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 in a unit. In this example I have mapped out the links between 4s. We are certain than ALL blues are the solution or ALL greens. Therefore where there are candidates that can see both colours they can be removed. By 'see' we mean any candidates that are in the same row, column or box as a green and blue candidate.

We can get just two links into box 3. They are from A2 to A8 (giving us a blue) and from G9 to C9. The other 4s cannot be linked but they can be removed.

Many thanks for Michael Wallis for the example. This rule is shared with 3D Medusa.

We can get just two links into box 3. They are from A2 to A8 (giving us a blue) and from G9 to C9. The other 4s cannot be linked but they can be removed.

Many thanks for Michael Wallis for the example. This rule is shared with 3D Medusa.

This rule is shared with 3D Medusa.

Having identified all the links for a candidate number, we can look elsewhere for 'off-chain' eliminations. Simply put, if you can spot a candidate X that can see an X of both colours - then it must be removed. In the same Sudoku puzzle as above, but one step before (if you run it through the solver), a Rule 5 example is to be found.

At this stage of the puzzle there are fewer links to be found on candidate 5. But if we colour them we find something interesting going on with the 5 in E3. It can see the blue 5 in E9 and the green 5 in G3. We know the either blue 5s will be the solution or the green 5s will be the solution - so that leaves no room for any other 5 that happens to see both colours.

Having identified all the links for a candidate number, we can look elsewhere for 'off-chain' eliminations. Simply put, if you can spot a candidate X that can see an X of both colours - then it must be removed. In the same Sudoku puzzle as above, but one step before (if you run it through the solver), a Rule 5 example is to be found.

At this stage of the puzzle there are fewer links to be found on candidate 5. But if we colour them we find something interesting going on with the 5 in E3. It can see the blue 5 in E9 and the green 5 in G3. We know the either blue 5s will be the solution or the green 5s will be the solution - so that leaves no room for any other 5 that happens to see both colours.

If you have looked at X-Cycles you'll spot how these two strategies overlap - if your colouring happens to form a loop.

The three blue lines connect pairs of 9s - along columns and rows and within box 7. These are the only units with just two 9s left. As we connect them we alternate between green and blue colours. It doesn't matter where we start or which colour we start with. In this configuration we spot an uncoloured 9 in C7 which can 'see' both a green 9 (C2) and a blue 9 (G7). Since we know that either green or blue will be the solution this 9 in C7 can't be a solution.

I have connected C7 using red lines. You will notice that there are four 9s in total in row C and three 9s in column 7. Simple Colouring can't connect these 9s as only pairs can make a connection.

The three blue lines connect pairs of 9s - along columns and rows and within box 7. These are the only units with just two 9s left. As we connect them we alternate between green and blue colours. It doesn't matter where we start or which colour we start with. In this configuration we spot an uncoloured 9 in C7 which can 'see' both a green 9 (C2) and a blue 9 (G7). Since we know that either green or blue will be the solution this 9 in C7 can't be a solution.

I have connected C7 using red lines. You will notice that there are four 9s in total in row C and three 9s in column 7. Simple Colouring can't connect these 9s as only pairs can make a connection.

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.

## Comments

Comments Talk## Saturday 24-May-2014

## ... by: Clheins

How do you know which candidate to link? In your example above for candidate 5, could you have used candidate 2 instead?## Sunday 15-Sep-2013

## ... by: Jackg

In the fourth example down J9 is linked into box 6 containing more than two 5'sG3 is not linked up into box 4 containing more than two 5's resulting in E3 seeing two colors. Any reason (besides giving an example of rule 5) that G3 couldn't be linked to E3 resulting in two colors (blue) in a unit therefore allowing for the removal of all blues (rule 2)?

## Thursday 5-Sep-2013

## ... by: Alan alansara@zoominternet.net

In my simple way of thinking do I have this right concerning single's chain?You can start with any number or color as long as there is only two of that number in any box,column or row. Once the chain is completed using alternate colors you simply look for any box, column or row that has two or more of the same color and then you simply eliminate all of that color in that box, column or row.

Did I cover everything in those two sentences?

alansara@zoominternet.net

## Tuesday 26-Feb-2013

## ... by: Steve Thier

I think rule 2 should include the condition that they be on the one chain.## Tuesday 22-Jan-2013

## ... by: td

andrew, great website. I have a suggestion and a question. Suggest the use of a tracing paper grid identical to the sudoku grid for coloring . The lines are easily erased leaving the sudoku uncluttered.. question: how do you go about choosing the candidate most likely to yeild results? Greatest number of candidates? If one has to check all of the possible candidates, it would be a very lengthy process. Any hints?## Thursday 25-Oct-2012

## ... by: Ian

Andrew, I'm puzzled by one aspect of the chaining and networking type strategies inasmuch as what sort of criteria do you use when choosing the starting tuple? In the example diagram 1 above, you start by saying; "Ive clicked on the 5 at C8 to highlight all the 5's in the puzzle...". But why 5? What determined that you looked at 5 to start with in that specific case? I'm trying to tie down - to define precisely - the differences between 'trial and error' strategies and 'logical' strategies - or what I like to think of as 'pattern recognition' strategies. At the time of asking this question, I put these single tuple type networking strategies in the 'trial and error' box but I desperately want for them to be moved to the other camp. However, for this to happen I need a 'pattern' type of reason. I hope this question makes sense. Thanks in advance.Great site! Fascinating!

Ian

## Friday 19-Oct-2012

## ... by: Mike Kerstetter

So just to clarify, am I correct in thinking that while the top diagram shows two unconnected 5 chains, I should not try to apply these coloring rules to multiple chains at the same time? I should apply them only to a SINGLE connected chain at a time?Multi-Coloring Strategy

But if you have two isolated chains you can use them individually to use Single Chains strategy - they shouldn't inhibit each other. Just don't reach across.

## Monday 3-Sep-2012

## ... by: Can I use singles chains in Windoku puzzles?

I am trying to get through a series on fiendish Hyper-Sudoku puzzles (from an Android app - Super Sudoku). I recently found they are also called Windoku. I find that I can use singles chains when I have exhausted easier strategies. I can check my board and the app will tell where i have made mistakes and let me back up, so I know when I have made a mistake. Hyper-Sudoku has a set of four shaded square areas overlaying the board (B2-D4, B6-D8, F2-H4, F6-H8) and they must be also filled in accordance with the normal rules.Anyway, I find that sometimes singles chaining will indicate a number can be eliminated from a cell and the program indicates that it cannot. I'm thinking that either I am missing something, but cannot use your solver to find what I've missed. Maybe I can use the killer solver, but I think it works differently. I just want to know if I should be using singles chaining for this type of Sudoku puzzle and if your solvers can be used to solve them. I really like your solvers and they are very helpful in learning the various strategies.

It is very dangerous to use the normal Sudoku solver for Windoku. Windoku has extra constraints (the four 3x3 blocks) that mean you can't assume certain things. All the normal Sudoku strategies apply but the normal Sudoku solver is not aware of the extra constraints so it will go down the wrong path eventually. It is also likely the solver will find multiple solutions as well.

I intend to provide a solver for Windoku just as I have a seperate solver for Sudoku X and Jigsaw.

## Tuesday 14-Aug-2012

## ... by: Ludek Frybort

I found a sudoku, which can demonstrate what I find an interesting twist to the colouring strategies:The interesting part comes if you run it through the solver to the point where it uses simple colouring on 9s.

You get this:

CLICK

(but if you load this in the solver, you'll have to go through some pointing pairs and an X-wing to eliminate some candidates, which were already eliminated in the original process, to get to the point I'm talking about)

At this point (actually right before this point but the order doesn't matter) I tried to use simple colouring on 2s - unsuccessfully, nothing was eliminated. But since I did it on paper, the "colors" remained on the sheet.

Now I correctly guessed it was the time to look for Y-wings. Before I could spot the D2 E3 E7 Y-wing, which the solver uses to solve the puzzle, I spotted another one at E4 F5 F9.

The latter Y-wing is useless itself, it doesn't eliminate anything, but I noticed that the 2s on both its ends have the same color from the previous attempt at colouring.

So the logic of Y-wing says one of them must be the solution and the logic of simple colouring says that if one of them is, the other must be, too. So I could conclude both must be the solution, quickly solve all the remaining 2s and the rest was trivial.

I'm posting this firstly to just share what I find an interesting puzzle, and secondly in the hope that maybe somebody with deeper understanding of the strategies might find a way to generalize it and maybe extend the power of the colouring strategies.

## Saturday 25-Feb-2012

## ... by: John Riddle

OK I get it...E3 was eliminated by rule 5 on the previous turn leaving E1 and D3 as the only cantidates in box 4.## Saturday 25-Feb-2012

## ... by: John Riddle

There are three 5's cantidates in box 4. So why is E1 and D3 listed as a valid link? Please advise - I don't understand.## Thursday 19-Jan-2012

## ... by: Luke

How do you choose which colour to start a particular chain with?## Thursday 22-Sep-2011

## ... by: Peter Foster

It appears to me that simple colouring and the 3D Medusa techniques are just another name for 'trial and error'.We colour a possibility and 'see where that takes us'. Maybe during the colouring it indicates that the number we started with can't be right or the same number in other parts of the grid can't be there.

But essentially it seems to me that we are 'trying' a number and finding out whether it is valid or not.

The only difference is that should we 'try' the number and it 'not' result in eliminations in other parts of the grid, then it can't be assumed that that number was correct. It's a negative rather than positive result.

Am I misunderstanding something ?

I use the 2 extensively, but it does seem like I'm getting short-changed when compared to X-wing etc.

I try and be more clear about this subject on this page

Crook's Algorithm

I hope that helps a bit

## Wednesday 17-Aug-2011

## ... by: Joe Badillo

I need help with example 2:Why is the 5 in A2 eliminated and not the 5 in J2

Why is the 5 kept in D3 and not the 5 in D6

Why is the 5 in A1 not colored at all and will eventually kept

Simple coloring is not so simple for me

Please help

## Tuesday 26-Jul-2011

## ... by: Michael Wallis

I was working on the this puzzle:Your sudoku solver returned the following when I clicked 'Take Step':

"SIMPLE COLORING (RULE 4): Two different colors for 4 appear in Box 3 with more than two occurences for 4. The uncolored candidates can be eliminated from B8 C7".

However, when I clicked on the explanation for 'Simple Colouring', I only saw references to Rules 2 and 5. What is 'Rule 4' and why isn't it in the explanation for 'Simple Colouring'. You did mention a Rule 4 for 3D-Medusa, but that seems to be a different strategy than simple colouring.

## Friday 3-Jun-2011

## ... by: Guy Renauldon

For Glisten and Donald WaddellThe candidates 5 are not shown in E3 in the introduction and in the second diagram because these diagrams represent a step after the third diagram (the third diagram is the first step if you prefer) First and second grid are identical.

To solve this sudoku two steps are necessary:

First step (third diagram): rule 5 allows you to remove the 5 in E3.

Second step (second diagram): the 5 being removed from E3 give a strong link between E1 and D3. Then you can use Rule 2 which allows you to remove all yellow candidates 5 in the second diagram.

Very good example of Single's chains: rule 5 followed by rule 2

## Monday 23-May-2011

## ... by: glisten

There seems to be a problem with the first three stages of the example sudoku. In the first two stages, E3 is shown only with candidates 1,2. But in the third stage, the candidate 5 appears. I believe candidate 5 should appear at stage 1 and 2. And if so, that would invalidate the links to the left and down from D3. You are way better at this than I am. What did I miss?## Monday 16-May-2011

## ... by: Donald Waddell

Love the Sudoku site and increasing my knowledge of its logic. In following the example for Single's Chain, is there a reason why the candidate 5 in cell E3 is not shown on the introduction, but is shown on a subsequent diagram for the same puzzle for "Rule 5"?## Thursday 12-May-2011

## ... by: Zophuko

Rule 5, two colors elsewhere , appears logically quite transparent to me but I am having trouble understanding Rule 2, twice in a unit, regarding removing all instances of the number that occurs twice in one unit. Would appreciate a more detailed explanation of the logic underlying Rule 2.Thanks for a wonderful wevbsite

## Friday 8-Apr-2011

## ... by: John

Thanks for the wonderful instruction. My question has to do with practicality/tactics for when one is doing a paper (not online) puzzle. Do you draw pencil lines between the numbers? Could get extraordinarily messy. What is the best way of applying this on in a pencil/paper mode?## Wednesday 30-Mar-2011

## ... by: Zophuko

Thank you Andrew. The linking rules now appear very clear.Another observation and question regarding your example next to the Rule 5 discussion above. It seems to me that, under the rules, the following connections could also be made: A4 to B5 to B1; B5 to H5 to H6 to D6 to D3; D6 to F4 to A4; F4 to F7 to G7. What is the reason you do not show those links?

## Tuesday 29-Mar-2011

## ... by: Zophuko

Thank you Andrew. It is now very clear. Also, it appears to me that if intra-box links are omitted, one obtains only possible X-wing or swordfish patterns.## Sunday 27-Mar-2011

## ... by: Zophuko

RE: constructing single chains.Referring to the example, I do not perceive clear and restrictive rules for not drawing links between the following: A2 and C3, D6 and E7, H6 and G7, D6 and F7, A4 and D3, A1 and E1; or B1 to E1, or A1 to D3 leaving B1 unlinked.

Someone please explain the rules that prevent these kinds of links but allow those shown.

## Saturday 5-Feb-2011

## ... by: Ed

I'm fairly new to Sudoku puzzles and have enjoyed reading through the explanations for the various strategies presented on this website. The step by step solver with the detailed explanations that you provide seems like an excellent way to approach figuring out the most difficult puzzles. Thanks for all of the hours you've put in to this website.From reading through the posts below it looks like there may be some confusion regarding single chains and the explanation above. Some of the user comments seem to prove the logic or point out alternative methods to solving the example cells in question, but it still looks like there may be a bit of confusion on, for example, where to start, what color to start with, etc.

Basically restating what you've presented above...

1) Pick a number and link all of the single chains: A single chain occurs in a row, column or box (9 cells) when there are only two occurrences of that number in the given row, column or box.

From the first example puzzle above you chose the number 5. Notice that box 1 has three 5s so there are no single chains (both starting AND ending) in that box. Column 2, however, has only two 5s in it and they are connected as a single chain (A2->J9: both starting AND ending in that column). The 5s in Row D get connected (D3->D6) because there are only two in that row. Notice that the 5s in Row E don't get connected because there are more than two in that row. This process is repeated until all of the single chains are completed and you end up with what you've shown in the first example puzzle above.

2) Add the alternating colors: Start anywhere in the puzzle with any the chains you've just laid out and color one of the cells that is pointed to by one of the chains. It doesn't matter where you start. From there, branch out, making sure to alternate colors as you move along the various connections.

3) Eliminate the numbers in the cells (if possible) using the rules provided above:

Rule 2 (Twice in a Unit): If you see a box, row or column with two cells that have the same color, remove all instances of the number that share that color.

In the example above, there are three instances of units that contain more than one 5 with the same color: Box 1: (A2 & B1), Row A (A2 & A4) and Column 1 (B1 & E1). In addition to the four cells (A2, A4, B1 and E1) that are colored the same in these three units, all of the cells of the same color can be removed.

Rule 5 (Two Colors Elsewhere): If there is a cell that can "see" two other chained cells with alternate colors, it can be eliminated.

The third example puzzle above shows the single chains drawn in blue (I think the G7->J9 link is supposed to be blue to show it is also a chain). Cell E3 sees a blue 5 (E9) and green 5 (G3) and therefore the 5 in cell E3 can be eliminated. Notice that the 5 in cell E3 was not part of the original network of single chains.

Basically, Rule 2 helps eliminate candidates that are part of the network of chains and Rule 5 helps eliminate candidates that the network of chains can "see".

Reasonable?

## Saturday 29-Jan-2011

## ... by: John

I believe the easiest way to state this rule isWe have a chain, between PAIRS of a DIGIT, which has an ODD number of links. Any cell which can see BOTH ends of the chain cannot have that DIGIT.

Why? See the diagram where we have a chain between pairs of 9. One end is C2, the other end is G7.

Case 1: Try C2=9, follow the chain, G7="not9".

Case 2: Try C2="not9", follow the chain, G7=9.

One of these cases must be right but we don't know which.

If Case 1 is true, and C2=9, then C7 cannot be 9 because it can see the 9 at C2.

If Case 2 is true, and C2="not9", then C7 cannot be 9 because it can see the 9 at G7.

So C7 cannot be 9 in either case - because it can see both ends of a chain with an odd number of links.

It does not matter "where we start".

If we have a chain of pairs of digits with more than 3 links, we can look at all the sub-chains which have an odd number of links. So, with 4 links p-q, q-r, r-s, s-t, we first look at p-q, q-r and r-s. Any cell which can see both p and s cannot have the digit. We also look at q-r, r-s and s-t. Any cell which can see both q and t cannot have the digit.

## Thursday 27-Jan-2011

## ... by: Lenni Nero

Best site I've found yet.Might I please ask, for those of us who are red-green colour-blind (about 10% of males), would please avoid using those colours together whenever you do a re-write of the examples. TIA.

## Tuesday 25-Jan-2011

## ... by: ozzy

I get stuck often at "SIMPLE COLORING (RULE 2)" where the only colour is green and (and yellow) not combined green & blue as in example above (where it is shown the combination beteween those two colours, plus yellow for the "unwanted candidates"). In my solutions are only green and yellow. Am I missing something?The second rule is very hard for me to understand.

Thank you for a wonderful site!

## Tuesday 4-May-2010

## ... 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 ?## Friday 26-Feb-2010

## ... 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?## Thursday 25-Feb-2010

## ... 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.## Wednesday 10-Feb-2010

## ... 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?

## Tuesday 5-Jan-2010

## ... by: Ben Wearn

The 11-link chain in the last example may be shortened to a 5-link chain by linking B5 to A6.## Tuesday 1-Dec-2009

## ... 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.

## Thursday 22-Oct-2009

## ... 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.

## Sunday 28-Jun-2009

## ... 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.## Thursday 18-Jun-2009

## ... 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?## Tuesday 12-May-2009

## ... 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.