Comments - Talk
... by: Shum
This isn't stated, but it seems to me a necessary condition is the length of the chain must be even (referring to the number of cells, not the number of links). All the examples here have length 4 and the alternate coloring seems to imply this restriction. Agree?
Andrew Stuart writes:
That is a good observation but I fail in my documentation to include longer examples. I think I was hung up on finding short simple ones and now I realise that I should show they can be several lengths. I also believe they will contain an even number of cells in total. I will have a hunt in my 2014 stock for longer ones and also try and find a crazy super length one, there has to be a 'longest'
... by: suneet shrotri
I have discovered a new method to eliminate a possibility by 2 chains with one common end & other 2 ends are part of a set then if common end is not X then if in some cases other ends come to have XY possibilities then possibility X can be eliminated from other remaining members of the set. I am sure that no other method on this website can do this.
... by: gg fuller
I'm unsure about the description as it mentions it is a more encompassing version of "Y-Wing Chains" and also refers to "Y-chains". Since there are no such strategies described on this site, I don't know if that refers to "Y-wing" or "X-chains" or both. If I just look for XY-chains will that catch all Y-Wings and X-chains, or is X-chains separate? X-chains seems to require a loop to make any eliminations.
Andrew Stuart writes:
Hi, yes there is a page, click here
but I subsumed this into the more general chaining strategies and that opening line needs to be edited. This goes way way back to the early days.
... by: Guy Renauldon
The Exclamation Mark Method
This is to answer to François Tremblay and John Robinson question.
Yes François, there is a trick to discover an XY chain.
Recently, I settled a kind of method which I name the “Exclamation Mark Method”
That is what it is:
At first I select a bi value cell. Not anyone, but rather one containing two candidates regarding the digits missing the most.
Let’s say the cell I select contain “a” and “b” as candidates.
Then I make a double bet.
First bet: “a” true,
Second bet: “b” true.
Regarding the first bet, I draw a small vertical line “I” just under the “a” of the bet cell. Then I consider the other bi values cells only and I mark all the candidates deducted by this first bet by an “I”, drawn just under the candidates.
Regarding the second bet “b” I do the same but the mark will be a small point “.”under the “b” in the bet cell.
Most of the time, soon or later, if a XY chain exists, I obtain a cell containing a candidate marked with an exclamation mark “!” (This is the reason of the name of my “method”, although the exclamation mark can be reverse, the point on the top)
So a candidate , let’s say “k”, will be marked with an exclamation mark. It means it is true in both bets. So “k” is a true digit. (Although I can’t say anything on “a” or “b”). Most of the time, this true digit will solve the Sudoku, when the Sudoku is not difficult too much.
But I do not consider the job as finished. I have to find my XY chain.
Let’s see the simplest case at first: the cell where we have the “!” is a bi value cell too. I consider the other candidate in that cell, the one without the “!”, let’s say “m”. “m” have to be eliminated, so “m” must see two other bi value cells containing an “m”, which will be the two ends of my XY chain (to understand that, you have to know what an XY chain is exactly). Then I consider each end of my chain and I follow the reverse itinerary from each end. I draw the links between the bi value cells really, with a pen. Doing so I’ll automatically reach my double bet cell. And that it is, I’ve drawn my XY chain! Very easily, without the headaches y suffered before I applied that method. And doing so, you can find very long XY chains, containing ten links for example, or more.
Note that you can find a shorter chain which does not go through the double bet cell, but with the same ends. The reason is that more than one XY chain exists on a same grid most of the time.
If the cell which sees the two ends of the chain is not a bi value cell, it is a little bit more complicated. But the “method” works too. If the cell contains three candidates, you will have two candidates marked with an exclamation mark. Then the third candidate can be removed. If a cell contains four candidates, you’ll find three exclamation marks in that cell. Generally speaking, if the cell contains N candidates, you’ll find N-1 candidates with an exclamation mark. The other one left without the “!” can be removed. This chain does not give a true digit directly.
Sorry, I’ve been little bit long in my explanation, and please excuse my English, being French (perhaps François Tremblay will understand me better, his name sounding very French …)
I’d like to add some remarks, if Andrew Stuarts give me some more place.
1/An xy chain is in fact a multi value X Cycle in three dimension, containing bi values cells only from one end to the other. The strong links are located inside the bi value cells. The other links can be either strong or weak, except the links between “m”, which must be weak. In fact it is a special AIC, with bi value cells only.
3/ This “method” is not mine…in fact it is the well known Forcing Chain Strategy. But the exclamation mark trick is mine, I think.
Many thanks to Andrew Stuart if he accepts to publish my comment, which maybe he can find a bit too long and not very clear (it is to me, but may be not to all the readers due to my bad English…).
... by: John Robinson
I have the same question as Francois Trembly on 28-Oct-2009
... by: S.Monta
First i would like to thank you very much for your great website witch made me dramatically improve my Sudoku skills.
I have a question : could the chain 56-67-67-78-83-89-96 be an XY-Chain?
... by: Marshal L. Merriam
I think I have another type of xy chain. In a sudoku I'm doing now (not one of these examples), I find the sequence 28-87-73-36-68-82 where the first pair and the last are the same cell! If its value were 8, then xy chain logic demands that it also be 2, a contradiction. No such contradiction prevents it being 2.
Alternatively we could argue that the xy chain demands that at least one of the endpoints be a 2. Since they are the same, the cell must be a 2.
... by: Marshal L. Merriam
I now understand Anton's confusion. It stems from the explanation of example 1. I would add one more bullet:
If C2 is not 5 then it must be 6. A1 cannot be 6, A5 cannot be 2 and A7 cannot be 9. Ergo, if C2 is not 5, then A7 must be 5. As noted in the first bullet, if A7 is not 5 then C2 must be 5. So either A7 or C2 must be 5.
When stated this way, the chain does not require locked sets. If C2 is not 5, then no other cell in box 1 can be 6, so even if there were a 6 at B3(say), all of the bullets would still hold.
... by: Matt Lala
I read something on a different site (if I understand this correctly) is that when you reach the end of the chain, you are using the "leftover" value to make your elimination. In the first example, the last green arrow is to a 6, and the leftover in that cell is a 5, which is what gets eliminated. If the last link had been to the 5, then the leftover is 6, and you cannot use that to end the chain and eliminate the 5's. Not sure what conditions are needed for the start. I may be wrong on this.
Francois, I don't think you need a solver, but it does help to have a program that highlights all of them. I glance at the 'busiest' groupings of bivalue cells and then pick one and just start driving. If a fork presents itself I'll choose whichever option seems to steer me back towards the start of the chain.
Anton, I don't think they have to be locked, they just need to be bivalue. Try plugging in a 2 at the start of that chain in the 2nd example, and follow the links, and you'll see how the green cell at the end becomes a 6 (despite the unlocked cells used). And plugging in a 6 at the start makes the eliminations pretty obvious.
... by: Francois Tremblay
I understand the logic but my problem is how do you spot this without the help of a solver? Visually, the start and end of such a chain are tough to find. Do you go through all the grid to start the chain at all possible bi-value cells? In the "simple" cases above, you have respectively 21 & 22 bi-value cells (even your book's example on page 62, figure 22.1 shows 21 of them) which would represent a lot of permutations that only a computer could run through. As a human solver, is there a trick to find those chains?
... by: Anton Delprado
Maybe I am missing something but it looks like 4E and 8E are not locked for the value 6 because 9E is potentially 6 as well.
An XY-Chain is possible from this although it would have the pivot chain below:
3A - 3C - 3H - 6H - 6F - 5F - 9F
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