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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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 Posted: Sun Apr 13, 2008 11:06 pm Post subject: competition #1008 |
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| Code: | 5..|.84|..1
...|.7.|.9.
...|...|478
---+---+---
.3.|...|2..
.7.|5.8|.4.
..1|...|.3.
---+---+---
32.|...|...
...|.5.|...
9..|12.|..7 |
w-wing on {3,6} to start |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Apr 13, 2008 11:23 pm Post subject: |
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Yes, but what to end? It has me stumped.
Keith |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Mon Apr 14, 2008 3:08 am Post subject: |
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I couldn't see a {36} W-Wing. This is the basic grid I got:
| Code: | +-------------+-------------+------------+
| 5 9 7 | 23 8 4 | 36 26 1 |
| 124 48 2348 | 236 7 126 | 5 9 23 |
| 12 6 23 | 239 19 5 | 4 7 8 |
+-------------+-------------+------------+
| 46 3 49 | 7 169 169 | 2 8 5 |
| 26 7 29 | 5 3 8 | 1 4 69 |
| 8 5 1 | 269 4 269 | 7 3 69 |
+-------------+-------------+------------+
| 3 2 5 | 8 69 7 | 69 1 4 |
| 7 1 68 | 4 5 369 | 3689 26 23 |
| 9 48 468 | 1 2 36 | 368 5 7 |
+-------------+-------------+------------+ |
In fact, rather than being a W-Wing, there is an AIC involving those remote {36}s that fixes r9c6 as <3>:
(3=6)r9c6-(6)r9c3=(6)r8c3-(6)r8c78=(6)r789c7-(6=3)r1c7-(3)r2c9
=(3)r8c9-(3)r8c6=(3)r9c6; r9c6=3
While that AIC is not too hard to see (try setting r1c7 to <3> and then to <6> and see that r9c6 is <3> in either case), the puzzle can be solved without it. The easiest route is some fairly simple Medusa Multi-coloring. However, without coloring, here's a path:
(1) 48 UR r29c23: r9c3<>4 and r2c3<>8
(2) 68 Type 4 UR r89c37: r89c7<>6
(3) XY Chain: r3c5-r3c1-r5c1-r4c1-r4c3; r4c5<>9
(4) XY Chain: r9c6-r9c3-r8c3-r8c8-r8c9; r8c6<>3
(5) XYZ Wing, pivot r4c6; r6c6<>6
(6) XY Chain: r6c6-r8c6-r8c8-r8c9-r2c9; r2c6<>2
(7) XY Wing, pivot r2c6; r7c5<>9 |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Apr 14, 2008 3:32 am Post subject: |
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| Quote: | | I couldn't see a {36} W-Wing. |
<36> in R9C6 and R1C7, with strong link on <6> in R7.
Keith |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Mon Apr 14, 2008 4:03 am Post subject: |
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| Thanks! Now I see it. |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Mon Apr 14, 2008 6:49 pm Post subject: |
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Edit! If you read some stuff I'd written here, please ignore: not for the first time, I didn't think carefully enough & wrote some rubbish.
Final statement still true: thanks S_N - really nice puzzle! |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Mon Apr 14, 2008 7:58 pm Post subject: another |
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Assellus,
Could you do the steps in your two UR's, 48 & 68.
It would help those of us not adept at complex UR's.
Earl |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Mon Apr 14, 2008 8:18 pm Post subject: Re: another |
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| Earl wrote: | Assellus,
Could you do the steps in your two UR's, 48 & 68.
It would help those of us not adept at complex UR's.
Earl |
First of all, that extra "s" is appropriate I hope you realize.
The 48 UR is the tricky one. Note that the <8>s in both r2 and c2 are strongly linked. And, note that the <4>s in in both c2 and r9. These are what allow the eliminations.
First, if r2c2 is <4>, then r2c8 must be <8> (due to the strong link), r9c2 is <8>, and r9c3 cannot be <4>. If r2c2 is not <4>, then r9c2 must be <4> (strong link) and r9c3 cannot be <4>. So, r9c3 cannot be <4> in either case.
Next, the <8> situation is done the same way. If r9c2 is <8>, r9c3 and r2c2 must be <4>, so r2c3 cannot be <8>. If r9c2 is not <8>, then r2c2 is <8>. So, r2c3 cannot be <8>.
This solves r9c2 as <4> and the 68 UR becomes a standard Type 4. R89c3 are both {68} bivalues. r89c7 are the "roof," both cells on that side of the UR having extra digits. The <8>s in the roof are strongly linked. Thus, the <6>s are eliminated from both roof cells (whichever of them is <8>, the other cannot be <6>). |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Tue Apr 15, 2008 10:54 am Post subject: |
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A one step solution.
| Code: | *-------------------------------------------------*
| 5 9 7 | 23 8 4 | 36 26 1 |
| 124 48 2348 | 236 7 126 | 5 9 23 |
|*12 6 -23 |#239 *19 5 | 4 7 8 |
|----------------+----------------+---------------|
| 46 3 49 | 7 169 169 | 2 8 5 |
|-26 7 *29 | 5 3 8 | 1 4 #69 |
| 8 5 1 |@269 4 269 | 7 3 @69 |
|----------------+----------------+---------------|
| 3 2 5 | 8 69 7 | 69 1 4 |
| 7 1 68 | 4 5 369 | 3689 26 23 |
| 9 48 468 | 1 2 36 | 368 5 7 |
*-------------------------------------------------*
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One of r3c4 and r5c9 must be 9:
r3c4<>9 => r6c4=9 => r6c9=6 => r5c9=9
r3c4=9 => r3c5=1 => r3c1=2
r5c9=9 => r5c3=2
==> r3c3<>2, r5c1<>2 |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Tue Apr 15, 2008 7:57 pm Post subject: |
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ravel,
That's nice... an otherwise useless Skyscraper activating <2> pincers. The same thing falls out quickly with Medusa multi-coloring. |
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