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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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 Posted: Tue May 06, 2008 9:00 pm Post subject: For DP experts? |
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M6400430 (52)
| Code: |
+----------------+--------------+--------+
| 1358 1358 158 | 4 7 358 | 2 6 9 |
| 356 2 4 | 1356 156 9 | 8 7 15 |
| 7 589 5689 | 1568 2 58 | 4 3 15 |
+----------------+--------------+--------+
| 138 138 18 | 2 4 7 | 5 9 6 |
| 569 59 2569 | 1358 15 358 | 7 4 28 |
| 4 7 25 | 58 9 6 | 3 1 28 |
+----------------+--------------+--------+
| 89 6 89 | 7 3 2 | 1 5 4 |
| 2 4 7 | 56 56 1 | 9 8 3 |
| 15 15 3 | 9 8 4 | 6 2 7 |
+----------------+--------------+--------+
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This is from my pile of unsolved puzzles. I wouldn't bother posting it if it weren't that there are so many potential DPs. I have the feeling that one should be able to make some smart deductions, but I can't see any.
Any thoughts about DPs? - or indeed anything else (preferably without 'Medusa' in the technique name). The sole elimination I can make is the 5 in r2c4. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Tue May 06, 2008 9:56 pm Post subject: |
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It doesn't help much, but you can eliminate <1> from r4c2 with a DP-based argument.
r4c3 is <1> or it is <8>. If <8>, then r4c12 is {13} and r1c3 is <1>, so r1c12 is {358}. To avoid the {35},{13},{15} DP in r149c12, one of r1c12 must be <8>. (In fact, it must be r1c2; but that doesn't matter.) Thus, r3c2 is {59}, producing a {59} locked pair in c2, so r9c2 is <1>.
So, r4c3 and/or r9c2 is <1> and r4c2 cannot be. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Tue May 06, 2008 10:10 pm Post subject: |
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You can also eliminate the <5> in r5c4. There is a potential {25},{58},{28} DP in r56c349. The <2>s are an X-Wing. If r5c3 is <2>, r5c4 cannot be <5> (to avoid the DP). If r6c3 is <2>, r6c4 is <5>. Either way, r5c4 is not <5>.
Again, this is not greatly useful. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Tue May 06, 2008 10:44 pm Post subject: |
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As it happens, a relatively simple AIC solves the puzzle without need of any of the DP eliminations:
(5)r6c3-(5={18})r14c3-(8=9)r7c3-(9=9)r3c23-(9=5)r5c2-(5)r6c3; r6c3<>5 |
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Myth Jellies
Joined: 27 Jun 2006
Posts: 64
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| Posted: Wed May 07, 2008 1:51 am Post subject: Re: For DP experts? |
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| Code: |
+--------------------+--------------+--------+
|*138+5 *138+5 *18+5 | 4 7 358 | 2 6 9 |
| 356 2 4 | 1356 156 9 | 8 7 15 |
| 7 589 5689 | 1568 2 58 | 4 3 15 |
+--------------------+--------------+--------+
|*138 *138 *18 | 2 4 7 | 5 9 6 |
| 569 59 2569 | 1358 15 358 | 7 4 28 |
| 4 7 25 | 58 9 6 | 3 1 28 |
+--------------------+--------------+--------+
| 89 6 89 | 7 3 2 | 1 5 4 |
| 2 4 7 | 56 56 1 | 9 8 3 |
| 15 15 3 | 9 8 4 | 6 2 7 |
+--------------------+--------------+--------+
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One of my favorite DPs, the MUG. Here we have a 138 MUG in r14c123 which implies one of the fives in r1c123 must be true. Eliminate the other fives in box 1 and row 1.
This exposes an xy-wing around pivot point r3c2 which solves a 5 and exposes a naked triple.
| Code: |
*-----------------------------------------------------------*
| 1358 1358 158 | 4 7 38 | 2 6 9 |
| 36 2 4 | 13 16 9 | 8 7 5 |
| 7 89 689 | 68 2 5 | 4 3 1 |
|-------------------+-------------------+-------------------|
| 138 138 18 | 2 4 7 | 5 9 6 |
| 569 59 2569 | 13 15 38 | 7 4 28 |
| 4 7 25 | 58 9 6 | 3 1 28 |
|-------------------+-------------------+-------------------|
| 89 6 89 | 7 3 2 | 1 5 4 |
| 2 4 7 | 56 56 1 | 9 8 3 |
| 15 15 3 | 9 8 4 | 6 2 7 |
*-----------------------------------------------------------*
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The 15-UR in r19c12 => r1c3 <> 8
Then I've sort of stalled out with the DP's |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Wed May 07, 2008 1:58 am Post subject: DP |
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Asellus,
I believe that many of us would like to have the specifics of your AIC, being unfamiliar with that technique. Thanks
Earl |
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Myth Jellies
Joined: 27 Jun 2006
Posts: 64
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| Posted: Wed May 07, 2008 5:49 am Post subject: |
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| Code: |
*-----------------------------------------------------------*
| 1358 1358 A158 | 4 7 358 | 2 6 9 |
| 356 2 4 | 1356 156 9 | 8 7 15 |
| 7 589 5689 | 1568 2 58 | 4 3 15 |
|-------------------+-------------------+-------------------|
| 138 138 A18 | 2 4 7 | 5 9 6 |
|C569 D59 2569 | 1358 15 358 | 7 4 28 |
| 4 7 25 | 58 9 6 | 3 1 28 |
|-------------------+-------------------+-------------------|
|C89 6 B89 | 7 3 2 | 1 5 4 |
| 2 4 7 | 56 56 1 | 9 8 3 |
| 15 15 3 | 9 8 4 | 6 2 7 |
*-----------------------------------------------------------*
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Essentially, Asellus has noticed...
A: two-thirds of a naked triple (an ALS)
B: half of a naked pair
C: half of an x-wing on nines, and
D: another half a naked pair
...and he has noticed that all these sub-patterns connect together to build a pattern called an Alternating Inference Chain with a 5 in r1c3 and a 5 in r5c2 as the AIC endpoints. Just like an xy-wing, one of those endpoints must be true.
the preferred notation would be more like...
(5=1&8)r14c3 - (8=9)r7c3 - (9)r7c1 = (9)r5c1 - (9=5)r5c2 => r13c2 & r56c3 <> 5 |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Wed May 07, 2008 5:59 am Post subject: |
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That MUG is nice! I'll have to file that away in the memory cells somewhere.
As for the AIC... Earl, I'm not sure exactly what aspect of the AIC needs to be elaborated. So, I'll err on the side of providing too much information. Also, I'm repeating the grid for ease of reference:
| Code: | +----------------+--------------+--------+
| 1358 1358 158 | 4 7 358 | 2 6 9 |
| 356 2 4 | 1356 156 9 | 8 7 15 |
| 7 589 5689 | 1568 2 58 | 4 3 15 |
+----------------+--------------+--------+
| 138 138 18 | 2 4 7 | 5 9 6 |
| 569 59 2569 | 1358 15 358 | 7 4 28 |
| 4 7 25 | 58 9 6 | 3 1 28 |
+----------------+--------------+--------+
| 89 6 89 | 7 3 2 | 1 5 4 |
| 2 4 7 | 56 56 1 | 9 8 3 |
| 15 15 3 | 9 8 4 | 6 2 7 |
+----------------+--------------+--------+ |
When I exhaust other methods, I start exploring ALS and strong links for chain possibilities. Here, the {158} ALS in r14c3 caught my attention. If it's not <5> (in r1c3), then it's a {18} Locked Pair. Continuing as if an XY Chain to r7c3, we shift to <9>. Then, we use the strong link on <9> in r3 to continue the chain to the <9> at r3c2, then XY again to the <5> at r5c2. The <5>s at the ends of the chain are pincers. They actually eliminate the <5>s at r13c2 and r56c3, and not just the one at r6c3. However, the puzzle fell apart with just the latter elimination so that's all I notated.
Any ALS works essentially the same as a bivalue (which is just the smallest ALS) in a chain. If r1c3 were a {58} bivalue, then the entire chain would be an XY Chain with the exception of the strong link step in r3. The {158} ALS in r13c3 works the same way in this case.
Any ALS can be seen as an "extra digit" that has a conjugate ("either/or") link to a Locked Set of the remaining digit(s). It is easy to see that the two digits in a bivalue cell ALS have such a link. But, think of it for a moment as an extra digit linked to a naked single locked set. In the 2-cell {158} ALS, we can think of it as (5={18}), or in other words as an "extra" <5> digit strongly linked with a locked {18} pair. And, since it is a conjugate link, we can also think of it as (5-{18}), or as the <5> weakly linked with the locked {18} pair. Sometimes, a chain requires that we exploit it as a weak link rather than as a strong link. The nice thing about conjugate links is that they serve both strong and weak functions.
A couple more things to note about working with multi-cell ALS links in a chain...
The digits can be (and frequently are) grouped in the link. The links to or from such a group must be able to "see" the entire group. Here, the grouped <8>s in r14c3 (as part of the {18} pair) both see the next <8> in the r7c3 bivalue.
Also, I can consider any of the digits as the "extra" one. For instance, the chain above can also be considered (a bit less obviously) with the ALS expressed as ({15}=8)r14c3. Examining the referenced cells shows that there is only one <5> (in r1c3) and that the "8" part of the link is a pair of grouped <8>s in r14c3. [Note: In my chain, the <1> doesn't play a role, other than as part of the ALS. So, the option (1={58}) wouldn't work in that chain. But, it would work in some other chain that continued via a weak link of <1>s, though that's not possible in this grid.]
Now, as another example of using a 2-cell ALS as part of a chain, consider the {569} ALS in r5c12. A partial chain could be:
-(1=5)r5c5-({59}=6)r5c12-(6=6)r35c3-
The nodes of this fragment are a bivalue, a 2-cell ALS, and a strong link on <6> in c3, respectively. (As far as I can see, this fragment doesn't lead to a useful chain. But, I didn't try to exhaust the possibilities.)
And, one isn't limited to 2-cell ALS. For instance, here's a fragment using the 3-cell {1358} ALS in r1c123:
-(3=3)r15c6-(3={158})r1c123-(8={59})r35c2-(5=1)r9c2-
The nodes are: a strong link on <3> in c6; the 3-cell ALS; another 2-cell ALS; and a bivalue. Note that the <8> weak link works because the <8> in r3c2 can see all of the grouped <8>s in r1c123.
As always, an AIC requires that the links alternate weak-strong-weak-strong-etc., with the victim(s) weakly linked at both ends of the chain. If you need help following that, then let's return to the original chain (but showing all victims):
(5)r13c2|r56c3-(5={18})r14c3-(8=9)r7c3-(9=9)r3c23-(9=5)r5c2-(5)r13c2|r56c3; r13c2|r56c3<>5
Reading from left to right, all 4 victim <5>s are weakly linked to the <5> in r1c3, which in turn has a conjugate link with the {18} pair in the r14c3 ALS. So, we consider that conjugate link as strong. The {18} pair has a weak link with the <8> in r7c3 (since there is another <8> in c3). Then, the {89} bivalue conjugate link is considered as strong, (8=9). The link to the <9> at r3c3 is weak (because of the extra <9> in r5c3). Then, there is the conjugate (strong) link between the <9>s in r3. Next, the link between the <9>s in r3c2 and r5c2 is also conjugate (they are the only 2 <9>s in c2). But, we now need a weak link so consider it as weak. Next is the conjugate bivalue link to <5>, considered as strong, then the weak links to the victims again, completing the chain and eliminating the victims.
Finally(!!), considering multi-cell ALS, as well as grouped single digit links, for constructing an AIC is often productive in difficult puzzles. Medusa Multi-coloring is a good technique for uncovering many AICs. But, it can be a bit tedious to carry out and it isn't so good at revealing AICs that are based on the grouped sorts of links that I have been discussing here. If you are creative (and get comfortable with the techniques) it is sometimes possible to find a useful AIC when Medusa Multi-coloring isn't leading readily to eliminations by extending an otherwise useless AIC from Medusa using one or two of these grouped links. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Wed May 07, 2008 6:08 am Post subject: |
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Myth Jellies,
I saw one thread on the "players Forum" about Mugs, but it was more about discussing whether the pattern should be considered a technique. I can't remember if there were any other threads anywhere else that had some of the rules discussed, maybe you would know?? |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Wed May 07, 2008 9:14 am Post subject: |
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| Thanks folks, that's all very impressive, and very interesting. Like Norm I'd value a reference or two to MUGs. |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Wed May 07, 2008 7:32 pm Post subject: |
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OK, I've been doing some homework, looking up some old threads about DPs. Now, I understand the principle behind BUG-lites (which isn't to say that I could find them), such as the two posted by Myth Jellies & Steve R in a recent thread started by Keith. You could say that every subset of the BUG-lite pattern must be locked in each house in which it appears. (And I can see the fun in finding them if you're good enough at it.)
And I value and understand the stuff Asellus gave us, even if again I couldn't find it myself.
But I don't really see how you can identify a MUG. The 138 MUG found by Myth Jellies doesn't satisfy the condition that all subsets be locked in their houses. As far as I can understand the deep & difficult discussions on this subject, the criterion seems to be something like this: if an omnipotent clever clogs somehow filled in all the cells outside the MUG, then you'd be left with either an impossible grid, or one with more than one solution. And if I've interpreted that correctly, well, I don't see how to actually see these things. |
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Steve R
Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England
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| Posted: Wed May 07, 2008 9:55 pm Post subject: |
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Victor
Suppose you have a grid like this
| Code: | +-----------------------------+
| xyz xyz xyz | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
-------------------------------
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
-------------------------------
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| xyz xyz xyz | . . . | . . . |
+-----------------------------+ |
By reassigning x, y and z, if necessary, to match the solution you can assume the top line is
| Code: | +-----------------------------+
| x y z | . . . | . . . | |
There are just two possibilities for the bottom line
| | Code: | y z x | . . . | . . . |
+-----------------------------+ |
and
| Code: | | z x y | . . . | . . . |
+-----------------------------+ |
The crucial point is that, whichever possibility applies, you will get a second solution by transposing the first and ninth rows in each of the first three columns.
So a puzzle with a unique solution cannot have only x, y and z as the candidates for the six cells shown.
Steve |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Thu May 08, 2008 10:48 am Post subject: |
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| Thanks Steve, lucid and concise as ever! I do get that but, since I missed this one, life is such that I may well never see another. (I noted your recent remark that these things are rare.) |
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Myth Jellies
Joined: 27 Jun 2006
Posts: 64
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