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ravel's weak or half M-wing

 
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keith



Joined: 19 Sep 2005
Posts: 3179
Location: near Detroit, Michigan, USA

PostPosted: Sat Jan 19, 2008 4:06 am    Post subject: ravel's weak or half M-wing Reply with quote

I decided to look for another of these, which ravel pointed out a week or two ago.
Code:

Puzzle: M5963221sh(15)
+-------+-------+-------+
| . . 1 | . . . | . 3 . |
| . . 8 | 5 . . | 1 9 6 |
| . . 2 | . 8 . | 5 . . |
+-------+-------+-------+
| . . . | . . 8 | . . . |
| 4 . . | 6 . . | . . . |
| . . 7 | . . . | . . 5 |
+-------+-------+-------+
| 3 . . | . 1 . | 6 . 7 |
| . . 5 | . 7 . | 8 . 3 |
| . . . | . 2 4 | . . . |
+-------+-------+-------+

An XYZ-wing and extreme basics get us to here:
Code:
+----------------+----------------+----------------+
| 59   45   1    | 47   6    79   | 2    3    8    |
| 7    34   8    | 5    34   2    | 1    9    6    |
| 69   36   2    | 1    8    39   | 5    7    4    |
+----------------+----------------+----------------+
| 156  156  39   | 247  34A  8    | 347B 16   29   |
| 4    128  39   | 6    5    1-37 | 37   18   29   |
| 126  1268 7    | 24   9    13D  | 34C  168  5    |
+----------------+----------------+----------------+
| 3    9    4    | 8    1    5    | 6    2    7    |
| 12   12   5    | 9    7    6    | 8    4    3    |
| 8    7    6    | 3    2    4    | 9    5    1    |
+----------------+----------------+----------------+

A and C have the same two candidates. They are not a remote pair, they are not a complementary pair. All you can say is: If A is <4>, C is <4>. (There is a weak link A-B, and a strong link, B=C, on <4>.)

But, there is also a strong link on <3> in C=D.

The logic for the elimination is:

Either:

1. A is <3>.

Or:

2. A is <4>, C is <4>, D is <3>.

Any cell that sees both A and D cannot be <3>. In particular, we can take out <3> from R5C6.

Which leads to an XY-wing, which solves the puzzle.

Bravo, ravel!
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Victor



Joined: 29 Sep 2005
Posts: 207
Location: NI

PostPosted: Sun Jan 20, 2008 7:49 pm    Post subject: Reply with quote

Nice! This is really an AIC I suppose? (Kind of thing I search for these days, mostly without success.)
The chain is 3R4C5 = 4R4C5 - ..... = 3R6C6, killing any 3s seen by both ends.
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Marty R.



Joined: 12 Feb 2006
Posts: 5166
Location: Rochester, NY, USA

PostPosted: Sun Jan 20, 2008 8:22 pm    Post subject: Reply with quote

Victor wrote:
Nice! This is really an AIC I suppose? (Kind of thing I search for these days, mostly without success.)
The chain is 3R4C5 = 4R4C5 - ..... = 3R6C6, killing any 3s seen by both ends.

I wouldn't know an AIC if I stepped on one. But it's also an XY-Chain by my definition. If A = 4, a 37 naked pair is created in box 6, making r6c7 = 4 and r6c6 = 3. However, Keith used logic whereas the XY-Chain is a little more guesswork.

I also have no idea of what "extreme basics" are, other than an oxymoron. Laughing
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keith



Joined: 19 Sep 2005
Posts: 3179
Location: near Detroit, Michigan, USA

PostPosted: Mon Jan 21, 2008 1:26 am    Post subject: Reply with quote

Quote:
I also have no idea of what "extreme basics" are,

The hidden triple <156> in R4. I have a real hard time finding these things. Maybe the naked quad <23479> is easier to see?

Edit: Should be "naked quint".

Keith
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Marty R.



Joined: 12 Feb 2006
Posts: 5166
Location: Rochester, NY, USA

PostPosted: Mon Jan 21, 2008 4:43 am    Post subject: Reply with quote

Quote:
The hidden triple <156> in R4. I have a real hard time finding these things. Maybe the naked quad <23479> is easier to see?

In my 2+ years, I've never found a hidden anything bigger than a single. I'll take a naked quint over a hidden pair anytime.
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Victor



Joined: 29 Sep 2005
Posts: 207
Location: NI

PostPosted: Mon Jan 21, 2008 9:13 pm    Post subject: Reply with quote

Marty, not sure that it's quite an XY-chain. I thought they were constructed purely from bivalue cells. You could solve this puzzle with an XY-chain: look at R4C359 & R5C4 - cells 29, 39, 34, 24 in that order. If either end is NOT a 2, then the other end must be - works like an XY-wing, in this case removing 2 from R4C4. (But Keith's method is nicer!)

Here are some general thoughts about AICs and things. I've come to a dead end sometimes when doing puzzles - fairly certain that there's no more colouring / skyscrapers / etc. to be found. So I went looking for other techniques and took a liking to ALS - elegant idea, easy to understand if not always easy to find. They've turned out to be quite powerful in the sense that they've cracked some puzzles I couldn't otherwise do, and they seem to turn up quite often in Menneske puzzles. (You could solve this puzzle with ALS.) BUT there are 2 cons:
(a) I can't be sure I've spotted them even if they're there, in the way that be with say XY-wings.
(b) I'm fairly sure that there are Menneskes - and others - that they won't crack.

So what else? Well, you've gone for Medusa colouring, and I can see how powerful that is, but I'm sort of mentally prejudiced against this technique - no sound reason! I've watched people crack puzzles with what are essentially AICs I think, like Steve doing the puzzle you posted in the Solving Techniques .. section. And Nataraj, Ravel & Asellus at work on this one: http://www.dailysudoku.co.uk/sudoku/forums/viewtopic.php?t=2290&start=0&postdays=0&postorder=asc&highlight=
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Victor



Joined: 29 Sep 2005
Posts: 207
Location: NI

PostPosted: Mon Jan 21, 2008 9:19 pm    Post subject: Reply with quote

Sorry, accidentally clicked Submit before I'd finished.
The point I'm making is that I need something beyond ALS to try to solve difficult puzzles, but I'm not sure how to go about it without doing Medusa.
(Not sure if Medusa solves all puzzles that can be solved by AICs?)
There are of course rare techniques that might solve a puzzle here or there, like Ravel's recent smart sort of repeating W-wing, or Sue de Coq. Nice to meet things like that, but they're icing on the cake, and won't help with many puzzles.
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Marty R.



Joined: 12 Feb 2006
Posts: 5166
Location: Rochester, NY, USA

PostPosted: Mon Jan 21, 2008 10:09 pm    Post subject: Reply with quote

Quote:
Marty, not sure that it's quite an XY-chain. I thought they were constructed purely from bivalue cells.

You're right, I take some liberties with the definition. If I can start a chain with a bivalue cell and end it with a value that creates a pincer situation, I figure if it involves a few trivalue cells, it's pretty much the same for my purposes.

Quote:
Sorry, accidentally clicked Submit before I'd finished.

Were you aware that you can edit your original post by clicking "Edit" at the upper right of that message?
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Mon Jan 21, 2008 11:26 pm    Post subject: Reply with quote

Victor,

There are two ways to see this AIC. The simplest (conforming to Keith's description) is:

(3)r5c6-(3=4)r4c5-(4)r4c7=(4-3)r6c7=(3)r6c6-(3)r5c6; r5c6<>3

Only slightly different (and conforming more, I suspect, to Marty's description) is:

(3)r5c6-(3=4)r4c5-(4={37})r45c7-(3)r6c7=(3)r6c6-(3)r5c6; r5c6<>3

The is almost an ALS Chain, which might be what Marty means by calling it an "XY Chain." Instead of the strongly linked <4>s in c7 exploited in the first example, it exploits the 2-cell {347} ALS in r45c7.

An XY Chain is just the simplest form of ALS Chain... one limited entirely to bivalue cells (the simplest form of ALS). But, the same concept applies if you use larger ALSs along the way, such as that 2-cell 3-digit ALS used here. In all cases, an ALS collapses to a locked set along the inference chain sequence and thus propogates the chain. (A naked single is the "locked set" when a bivalue collapses.)

The second example deviates from a true ALS Chain by exploiting the strongly linked <3>s in r6. So, we probably have to say that the more generic term "AIC" is the most accurate description.

By the way, the key to understanding the Eureka notation for an ALS chain node is to realize that in an ALS each digit is strongly linked to the alternate locked set. So here, the <4> in the <347> ALS is strongly linked to the {37} locked set: (4={37})r45c7. Had we needed, we could also have exploited (3={47}) or (7={34}).

And for "extra credit": The "xz" ALS technique is simply a 2-node ALS Chain:

(x)victims - ({ax}=z)ALS1 - (z={bx})ALS2 - (x)victims; victims<>x

where a and b are any other digits that form the ALSs. The "shared exclusive" z is weakly linked (cannot both be true) between the two ALSs.
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Victor



Joined: 29 Sep 2005
Posts: 207
Location: NI

PostPosted: Thu Jan 24, 2008 8:52 pm    Post subject: Reply with quote

Thanks Marty - no I didn't know that I could edit a post - thanks for telling me.

And thanks Asellus. You're turning me into a proper sudoku-er! I do appreciate it.
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