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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Mon Jan 19, 2009 4:55 am Post subject: BB 51784 |
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From April 14, 2008. If you like VH puzzles, this one's for you!
| Code: |
+-------+-------+-------+
| . 3 . | . . . | . 6 . |
| 1 . . | 9 . 3 | . . 8 |
| . . 8 | . 1 . | 3 . . |
+-------+-------+-------+
| . 8 . | 1 5 4 | . 3 . |
| . . 6 | 7 . 8 | 5 . . |
| . 1 . | 6 3 9 | . 7 . |
+-------+-------+-------+
| . . 7 | . 4 . | 1 . . |
| 4 . . | 5 . 1 | . . 7 |
| . 6 . | . . . | . 9 . |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
Keith |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Jan 19, 2009 5:27 am Post subject: |
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What is this??
In looking for creative ways to solve this one, after basics and a UR I get here:
| Code: | +----------------+----------------+----------------+
| 259 3 2459 | 248 78 257 | 279 6 1 |
| 1 457 245 | 9 6 3 | 27 245 8 |
| 6 4-579 8 | 24 1 257 | 3 245 259a |
+----------------+----------------+----------------+
| 7 8 29 | 1 5 4 | 29 3 6 |
| 3 49c 6 | 7 2 8 | 5 1 49b |
| 25 1 245 | 6 3 9 | 8 7 24 |
+----------------+----------------+----------------+
| 589 59d 7 | 38 4 6 | 1 25 235 |
| 4 2 3 | 5 9 1 | 6 8 7 |
| 58 6 1 | 238 78 27 | 4 9 35 |
+----------------+----------------+----------------+ |
abcd sure looks like an almost W-wing and, sure enough, R3C2 cannot be <5>. But, I don't quite see it.
It seems to me this is not an M-wing or W-wing variant, but simply a backwards way of looking at the X-wing on <9> that solves d as <5>.
(You can argue that if R3C2 is <5>, so is d. But that depends on the strong link on <9> in R3. a and d are not "pincers".)
Any opinions?
Keith |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Mon Jan 19, 2009 7:43 am Post subject: |
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keith, don't stop at r3c2. after that 5 is eliminated, then you can keep that chain going through the 9 in r1c1, then down to the 9 in r7c1 and the 5 is eliminated from that cell as well.
and then the 8 in r7c1 continues the chain to r9c1 and eliminates the 5 in that cell too.
so your chain goes through three cells that sees the 5 in r7c2 and eliminates all of them.
these eliminations compliment the damage done by the x-wing on 9 and bring you to the same point.
this chain will get the job done after basics
| Code: | .------------------.------------------.------------------.
| 259 3 2459 | 248 78 257 | 279 6 1 |
| 1 *457 245 | 9 6 3 |*27 245 8 |
| 6 *4579 8 | 24 1 257 | 3 245 *259 |
:------------------+------------------+------------------:
| 7 8 29 | 1 5 4 |*29 3 6 |
| 3 49 6 | 7 2 8 | 5 1 4-9 |
| 25 1 245 | 6 3 9 | 8 7 24 |
:------------------+------------------+------------------:
| 589 59 7 | 238 4 6 | 1 25 235 |
| 4 2 3 | 5 9 1 | 6 8 7 |
| 58 6 1 | 238 78 27 | 4 9 235 |
'------------------'------------------'------------------' |
(9=2)r4c7 - (2=7)r2c7 - (7)r2c2 = (7-9)r3c2 = (9)r3c9; r5c9 <> 9
don't know what to call it tho. |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Mon Jan 19, 2009 9:29 am Post subject: |
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| keith wrote: | What is this??
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Keith, it looks to me like you hooked up with a technique that I call Wishful Thinking. Unlike me, you actually ended up with a correct elimination. As for [r3c2]<>5, my solver ties into the X-Wing on <9> to get it.
| Code: | (5=9)r7c2 - (9)r5c2=r5c9-r3c9=r3c2 => [r3c2]<>5
+--------------------------------------------------------------+
| 259 3 2459 | 248 78 257 | 279 6 1 |
| 1 457 245 | 9 6 3 | 27 245 8 |
| 6 479-5 8 | 24 1 257 | 3 245 259 |
|--------------------+--------------------+--------------------|
| 7 8 29 | 1 5 4 | 29 3 6 |
| 3 49 6 | 7 2 8 | 5 1 49 |
| 25 1 245 | 6 3 9 | 8 7 24 |
|--------------------+--------------------+--------------------|
| 589 59 7 | 38 4 6 | 1 25 235 |
| 4 2 3 | 5 9 1 | 6 8 7 |
| 58 6 1 | 238 78 27 | 4 9 35 |
+--------------------------------------------------------------+
# 50 eliminations remain
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Mon Jan 19, 2009 4:05 pm Post subject: |
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More conventional for me.
x-wings on <9> and <2>.
xyz for <245>
xy for <479> |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Jan 19, 2009 4:57 pm Post subject: |
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A Finned W-Wing  |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Mon Jan 19, 2009 9:55 pm Post subject: |
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Here’s a possible explanation.
| Code: | +----------------+----------------+----------------+
| 259 3 2459 | 248 78 257 | 279 6 1 |
| 1 457 245 | 9 6 3 | 27 245 8 |
| 6 4-579 8 | 24 1 257 | 3 245 259a |
+----------------+----------------+----------------+
| 7 8 29 | 1 5 4 | 29 3 6 |
| 3 49c 6 | 7 2 8 | 5 1 49b |
| 25 1 245 | 6 3 9 | 8 7 24 |
+----------------+----------------+----------------+
| 589 59d 7 | 38 4 6 | 1 25 235 |
| 4 2 3 | 5 9 1 | 6 8 7 |
| 58 6 1 | 238 78 27 | 4 9 35 |
+----------------+----------------+----------------+ |
The conjugates b and c with respect to 9 are potentially the backbone of a W-wing. If 2 were not a candidate for a, the pattern would be a W-wing with pincers a and d giving Keith’s elimination.
When you think about it, though, the pincers are not required to be bivalent cells for the structure to work. One or both could be almost locked sets provided only that the backbone eliminates a candidate from the als.
From this point of view the upper pincer here is r3c489 with candidates {2, 4, 5, 9}. Once 9 is eliminated, 2, 4 and 5 are left to support eliminations in the rest of row 3. Of course, in this case, the other pincer, d, only supports the elimination of 5 but I think it justifies Keith’s approach and perhaps Marty’s comment.
This sort of thing can be useful in building chains and is clearly associated with the W-wing. I hesitate to call it a W-wing, however, because it lacks the readily spotted pair of identical bivalent pincers.
Steve |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Mon Jan 19, 2009 11:14 pm Post subject: |
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| Steve R wrote: | From this point of view the upper pincer here is r3c489 with candidates {2, 4, 5, 9}. Once 9 is eliminated, 2, 4 and 5 are left to support eliminations in the rest of row 3. Of course, in this case, the other pincer, d, only supports the elimination of 5 but I think it justifies Keith’s approach and perhaps Marty’s comment.
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Once <9> is eliminated in [r3c9], the strong link on <9> forces [r3c2]=9 and <>5.
Whether you use the strong link on <9> or the ALS or something else, the cells abcd are going to need outside help in order to produce an elimination. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Jan 19, 2009 11:27 pm Post subject: |
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| Marty R. wrote: |
A Finned W-Wing |
Marty,
| Quote: | At the risk of my being abused by the supposed (and self-anointed) fin cognoscenti, I say:
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You are correct!
This is a finned W-wing. Either the W-wing <59> in R3C9 is true, or the fin, R3C9 is <2>. Either way, R3C2 is not <5>.
If the W-wing is true, it also eliminates <5> in R7C9. I do not see that the fin helps to confirm that elimination.
Keith
(edited for grammar.) |
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