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ravel · Joined: 21 Apr 2006 Posts: 536

By TTHsieh

Marty R. · Posted: Sat Jan 12, 2008 4:44 pm Post subject:

ravel · Joined: 21 Apr 2006 Posts: 536

Johan · Joined: 25 Jun 2007 Posts: 206 Location: Bornem Belgium

I didn't find any quad either, or could solve this with only basic moves, probably too much Duvel(strong Belgian beer) gave me a fuzzy look on the grid, and

needed two non basic moves to solve this one. The first is when R1C2 is <5> results in a contradiction in C2 : R1C2=5 => R2C3=8 => R9C2=8 => digit <6> is

lost in C2 => R1C2 ≠ 5. After that an [25-26-56] xy-wing to finish it.

Cheers

nataraj · Posted: Sun Jan 13, 2008 6:26 pm Post subject:

basic moves? Marty you must be kidding?

basic moves got me here (edit - oops I see this is Johan's position exactly)

Marty R. · Posted: Sun Jan 13, 2008 10:39 pm Post subject:

ravel · Joined: 21 Apr 2006 Posts: 536

Nice solutions, Johan and nataraj.

I found the same elimination as nataraj in a different way:

Marty R. · Posted: Mon Jan 14, 2008 12:07 am Post subject:

nataraj · Posted: Mon Jan 14, 2008 6:32 am Post subject:

ravel · Joined: 21 Apr 2006 Posts: 536

Yes, its just like nataraj said.

Looking at it as a pattern, the strong link of a w-wing is replaced by a skyscraper (could be a kite also). It is sufficient, that we know, that at least in one of the ends of the skyscraper the number must be true (in this case possibly in both) to conclude that at least one of the pairs must hold the other number then.

This is what we always do with solving methods to be able to solve a bit harder puzzles with them - generalize it.

In a skysraper we can add another weak and strong link at one end to get 3 strong links, where the ends have the same property, that at least one end has to hold the number.

In an xy-wing. that does not lead to an elimination, we also can add a weak and strong link at each pincer - or a bivalue chain in order to get an xy-chain.

All this can be looked at as adding links at an end of an AIC until hopefully you can make eliminations with the end's assignment.

btw Johans solution above for me is a "strong link chain".
There is a strong link for 6 in column 2.
Either r1c2=6 or (r9c2=6 => r9c3=8 => r2c3=5). In both cases r1c2<>5.

Marty R. · Posted: Mon Jan 14, 2008 4:58 pm Post subject:

Thanks to both of you. I'd be lying if I said I understood all the notation and AICs and the like, but I see clearly now that it works with the two strong links based in column 2.

Asellus · Posted: Tue Jan 15, 2008 1:10 am Post subject:

I once suggested that the best definition of a W-Wing is a remote pair of matching bivalues that each "sees" the opposite ends of an external strong link on one of their digits.

Here, the external strong link is the result of an otherwise useless "color wing" or "multi-coloring" on <5>:

Victor · Joined: 29 Sep 2005 Posts: 207 Location: NI

Nice puzzle, smart answers. Particularly like Johan's punchy method. I found ALS - some of C2, & R2C3 on its own: x = 8, z = 5, killing the 5 in R1C2.
Nervous about my own ideas. Is this OK? -
Consider the two 56s in R1 & 2. Now, in colouring terms the 6s are the same colour & therefore so are the 5s, say Red. (Basis of Keith's M-wings.) Colour the 5s clockwise from R1C8: R7C8 = NOT-Red, R7C2 = Red. But that gives a contradiction: impossible to colour the 5s in R5, since each is seen by a Red. So Red is False, etc.

ravel · Joined: 21 Apr 2006 Posts: 536

Victor · Joined: 29 Sep 2005 Posts: 207 Location: NI

Thanks Ravel. When I started to get bogged down with some Menneskes before Xmas, I looked for some new techniques & took a particular fancy to ALS - elegant, & capable of breaking some puzzles when I got used to spotting them. I've glanced at APEs but never used them - seems a nice technique.

I've got confused about the different ways of doing this puzzle, but let me explain again what I meant.

Say you've two cells A (5,6) & D also (5,6) connected via B & C which contain 6s but not 5s. Now if this is say a skyscraper with 6A=6B-6C=6D, then the 6s in A and D are strongly connected in the sense that at least one is True, maybe both. So the best you can say of the 5s is they are weakly connected, at most one being True, maybe both False.

But suppose that A=B=C=D is a colouring chain for 6s, i.e. with every link conjugate. Then you can say that A=6 & D=6 are conjugate - precisely one True. Therefore that's also the case for the 5s - conjugately linked, precisely one True. So A & D are remote pairs, and you can do with them whatever you can with natural remote pairs, e.g. swapping over & colouring on with the 5s.

Now consider the case when A & D are connected by just two conjugate links in 6, 6A=6B=6D In colouring terms, they're the same colour - they're both True, or both False. That must apply also to the 5s - TT or FF. And, this is the point, you can xx over and continue colouring with the 5s, which was what I tried to explain above. (And as I said, it's the basis for Keith's M-wings.) Start with R2C1 = Yellow say. Then R1C8 is Y for the reason just given (the 6s are TT or FF & therefore the 5s also), R7C8 is NOT-Y, R7C2 is Y. So far so good - I'm certain about that. Here's the punchline: each of the 5s in R5 is seen by a Y, which is a contradiction I think: so Y is False. & that solves the puzzle.

ravel · Joined: 21 Apr 2006 Posts: 536

Victor,

i agree to all, but in the grid i looked at (johan's and nataraj's starting grid) there were no two conjugate links in 6 to connect the cells with the 56-pairs.

Instead i found 2 chains (with grouped links), that lead to the same conclusion: r1c8=6 <=> r2c1=6.

Victor · Joined: 29 Sep 2005 Posts: 207 Location: NI

Ravel, thanks, didn't notice the other 6s in box 3 in your grid. Two possible explanations for me:
(a) I had in fact found some previous way of killing those other 6s on my own grid.
(b) The 6s were there but were I just didn't see them. (I.e. silly mistake, which I'm easily capable of making.)
But I've thrown away my sheet of paper & will now never know!