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Learning W-Wings

 
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leocar



Joined: 04 Mar 2010
Posts: 12

PostPosted: Thu Jun 16, 2022 3:17 am    Post subject: Learning W-Wings Reply with quote

Hi, I'm learning how to find and use W-Wings. Today's daily sudoku (16/06/2022) I think may have had one even though it has nothing to do with solving the puzzle. Can someone please let me know if I have found a W-Wing or if I am way off. From what I can see the 1,9's R2C7, R5C8 (*) are linked by 3,9 R7C7 and 2,9 R8C8(#) meaning 1 is removed from R5C7 is that correct?

Code:

+----------+------------+-------------+
| 2  5 1   | 4   9    3 | 7    6   8  |
| 4  7 8   | 6   5    2 | 19*  3   19 |
| 6  9 3   | 1   7    8 | 5    4   2  |
+----------+------------+-------------+
| 39 6 59  | 235 1238 4 | 28   7   139|
| 8  2 459 | 35  13   7 | 1349 19*  6 |
| 37 1 47  | 9   28   6 | 28   5   34 |
+----------+------------+-------------+
| 1  4 29  | 23  6    5 | 39#  8    7 |
| 79 3 279 | 8   4    1 | 6    29#  5 |
| 5  8 6   | 7   23   9 | 134  12  134|
+----------+------------+-------------+


Thanks
Leo
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Mogulmeister



Joined: 03 May 2007
Posts: 1151

PostPosted: Thu Jun 16, 2022 3:08 pm    Post subject: Reply with quote

Hi Leo and welcome

It’s all about strong links for these various wings to work.

Looking for wings is helpful but also consider things from a fundamental level.

You have described something that has two ends or pincers - the remote pairs of 19’s in box 3 and box 6. One of these must be 1. It doesn’t matter which.

If you set one of the pairs to 9 and follow the chain the other pair becomes 1 making your elimination.
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Mogulmeister



Joined: 03 May 2007
Posts: 1151

PostPosted: Thu Jun 16, 2022 3:14 pm    Post subject: Reply with quote

Mogulmeister wrote:
Hi Leo and welcome

It’s all about strong links for these various wings to work.

Looking for wings is helpful but also consider things from a fundamental level.

You have described something that has two ends or pincers - the remote pairs of 19’s in box 3 and box 6. One of these must be 1. It doesn’t matter which.

If you set one of the pairs to 9 and follow the chain the other pair becomes 1 making your elimination.


So the definition of W wings:

“They consist of two bivalue cells with the same candidates, that are connected by a strong link on one of the candidates.”

Which you have.
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TomC



Joined: 30 Oct 2020
Posts: 349
Location: Wales

PostPosted: Thu Jun 16, 2022 3:35 pm    Post subject: Reply with quote

Very good explanation.

For years I have always thought W wings as contradictory (as well as other stuff)

In this puzzle, r5c7 <> 1 as by using the <19> pair this would lead to no <9> in box 9 or in column 9
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Mogulmeister



Joined: 03 May 2007
Posts: 1151

PostPosted: Thu Jun 16, 2022 4:39 pm    Post subject: Reply with quote

Excellent !

So this Leo is a really good way to get into the guts of a puzzle. Same solution but arrived at in a very different way. Instead of starting out looking for a wing of some sort, consider the effect of a placement.

Tom has shown what happens if you put a 1 in r5c7 ?

It is very elegant because we are looking to disrupt that pair of 19’s and how!

This causes contradictions on a massive scale - no 9’s in column 9 and no 9 in box 9. So r5c7 can never be 1.

This method may seem inside out but, if anything it is even more powerful than the w wing search. Definitely worth adding to your armoury.
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TomC



Joined: 30 Oct 2020
Posts: 349
Location: Wales

PostPosted: Thu Jun 16, 2022 5:18 pm    Post subject: Reply with quote

A pair of cells containing <19> can go four ways 1:1 1:9 9:1 9:9

W wings prove that if one cell is 1 the other is 9 and vice versa, but always consider what would if happen if both were 1 or both were 9

Here, I can see the effect if both were 9 and just look to cause maximum damage by sticking 1 in r5c7
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leocar



Joined: 04 Mar 2010
Posts: 12

PostPosted: Thu Jun 16, 2022 10:41 pm    Post subject: Reply with quote

Thank you both for the replies. It has helped me understand the method better as well as proving it and having another potential solving method.
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leocar



Joined: 04 Mar 2010
Posts: 12

PostPosted: Thu Jun 30, 2022 3:01 am    Post subject: Reply with quote

Hi, while I was waiting for another VH, I looked at one from a few years back (I think it was from Tue 3-Jun-2014). After completing a X-Y wing I was left with the puzzle below which is graded hard.

I think I have found a W-Wing on 19's R8C1 and R9C7 with a few links eg R89C4 with both 1 and 9.

Does this mean R9C1 and R8C7 cannot be 1 or 9 i.e. R9C1=5 and R8C7=8?

Code:

+------------+--------------+-----------+
| 2   8   4  | 6   13   13  | 79  79  5 |
| 7   6   9  | 5   4    2   | 3   1   8 |
| 3   5   1  | 8   9    7   | 6   4   2 |
+------------+--------------+-----------+
| 159 2   8  | 7   15   19  | 4   6   3 |
| 4   13  35 | 2   1358 6   | 78  578 9 |
| 6   39  7  | 39  58   4   | 2   58  1 |
+------------+--------------+-----------+
| 8   139 2  | 4   7    139 | 5   39  6 |
| 19  4   6  | 139 2    5   | 189 389 7 |
| 159 7   35 | 139 6    8   | 19  2   4 |
+------------+--------------+-----------+

Play this puzzle online at the Daily Sudoku site
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glesco



Joined: 12 May 2022
Posts: 35

PostPosted: Thu Jun 30, 2022 3:03 pm    Post subject: Reply with quote

A 9 in column 4 is also possible in R6. 1s in column 4 are restricted to R89, so the 19 W-Wing in R8C1 & R9C7 removes 9s in R9C1 & R8C7
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Pat



Joined: 23 Feb 2010
Posts: 207

PostPosted: Thu Jul 07, 2022 2:12 am    Post subject: Reply with quote

glesco wrote:


1s in column 4 are restricted


— to box 8,
this solves the 1 for row 7.
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leocar



Joined: 04 Mar 2010
Posts: 12

PostPosted: Mon Aug 08, 2022 10:33 pm    Post subject: Reply with quote

Code:

+------------+----------+----------+
| 28 1  3    | 28  7  5 | 9  4  6  |
| 49 7  2489 | 3   89 6 | 1  28 5  |
| 6  5  289  | 289 1  4 | 3  7  28 |
+------------+----------+----------+
| 49 3  49   | 7   2  8 | 6  5  1  |
| 7  6  28   | 1   5  9 | 4  28 3  |
| 1  28 5    | 6   4  3 | 7  9  28 |
+------------+----------+----------+
| 28 4  7    | 58  6  1 | 25 3  9  |
| 3  28 1    | 589 89 7 | 25 6  4  |
| 5  9  6    | 4   3  2 | 8  1  7  |
+------------+----------+----------+

Play this puzzle online at the Daily Sudoku site

Hi, I was looking at this puzzle from yesterday and noticed a W-Wing with 28 in R5C3 and R3C9 with links to 2 and 8 in R6C29, from what I can see that means R3C3 cannot be either 2 or 8 therefore it has to be 9, is that correct? In this situation both numbers 2 and 8 are removed and not just one of them?

Cheers
Leo
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storm_norm22



Joined: 24 Oct 2020
Posts: 15

PostPosted: Wed Dec 21, 2022 4:09 pm    Post subject: Reply with quote

leocar wrote:
Code:

+------------+----------+----------+
| 28 1  3    | 28  7  5 | 9  4  6  |
| 49 7  2489 | 3   89 6 | 1  28 5  |
| 6  5  289  | 289 1  4 | 3  7  28 |
+------------+----------+----------+
| 49 3  49   | 7   2  8 | 6  5  1  |
| 7  6  28   | 1   5  9 | 4  28 3  |
| 1  28 5    | 6   4  3 | 7  9  28 |
+------------+----------+----------+
| 28 4  7    | 58  6  1 | 25 3  9  |
| 3  28 1    | 589 89 7 | 25 6  4  |
| 5  9  6    | 4   3  2 | 8  1  7  |
+------------+----------+----------+

Play this puzzle online at the Daily Sudoku site

Hi, I was looking at this puzzle from yesterday and noticed a W-Wing with 28 in R5C3 and R3C9 with links to 2 and 8 in R6C29, from what I can see that means R3C3 cannot be either 2 or 8 therefore it has to be 9, is that correct? In this situation both numbers 2 and 8 are removed and not just one of them?

Cheers
Leo


sometimes its better to see a picture of it.



obviously, the W-wing is one of the more sought after patterns because its fairly easy to spot the bi-value cells. in this case, form your grid, we have a multitude.
in the picture, we have (2=8) in r5c3 and (2=8) in r3c9.
easy to spot, easy to see.
what gives these two cells its power to eliminate the 2 in r3c3 takes a bit more logic.

what connects these cells are the 8's in r5c8 and r6c9.
these 8's being the only 8's in BOX 6 are strongly inferenced.
we denote this strong inference like so (8)r5c8 = (8)r6c9

in the image above, you can put the chain together

(2=8)r3c9 - (8)r6c9 = (8)r5c8 - (8=2)r5c3

notice how the 2's are now strongly inferenced and are at the ends of the chain. if r5c3 is NOT a 2, then r3c9 is a 2

if r3c9 is NOT a 2 then r5c3 is a 2.

just like the 8's in BOX 6.
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