... by: Robert

You say it doesn't matter where you start with the colours, so what if you start with the one you have as yellow (c7) and made it blue the one in (c2) green and so on you end up with the one in (g7) as the yellow one

... by: Robin

Hi, Many thanks for the simple colouring explanation; that helped with a puzzle I couldn't solve; however the "simple colouring Rule 5 example" appears to be a very simple puzzle that does not require anything beyond very basic skill to solve. This must be an error ?

... by: Mike

X and A form a conjugate pair, so why can't X be one end of the chain and D the exlusionary square?

... by: Marge Falconer

Agree that this explanation remains confusing. However, using the basic premise of the chain, I've just tried to replace at any point in the chain the number I'm dealing with and one answer soon becomes apparent. Unfortunately, I need more practice with this strategy and only come across it rarely. I would love a site that explained a strategy and then gave you three or four puzzles which would require you to use it.

... by: soooconfused

Why does the solver use coloring when there are 3+ candidates in a unit? I thought this was not allowed for singles chains. http://i46.tinypic.com/2akfnnm.jpg

... by: Lea Hayes

Thank you for your interesting descriptions on various Sudoku strategies.

What is the difference between single's chains and x-cycles? Do x-cycles always cover the same ground, or are they completely different?

... by: Ben Wearn

The 11-link chain in the last example may be shortened to a 5-link chain by linking B5 to A6.

... by: Steve

Colouring example 1 in fact shows rule 2, not rule 1: A and X are a conjugate pair, and there is no justification for arbitrarily ending the chain at A instead of continuing it to X. If you add 5 as a candidate in F3, then you will have a valid example of rule 1.

Interestingly, though, example 1 also has an X-cycle with a weak link discontinuity at X, which provides a different reason for eliminating 5 at X.

... by: gerryfromktown

Yeah, this explanation is a mess. It begins with this reference to non-existent text:

"Another way of looking at this is the popular technique of Colouring."

It appears the explanation is a cut-and-paste from here:

http://eric4ever.sudoku.googlepages.com/strategies_advanced.htm#SC

where you can find the missing text.

... by: Mike Wallis

"Another way of looking at this is the popular technique of Colouring". What does "this" refer to?

The explanation for Rule 1 states that 5 can be removed from X (G3) because it's outside the chain and points to A (D3) and D (G8).

However, the explanation for Rule 2 states that X (G3) and D (G8) are the "false' color because they're in the chain, both are blue, and both in the same row.

How can a square be both in a chain and outside of it at the same time? Moreover, Example 1 shows X (G3) and D (G8) in two separate chains - not in "a' chain.

There is another chain that runs from F to C, C to D, D to E, E to B, B to A, and A to X. However the example for Rule 2 doesn't show that.

In fact, none of the examples on this page show Rule 2 in action. They only show Rule 1.

The disconnect between Rule 2 and the examples has been very confusing to me and has caused me difficulty in solving puzzles using this technique. For this reason, I think the article for Singles Chains needs to be revised with new examples that more accurately reflect Rule 2.

... by: Don

Cell X cannot be green because cells D and X do not form a conjugate pair. That's because two other cells (G1 and G2) in row G contain 5 as a candidate. In column 3, however, cells A and X are the only two that contain 5 as a candidate; hence they do form a conjugate pair, making cell X blue.

... by: Clark

If you start with E=green, then B=blue, C=green, D=blue, X=green. 'A' now looks like the exclusion. Can you explain this better? Why is X the exclusion and not A?

... by: Semax

Every part of the chain links two cells which share one or two units, but these and only these shared units have to fulfill the condition that there can't be any other cells containing the candidate.

So in the first example you can link D7 to E8 because in their shared unit (box 6) there is no other cell containing a 5. In other words, if D7 is 5 then E8 can't be 5, and vice versa. Therefore it doesn't matter what happens in rows D and E or in columns 7 and 8.

... by: Stephen P. Byers

The Single's Chain business is very confusing. In your article on the subject you write, "... we are looking at candidate 5 and units with two 5's in them ( called conjugate pairs)." In the first example, however, you link to Cell E8 that is one of two 5's in Box 6, but appears disqualified to be in the chain because it is one of three 5's in Row E. Is this viable?

A similiar problem exists in your example titled "Coloring Example 1" at the end of the article. There are three 9's in Colimn 1 so how can two of them be included in the chain? Please explain.