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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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 Posted: Sun Feb 10, 2008 9:21 am Post subject: LAT 08-Feb-08 |
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No one posted this week's 5-star LAT/Freep puzzle. So, here it is. It's not so easy.
| Code: |
. . . | 1 . . | . . .
. 6 . | . . . | 1 4 9
1 8 . | . . . | . . .
------+-------+------
. 1 . | 8 . 6 | . . 3
3 . . | 4 7 9 | . . 8
6 . . | 5 . 3 | . 7 .
------+-------+------
. . . | . . . | . 6 2
2 5 4 | . . . | . 9 .
. . . | . . 4 | . . .
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Sun Feb 10, 2008 1:05 pm Post subject: |
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| Code: | *--------------------------------------------------*
| 57 4 39 | 1 69 257 | 378 238 56 |
| 57 6 2 | 37 358 578 | 1 4 9 |
| 1 8 39 | 69 4 257 | 37 23 56 |
|----------------+----------------+----------------|
| 4 1 7 | 8 2 6 | 9 5 3 |
| 3 2 5 | 4 7 9 | 6 1 8 |
| 6 9 8 | 5 1 3 | 2 7 4 |
|----------------+----------------+----------------|
| 89 3 1 | 79 589 578 | 4 6 2 |
| 2 5 4 | 36 368 1 | 38 9 7 |
| 89 7 6 | 2 389 4 | 5 38 1 |
*--------------------------------------------------* |
I tried to use the 37 pair.
r2c4=7 => r3c7=7 did not help.
Then i found r3c7=3 => r3c3=9 => r3c4=6 => r8c4=3 => r2c4=7
Transporting r2c4=7 => r7c4<>7 => r7c6=7 then elimiates 7 from r3c6.
This only gives one number, but also opens an xy-chain to eliminate 3 from r1c3. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Feb 10, 2008 6:12 pm Post subject: |
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| I needed a Medusa trap because I couldn't see anything else. That led to a Type 1 UR and a W-Wing, after which another Medusa trap finished it off. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Sun Feb 10, 2008 7:35 pm Post subject: |
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Here's what I did:
From ravel's position, the xy-wing 39-69-36 is flightless, but extend the 36 in r8c4 via 38 - 38 in box 9 and the 3 in r3c8 is toast.
Unfotunately, this only solves r3c8 and r1c6.
But ... there is an xy chain now
'56' (r3c9) ... r3c6,r2c4...'36'(r8c4) that removes 6 from cell(s) r3c4 and the puzzle is solved |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Feb 10, 2008 10:03 pm Post subject: |
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| Quote: | | From ravel's position, the xy-wing 39-69-36 is flightless, but extend the 36 in r8c4 via 38 - 38 in box 9 and the 3 in r3c8 is toast. |
Nataraj,
How about an explanation for that extension? It has to be done via means that I'm unfamiliar with. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Feb 10, 2008 10:36 pm Post subject: |
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Marty,
The XY Wing induces a strong (inferential) link between the <3>s in the pincers, the <3>s in r8 are weakly linked, and those in b9 are strongly linked. Thus:
(3)r3c8-(3)r3c3=(3)r8c4-(3)r8c7=(3)r9c8-(3)r3c8; r3c8<>3
Transporting a pincer does not depend upon having only conjugate links. A short-cut way to consider it is: if r8c4 is <3>, then r8c7 is not <3>, so r9c8 must be <3>. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Feb 10, 2008 10:58 pm Post subject: |
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In Ravel's grid, the {89}r79c15 UR allows us to remove <8> from r7c5. This exposes a flightless XY Wing where the <5> in the r1c9 pincer can be transported to r3c6, eliminating <5> from r2c5 and r7c6. This gets us here:
| Code: |
*---------------------------------------------------*
| 57 4 39 | 1 69 (2)57 | 378 238 56 |
| 57 6 2 | 37 [3]8 57[8]| 1 4 9 |
| 1 8 39 | 69 4 257 | 37 [(2)]3 56 |
|----------------+----------------+-----------------|
| 4 1 7 | 8 2 6 | 9 5 3 |
| 3 2 5 | 4 7 9 | 6 1 8 |
| 6 9 8 | 5 1 3 | 2 7 4 |
|----------------+----------------+-----------------|
| 89 3 1 | 79 5 78 | 4 6 2 |
| 2 5 4 |[3]6 368 1 | 38 9 7 |
| 89 7 6 | 2 389 4 | 5 [3]8 1 |
*---------------------------------------------------*
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To avoid the {57}r12c16 UR, r1c6=2 and/or r2c6=8.
IF r1c6=2, then r3c8=2.
IF r2c8=8, r2c5=3, r8c4=3, r9c7=3, and r3c8=2.
Either way, r3c8=2.
This doesn't eliminate the need for an XY Chain. But, it is interesting. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sun Feb 10, 2008 11:21 pm Post subject: |
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By the way, basic Medusa alone (without extensions) can solve the puzzle completely. First:
| Code: |
*-----------------------------------------------------*
| 57 4 3a9A | 1 6A9a 257 | 378 238 5A6a |
| 57 6 2 | 3A7a 358 578 | 1 4 9 |
| 1 8 3A9a | 6a9A 4 25A-7 | 37 23 5a6A |
|----------------+------------------+-----------------|
| 4 1 7 | 8 2 6 | 9 5 3 |
| 3 2 5 | 4 7 9 | 6 1 8 |
| 6 9 8 | 5 1 3 | 2 7 4 |
|----------------+------------------+-----------------|
| 89 3 1 | 7A9a 589 -57a8 | 4 6 2 |
| 2 5 4 | 3a6A 36a8 1 | 38 9 7 |
| 89 7 6 | 2 389 4 | 5 38 1 |
*-----------------------------------------------------*
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There are two eliminations in c6. Continuing:
| Code: |
*-----------------------------------------------------*
| 57 4 3a9A | 1 6A9a 2A5-7 | 38 2a38 5A6a |
| 57 6 2 | 3A7a 3a8A -578a | 1 4 9 |
| 1 8 3A9a | 6a9A 4 2a5A | 7 2A3a 5a6A |
|----------------+------------------+-----------------|
| 4 1 7 | 8 2 6 | 9 5 3 |
| 3 2 5 | 4 7 9 | 6 1 8 |
| 6 9 8 | 5 1 3 | 2 7 4 |
|----------------+------------------+-----------------|
| 8a9A 3 1 | 7A9a 5 7a8A | 4 6 2 |
| 2 5 4 | 3a6A 36a-8 1 | 38 9 7 |
| 8A9a 7 6 | 2 -389A 4 | 5 38 1 |
*-----------------------------------------------------*
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These four eliminations finish it off. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Feb 11, 2008 1:57 am Post subject: |
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| Asellus wrote: | Marty,
The XY Wing induces a strong (inferential) link between the <3>s in the pincers, the <3>s in r8 are weakly linked, and those in b9 are strongly linked. Thus:
(3)r3c8-(3)r3c3=(3)r8c4-(3)r8c7=(3)r9c8-(3)r3c8; r3c8<>3
Transporting a pincer does not depend upon having only conjugate links. A short-cut way to consider it is: if r8c4 is <3>, then r8c7 is not <3>, so r9c8 must be <3>. |
Thanks. Undoubtedly easier for me to understand than to actually spot, but I'll keep my eyes open. |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Mon Feb 11, 2008 2:13 am Post subject: LA |
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Two xy-chains eliminating 7 from R1C7, and 3 from R9C5 will solve the puzzle.
Earl |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Mon Feb 11, 2008 2:48 am Post subject: |
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| wow, the people reading this paper are getting some hard puzzles to solve. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Feb 11, 2008 7:49 am Post subject: |
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| Marty R. wrote: | | Quote: | | From ravel's position, the xy-wing 39-69-36 is flightless, but extend the 36 in r8c4 via 38 - 38 in box 9 and the 3 in r3c8 is toast. |
Nataraj,
How about an explanation for that extension? It has to be done via means that I'm unfamiliar with. |
Thanks to Asellus for giving the explanation how this works - I'd just add a little more about how I work:
After reading about "transport" in this forum last year (some time late November early December, if I remember correctly), I've made it a habit to check every "useless" xy-wing for possible extensions into an xy-chain.
In order to be useful, an xy-chain must have the same candidate in both end cells, so for the xy-wing to xy-chain extension I first look for "two more" cells {a,b} that contain:
a - the pincer candidate from the xy-wing (in our case "3")
b - any other candidate
either: those two cells "see" each other and one of the two must see a pincer cell of the xy-wing
or: those two cells "see" different ends of the xy-wing (in that case it is "b" that gets eliminated by the resulting xy-chain)
example: our xy-wing 39-69-36 could be extended to:
39-96-63-38-83 like here in this puzzle, or:
53-39-96-63-35 in some other hypothetical puzzle
Of course, there is no guarantee to find a useful xy-chain, but it is extremely easy to do - so I made that check part of my routine (ahem, if I am in the mood for routine, that is).
And, yes, there are other transports but less easy to spot so I usually do not find them. My limit for xy-chains seems to be 4-5 chainlinks, maybe 6.
___
P.S. The way I look at xy-wings is like
"If r3c8 is not 3, then ... r8c4 is 3" - look for cells that see both ends and have 3
This extends naturally (for me) into
"...then r8c7 is not 3 then r9c8 is 3" - again look for cells that see both ends and have 3
edited 1217GMT (after Asellus' post) to include transport from both ends
Last edited by nataraj on Mon Feb 11, 2008 12:22 pm; edited 1 time in total |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Mon Feb 11, 2008 8:49 am Post subject: |
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| Marty wrote: | | Undoubtedly easier for me to understand than to actually spot, but I'll keep my eyes open. |
Actually, transporting pincers is exactly the same as "extending" Medusa coloring. You have two <3> pincers, A and B, say. Any <3> that must be true if the pincer A <3> is true can serve as a transported pincer A. And, any <3> which must be true if pincer B is true can serve as a transported pincer B. Seen this way, you can transport both pincers at the same time safely.
You can even change digits. For instance, say there is a <7> that must be true if the A <3> is true and also a <7> which must be true if the B <3> is true. Then, these can be transported pincers for <7> eliminations from a wing with pincers in <3>.
This "if this is true then" approach is not elegant and so is objectionable to some. But... we won't go down that road. Let's just say that it's the "poor man's approach" to finding AICs. |
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