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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Wed Mar 31, 2010 9:12 pm Post subject: Puzzle 10/03/31 BBDB/Advanced (I Think) |
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| Code: | +-----------------------+
| 9 4 . | . . . | 6 . . |
| 1 6 . | . . 7 | . 9 . |
| . . 5 | . . . | . . . |
|-------+-------+-------|
| . . . | . 9 . | . 4 . |
| . . . | 4 2 6 | 5 8 . |
| . 7 . | . 8 5 | . . 9 |
|-------+-------+-------|
| 4 . . | . 5 . | 1 . . |
| . 5 . | 2 6 . | . . 4 |
| . . . | . . 4 | . 5 . |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Thu Apr 01, 2010 8:20 am Post subject: |
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#1...x-wing on 7's
then this grid.
| Code: | +----------------------+-------------------+-----------------+
| 9 4 38(7) | 58 13 2 | 6 (17) 1358 |
| 1 6 238 | 58 34 7 | 48 9 2358 |
| 28(7) 238 5 | 6 134 9 | 478 12 1238 |
+----------------------+-------------------+-----------------+
| 5 28 268 | 7 9 13 | 23 4 16 |
| 3 19 19 | 4 2 6 | 5 8 7 |
| 26 7 4 | 3(1) 8 5 | 23 6(1) 9 |
+----------------------+-------------------+-----------------+
| 4 2389 236789 | 39 5 38 | 1 267 268 |
| (78) 5 189 | 2 6 -8(1) | 789 3 4 |
| 268 12389 123689 | 39(1) 7 4 | 89 5 268 |
+----------------------+-------------------+-----------------+ |
#2...(8=7)r8c1 - (7)r3c1 = (7)r1c3 - (7=1)r1c8 - (1)r6c8 = (1)r6c4 - (1)r9c4 = (1)r8c6; r8c6 <> 8
#3... xyz-wing {4,7,8} removes 8 from r3c9 |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Thu Apr 01, 2010 8:35 am Post subject: |
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after the x-wing here is a totally sick, scratch my eyes out move, but doesn't do much.
| Code: | +---------------------+-----------------+-------------------+
| 9 4 378 | 58 13 2 | 6 17 8-1(35) |
| 1 6 238 | 58 34 7 | 48 9 28(35) |
| 278 (238) 5 | 6 134 9 | 478 12 128(3) |
+---------------------+-----------------+-------------------+
| 5 (28) 268 | 7 9 (13) | 23 4 6(1) |
| 3 19 19 | 4 2 6 | 5 8 7 |
| 26 7 4 | 13 8 5 | 23 16 9 |
+---------------------+-----------------+-------------------+
| 4 (2389) 236789 | (39) 5 8(3) | 1 267 268 |
| 78 5 189 | 2 6 18 | 789 3 4 |
| 268 12389 123689 | 139 7 4 | 89 5 268 |
+---------------------+-----------------+-------------------+ |
(35)r12c9 = (3)r3c9 - (3=28)r34c2 - (28=39)r7c24 - (3)r7c6 = (3-1)r4c6 = (1)r4c9; r1c9 <> 1
[daj Edit: corrected typo in last strong link of Norm's chain.] |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Apr 02, 2010 5:06 am Post subject: |
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| I played the Type 4 UR (58), five ERs, three XY-Chains and finished with the XYZ-Wing (478). |
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wapati
Joined: 10 Jun 2008
Posts: 472
Location: Brampton, Ontario, Canada.
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| Posted: Fri Apr 02, 2010 12:02 pm Post subject: |
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I dropped this into Ruud's program and, though this isn't what it found, I saw that a UR4, x-wing, xy-wing and an xyz did it all, in some order.
Given how many are available finding the right 4 is indeed BBDB ! |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Fri Apr 02, 2010 2:40 pm Post subject: |
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| wapati wrote: | I dropped this into Ruud's program and, though this isn't what it found, I saw that a UR4, x-wing, xy-wing and an xyz did it all, in some order.
Given how many are available finding the right 4 is indeed BBDB ! |
All of the permutations my solver tried used at least one turbot fish class move trying to bypass the XYZ-Wing. Of course, it failed. With turbot fish moves always showing up in my solutions, it made it "5" steps -- my criteria for BBDB. In any case, there were technique choices to suit almost everyone.
Regards, Danny |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Fri Apr 02, 2010 3:59 pm Post subject: |
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Danny et al,
I'm still working on it (and may be wrong!!!) but I think that there is a loop which can prove that the 7 in r7c8 can't exist.
This means that r7c3 must be 7 doesn't it ? If it is the puzzle (I think!!) is done. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Fri Apr 02, 2010 7:13 pm Post subject: |
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| Mogulmeister wrote: | Danny et al,
I'm still working on it (and may be wrong!!!) but I think that there is a loop which can prove that the 7 in r7c8 can't exist.
This means that r7c3 must be 7 doesn't it ? If it is the puzzle (I think!!) is done. |
After Basics and before the X-Wing: r7c8<>7 does crack puzzle, but ...
It takes an advanced chain like Norm uses. Otherwise, you'll need an ALS XY-Wing for the elimination.
Note: there is a chain that attacks r1c8<>1 to crack the puzzle.
Regards, Danny |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Sat Apr 03, 2010 7:55 am Post subject: |
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As you say Danny, Norm would put this in a much tidier package but what I'm trying to say is that r7c8 can never be 7.
1)If r7c8 is not 7 then r7c3 is and puzzle solved.
2)However, If r7c8 is 7 then:
7(r8c7)=(7-4)r3c7=(4-1)r3c5=r1c5-r1c8
At which point we're saying r1c8 is not 1. This means we have a locked pair at r18c8 of <37> which eliminates the 7 at our start cell of r7c8
7(r8c7)=(7-4)r3c7=(4-1)r3c5=r1c5-r1c8(1=7)[r1c8|r8c8]-r7c8
SO r7c8 <> 7 therefore r7c3 = 7 and puzzle solved
I'm afraid I can't diagram this yet. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sat Apr 03, 2010 8:33 am Post subject: |
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| Mogulmeister wrote: | | I'm afraid I can't diagram this yet. |
Okay, you have the advanced chain that I mentioned because you used the <37> pair as part of a strong inference. Here's how I would write it.
| Code: | (7)r8c7 = (7-4)r3c7 = (4-1)r3c5 = (1)r1c5 - (1=37)r18c8 => r7c8<>7
+--------------------------------------------------------------------------------+
| 9 4 378 | 58 d13 2 | 6 e137 1358 |
| 1 6 238 | 58 34 7 | 48 9 2358 |
| 278 238 5 | 6 c134 9 | b478 1237 1238 |
|--------------------------+--------------------------+--------------------------|
| 5 28 268 | 7 9 13 | 23 4 16 |
| 3 19 19 | 4 2 6 | 5 8 7 |
| 26 7 4 | 13 8 5 | 23 16 9 |
|--------------------------+--------------------------+--------------------------|
| 4 2389 236789 | 39 5 38 | 1 236-7 2368 |
| 78 5 13789 | 2 6 138 | a789 e37 4 |
| 268 12389 123689 | 139 7 4 | 89 5 2368 |
+--------------------------------------------------------------------------------+
# 86 eliminations remain
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My chain is equivalent to the forcing chain w/ streams:
| Code: | *) r8c7= 7 => r7c8<>7
*) r8c7<>7 ... r18c8=37 => r7c8<>7
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Last edited by daj95376 on Sat Apr 03, 2010 8:40 am; edited 1 time in total |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Sat Apr 03, 2010 8:38 am Post subject: |
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| Thanks !! |
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