1
B. r8c1 = 9 ==> r8c4 = 6 ==> r8c9 = 1
And setting r8c9 = 1 is enough to crack it wide open. dcb |
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AZ Matt
Joined: 03 Nov 2005
Posts: 63
Location: Hiding under my desk in Phoenix AZ USA
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| Posted: Fri Jul 07, 2006 3:51 pm Post subject: Oops! |
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Sorry about the typo in the grid  : r7c8=4
Great puzzle, isn't it? |
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Myth Jellies
Joined: 27 Jun 2006
Posts: 64
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| Posted: Fri Jul 07, 2006 5:35 pm Post subject: |
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There is a pattern based method for eliminating the 3 in r1c1 called the finned sashimi x-wing.
| Code: |
-37 6 *9 5 4 8 *137 137 2
347 2 #34 1 6 9 5 8 37
8 1 5 7 2 3 6 9 4
1 359 2 369 79 67 4 356 8
349 389 *34 3689 5 2 *137 1367 137
6 358 7 38 1 4 2 35 9
359 379 1 2 79 567 8 4 367
59 4 8 69 3 567 17 2 167
2 37 6 4 8 1 9 37 5 |
Either the fin in r2c3 is true or the sashimi x-wing in r15c37 is true (sashimi because the vertex of the x-wing in the same box as the fin is missing which does not hurt the x-wing at all). In either case r1c1 cannot be a 3. This reduction can also be shown via two connected strong links in columns 3 and 7...
| Code: |
3xx xxx 33x
xx3 xxx |x3
xx| xxx |xx
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x3| 3xx |3x
x33-3---333
x3x 3xx x3x
33x xxx xx3
xxx xxx xxx
x3x xxx x3x
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...which wipes out any 3 that sees both r1c7 and r2c3 (which in this case wipes out the 3s in r1c1 and r2c9.) |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Jul 07, 2006 8:14 pm Post subject: Another UR in the puzzle |
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There is also another UR in the puzzle:
| Code: |
7 6 9 5 4 8 13 13 2
34 2 34 1 6 9 5 8 7
8 1 5 7 2 3 6 9 4
1 359 2 369 79 67 4 356 8
349 389a 34 3689c 5 2 137 1367 13
6 358b 7 38d 1 4 2 35 9
359 379 1 2 79 567 8 4 36
59 4 8 69 3 567 17 2 16
2 37 6 4 8 1 9 37 5
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The cell marked "a" cannot be <3>, for that would force a non-unique pattern in "abcd". Note the strong links (an X-wing) on <8>.
Keith |
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TKiel
Joined: 22 Feb 2006
Posts: 292
Location: Kalamazoo, MI
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| Posted: Fri Jul 07, 2006 9:13 pm Post subject: |
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From David Bryant's ammended grid, after the eliminations from the swordfish on 3's you get this:
| Code: |
*--------------------------------------------------*
| 37B 6 9 | 5 4 8 | 137b 17 2 |
| 347 2 34 | 1 6 9 | 5 8 37B |
| 8 1 5 | 7 2 3 | 6 9 4 |
|----------------+----------------+----------------|
| 1 359 2 | 369 79 67 | 4 356 8 |
| 349 89 34 | 689 5 2 | 137 167 137 |
| 6 358 7 | 38 1 4 | 2 35 9 |
|----------------+----------------+----------------|
| 359A 79 1 | 2 79 567 | 8 4 367a |
| 59 4 8 | 69 3 567 | 17 2 167 |
| 2 37a 6 | 4 8 1 | 9 37A 5 |
*--------------------------------------------------*
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which leaves us with a colouring pattern similar to AZ Matt's original one, which is usually referred to as multi-colouring (more that one chain based on the same number). (B) shares a group with both (A) and (a), so (b) must be true. |
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dejsmith
Joined: 23 Oct 2005
Posts: 42
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| Posted: Sat Jul 08, 2006 6:07 pm Post subject: Does This Simple Coloring Work? |
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After I found the Type 4 UR in R25C13 & eliminated the 3s at R25C1, I noticed a potentially simple coloring on 3 proceeding from from R7C1 to R1C1 to R2C3 to R2C9 that appeared to eliminate 3 from R7C9. Even though that solved the puzzle, I became concerned if the 3 at R7C2 invalidated that logic. Does it?
37 6 9 5 4 8 137 137 2
47 2 34 1 6 9 5 8 37
8 1 5 7 2 3 6 9 4
1 359 2 369 79 67 4 356 8
49 389 34 3689 5 2 137 1367 137
6 358 7 38 1 4 2 35 9
359 379 1 2 79 567 8 4 367
59 4 8 69 3 567 17 2 167
2 37 6 4 8 1 9 37 5
Dave |
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TKiel
Joined: 22 Feb 2006
Posts: 292
Location: Kalamazoo, MI
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| Posted: Sun Jul 09, 2006 6:05 pm Post subject: |
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| No, the logic from the simple colouring chain tells you that either r2c9 or r7c1 must be a 3. Since r7c9 sees both colours of the chain, it can't be a 3 because either column 1 or row 2 would end up not having a 3. The fact that r7c2 could be a 3 has no bearing on the colouring logic, but if it were a 3 then r7c9 couldn't be a 3 either. |
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AZ Matt
Joined: 03 Nov 2005
Posts: 63
Location: Hiding under my desk in Phoenix AZ USA
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| Posted: Tue Jul 11, 2006 4:40 pm Post subject: Hmm... |
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| I saw the 3s the way Myth Jellies describes, which I now see is a form of x-wing, but doesn't that also mean the 3 must be in either r5c3 or r5c7, and therefore also eliminate all the other 3 candidates in r5? |
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AZ Matt
Joined: 03 Nov 2005
Posts: 63
Location: Hiding under my desk in Phoenix AZ USA
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| Posted: Tue Jul 11, 2006 5:23 pm Post subject: Oops! |
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| Never mind my previous post. Doh... |
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