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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Tue Nov 18, 2008 8:34 am Post subject: |
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Hm, I don't think this has a lot to do with w-wings at all. Although, I can see that with a w-wing, the pincers usually don't share a house (don't know why, just from my experience) so another means of closing the loop is necessary.
With whatever type of m-wing, it is more common that both pincers see each other and the loop can be closed.
To find another example, I looked at today's Brainbasher's Nov. 18 super hard
This is the position after basics:
Code: |
+--------------------------+--------------------------+--------------------------+
| 5 239 28 | 4 89 1 | 36 369 7 |
| 89 6 7 | 39 289 238 | 4 5 1 |
| 39 4 1 | 5 7 6 | 2 8 39 |
+--------------------------+--------------------------+--------------------------+
| 1 5 4 | 39 6 38 | 378 379 2 |
| 389 239 6 | 7 289 4 | 5 1 39 |
| 7 239 28 | 1 5 238 | 368 369 4 |
+--------------------------+--------------------------+--------------------------+
| 6 7 9 | 2 3 5 | 1 4 8 |
| 4 8 3 | 6 1 7 | 9 2 5 |
| 2 1 5 | 8 4 9 | 37 37 6 |
+--------------------------+--------------------------+--------------------------+
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There are a number of coloring eliminations in (8), plus an x-wing in (3),
but this time I'd like to concentrate on the m-wing pattern, or rather (in order to avoid naming discussions at this point)
"The Pattern":
The Pattern is characterized by
- bi-value cell X:{a,b} (<2,8> in r6c3)
- "sees" (weak link in one of the two candidates, "a": in this case "8" in box 4)
- one end (cell V: r5c1) of a strong link in that candidate a ( (8)r5c1=r5c5 ).
- The other end of that strong link shares a cell W (r5c5) with one end of
- a strong link in the other candidate b ( (2)r5c5=r5c2 )
- the other end of that strong link ( cell Y: r5c2 ) and our first cell X form a new strong link ("pincers") that remove b from all cells that see both pincers.
Pattern: -(b=a)X-V=(a-b)W=Y- (sorry if the notation is wrong, but I hope you get the meaning)
Visually, I picture this situation as a hockey stick:
two strong links in different candidates meet in one cell, one of the ends "sees" the puck.
Sometimes I think of it as a flail: whereas the nunchuck has a piece of chain between the sticks, the parts of a flail are more closely connected (with a hinge or something. not sure, I've never used such a thing).
In this picture, the hockey stick is rather deformed:
In the example, this would allow us to remove 2 from r6c2.
But there is more (as re'born pointed out):
since the first and the last cell "see" each other, the whole chain now becomes a continuous loop and all the weak links act as strong links.
For example, let's look at the cell common to both strong links, called W in the above chain (r5c5):
we can re-write the loop (and "open" it at this new position)
-(b)W=Y-(b=a)X-V=W(a)-
What we get is an AIC that effectively says one of b and a must be true in cell W.
This means r5c5<>9
So, in addition to the "main" attraction (the elimination of 2 in r6c2), we get an elimination in a totally different candidate. Nice.
Same with the initial weak link (between the bi-value cell and the first strong link). We could remove all other "8"s from box 4 (but there aren't any in this example).
----
Edited to add picture of a flail:
Last edited by nataraj on Tue Nov 18, 2008 7:07 pm; edited 1 time in total |
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daj95376
Joined: 23 Aug 2008 Posts: 3854
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Posted: Tue Nov 18, 2008 10:48 am Post subject: |
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Nice presentation everyone
With any continuous loop, secondary eliminations (through weak links/inferences) are a bonus!
Myth Jellies wrote: | If the two endpoints candidates are weakly linked, then you have an (continuous) AIC loop.
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(Asellus) FWIW: The concept of an AIC loop seemed to be an oxymoron unless the AIC actually returned to the starting cell. Now, I realize that an AIC loop is also defined as an extension to an AIC.
Please tell me that it's a discontinuous AIC loop when an AIC returns to the starting cell for the same candidate. Otherwise, storm_norm and I need to update a recent post on AUR. |
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re'born
Joined: 28 Oct 2007 Posts: 80
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Posted: Tue Nov 18, 2008 2:39 pm Post subject: |
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keith wrote: | re'born wrote: |
Quote: | cells {a b}, {a b} that are connected by w-wings on both digits | In that case, the cells {a b} are a remote pair. Obviously. Why this is remarkable, I do not know. The first W-wing destroys the first candidate, the second the second. Knowing both of them together does not add anything to knowing them separately.
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Here is the puzzle from Bud's thread:
Code: |
*-----------------------------------------------------------*
| 5 67 3 | 9 2 68 | 4 178 16 |
| 4 167 168 | 5 67 3 | 678 9 2 |
| 89 2 689 | 17 4 168 | 678 5 3 |
|-------------------+-------------------+-------------------|
| 3 4 7 | 2 19 19 | 5 6 8 |
| 6 9 2 | 8 5 7 | 1 3 4 |
| 1 8 5 | 6 3 4 | 9 2 7 |
|-------------------+-------------------+-------------------|
| 2 5 4 | 3 1679 169 | 678 178 169 |
|B789 A16 A1689 |*17 1679 2 | 3 4 5 |
|b79 3 a169 | 4 8 5 | 2 *17 169 |
*-----------------------------------------------------------*
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Let's focus on the w-wing between r8c4<17> and r9c8<17> that goes through the 1's in box 7. When I first saw a w-wing (a long time ago...before it was called w-wing, y-wing style or semi-remote naked pair), I would have thought that this gave no eliminations on 7. Then, I learned to take the 7 in r9c8, transport it around to r8c1 and eliminate 7 from r8c5. Being a little older and a little wiser, I now see that in transporting, I created another w-wing, this time producing a strong link between the 1's. Again, there is no immediate deduction, but transporting the 1 in r8c4 to r9c3 or the 1 in r9c8 to r8c23 gives the deductions r9c9<>1 and r8c5<>1, respectively.
So in a very precise sense, you are correct, Keith. Individually, you get all of the deductions you get by taking the w-wings together and forming a continuous loop. However, it is, perhaps, quicker to see the continuous loop and make the deductions than it is to find all of the transport opportunities. On the other hand, one should recognize that any time you have a w-wing, you automatically can transport the end points along the w-wing path. |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Wed Nov 19, 2008 11:03 pm Post subject: |
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re'born wrote: | cells {a b}, {a b} that are connected by w-wings on both digits | keith wrote: | In that case, the cells {a b} are a remote pair. Obviously. Why this is remarkable, I do not know. The first W-wing destroys the first candidate, the second the second. Knowing both of them together does not add anything to knowing them separately. |
re'born wrote: | So in a very precise sense, you are correct, Keith. |
re'born is too kind.
If two cells {a b}, {a b} are connected as w-wings on both digits, then we can say:
One or both are {a} AND one or both are {b}.
But, if we put these together, there can only be: One is {a} and one is {b}.
Maybe the eliminations of this remote pair are the same as those made by the W-wings, but if you were to embed the W-wing pincers in some other pattern, I think the remote pair (W-wings together) is more powerful (and easier) than arguing the W-wings sequentially.
Keith |
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