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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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 Posted: Thu Dec 06, 2007 5:44 pm Post subject: another nightmare |
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I kept looking for UR solutions in today's nightmare from http://www.sudocue.net/daily.php, but no such luck (I am not too experienced in URs so I mostly depend on luck there).
Instead I found a w-wing (at least that's what I think it is) that works with a group of cells, for both candidates. Look at {4,5} in r3c3:
| Code: |
+--------------------------+--------------------------+--------------------------+
| 2 145679 3 | 57 8 47 | 169 14569 16 |
| 1459 14589 14589 | 2 345 6 | 7 13459 13 |
| 456 4567 45 | 1 3457 9 | 8 3456 2 |
+--------------------------+--------------------------+--------------------------+
| 139 1239 19 | 68 79 1237 | 4 68 5 |
| 8 12345 145 | 356 45 1234 | 1236 7 9 |
| 7 123459 6 | 358 459 1234 | 123 138 138 |
+--------------------------+--------------------------+--------------------------+
| 13 13 7 | 4 2 5 | 69 689 68 |
| 45 45 2 | 9 6 8 | 13 13 7 |
| 69 689 89 | 37 1 37 | 5 2 4 |
+--------------------------+--------------------------+--------------------------+
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if r3c3=4 then r23c1<>4 then r8c1=4 then r8c2=5
if r3c3=5 then r23c1<>5 then r8c1=5 then r8c2=4
the first line means that (if r3c3<>5 then r8c2=5) we can remove 5 from r123c2, the second line means we can remove 4 from those same cells.
Another try at Eureka:
(5=4)r3c3-(4)r23c1=(4)r8c1-(4=5)r8c2;r123c2<>5
After this, stuck again.
But, another grouped chain in boxes 4 and 7 helps to eliminate 1 from r5c3:
(5)r5c3=(5)r56c2-(5=4)r7c2-(4)r56c2=(4)r5c3;r5c3<>1
in words: if r5c3 is not 5, then one of r56c2 must be 5, then r7c2 is not 5.
but then r7c2 must be 4. Which means that none of r56c2=4 but then r5c3=4.
That means r5c3 is either 4 or 5 but not 1.
This solves the puzzle. Not quite the orthodox solution, I think. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Dec 06, 2007 10:19 pm Post subject: Re: another nightmare |
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| nataraj wrote: | I kept looking for UR solutions in today's nightmare from http://www.sudocue.net/daily.php, but no such luck (I am not too experienced in URs so I mostly depend on luck there).
Instead I found a w-wing (at least that's what I think it is) that works with a group of cells, for both candidates. Look at {4,5} in r3c3:
| Code: |
+--------------------------+--------------------------+--------------------------+
| 2 145679 3 | 57 8 47 | 169 14569 16 |
| 1459 14589 14589 | 2 345 6 | 7 13459 13 |
| 456 4567 45 | 1 3457 9 | 8 3456 2 |
+--------------------------+--------------------------+--------------------------+
| 139 1239 19 | 68 79 1237 | 4 68 5 |
| 8 12345 145 | 356 45 1234 | 1236 7 9 |
| 7 123459 6 | 358 459 1234 | 123 138 138 |
+--------------------------+--------------------------+--------------------------+
| 13 13 7 | 4 2 5 | 69 689 68 |
| 45 45 2 | 9 6 8 | 13 13 7 |
| 69 689 89 | 37 1 37 | 5 2 4 |
+--------------------------+--------------------------+--------------------------+
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if r3c3=4 then r23c1<>4 then r8c1=4 then r8c2=5
if r3c3=5 then r23c1<>5 then r8c1=5 then r8c2=4
the first line means that (if r3c3<>5 then r8c2=5) we can remove 5 from r123c2, the second line means we can remove 4 from those same cells.
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It's a W-Wing, but not worded in the way that I learned the technique. Consider the 45s in r3c3 and r8c2. They are connected by the 4s and 5s in column 1, boxes 1 and 7. Since the connection is with both 4 and 5, both numbers can be eliminated from cells that see both of the pairs. Notation and testing are not necessary. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Fri Dec 07, 2007 8:46 am Post subject: |
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nataraj,
Your {45}s are a remote naked pair: both candidates are strongly linked and thus conjugate. (That's why re'born refers to a W-Wing as a "semi-" remote naked pair.)
Your <1> elimination in Box 4 is the "famous Sue" de Coq again! Take another look. There are 6 digits in 6 cells that can be grouped into two overlapping Locked Sets: {1239} in Box 4 and {45} in Column 2. The cells involved are R4C123|R568C2. There can be no other <4> or <5> in C2 and no other <1>, <2>, <3> or <9> in B4. Notice that this means that those remote pair eliminations would have been accomplished with this Sue de Coq alone.
I'm not sure how this completed the puzzle without some further advanced step. I couldn't. Next, I got to here:
| Code: | +----------------------+-----------------+------------------+
| 2 1679 3 | 57 8 47 | 169 14569 16 |
| 1459 189 189 | 2 345 6 | 7 13459 13 |
| 456 67 45 | 1 3457 9 | 8 3456 2 |
+----------------------+-----------------+------------------+
| 139 1239 19 | 68 #7-9 1237 | 4 68 5 |
| 8 123 45 | 36 @45 123 | 1236 7 9 |
| 7 #-1-2-345-9 6 |@358 @459 @1234 |@123 @138 @138 |
+----------------------+-----------------+------------------+
| 13 13 7 | 4 2 5 | 69 689 68 |
| 45 45 2 | 9 6 8 | 13 13 7 |
| 69 689 89 | 37 1 37 | 5 2 4 |
+----------------------+-----------------+------------------+ |
It's another Sue de Coq! There is a {1238} Locked Set in R6 overlapped with a {459} Locked Set in B5, marked @: 7 digits in 7 cells. Five eliminations occur in the two cells marked #. (<9> is removed from R6C2 since the only place for <9> in the Sue de Coq cells can see both houses.)
But this didn't solve it for me, either. There is something else interesting that was there from the beginning. So, I'll go back to the first grid:
| Code: | +--------------------+---------------+---------------------+
| 2 #145-679 3 | 57 8 47 | B169 #145-69 B16 |
| 1459 14589 14589 | 2 345 6 | 7 13459 13 |
| 456 4567 45 | 1 3457 9 | 8 #345-6 2 |
+--------------------+---------------+---------------------+
| 139 1239 19 | 68 79 1237 | 4 A68 5 |
| 8 12345 145 | 356 45 1234 | 1236 7 9 |
| 7 123459 6 | 358 459 1234 | 123 138 138 |
+--------------------+---------------+---------------------+
| 13 13 7 | 4 2 5 |AB69 A689 68 |
| 45 45 2 | 9 6 8 | 13 13 7 |
| 69 689 89 | 37 1 37 | 5 2 4 |
+--------------------+---------------+---------------------+ |
There are two otherwise useless XYZ Wings, marked A and B. Both would eliminate <6>. And, the two <6>s in Box 6 are strongly linked. Looking at the "A" Wing, if R7C7 is <6>, then R4C8 is <6> and the cells marked # cannot be <6>. If R7C7 is not <6>, then one of R47C8 is <6> and the cells marked # again cannot be <6>. (The same reasoning works with the "B" Wing. We only need one of them.) Essentially, the strong link in Box 4 transports R7C7 to R4C8 and allows either Wing to make the eliminations.
But, even when I used this in addition to the two Sue de Coqs, the puzzle wasn't solved!
I had to color <4> and then <5> to solve the puzzle. (The interesting bit with the <6>s wasn't required for the solution.)
[Edit to add a missing parenthesis.] |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Fri Dec 07, 2007 6:15 pm Post subject: |
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| Asellus wrote: | Next, I got to here...
But this didn't solve it for me, either. |
Asellus, in your first grid there is a naked quad in box 4 (1239) , which makes r6c2={45} and then leaves a naked single 9 in r6c5.
But I had to do some coloring, too, in order to remove 4 from r3c8 and 5 from r3c8, plus again one more time to remove 4 from r3c8. Sorry, I was so exhilarated after those 2 grouped exclusions (and would have been more so had I known I'd finally encountered a Sue de Coq!) that I didn't mention those coloring eliminations.
That was a tough one, indeed! |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Fri Dec 07, 2007 6:21 pm Post subject: |
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| Asellus wrote: | | Your {45}s are a remote naked pair: both candidates are strongly linked and thus conjugate. |
Ah, revelation!
I get it:
- both candidates in both cells strongly linked -> remote naked pair
- one candidate strongly linked -> W-wing
and now I finally understand the "semi". (That was Nov 1st when we had that long discussion about proper behaviour of pairs, remote or otherwise, wasn't it?)
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Fri Dec 07, 2007 10:11 pm Post subject: |
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| nataraj wrote: | | Asellus wrote: | | Your {45}s are a remote naked pair: both candidates are strongly linked and thus conjugate. |
Ah, revelation!
I get it:
- both candidates in both cells strongly linked -> remote naked pair
- one candidate strongly linked -> W-wing
and now I finally understand the "semi". (That was Nov 1st when we had that long discussion about proper behaviour of pairs, remote or otherwise, wasn't it?)
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Not quite, I think:
1. There is the classic remote pair
WX=WX=WX=WX
where the same pair occurs in all cells in the chain. = is a strong link, obviously in both candidates. One end must be W, the other must be X.
2. There is the other remote pair:
WX=Xy=Xz=WX
where y and z are any candidates (and each may be more than one candidate). = is a strong link on X only. One end must be X, the other must be W. You do NOT need strong links on both candidates, only on one.
3. There is the semi-remote pair (W-wing or remote semi-pair):
WX-Xy=Xz-WX
- is a weak link on X, = is a strong link. One end must be W.
4.? There was an interesting one last weekend, which I did not point out: Two cells WX which were both a W-wing on W, and a W-wing on X. (The strong link on X was in different cells than the one for W.) The net result is a naked pair. (I'll post that remark in the appropriate thread.)
Keith
Last edited by keith on Sat Dec 08, 2007 6:40 am; 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: Fri Dec 07, 2007 11:50 pm Post subject: |
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Marty,
First, in your item 3 for the W-Wing, you should say "one or both ends must be W."
In the case of the {45} pair in the puzzle above, it is a remote naked pair, though not with a pattern we may have seen before. The {45} pair can be seen as two possible W-Wings: one that eliminates <4>s and one that eliminates <5>s. By W-Wing logic, one or both cells must be <4> (strong inference). But also, one or both cells must be <5>. Because of this double strong inference in the same two cells, the "both can be true" part of the strong inference collapses and we are left with strong conjugate links: one cell must be <4> and the other must be <5>.
(It doesn't matter that the two alternate W-Wings involve weak links. Once we have those two strong inferences, the other links don't matter.) |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Sat Dec 08, 2007 12:01 am Post subject: |
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Marty,
On careful re-reading, I realize that you are not necessarily disagreeing that the current case is a remote naked pair, since your item 4 is another example of the same thing. So, perhaps it is nataraj's "strong link" wording that is the trouble.
The first statement | Quote: | | - both candidates in both cells strongly linked -> remote naked pair |
is fine, it seems to me.
The second | Quote: | | - one candidate strongly linked -> W-wing |
should say "- one candidate with a strong inferential link -> W-Wing."
These statements refer only to the results within the two remote bivalue cells, not how we got there. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat Dec 08, 2007 1:43 am Post subject: |
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| Quote: | Marty,
On careful re-reading, I realize that you are not necessarily disagreeing that the current case is a remote naked pair, since your item 4 is another example of the same thing. So, perhaps it is nataraj's "strong link" wording that is the trouble. |
Asellus, I think you might have meant to address Keith since I had no numbered items. But as long as I'm typing, I might add that, being the non-theorist that I am, I wouldn't know what to do with the distinctions between W-Wing, Remote Naked Pairs and Semi-Remote Naked Pairs other than to get more confused than I already am.
If there are identical pairs in different houses and they're connected with strong links, it's a W-Wing to me whether they're connected by one or both of the numbers. |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Sun Dec 09, 2007 10:28 pm Post subject: |
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Mama mia! This is the most mind-bogglingly confusing & difficult discussion I've encountered since I began studying sudokus more seriously a week or two ago. Guess I'll have to leave it and come back later & hope it makes sense then. Meanwhile ....
.... it doesn't do much, but there is at least one UR in there. Look at the 13s in R4C12, R7C12. You can eliminate the 1 in R4C2. Try running clockwise round this rectangle starting with that 1 and you'll produce a DP. I don't know which type this is, but it depends on the strong link in 3s in C1. |
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