Hybrid Wing - or is there a better way?

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Hybrid Wing - or is there a better way?

Postby udosuk » Fri Jan 09, 2009 11:37 pm

With this puzzle:

Code: Select all
3...51.7.
..6.....1
.9..2.5..
....4...3
9.43.56.7
1...7....
..3.6..2.
8.....7..
.1.58...9


Easy moves get you to here:

Code: Select all
+----------------------+----------------------+----------------------+
| 3      48    *28     | 468    5      1      | 9      7     *2468   |
| 2457   4578   6      | 478    9      3478   | 2348   34     1      |
| 47     9      1      | 4678   2      34678  | 5      346    468    |
+----------------------+----------------------+----------------------+
| 67     678    578    | 289    4      289    | 12     159    3      |
| 9      2      4      | 3      1      5      | 6      8      7      |
| 1      3     *58     | 2689   7      2689   | 24     59    -245    |
+----------------------+----------------------+----------------------+
| 457    457    3      | 1479   6      479    | 148    2      458    |
| 8      456    9      | 124    3      24     | 7      15     456    |
| 2467   1      27     | 5      8      47     | 34     346    9      |
+----------------------+----------------------+----------------------+

Then I'll use what looks like a "Hybrid Wing" to crack it:

r1c3 from {28}
=> r1c9+r6c3 can't be [28]
But there is a strong link of 2 @ r16c9
=> r6c9 can't be 5 (otherwise it will force r1c9+r6c3=[28])

Or, put it in another way:

r16c3=[25|28|85] must have 2|5
=> r16c9 can't be [25]
=> With the strong link of 2 @ r16c9, r6c9 can't be 5

Is there a better (i.e. more elegant) way to solve the puzzle, such as a critical move involving fewer cells or simpler logic?:?:
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Postby Steve R » Sat Jan 10, 2009 12:31 am

Code: Select all
+---------------------+----------------------+---------------------+
| 3      48    A28    | 468    5      1      | 9      7     B2468  |
| 2457   4578   6     | 478    9      3478   | 2348   34     1     |
| 47     9      1     | 4678   2      34678  | 5      346   B468   |
+---------------------+----------------------+---------------------+
| 67     678    578   | 289    4      289    | 12     159    3     |
| 9      2      4     | 3      1      5      | 6      8      7     |
| 1      3     A58    | 2689   7      2689   | 24     59     24-5  |
+---------------------+----------------------+---------------------+
| 457    457    3     | 1479   6      479    | 148    2     B458   |
| 8      456    9     | 124    3      24     | 7      15    B456   |
| 2467   1      27    | 5      8      47     | 34     346    9     |
+---------------------+----------------------+---------------------+


The sets A and B are almost locked, with 2 as a restricted common candidate.

Hardly elegant.

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Postby udosuk » Sat Jan 10, 2009 12:53 am

Thanks Steve, your move is fine. It involves more cells, but the ALS-xz logic is arguably simpler. I always regard moves involving the spotting of strong links as less preferable (a personal taste thing).

Of course, your elimination is essentially the same as mine, just making use of the ALS @ r1378c9 instead of the strong link @ r16c9.:)

I was sort of expecting someone to find some uniqueness-based moves to crack it.:?:
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Postby daj95376 » Sat Jan 10, 2009 4:42 am

Simpler is subjective. What you're describing is a simple forcing chain. What's wrong with it?

Code: Select all
[r1c3]=2 [r1c9]<>2 [r6c9]=2 [r6c9]<>5
[r1c3]=8           [r6c3]=5 [r6c9]<>5

Or the short chain ...

Code: Select all
[r6c9]=2=[r1c9]-2-[r1c3]-8-[r6c3]-5-[r6c9]           =>  [r6c9]<>5

(2)r6c9 = (2)r1c9 - (2=8)r1c3 - (8=5)r6c3 - (5)r6c9  =>  [r6c9]<>5   (practicing Eureka)
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Postby DonM » Sat Jan 10, 2009 7:11 am

daj95376 wrote:Simpler is subjective. What you're describing is a simple forcing chain. What's wrong with it?

Code: Select all
[r1c3]=2 [r1c9]<>2 [r6c9]=2 [r6c9]<>5
[r1c3]=8           [r6c3]=5 [r6c9]<>5

Or the short chain ...

Code: Select all
[r6c9]=2=[r1c9]-2-[r1c3]-8-[r6c3]-5-[r6c9]           =>  [r6c9]<>5

(2)r6c9 = (2)r1c9 - (2=8)r1c3 - (8=5)r6c3 - (5)r6c9  =>  [r6c9]<>5   (practicing Eureka)


What's wrong with it? Purely IMO, that type of forcing chain is maximally assumptive & the procedure somewhat random (for instance, there is no logical reason for choosing r1c3 other than the fact that it is a bivalue cell.) On the other hand, again, purely IMO (though supported by many others with my same belief system:) ), the 2nd method using the NL/AIC, though somewhat assumptive as any chain may be to a greater or lesser extent, is a far more elegant proof of the elimination. (Ordinarily, I wouldn't have commented, but since the question was asked....)
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Postby aran » Sat Jan 10, 2009 12:58 pm

DonM wrote:
daj95376 wrote:Simpler is subjective. What you're describing is a simple forcing chain. What's wrong with it?

Code: Select all
[r1c3]=2 [r1c9]<>2 [r6c9]=2 [r6c9]<>5
[r1c3]=8           [r6c3]=5 [r6c9]<>5

Or the short chain ...

Code: Select all
[r6c9]=2=[r1c9]-2-[r1c3]-8-[r6c3]-5-[r6c9]           =>  [r6c9]<>5

(2)r6c9 = (2)r1c9 - (2=8)r1c3 - (8=5)r6c3 - (5)r6c9  =>  [r6c9]<>5   (practicing Eureka)


What's wrong with it? Purely IMO, that type of forcing chain is maximally assumptive & the procedure somewhat random (for instance, there is no logical reason for choosing r1c3 other than the fact that it is a bivalue cell.) On the other hand, again, purely IMO (though supported by many others with my same belief system:) ), the 2nd method using the NL/AIC, though somewhat assumptive as any chain may be to a greater or lesser extent, is a far more elegant proof of the elimination. (Ordinarily, I wouldn't have commented, but since the question was asked....)


Udusok is saying with a minimum of fuss that if r6c9 is 5 then there is no room for 2 in column 9.
When seen this is immediately obvious and no orthodoxy (eg why on earth did you ?) can override it.
BTW I do realise that Don is not suggesting that.
Steve R's ALS alternative is of course fine.
Personally I don't like (looking for) ALS's but nobody's perfect.
A much worse alternative to all of the above is to look at the hidden pair 59r46c8.
Its sole merit in a somewhat lengthy journey is to target 5r6c9/5r8c8 from the outset.
59r46c8=1r4c8-(1=2)r4c7-(2=89)r4c46-(8=67)r4c12-(78=5)r4c3-(5=8)r6c3-(8=2)r1c3-2r1c9=(2-5)r6c9 : =><5>r6c9
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Postby daj95376 » Sat Jan 10, 2009 3:14 pm

DonM wrote:
daj95376 wrote:Simpler is subjective. What you're describing is a simple forcing chain. What's wrong with it?

What's wrong with it? Purely IMO, that type of forcing chain is maximally assumptive & the procedure somewhat random (for instance, there is no logical reason for choosing r1c3 other than the fact that it is a bivalue cell.)

Uh, that was my point. udosuk's description had no more foundation than a forcing chain. It started with bivalue cell [r1c3] and generated two short inference streams. The commonality/verity was [r6c9]<>5. If [r6c9]=5, then backtracking on each stream resulted in [r1c3] being empty.

I was asking what's wrong with using it -- a forcing chain -- to represent what udosuk described in words.
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Postby DonM » Sat Jan 10, 2009 3:45 pm

daj95376 wrote:
DonM wrote:
daj95376 wrote:Simpler is subjective. What you're describing is a simple forcing chain. What's wrong with it?

What's wrong with it? Purely IMO, that type of forcing chain is maximally assumptive & the procedure somewhat random (for instance, there is no logical reason for choosing r1c3 other than the fact that it is a bivalue cell.)

Uh, that was my point. udosuk's description had no more foundation than a forcing chain. It started with bivalue cell [r1c3] and generated two short inference streams. The commonality/verity was [r6c9]<>5. If [r6c9]=5, then backtracking on each stream resulted in [r1c3] being empty. I was asking what's wrong with using it -- a forcing chain -- to represent what udosuk described in words.


Sorry, I misunderstood what the 'What's wrong with it?' was referring to.
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Postby Luke » Sat Jan 10, 2009 3:55 pm

Doesn't this accomplish the same thing?

5r6c3-8-r1c3-2-r1c9=2=r6c9 =>r6c9 <> 5.
Or (5=8)r6c3-(8=2)r1c3-(2)r1c9=(2-5)r6c9 => <5> r6c9.

Or maybe the idea is to avoid chains altogether.....
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Postby udosuk » Sun Jan 11, 2009 3:29 am

Well, thanks for all the replies mates. It's a good way to know a lot about solving habits among ourselves.:)

As for the arguments about chains, like I always say, it's all about presentation. Every move can be expressed as a forcing chain, so it's not about the question "what's wrong with the forcing chain?", but rather "is this forcing chain equivalent to something with a different name?".:idea:

Like DonM suggested, a random forcing chain, no matter how short, is not as elegant as a more structured one, e.g. one only involving bivalue cells (xy-chain) or only a couple of almost-locked-subsets (ALS-xz). But of course when we have no choice, a shorter chain is always better than a long one (such as the one aran showed).

So the point is, when do we consider a move "well-structured", and are there any objective criteria? Just take the "hybrid wing" I cited as an example, even though it involves only 4 cells, because it uses both weak & strong inferences, I can't say for sure it's an elegant move, compare to the following 2 "established" moves:

Code: Select all
 .  .  28 |  .  .  .  |  .  .  25
 .  .  .  |  .  .  .  |  .  .  .
 .  .  .  |  .  .  .  |  .  .  .
----------+-----------+----------
 .  .  .  |  .  .  .  |  .  .  .
 .  .  .  |  .  .  .  |  .  .  .
 .  .  58 |  .  .  .  |  .  .  24-5
----------+-----------+----------
 .  .  .  |  .  .  .  |  .  .  .
 .  .  .  |  .  .  .  |  .  .  .
 .  .  .  |  .  .  .  |  .  .  .

XY-Wing:
r1c3 must be 2 or 8
=> r1c9+r6c3 can't be [28]
=> r6c9 can't be 5


 .  .  13 |  .  .  .  |  .  .  12
 .  .  .  |  .  .  .  |  .  .  .
 .  .  .  |  .  .  .  |  .  .  .
----------+-----------+----------
 .  .  .  |  .  .  .  |  .  .  .
 .  .  .  |  .  .  .  |  .  .  .
 .  .  34 |  .  .  .  |  .  .  24-5
----------+-----------+----------
 .  .  .  |  .  .  .  |  .  .  .
 .  .  .  |  .  .  .  |  .  .  .
 .  .  .  |  .  .  .  |  .  .  .

1 @ r1 locked @ r1c39
4 @ r6 locked @ r6c39
3 @ c3 locked @ r16c3
2 @ c9 locked @ r16c9

Strong-Wing:
r1c3 can't be 1 & 3 simutaneously
=> r1c9+r6c3 can't be [24]
=> r6c9 can't be 5
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Re: Hybrid Wing - or is there a better way?

Postby Luke » Sun Jan 11, 2009 6:54 pm

udosuk wrote:Is there a better (i.e. more elegant) way to solve the puzzle, such as a critical move involving fewer cells or simpler logic?:?:

I say, "hardly," leastways not by me. I thought I'd try to get somewhere without hunting for chains, which was tough enough because I'm really into them these days.

I'm also into rectangles with strong links. How about a "strong wing minus one" (non-UR with three strong links instead of four)?
Code: Select all
+----------------------+----------------------+----------------------+
| 3      48     28     | 468    5      1      | 9      7      2468   |
| 2457   4578   6      | 478    9      3478   | 2348   34     1      |
| 47     9      1      | 4678   2      34678  | 5      346    468    |
+----------------------+----------------------+----------------------+
|*67     678   *578    | 289    4      289    | 12     159    3      |
| 9      2      4      | 3      1      5      | 6      8      7      |
| 1      3      58     | 2689   7      2689   | 24     59     245    |
+----------------------+----------------------+----------------------+
| 457    457    3      | 1479   6      479    | 148    2      458    |
| 8      456    9      | 124    3      24     | 7      15     456    |
|*2467   1     *27     | 5      8      47     | 34     346    9      |
+----------------------+----------------------+----------------------+

Strong links on [267] mean r4c1<>7. I know there's a chain behind this, but it's pretty easy to spot without it (well, here it is anyway: r4c1=6=r9c1=2=r9c3=7=r4c3.)

After tidying up there's this:
Code: Select all
 *--------------------------------------------------------------------*
 | 3      48     28     | 468    5      1      | 9      7      268    |
 | 2457  *58     6      | 478    9      3478   |*28     34     1      |
 | 47     9      1      | 4678   2      34678  | 5      34     68     |
 |----------------------+----------------------+----------------------|
 | 6      78     578    | 289    4      289    | 12     159    3      |
 | 9      2      4      | 3      1      5      | 6      8      7      |
 | 1      3      58     | 2689   7      2689   | 24     59     245    |
 |----------------------+----------------------+----------------------|
 | 45     4-57   3      | 1479   6      479    |*148    2     *48     |
 | 8      6      9      | 124    3      24     | 7      15     45     |
 | 24     1      27     | 5      8      47     | 3      6      9      |
 *--------------------------------------------------------------------*

Strong link of 8 @ r27c7 => r2c2 + r7c9 <> [54]. If 5 is in r7c2 it will force exactly that with the help of [45] in r7c1.
=> r7c2 <> 5. One xy cycle brings it home. Wait, that's a chain. Dang.:)
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Re: Hybrid Wing - or is there a better way?

Postby aran » Sun Jan 11, 2009 10:10 pm

Luke451 wrote:
udosuk wrote:Is there a better (i.e. more elegant) way to solve the puzzle, such as a critical move involving fewer cells or simpler logic?:?:

I say, "hardly," leastways not by me. I thought I'd try to get somewhere without hunting for chains, which was tough enough because I'm really into them these days.

I'm also into rectangles with strong links. How about a "strong wing minus one" (non-UR with three strong links instead of four)?
Code: Select all
+----------------------+----------------------+----------------------+
| 3      48     28     | 468    5      1      | 9      7      2468   |
| 2457   4578   6      | 478    9      3478   | 2348   34     1      |
| 47     9      1      | 4678   2      34678  | 5      346    468    |
+----------------------+----------------------+----------------------+
|*67     678   *578    | 289    4      289    | 12     159    3      |
| 9      2      4      | 3      1      5      | 6      8      7      |
| 1      3      58     | 2689   7      2689   | 24     59     245    |
+----------------------+----------------------+----------------------+
| 457    457    3      | 1479   6      479    | 148    2      458    |
| 8      456    9      | 124    3      24     | 7      15     456    |
|*2467   1     *27     | 5      8      47     | 34     346    9      |
+----------------------+----------------------+----------------------+

Strong links on [267] mean r4c1<>7. I know there's a chain behind this, but it's pretty easy to spot without it (well, here it is anyway: r4c1=6=r9c1=2=r9c3=7=r4c3.)

After tidying up there's this:
Code: Select all
 *--------------------------------------------------------------------*
 | 3      48     28     | 468    5      1      | 9      7      268    |
 | 2457  *58     6      | 478    9      3478   |*28     34     1      |
 | 47     9      1      | 4678   2      34678  | 5      34     68     |
 |----------------------+----------------------+----------------------|
 | 6      78     578    | 289    4      289    | 12     159    3      |
 | 9      2      4      | 3      1      5      | 6      8      7      |
 | 1      3      58     | 2689   7      2689   | 24     59     245    |
 |----------------------+----------------------+----------------------|
 | 45     4-57   3      | 1479   6      479    |*148    2     *48     |
 | 8      6      9      | 124    3      24     | 7      15     45     |
 | 24     1      27     | 5      8      47     | 3      6      9      |
 *--------------------------------------------------------------------*

Strong link of 8 @ r27c7 => r2c2 + r7c9 <> [54]. If 5 is in r7c2 it will force exactly that with the help of [45] in r7c1.
=> r7c2 <> 5. One xy cycle brings it home. Wait, that's a chain. Dang.:)

Luke
Your first deduction points to this rectangular general pattern :
rectangle : cells C1,C2,C3,C4 : ax abx bcx cx
(a,a) (b,b) (c,c) : conjugates
x : any number of extra candidates.
Then :
C1 <c> and C4 <a>.
So in your instructive example, had there been a 6 in r4c3, then it could have been eliminated as well.

A minor misprint in your second point, this
Strong link of 8 @ r27c7 => r2c2 + r7c9 <> [54] should read
Strong link of 8 @ r27c7 => r7c2 + r7c9 <> [54]
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Postby Luke » Mon Jan 12, 2009 2:32 am

Sorry, I meant to say:

"Strong link of 8 @ r27c7 => r2c2 + r7c9 <> [88]. If 5 is in r7c2 it will force exactly that with the help of [45] in r7c1.
=> r7c2 <> 5."

I'm not really sure this is the best way to write it either. It took me two minutes to spot it and half an hour to figure out a way to express it (incorrectly)!
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Postby aran » Mon Jan 12, 2009 2:54 am

Luke451 wrote:Sorry, I meant to say:

"Strong link of 8 @ r27c7 => r2c2 + r7c9 <> [88]. If 5 is in r7c2 it will force exactly that with the help of [45] in r7c1.
=> r7c2 <> 5."

I'm not really sure this is the best way to write it either. It took me two minutes to spot it and half an hour to figure out a way to express it (incorrectly)!

That was a nice way to see the elimination.
Whether there is then any need to express as a chain is another matter although chains do help to standardise things.
Anyway, since any elimination (other than a direct uniqueness deduction) can be transcribed into a discontinuous loop starting with the elimination itself as "true", you could always do that :
So here :
5r7c2-(5=4)r7c1-(4=8)r7c9-8r7c7=8r2c7-(8=5)r2c2-5r7c2 : <5>r7c2.
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Re: Hybrid Wing - or is there a better way?

Postby udosuk » Mon Jan 12, 2009 8:30 am

Luke451 wrote:I'm also into rectangles with strong links. How about a "strong wing minus one" (non-UR with three strong links instead of four)?
Code: Select all
+----------------------+----------------------+----------------------+
| 3      48     28     | 468    5      1      | 9      7      2468   |
| 2457   4578   6      | 478    9      3478   | 2348   34     1      |
| 47     9      1      | 4678   2      34678  | 5      346    468    |
+----------------------+----------------------+----------------------+
|*67     678   *578    | 289    4      289    | 12     159    3      |
| 9      2      4      | 3      1      5      | 6      8      7      |
| 1      3      58     | 2689   7      2689   | 24     59     245    |
+----------------------+----------------------+----------------------+
| 457    457    3      | 1479   6      479    | 148    2      458    |
| 8      456    9      | 124    3      24     | 7      15     456    |
|*2467   1     *27     | 5      8      47     | 34     346    9      |
+----------------------+----------------------+----------------------+

Strong links on [267] mean r4c1<>7. I know there's a chain behind this, but it's pretty easy to spot without it (well, here it is anyway: r4c1=6=r9c1=2=r9c3=7=r4c3.)

Nice move but I'm not sure if your move is much better than my "hybrid-wing". Mind you, my hybrid-wing uses 3 weak links + 1 strong link while your move uses 1 weak link + 3 strong links so in a sense they're on the same level of "elegancy".

To present your move "my way" it'll go like this:

r9c3 can't be 2 & 7 simultaneously
=> r4c3+r9c1 can't be [56|86] (strong links of 7 @ r49c3 & 2 @ r9c13)
=> r4c1 can't be 7 (strong link of 6 @ r49c1)

Luke451 wrote:After tidying up there's this:
Code: Select all
 *--------------------------------------------------------------------*
 | 3      48     28     | 468    5      1      | 9      7      268    |
 | 2457  *58     6      | 478    9      3478   |*28     34     1      |
 | 47     9      1      | 4678   2      34678  | 5      34     68     |
 |----------------------+----------------------+----------------------|
 | 6      78     578    | 289    4      289    | 12     159    3      |
 | 9      2      4      | 3      1      5      | 6      8      7      |
 | 1      3      58     | 2689   7      2689   | 24     59     245    |
 |----------------------+----------------------+----------------------|
 | 45     4-57   3      | 1479   6      479    |*148    2     *48     |
 | 8      6      9      | 124    3      24     | 7      15     45     |
 | 24     1      27     | 5      8      47     | 3      6      9      |
 *--------------------------------------------------------------------*

Strong link of 8 @ r27c7 => r2c2 + r7c9 <> [88]. If 5 is in r7c2 it will force exactly that with the help of [45] in r7c1.
=> r7c2 <> 5. One xy cycle brings it home. Wait, that's a chain. Dang.:)

An xy cycle being a chain is alright, since it's an "elegant" chain.:)

However, I'm curious about your "tidying up".:?: From your previous position, after singles, pairs, locked candidates and an xy-wing I can almost reach that position barring an elimination of 4 from r2c2. It will be much appreciated if you can elaborate how you eliminated that 4 from r2c2.:?:
udosuk
 
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