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

Posted: Tue Jan 15, 2008 4:48 pm Post subject: Mistake! 


Here's M3504165, which I've posted as a puzzle, and here's the story of my mistake. (Yes, I know many people have been here before me  these are my musings.) Grid post basics:
Code: 
++++
 3 26 8  9 7 4  5 1 26 
 4 5 27  6 1 8  29 279 3 
 19 17 1679  35 35 2  8 4 67 
++++
E125 3 4  8 A25 7 B12 6 9 
 6 17 1257  245 9 C15  1234 23 8 
 8 9 12  234 236 F16  7 5 124 
++++
 1259 8 123569  7 2456 56  123469 239 124 
 7 26 236  1 246 9  2346 8 5 
 1259 4 12569  25 8 3  1269 279 127 
++++

Forget the nice Wwings etc. and look at the XYwing pivoted on A, r4c5. When I'd eliminated the 1 in r5c7, I tried to continue with colouring on 1s: on to E from B, on to F from C, i.e. a 3link colouring chain in 1s from E to F, killing the 1 in r6c3. But that led to a contradiction in due course. (r6c3 is in fact 1.) Why?  the answer's easy, that both of B & C might be 1, so that neither of E nor F need be. And that's led me to read up / think more clearly about the nature of strong links etc. Yes, I know, I could have read Asellus's explanation in a thread started by Keith, http://www.dailysudoku.co.uk/sudoku/forums/viewtopic.php?t=2028, or elsewhere, and I know I could just have a figured it out, but I hadn't done any of those things.
The answer as far as colouring on from an XYchain, or Wwings or any such construct is easy enough to state: you need an even number (inc.0) of colouring links from each pincer. But it's prompted me to read up a bit more about strong links & I have found some difference in definition.
Everybody agrees what a weak link is, but e.g. Sudopedia defines a strong link as being the same as a conjugate link (often known as XOR  eXclusive Or) whereas others define it as being Either/Both (often known as OR). I'm finding it easier to think of strong as ORlinks (truth table TT,TF,FT) rather than XOR (TF,FT). If you so define them, then it's easy to say that a conjugate link is weak or strong as needed. And, if you have a chain A=BC=D then clearly this condenses to A=D, just another strong link, extensible to any length chain. And AB=CD condenses to AD.
However, I like really to think 'practically' and that's involving me in difficulties. Take the XYwing above. As an AIC you could write that 1C=5C5A=2A2B=1B, which leads to 1C=1B, killing any 1s they both see. More practically, you can write it C ≠ 1 ==> C=5 ==>A =2 ==> B =1, and B ≠ 1 ==> ... ==> C=1. (You can't start C=1, since that won't propagate along the chain.) That's OK, but how about the sort of AIC that starts and finishes in the same set (house) but with different numbers? E.g. say cell G contains 1,4,6 and cell H in the same column contains 4,7. Say you can find a chain 1G= ...... = 4H. Then clearly you can say that G can't contain 4, since Either it contains 1 Or a cell that sees it contains 4, but I find that difficult to explain in practical terms, since I see the chain as saying G ≠ 1 == > H=4 and H ≠ 4 ==> G = 1 which doesen't seem to tell us much.
Anyway, does anybody know the answer to this question? I found a thread a couple of years ago, in which the theorist Myth Jellies suggested that as well as the ends of AICs being strongonly, as he put it, some links in URs might be like this  i.e. strong but not conjugate. Was this ever (dis/)proven?
A final question. A simple colouring chain on one digit is composed entirely of conjugate links, and cells separated by an odd number of links are conjugatelinked. (Note that if you want to write it in simple language you can start A = x ==> ... as well as A ≠ x ==> .)This isn't true of most chains such as an XYchain. Is it in fact true of ANY other kind of chain? 

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nataraj
Joined: 03 Aug 2007 Posts: 1045 Location: Vienna, Austria

Posted: Tue Jan 15, 2008 7:31 pm Post subject: Re: Mistake! 


Victor wrote:  E.g. say cell G contains 1,4,6 and cell H in the same column contains 4,7. Say you can find a chain 1G= ...... = 4H. Then clearly you can say that G can't contain 4, since Either it contains 1 Or a cell that sees it contains 4, but I find that difficult to explain in practical terms, since I see the chain as saying G ≠ 1 == > H=4 and H ≠ 4 ==> G = 1 which doesen't seem to tell us much.

Victor, I shall try and shed some light on this one.
"...difficult to explain in practical terms". But you just did. Your explanation of the consequence ("Then clearly you can say ...") is quite practical.
To express the same logic in terms of an AIC it is necessary to add the information that one cell (G) contains 1 and 4, and that both cells are in the same house, which we do via weak links  just add one weak link to each end of your chain and voilą, what you get now is a "discontinuous nice loop". The discontinuity consists of a node (4)G connected to the loop at both sides through a weak link. Such a node must be false.
(4)G(1)G=[here is your alternating chain with strong links at the ends]=(4)H
The first weak link (4)G(1)G comes from the cell {1,4,6} ("G cannot be both 1 and 4"), the second weak link (4)H(4)G from the house ("both cells cannot be 4 at the same time"). 

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