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Brainbashers Super Hard July 5: Colouring?

 
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AZ Matt



Joined: 03 Nov 2005
Posts: 63
Location: Hiding under my desk in Phoenix AZ USA

PostPosted: Thu Jul 06, 2006 11:30 pm    Post subject: Brainbashers Super Hard July 5: Colouring? Reply with quote

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

PostPosted: Fri Jul 07, 2006 11:57 am    Post subject: This technique is called "Nishio" Reply with quote

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: 3179
Location: near Detroit, Michigan, USA

PostPosted: Fri Jul 07, 2006 1:37 pm    Post subject: Valid puzzle? Reply with quote

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

PostPosted: Fri Jul 07, 2006 2:20 pm    Post subject: Correcting the grid Reply with quote

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

PostPosted: Fri Jul 07, 2006 3:51 pm    Post subject: Oops! Reply with quote

Sorry about the typo in the grid Embarassed : r7c8=4

Great puzzle, isn't it?
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Myth Jellies



Joined: 27 Jun 2006
Posts: 64

PostPosted: Fri Jul 07, 2006 5:35 pm    Post subject: Reply with quote

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
  |     |
x3| 3xx |3x
x33-3---333
x3x 3xx x3x

33x xxx xx3
xxx xxx xxx
x3x xxx x3x

...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: 3179
Location: near Detroit, Michigan, USA

PostPosted: Fri Jul 07, 2006 8:14 pm    Post subject: Another UR in the puzzle Reply with quote

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

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

PostPosted: Fri Jul 07, 2006 9:13 pm    Post subject: Reply with quote

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    |
 *--------------------------------------------------*


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

PostPosted: Sat Jul 08, 2006 6:07 pm    Post subject: Does This Simple Coloring Work? Reply with quote

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

PostPosted: Sun Jul 09, 2006 6:05 pm    Post subject: Reply with quote

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

PostPosted: Tue Jul 11, 2006 4:40 pm    Post subject: Hmm... Reply with quote

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

PostPosted: Tue Jul 11, 2006 5:23 pm    Post subject: Oops! Reply with quote

Never mind my previous post. Doh...
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