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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Wed Oct 15, 2008 3:40 pm Post subject: Vanhegan Extreme |
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Random 6-627799, rated 2.1.0.1.1.
Code: |
+-------+-------+-------+
| 9 3 . | 5 . 6 | . 4 7 |
| . . . | 9 1 7 | . . . |
| . . 7 | 3 . 4 | 9 . . |
+-------+-------+-------+
| . 9 . | . 5 . | . 6 . |
| 7 8 . | . . . | . 2 9 |
| . 2 . | . 9 . | . 3 . |
+-------+-------+-------+
| . . 5 | 2 . 1 | 8 . . |
| . . . | 8 3 9 | . . . |
| 8 1 . | 6 . 5 | . 7 2 |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
I had a couple of ERs and a seldom-seen Type 5 UR, but my arsenal isn't big enough to go further without my Medusa crutch. I suspect there are other techniques which will crack it. |
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Wed Oct 15, 2008 6:12 pm Post subject: |
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kite (1) row 1, col 1 => r4c7<>1
kite (6) row 7, col 7 => r2c2<>6
Here:
Code: |
+--------------------------+--------------------------+--------------------------+
| 9 3 128# | 5 -28 6 | 12* 4 7 |
| 246 45 248 | 9 1 7 | 26 58 3 |
| 126# 56 7 | 3 28* 4 | 9 158 156 |
+--------------------------+--------------------------+--------------------------+
| 14 9 3 | 147 5 2 | 47 6 8 |
| 7 8 14 | 14 6 3 | 5 2 9 |
| 5 2 6 | 47 9 8 | 147 3 14 |
+--------------------------+--------------------------+--------------------------+
| 3 46 5 | 2 7 1 | 8 9 46 |
| 246 7 24 | 8 3 9 | 146 15 1456 |
| 8 1 9 | 6 4 5 | 3 7 2 |
+--------------------------+--------------------------+--------------------------+
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(generalized) m-wing (2)r1c7=r3c5 via sl (1) box 1: r1c5<>2
This solves the puzzle. |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Wed Oct 15, 2008 9:44 pm Post subject: |
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From nataraj's PM:
Code: | sl (8) [r1],[c5]
sl (2) [r3],[c5]
sl (1) [b1]
If [r1c3]=8 then [r3c1]=2 and [b1]~1 => [r1c3]<>8
+--------------------------------------------------------------+
| 9 3 12-8 | 5 28 6 | 12 4 7 |
| 246 45 248 | 9 1 7 | 26 58 3 |
| 126 56 7 | 3 28 4 | 9 158 156 |
|--------------------+--------------------+--------------------|
| 14 9 3 | 147 5 2 | 47 6 8 |
| 7 8 14 | 14 6 3 | 5 2 9 |
| 5 2 6 | 47 9 8 | 147 3 14 |
|--------------------+--------------------+--------------------|
| 3 46 5 | 2 7 1 | 8 9 46 |
| 246 7 24 | 8 3 9 | 146 15 1456 |
| 8 1 9 | 6 4 5 | 3 7 2 |
+--------------------------------------------------------------+
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storm_norm
Joined: 18 Oct 2007 Posts: 1741
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Posted: Wed Oct 15, 2008 9:57 pm Post subject: |
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or do the generalized m-wing first then a xy-wing finishes it. |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Thu Oct 16, 2008 12:29 am Post subject: |
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Quote: | (generalized) m-wing (2)r1c7=r3c5 via sl (1) box 1: r1c5<>2 |
I see a strong link on 1 in box 1 and I see a couple of asterisks and a minus sign, but I don't have a clue what a generalized M-Wing is. |
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Thu Oct 16, 2008 5:45 am Post subject: |
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Marty,
what I call a "generalized" m-wing, is
- what Keith called "Appendix: ravel's half M-wing" in this thread
- without the restriction that the first and third cell in the chain must be identical.
The "classic" m-wing would look like this:
12-12-12=(2)=28
It forms two "pincers" - the cells containing 12 and 28 - and no cell that sees both pincers can contain 2. (I mark the pincer cells with "*").
The half m-wing is very similar, except that the second cell doesn't have to be identical to the first and third cells. BUT, there has to be a strong link on 1 between second and third cell. It would look like this:
12-128=(1)=12=(2)=28
It works exactly the same way as the classic m-wing
The "generalized" m-wing removes the restriction that first and third cell must be identical. The wing still works exactly the same way: if the first cell is "1", then the third cell is "1", then the fourth cell must be "2"
12-128=(1)=126=(2)=28
I mark the two "helper" cells (the ones with the strong link on 1) "#"
For a somewhat longer explanation, see this thread |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Thu Oct 16, 2008 3:53 pm Post subject: |
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[Withdrawn: After reading keith's reply, I realize that I was mistaken in thinking that the M-Wing was limited to four cells.]
Last edited by daj95376 on Fri Oct 17, 2008 12:37 pm; edited 6 times in total |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Thu Oct 16, 2008 3:59 pm Post subject: |
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Thanks Nataraj and Dan. |
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storm_norm
Joined: 18 Oct 2007 Posts: 1741
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Posted: Thu Oct 16, 2008 6:04 pm Post subject: |
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I'll provide the picture this time. you will see in the generalized M-wing that the third cell ( the cell that would normally house the same pair in the first cell) does have both candidates present. the difference is that because the link is weak within the cell that the cell can have more than just the two original candidates.
because you would think in your head
first cell: is not 2, must be 1
second cell: is not 1
third cell: has to be 1, cannot be 2 (and cannot be any other candidate in the cell)
fourth cell: has to be 2 (hence why the 2 in r1c5 can be eliminated)
so in the image above you can see the strong links in green on candidate 1.
the blue link is the strong link on 2 needed from r3c1 because if r3c1 is 1, then it can't be 2... that means r3c5 has to be 2.
and going in reverse, you would think in your head if r3c5 is not 2, then you would find that r1c7 has to be 2.
and just to go a little further... if that 6 in r3c1 were strongly linked anywhere else in the puzzle, it would be tempting to see what other eliminations could be made by seeing where the sprawling arms lead. |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Thu Oct 16, 2008 8:56 pm Post subject: |
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[Withdrawn: After reading keith's reply, I realize that I was mistaken in thinking that the M-Wing was limited to four cells.]
Aaaaaagh!!!
Last edited by daj95376 on Fri Oct 17, 2008 12:36 pm; edited 1 time in total |
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Thu Oct 16, 2008 9:46 pm Post subject: |
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Let's leave the "name"-topic for a second and look at the underlying pattern:
a) Ingredients:
- one strong link in candidate a
- one strong link in candidate b
(That shouldn't be too hard now, I think we might find a few in the cupboard of our puzzle ??...)
b) Preparation:
- combine the two links in a common cell (with possibly other candidates added), forming a hockey-stick pattern ( _/ )
- add a a bi-value cell {a,b}, (the puck) that sees one end of the hockey stick.
If done right, the whole pattern looks like this:
o..._/
c) Effects:
The puck and the grip form pincers to eliminate a candidate.
______
This pattern is both
- very common (although often duplicated by simple coloring)
- pretty effective in moderately difficult (vh+) puzzles
plus: relatively easy to spot (provided one looks for any sort of coloring elimination first - just combine two "useless" strong links)
Coming back to the "name" topic, actually I don't care a lot how this thing is called. It would be nice to have a name since it occurs quite often. Since you-know-what (a.k.a m-wing) is so no-no, we might call it a Voldemort just as well, or - returning full circle to the beginning of this post -
"Harvey Wallbanger"
Why? Because of the "Aaaaaagh !!!", of course
_______
Yes, it is a 4-cell chain / 5-link AIC.
With
G...grip
H...heel
T...toe
P...puck
the chain looks like this
G<a> - T<a> =(a)= H<a> =(b)= G<a>
and the AIC, I think, might be written somewhat like this:
-(b=a)P-(a)T=(a-b)H=(b)G-
but - for me, at least - neither chain nor AIC carry that visual image (and thus the search pattern) as well as the puck+hockey stick ... |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Thu Oct 16, 2008 10:36 pm Post subject: |
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[Withdrawn: After reading keith's reply, I realize that I was mistaken in thinking that the M-Wing was limited to four cells.]
Last edited by daj95376 on Fri Oct 17, 2008 4:52 am; edited 1 time in total |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Thu Oct 16, 2008 11:35 pm Post subject: I disagree, I think |
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I only have a few minutes, but let me say a couple of things.
(Strong link =, weak link -)
An M-wing starts with a complementary pair. Two cells that have the same two candidates, plus the knowledge that they have the same value.
XY ..?..XY
If the first and last cells are A and B: The value in A implies the same value in B, and vice-versa.
Now, suppose we can append a strong link on either X (or Y) at one of the ends:
XY ..?..XY = aX
where a is anything. Now we have A, B, C. If A is X, the value of C is immaterial. If A is Y, C must be X.
So, A and C are pincers. Any cell that sees both cannot be X.
So far, so good?
How are A and B connected to know they are the same? My original idea was that it does not matter. In the simplest case there are strong links on Y:
XY = bY = XY = aX
where b is anything. If you think about this long enough, you will realize that Medusa coloring will ALWAYS find this chain, hence the name "M-wing".
Two points:
1. XY = XY = XY = aX is, to me, not the simplest case of an M-wing. It is coloring on X.
2. Ravel's half-wing is a very astute observation that you may have A implies B, but B does not imply A. So,
XY - bY = XY = aX
makes the same eliminations as an M-wing.
Keith |
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