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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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 Posted: Sun Feb 28, 2010 3:06 am Post subject: M-Wing question |
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| Code: |
+--------------+-----------+--------------+
| 7 3 2456 | 469 12 46 | 125 159 8 |
| 25 15 9 | 7 3 8 | 125 6 4 |
| 24 148 2468 | 469 12 5 | 3 7 29 |
+--------------+-----------+--------------+
| 9 45 1 | 3 8 7 | 456 2 56 |
| 8 2 7 | 46 5 46 | 9 3 1 |
| 3 6 45 | 1 9 2 | 8 45 7 |
+--------------+-----------+--------------+
| 245 7 3 | 8 6 9 | 1245 145 25 |
| 6 9 58 | 2 4 1 | 7 58 3 |
| 1 48 248 | 5 7 3 | 246 489 269 |
+--------------+-----------+--------------+
|
Play this puzzle online at the Daily Sudoku site
This is the 2-28 VH which is solvable by an XY-Wing or UR. Note the 25 cells in boxes 19. Is that a valid M-Wing connected by 5's? The connection isn't the usual two simple strong links, but the effect seems to be the same, as if the extra 5's weren't present in r7c78. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Feb 28, 2010 3:33 am Post subject: |
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Marty,
| Code: | +----------------+----------------+----------------+
| 7 3 2456 | 469 12 46 | 125 159 8 |
| 25a 15 9 | 7 3 8 | 125c 6 4 |
| 24 148 2468 | 469 12 5 | 3 7 -29 |
+----------------+----------------+----------------+
| 9 45 1 | 3 8 7 | 456 2 56 |
| 8 2 7 | 46 5 46 | 9 3 1 |
| 3 6 45 | 1 9 2 | 8 45 7 |
+----------------+----------------+----------------+
| 245 7 3 | 8 6 9 |1-245 145 25b |
| 6 9 58 | 2 4 1 | 7 58 3 |
| 1 48 248 | 5 7 3 |-246 489 269 |
+----------------+----------------+----------------+ |
Well, you have the basis of an M-wing.
In an astounding case of grouped coloring on 5, you can show that if a is 5, b is 5. Also, by a separate path, if b is 5, a is 5. So, a and b are the basis of an M-wing, which is completed by the strong link ac in R2.
Making the three eliminations shown.
By the way, you do not need to show if a is 5, b is 5. You only need the half wing, if b is 5 a is 5.
Logic:
b is 2, or
b is 5, a is 5, c is 2.
b and c are pincers on 2.
Keith  |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Feb 28, 2010 4:51 am Post subject: |
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Thanks Keith. Hopefully, my M-Wings can expand if I don't dismiss them because of the lack of two simple strong links and start looking more closely.
In this case, I established that if b=5, then a=5. However, I'm not seeing how b=5 if a=5.
| Quote: | Logic:
b is 2, or
b is 5, a is 5, c is 2.
b and c are pincers on 2. |
I think in different terms. If b=5, then r7c1<5>r2c7. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Feb 28, 2010 4:52 am Post subject: |
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Thanks Keith. Hopefully, my M-Wings can expand if I don't dismiss them because of the lack of two simple strong links and start looking more closely.
In this case, I established that if b=5, then a=5. However, I'm not seeing how b=5 if a=5. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Feb 28, 2010 5:23 am Post subject: |
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| Code: | +----------------+----------------+----------------+
| 7 3 2456 | 469 12 46 | 125f 159 8 |
| 25a 15 9 | 7 3 8 | 125c 6 4 |
| 24 148 2468 | 469 12 5 | 3 7 -29 |
+----------------+----------------+----------------+
| 9 45d 1 | 3 8 7 | 456 2 56 |
| 8 2 7 | 46 5 46 | 9 3 1 |
| 3 6 45 | 1 9 2 | 8 45e 7 |
+----------------+----------------+----------------+
| 245 7 3 | 8 6 9 |1-245 145 25b |
| 6 9 58d | 2 4 1 | 7 58 3 |
| 1 48 248 | 5 7 3 |-246 489 269 |
+----------------+----------------+----------------+ |
If a is 5 (true), so are the cells d. If d are true, so is e. If e is true, so is f. If f is true, so is b.
(As you go through this, you need to mark the cells that are not true, if a is true.)
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Feb 28, 2010 6:15 am Post subject: |
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| Thanks, I can follow it, but it's way too convoluted for me to have found it. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Feb 28, 2010 2:10 pm Post subject: |
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| Marty R. wrote: | | Thanks, I can follow it, but it's way too convoluted for me to have found it. |
Marty,
In many situations, I find that following implications is very similar to your original question about the extra 5s in r7c78 possibly influencing the link between the two 25s in r2c1 and r7c9.
In this specific case, I would walk the path as:
If r2c1=5, then r2c2<>5, then r4c2=5 due to the strong link in col2,
If r4c2=5, then r4c9<>5, then r7c9=5 due tot he strong link in col9.
Going the other direction, I get:
If r7c9=5, then r7c1<>5, then r2c1=5 due to the strong link in col1.
Viewed in this manner, I do not find the path as convoluted.
Ted |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Feb 28, 2010 3:34 pm Post subject: |
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FWIW: Ted's cells are the vertices of a (222) Swordfish, with the base set being the three columns with strong links.
| Code: | Swordfish c129\r247 => r247c7,r7c8 <> 5
+-----------------------------------+
| . . 5 | . . . | 5 5 . |
| *5 *5 . | . . . | -5 . . |
| . . . | . . 5 | . . . |
|-----------+-----------+-----------|
| . *5 . | . . . | -5 . *5 |
| . . . | . 5 . | . . . |
| . . 5 | . . . | . 5 . |
|-----------+-----------+-----------|
| *5 . . | . . . | -5 -5 *5 |
| . . 5 | . . . | . 5 . |
| . . . | 5 . . | . . . |
+-----------------------------------+
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Since it's a (222) Swordfish, it can also be described as a continuous loop with the same eliminations.
| Code: | (5): r2c2 = r4c2 - r4c9 = r7c9 - r7c1 = r2c1 - loop => r247c7,r7c8 <> 5
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Notice in the loop that r2c1, r4c2, and r7c9 are treated as true. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Feb 28, 2010 4:31 pm Post subject: |
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| Quote: | In this specific case, I would walk the path as:
If r2c1=5, then r2c2<>5, then r4c2=5 due to the strong link in col2,
If r4c2=5, then r4c9<>5, then r7c9=5 due tot he strong link in col9.
Going the other direction, I get:
If r7c9=5, then r7c1<>5, then r2c1=5 due to the strong link in col1.
Viewed in this manner, I do not find the path as convoluted. |
Ted, starting with r7c9, that's exactly my path. As to going the other way, it's really not convoluted; that was a poor choice of words. All I meant was that I probably wouldn't have thought to try that train of thought.
I have one (hopefully) final question for anybody. In my previous M-Wings I used only simple strong links so it could always be proven that one cell forced the other and vice versa and either cell could be extended by the other number to form the pincers.
In this actual case it was shown that if a=5, then b=5 and vice versa. However, if I can prove that if a=5, then b=5, but can't prove it going the other way, does that mean that only b can be extended to form pincers? |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Feb 28, 2010 4:38 pm Post subject: |
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[Withdrew: Top part of post redundant.]
Marty: In your original description, you seemed fixated on the chain for <5> and ignored any chain on <2>. To me, it seemed that you missed the M-Wing completely. But, I've been wrong many other times.
[Edit: changed wording to better describe my interpretation -- which turned out to be wrong, again!]
Last edited by daj95376 on Sun Feb 28, 2010 6:30 pm; edited 2 times in total |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Feb 28, 2010 5:55 pm Post subject: |
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| Quote: | | Marty: In your original description, you were fixated on the chain for <5> and completely ignored any chain on <2>. To me, it seemed that you missed the M-Wing completely. But, I've been wrong many other times. |
Unlike you, I've never been wrong. 
I don't know if I was "fixated" on 5, but that's the connection I saw and the extension of the 2 solved the puzzle in one step. I just wanted to make sure that my method of connection was valid, as I was used to forming M-Wings only via simple strong link connections. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Feb 28, 2010 9:55 pm Post subject: |
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| Quote: | | In this actual case it was shown that if a=5, then b=5 and vice versa. However, if I can prove that if a=5, then b=5, but can't prove it going the other way, does that mean that only b can be extended to form pincers? |
Marty, YES!!
And, there is a little more:
1. Suppose a is 25 and b is 25, and you can prove that 5 in a forces 5 in b. You can add the strong link in 2 to b to make the M-wing.
2. Suppose a is 25 and b is 2567, and you can prove that 5 in a forces 5 in b. You can add the strong link in 2 to b to make the M-wing.
Something like this: a(25) - b(25x) = c(2y)
5 in a forces 5 in b. Not 2 in b forces 2 in c. a and c are pincers in 2. (x and y can be any candidates.)
Case 1 above is a half M-wing. Case 2 is a generalized M-wing.
Remember to check whether a and c are in the same house (row, column, box). If they are, you have an M-cycle, and additional eliminations are possible.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Feb 28, 2010 10:09 pm Post subject: |
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Thank you!!! 
| Quote: | | Case 1 above is a half M-wing. |
If you used one and posted your solution, would you list that step as a half M-Wing, as opposed to M-Wing?
| Quote: | | Case 2 is a generalized M-wing |
I think this might be a 2nd definition I've seen of a gM-Wing. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Feb 28, 2010 10:25 pm Post subject: |
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| Quote: |
If you used one and posted your solution, would you list that step as a half M-Wing, as opposed to M-Wing? |
I would call it an M-wing, but then I always show the cells involved when I post a solution.
| Quote: | | I think this might be a 2nd definition I've seen of a gM-Wing. |
| Quote: | | `When I use a word,' Humpty Dumpty said, in rather a scornful tone, `it means just what I choose it to mean -- neither more nor less.' |
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Mar 01, 2010 12:04 am Post subject: |
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I'm too old to have remembered anything about Humpty Dumpty, so I'll take your word on his wisdom.
Thanks again, I have no further questions for this witness. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Mar 01, 2010 4:24 am Post subject: |
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I saw that but don't understand one iota of it. The two cells are 17 and 159? |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Mon Mar 01, 2010 4:44 am Post subject: |
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Marty,
Starting from the beginning: R8 is the only cell in C3 that can be 4. | Code: | +----------------------------+----------------------------+----------------------------+
| 7 25 1 | 46 2456 3 | 8 56 9 |
| 25689 2359 23569 | 14679 1245679 124567 | 2457 3567 246 |
| 2569 4 23569 | 679 25679 8 | 257 1 26 |
+----------------------------+----------------------------+----------------------------+
| 4 379 3679 | 13678 13678 167 | 179 2 5 |
| 256 2357 23567 | 134678 12345678 9 | 147 678 1468 |
| 1 8 25679 | 467 24567 24567 | 479 679 3 |
+----------------------------+----------------------------+----------------------------+
| 29 6 279 | 5 1789 17 | 3 4 128 |
| 259 12579 4 | 136789 136789 167 | 1259 589 128 |
| 3 159 8 | 2 149 14 | 6 59 7 |
+----------------------------+----------------------------+----------------------------+ |
1. R7C6 is 1. Or,
2. R7C6 is 7. In which case, R8C2 is 7 (and not 1). In which case, R9C2 in C2 is 1.
R7C6 and R9C2 are pincers on 1. Puzzle solved.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Mar 01, 2010 6:27 am Post subject: |
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| keith wrote: | Marty,
Starting from the beginning: R8 is the only cell in C3 that can be 4. | Code: | +----------------------------+----------------------------+----------------------------+
| 7 25 1 | 46 2456 3 | 8 56 9 |
| 25689 2359 23569 | 14679 1245679 124567 | 2457 3567 246 |
| 2569 4 23569 | 679 25679 8 | 257 1 26 |
+----------------------------+----------------------------+----------------------------+
| 4 379 3679 | 13678 13678 167 | 179 2 5 |
| 256 2357 23567 | 134678 12345678 9 | 147 678 1468 |
| 1 8 25679 | 467 24567 24567 | 479 679 3 |
+----------------------------+----------------------------+----------------------------+
| 29 6 279 | 5 1789 17 | 3 4 128 |
| 259 12579 4 | 136789 136789 167 | 1259 589 128 |
| 3 159 8 | 2 149 14 | 6 59 7 |
+----------------------------+----------------------------+----------------------------+ |
1. R7C6 is 1. Or,
2. R7C6 is 7. In which case, R8C2 is 7 (and not 1). In which case, R9C2 in C2 is 1.
R7C6 and R9C2 are pincers on 1. Puzzle solved.
Keith |
OK, that's easy enough to follow. But what's M-Wingish about it? I'm assuming this is not trial and error. So what's there to tell the player to look at r7c6? |
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strmckr
Joined: 18 Aug 2009
Posts: 64
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| Posted: Mon Mar 01, 2010 8:46 am Post subject: |
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| Quote: | | 1 r8c2 -1- r8c9 =1= r7c9 -1- r7c6 -7- r7c3 =7= r8c2 => r8c2<>1 |
losely described above doesn't match any of the m-wings which covers all variations.
but it does match a "s - wing" which i have covered here
i also did post a copy of both threads to this site in (about 4 thread topics down u can find them...)
(these are an aic chain of very short length always discontinuous )
| Quote: |
So what's there to tell the player to look at r7c6? |
the bivalve digits 1 & 7
look for a strong link for digits "1" in a space.
&
look for a strong link for digits "7" in a space.
one cell from both 1 and 7 strongly linked cells must see the bivalve cell
if both 1 & 7 (end points) can see each other
then end points
1 <> 7 ,
7 <> 1
| Code: |
-------------------------------------------------
| . . . | . . . | . . / |
| . . . | . . . | . . / |
| . . . | . . . | . . / |
-------------------------------------------------
| \ \ 7 | . 17 . | . . 1 |
| \ 7-1 \ | . . . | . . 1-7 |
| \ \ \ | . . . | . . / |
-------------------------------------------------
| . . . | . . . | . . / |
| . . . | . . . | . . / |
| . . . | . . . | . . / |
-------------------------------------------------
reference:
\ <> 7
/ <> 1
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