Single's Chains, also known as **Simple Colouring** is a chaining strategy and part of a large family of such strategies. 'Simple' refers to the idea that one candidate number is considered - in contrast to 'multi-colouring' which is the basis of 3D Medusa. Single's Chains also related to X-Cycles.

A 'chain' is a series of links hopping from one candidate to another following very simple rules. A candidate can either be ON or OFF. That is, we either think it is a possible solution to that cell, or we do not. There are consequences to the rest of the board when we 'link' these two states. When we are starting out we don't know which will be ON or OFF so any two colours will do.

A 'chain' is a series of links hopping from one candidate to another following very simple rules. A candidate can either be ON or OFF. That is, we either think it is a possible solution to that cell, or we do not. There are consequences to the rest of the board when we 'link' these two states. When we are starting out we don't know which will be ON or OFF so any two colours will do.

On this board I have clicked on 5 in C8 to highlight all the 5s in the puzzle. From this it is possible to map all the possible chains. These are between 5s where there are only two remaining 5s left in any row, column or box. The technical term for these are 'bi-location' links. Where there are three or more 5s - for example in box 1 and box 6, no links are possible within those boxes. But there are plenty about.

Now, the Colouring aspect which sometimes gives this strategy its name, is illustrated in the Rule 2 diagram below. Each end of each link can be assigned a colour. You can start in any position, taking any 5 on the board and give it a colour. If you follow each chain link, alternate the colours. The strategy is all about recognising that one of those colours will be the solution and the other not. The rules that follow identify the contradictions that allow us to eliminate candidates or decide which colour (which end of every link) is the solution.

Now, the Colouring aspect which sometimes gives this strategy its name, is illustrated in the Rule 2 diagram below. Each end of each link can be assigned a colour. You can start in any position, taking any 5 on the board and give it a colour. If you follow each chain link, alternate the colours. The strategy is all about recognising that one of those colours will be the solution and the other not. The rules that follow identify the contradictions that allow us to eliminate candidates or decide which colour (which end of every link) is the solution.

Just a note on rule numbers: The solver uses the same search algorithm for both Singles Chains and 3D Medusa so I have synchronised the rule numbers that are returned. Rules 1 and 3 apply only to 3D Medusa (Multi-colouring) since they extend chains into multiple candidates. However, the solver needs to look for Singles Chains first because I deem it to be a simpler strategy that's easier to search for.

This rule is shared with 3D Medusa.

Taking the example we started with - and identifying all the chain links, we find that in three units there are 5s highlighted in the same colour. The top row has two yellow 5s in A2 and A4. Box 1 has yellow 5s in A2 and B1 and finally, column 1 has yellow fives in B1 and E1. This Rules says that if any unit has the same colour twice ALL those candidates which share that colour must be OFF. The alternative colour will be ON and the solution for that cell.

(Actually yellow is the colour I use to show eliminated candidates. The solver will return Green and Blue for the colouring but then switch one or other to yellow if the candidates are to be eliminated).

Taking the example we started with - and identifying all the chain links, we find that in three units there are 5s highlighted in the same colour. The top row has two yellow 5s in A2 and A4. Box 1 has yellow 5s in A2 and B1 and finally, column 1 has yellow fives in B1 and E1. This Rules says that if any unit has the same colour twice ALL those candidates which share that colour must be OFF. The alternative colour will be ON and the solution for that cell.

(Actually yellow is the colour I use to show eliminated candidates. The solver will return Green and Blue for the colouring but then switch one or other to yellow if the candidates are to be eliminated).

If we can eliminate "off chain" in a cell we can certainly do so off-chain. This is a great example puzzle for Rule 4 - there are three in succession and I'm showing the first and third. This first cannot be simpler. Either B9 is 3 or - follow the chain - F5 is 3. One or the other. Since the 3 at B5 can see both ends of the chain it can be removed. You've heard that story before in earlier strategies. With Simple Coloring the 3 in B5 can 'see' two different colors elsewhere.

If you have looked at X-Cycles you'll spot how these two strategies overlap - if your colouring happens to form a loop, as it does here. Also, if you've read as far as AICs you may recognize the pattern of 3s as a classic double alternating Nice Loop with a discontinuity in B5. But that's another story.

Michael Wallis is an early pioneer of Simple Coloring and this rule family.

This rule is shared with 3D Medusa.

If you have looked at X-Cycles you'll spot how these two strategies overlap - if your colouring happens to form a loop, as it does here. Also, if you've read as far as AICs you may recognize the pattern of 3s as a classic double alternating Nice Loop with a discontinuity in B5. But that's another story.

Michael Wallis is an early pioneer of Simple Coloring and this rule family.

This rule is shared with 3D Medusa.

Rule 4 is simply put: if you can spot a candidate X that can see an X of both colours - then it must be removed. The third instance of this strategy concerns 8s and I show it reveal a more complicated network of 8s that the 3s above. However, it is easy to pick out the blue and green 8s that point to the eliminations.

The documentation on this page has changed in February 2015 when a reader called

The strategy which naturally follows on from Singles Chains is 3D Medusa, but you should also read up on the article Introducing Chains and Links.

All contain only one Naked Pair in addition to Single Chains They make good practice puzzles.

## Comments

Talk Subject CommentsComments here pertain to corrections to the text, not the subject itself## Sunday 25-Mar-2012

## ... by: Pieter, Newtown, Oz

Andrew, since you've added Rule 4 and Michael Wallis's example here in Simple Colouring, you need to amend "Just a note on rule numbers: ... Rules 1 and 3 (not 1,3 & 4) apply only to 3D Medusa" (just before the explanation for Rule 2)## Thursday 1-Mar-2012

## ... by: Roman Malyniak

Some suggestions about the "All 5 links " top board:- the drawn 5-bi-locations links D3-E1 and D3-G3 are not valid.

- the 5-bi-locatin link G3-G7 is missing. Also this strong

link forms an ER example with boxes4 and 6,removing 5

from E3 and E7 and making the D3-E1 and D3-G3 links valid.

## Saturday 26-Feb-2011

## ... by: peter

Thanks for the clear explanation. If I may point out a small typo. In the description of the "Twice in a Unit" rule, you say: "The top row has two yellow 5s in A2 and A5". I think you mean "A2 and A4".## Friday 28-Jan-2011

## ... by: Bill Howard

This is very confusing to me. Where do you start? No hint about where to start. You say it will be rewritten, but that statement was long ago. Is this site still active?A better description is very necessary.

## Saturday 19-Sep-2009

## ... by: jO

The problem I have following your example starts with the beginning:"Another way of looking at this...." Please what is the first way of looking at this? I am looking for a different explanation since this one is not clear.

## Thursday 7-May-2009

## ... 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.