SudokuWiki.org
Strategies for Popular Number Puzzles

# Single's Chains

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.

On this board I have clicked on 5 in C8 to highlight all the 5s in the puzzle. You can also use the little tool under the list of strategies to focus on a specific number and view all the chains. Tick the "rows", "columns" and "boxes" and un-tick the other numbers. This gives us all the possible chains 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 rules below. Each end of each link can be assigned one of two colours. You can start in any position, taking any 5 on the board and give it one colour. Then follow each chain link alternating the colour. 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 to different numbers. 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.

## Rule 2 - Twice in a Unit

This rule is shared with 3D Medusa.

This puzzle has a series of Rule 4 Single Chains culminating in this excellent Rule 2 example. Mapping all the chain links for number 7 we find something interesting in row C. There are two yellow 7s in C1 and C4. This Rule 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).

## Rule 4 - Two colours 'elsewhere'

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.

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 we can see a more complicated network of 8s than 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 FallsOffRocks wrote to me to point out the algorithm for the old Rule 5 could always get all Rule 4 eliminations. So have rolled Rule 5 into Rule 4. Both rules are examples of off-chain eliminations looking at candidates in different cells. The simplification also affects 3D Medusa.

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.

## Singles Chains Exemplars

These puzzles require Singles Chains strategy at some point but are otherwise trivial.
All contain only one Naked Pair in addition to Single Chains They make good practice puzzles.

 Go back to X-Wings Continue to Y-Wings

## ... by: Tony

Friday 6-Oct-2023
Andrew. This is my go-to site to learn but this page has a huge problem. You start with an example for a basic discussion of simple coloring but 3 paragraphs later start a discussion of 3D Medusa and notes on Rules 1 and 3, synchronization, historic solver information , etc. All very advanced and confusing to the step by step learner.
You start with Rule 2 where you tell us the 5 in E3 is eliminated by Rule 4. Where is this explained? This reader is lost. I know I'm slow but it seems this page could be a little gentler. Thanks
Andrew Stuart writes:
I've just re-read the page. Yes it could be a bit clearer.
I was thinking of someone who might be stepping through the example puzzle in the solver. But Having done that myself I get a series of Rule 4 and one Rule 2 and it didn't follow. With my new string format I can take the user to the exact point in solver now, so I've re-written the rule 2 text and changed the diagram
Hope you think it's a bit better

Appreciate the feedback!
- [Del]

## ... by: Evelyn

Thursday 9-Mar-2023
I get the idea of simple coloring, is actually quite easy to understand, but would there normally be any cues for us to use the strategy? Sometimes I'm even well aware that I'd need to use this strategy but I just can't find anywhere I can start with.

## ... by: M.A.Kulkarni

Wednesday 25-Jan-2023
Illustration (4): is a particular case of the more general case of Singles with a mix of Strong links and Weak links, subject to the condition that the 1st and last links of the chain are each a Strong link.

## ... by: Molly

Wednesday 26-Oct-2022
Examplar 2 solved with basics and didn't require simple coloring. So I'm surprised that the solver used that strategy. BTW this website is great for me! I am using it to try to learn the tough strategies. I'm far from mastering them but enjoy trying with the practice puzzles.
Andrew Stuart writes:
Shows up for me stepping through the solver. But maybe another route to avoid - [Del]

## ... by: Paul Philbin

Wednesday 6-Apr-2022
I want to learn simple colouring. I don't get why specific numbers are ignored in the chains. In the examples with 5's, why are some 5's bypassed, and I don't get the logic of when to remove them. Thank you. Paul

## ... by: Picks

Tuesday 29-Mar-2022
I have a board that contains a simple coloring chain that has one cell that has to be both ON and OFF (that is, if you follow the chain it is ON at an early point in the chain which then loops back and subsequently indicates it should be OFF).
What does this indicate for the chain as a whole, or does it simply indicate that I've made an error somewhere?

## ... by: Greg

Monday 2-Mar-2020
I tried exemplar 5 and it can be solved with only one chain.

The solver uses two chains of 4s (rule 4) followed by a chain of 5s (rule 2). However, if you use the chain of 5s first you don't need the chains of 4s.

I found that chain first and was surprised because the puzzle was easier than I was expecting.

I guess the solver's algorithm picks the first chain it finds instead of checking which is more efficient.

## ... by: jon

Tuesday 30-Apr-2019
I really don't understand rule 2, what does it mean if two of the same colors see each other?
Andrew Stuart writes:
That two candidates with the same color are in the same row, box or column. - [Del]

## ... by: Ally

Monday 11-Mar-2019
I only see rules 2 and 4. Am I missing 1 and 3 or is that intentional?
Andrew Stuart writes:
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 to different numbers. - [Del]

## ... by: Kent

Sunday 24-Feb-2019
This colouring trick is great/cool/welcome, but if you don't happen to have "Sudoku-Solver's" help available for a given puzzle, a transparency sheet and dry-erase markers might also work. All these lines and colours on a sudoku puzzle would make a mess.

## ... by: Basil Dalie

Tuesday 27-Mar-2018
Hello.
Could you explain me what to do in the situation in which, during the construction of the chain, we find a strong link between two cells that have already been colored with the same color?
Andrew Stuart writes:
Sounds like a discontinuity. They cannot both be the same colour since that implies they same same number twice. It's not necessarily a mistake in your chaining but a highlighting of a contradiction to be taken advantage of. - [Del]

## ... by: Steve

Friday 26-Jan-2018
I think I've found an extension to rule 4: if you are discarding a candidate that is part of a different chain, then you can also discard other candidates in the chain that share that colour.

The image at
https://i.imgur.com/m91sfxS.png
shows an instance with multiple chains where rule 4 is eliminating a candidate from the second chain; if that happens, then the next element in chain 2 is the solution and so on.
Andrew Stuart writes:
Correct. You are chaining your chains! The solver (and I) would break that down into two steps but yes you get a cascade there. - [Del]

## ... by: Hound99

Wednesday 17-Jan-2018
in the Rule 4 example, why would you eliminate B5 but NOT F5. If you had started the chain at B5 you would have eliminated F5 instead.

In this case, shouldn't BOTH be eliminated because you have two of the same color on a single unit (row)?
Andrew Stuart writes:
There is no chain starting with 3 on B5. Too many 3s in the row, column and box. - [Del]

## ... by: Alpo1

Thursday 9-Nov-2017
Hi there and big thanks for the site.

I have a question about linking pairs. I was doing an Y-wing example and came across a Singles Chain solution. In that solution, for some reason a pair on the same row (7,8 in H1 and 2,7 in H7) wasn't linked.
The situation is quite similar as it is in the first example on this page, linking fives in J2 and J9.

Actually here's a screen capture: https://imgur.com/QolT21N

Why aren't the pair in question linked?

Thanks
Andrew Stuart writes:
Because there are three 7s in row H, no link possible - [Del]

## ... by: William

Tuesday 12-Sep-2017
I have read this for past one year when I get one that need this technique, and I believe that I will understand it eventually. I read it again these days, and make feel like the "hard" or level is as much I can go on sudoku......

## ... by: Harley

Saturday 28-Jan-2017
I have been using your solver for many months and love it.

In trying to solve single chains, it seems there should be a way to either turn a candidate on or off but I can't figure out how. Am I missing something?

## ... by: t.l. shroeger

Thursday 7-Apr-2016
Regarding Rule 2 - Twice in a unit. In your example in box 1, A2 and B1 are both used to start a chain. They are both colored yellow. Since the color choice to start a chain is arbitrary, how was the color yellow (for both) chosen? Obviously, if one of the locations starting the chain was given a blue color the outcome of the puzzle is changed.

The question is: when two or more chains are in a puzzle, how are the colors used to start a chain determined?

Thanks,
Tom
Andrew Stuart writes:
Well that’s the solver getting ahead of itself. It want to show both the pattern - the two colors (green and blue) and the eliminated cells - which are all the green ones in this case. My color code for elimination is yellow with red text. Actually A2 and B1 are not the starts of the same change net. The 5 in A1 prevents you from directly connecting A2 and B1, so you have to choose one to start with - if you trace A2 round to B1 you find it has the same color. - [Del]

## ... by: RD

Sunday 27-Mar-2016
I'm pretty sure that any eliminations found by applying Rule 2 can also be found by applying Rule 4, simply by starting at different points and following different paths through the network of connections. (But I can't provide a mathetmatical proof). If this is the case, then Rule 2 is redundant, and there would be no need for the 'off-chain' vs. 'off chain' distinction, which is perhaps more confusing than enlightening?

Having noted this, even greater simplification of strategies can be achieved by noting that all instances of Rule 4 Singles Chains are in fact examples of X-cycles with an odd number of nodes - the candidate that sees both ends of a chain that begins and ends on strong links will be the odd one out, linked by a W-W connection. In which case the strategy of 'Simple Colouring/Singles Chains' becomes entirely redundant.

From the humanities side, I would point out that the apostrophe in "Single's" isn't needed, you're trying to mark plurality not possession.

Thanks for the challenging and interesting site.

## ... by: Tom

Monday 12-Oct-2015
In your first example under Rule 4, you show two separate chains, B9 to F5 and J2 to H5. The J2 chain has not been assigned any colors. Can you assign colors to the J2 chain and have it work in concert with the B9 chain? I have used two chains together but I obtain different results depending on how the colors are assigned to the second chain. My question is, if two chains are used together to gain solutions, how is it determined which color is used to start the second chain as well as which end of the second chain is the starting end?

Enjoy your site and use it often, keep up the good work.

Thanks,
Tom
Andrew Stuart writes:
The two chains cannot be used together and cant be linked with colors. Eliminations will be made independently, but there is no guarantee either will produce a result. They are isolated form each other.
If you could link them then they would become one chain. That may not be possible with this number but there are several ways to extend chains, eg through cells (see 3D Medusa).
- [Del]

## ... by: Jason

Friday 26-Jun-2015
Is it possible to rephrase your rule explanation for rule 2? It is confusing and incomplete. If I understand your rule 2 correctly, do you mean the following .....
"Rule 2 states that if a row or a column or a cell contains 2 candidates that share the same color, they are contradicting to each other, and they can be removed. Therefore, the candidates circled in red can all be removed. "

## ... by: Volker Schaefer

Thursday 28-May-2015
I've a question about rule 4. There are two chains at the first grid shown on this page.
One of them starts at b1 to b5 to a4 to f4 to d6 to d3 etc.
Colouring this chain means that b1 gets green and d3 gets blue.
Why isn't it possible to remove the 5 of e1, which sees both ends of the chain?
Andrew Stuart writes:
Your idea is valid, you can remove the 5 form E1 for that reason. The solver has to do things in a blind order, top left to bottom right, and currently it only returns the first instance of a result in one strategy. So it gets the example one first.

Note, there are two chains and they are separate when considering this strategy
- [Del]

## ... by: Trevor

Sunday 8-Mar-2015
I'm sure that you know what you mean and not what you say in your reply to my comment of 2 March - you're not going to "remove all candidates on the ON cells" are you? What I would like to see is all ON candidates become solutions either in the same or the next step of the solver. The beauty of Simple Colouring Rule 2 is that we know immediately the fate of every candidate in the chain. In the example for Rule 2 there are seven ON candidates. Six of them become Naked Singles and become solutions in the next step because they are singles. But the 5 in E9 becomes a Hidden Single and doesn't get promoted to a solution until two steps later, i.e. its membership of the chain is forgotten and its fate is decided by later scanning. The fundamental problem is in the thinking - the message in the results window says "all yellow coloured candidates can be removed". This needs to be enhanced with "and all blue coloured candidates become solutions". This would reflect the full meaning of Rule 2 and not just half of it as now.
Andrew Stuart writes:
I've added some words as I wasn't clear

" I should remove all other candidates on the ON cells (leaving the one correct candidate behind)."

My point is that the solver is subtractive. I can't set a cell to a certain number. I have to remove candidates leaving one behind. The point where a single candidate is turned into a solution is in the solver "Check for solved cells" and must be there only. Might seem a pedantic point but its consistent behaviour I'm looking for.

It's true the current implementation only does the off side. I'll release an update soon as soon as I've done more testing
- [Del]

## ... by: Trevor

Monday 2-Mar-2015
In the description of Rule 2 it is stated uneqivocally that all candidates that share the colour with a pair that have the same colour will be OFF and all those with the alternative colour will be ON and THE SOLUTION FOR THE CELL. However, the solver only executes the OFF side of this and erases those candidates. The ON candidates are left to be dealt with by another technique. This not only counters the stated rule, it spoils the wonderful power of the rule that allows ALL the candidates in the chain to be dealt with according to their colour before moving on. Is there going to be an amendment to the solver to correct this omission?
Andrew Stuart writes:
That's a good observation. I'm so used to thinking in terms of removing candidates and then letting the solver find singles - this two part step is necessary to explain what is going on - if too much happens in one step it becomes difficult to see what happened. But logically you are right, I should remove all other candidates on the ON cells (leaving the one correct candidate behind). I will test this and push to the next update if successful.

I do note that Rule 2 usually gives a lot of singles anyway, so its not holding it back too much I believe, but I'll need to check quite a few examples.
- [Del]

## ... by: Clheins

Saturday 24-May-2014
How do you know which candidate to link? In your example above for candidate 5, could you have used candidate 2 instead?
Andrew Stuart writes:
Yes, each number could create an independent Singles Chain set, so each number should be checked, but I'd go with the one that looks like a nice spread with lots of links first. We know that 2 did not result in any Singles Chains in this example because the solver is exhaustive and tries numbers in order. - [Del]

## ... by: Jackg

Sunday 15-Sep-2013
In the fourth example down J9 is linked into box 6 containing more than two 5's
G3 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)?
Andrew Stuart writes:
J9 is not linked to E9 buy boxes. It is linked by their common column - where only two 5s exist. There are three 5s in column3 so there is no link between G3 and E3. Same with row E, four fives, so we can't connect to E3 and apply another rule. - [Del]

## ... by: Alan

Thursday 5-Sep-2013
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
Andrew Stuart writes:
Not everything, but mostly ;). It sounds like a handy rule when scanning with pen and paper. Just being aware of the different types and they'll get spotted. - [Del]

## ... by: Steve Thier

Tuesday 26-Feb-2013
I think rule 2 should include the condition that they be on the one chain.
Andrew Stuart writes:
Good idea, I've added a line at the top before the first rule. - [Del]

## ... by: td

Tuesday 22-Jan-2013
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?

## ... by: Ian

Thursday 25-Oct-2012
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

## ... by: Mike Kerstetter

Friday 19-Oct-2012
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?
Andrew Stuart writes:
You can mix two isolated sets of chains but it's quite complicated. Have a look at:
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.
- [Del]

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

Monday 3-Sep-2012
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.
Andrew Stuart writes:
Hi Dan
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. - [Del]

## ... by: Ludek Frybort

Tuesday 14-Aug-2012
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.

## ... by: John Riddle

Saturday 25-Feb-2012
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.
Andrew Stuart writes:
Yes but I used a screen shot of the previous step incorrectly. I have updated the diagram with 5 removed from E3. - [Del]

## ... by: John Riddle

Saturday 25-Feb-2012
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.

## ... by: Luke

Thursday 19-Jan-2012
How do you choose which colour to start a particular chain with?
Andrew Stuart writes:
It shouldn't matter which colour your start with or where you start. You should more or less get the same result. - [Del]

## ... by: Peter Foster

Thursday 22-Sep-2011
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.
Andrew Stuart writes:
I know where you are coming from. And there are degrees of "trail and error". I distinguish between the pattern found that leads to an elimination and searching for a pattern. The later can seem arbitrary long but the cut off between a logical strategy and trial and error is - is the logic based on a static pattern that exists on the board now - or have you modified the board by inserting and removing numbers?

Crook's Algorithm

I hope that helps a bit
- [Del]

Wednesday 17-Aug-2011
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
Andrew Stuart writes:
The 5 in A1 is critical to upsetting the neat array of pairs of 5 in rows, columns and boxes. It means there are three 5s in row A and column 1. That means no links can be made as part of the 'simple chains' web. But 5's across the board can be connected. It so happens that when they are connected with alternating colours (starting from any 5) the ringed 5s have the same colour. Since ALL of one colour is true and ALL of the other colour is false, the colour found twice in the row, column or box will be the false one, and all yellow can be eliminated. - [Del]

## ... by: Michael Wallis

Tuesday 26-Jul-2011
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.
Andrew Stuart writes:
Correct, I hadn't provided an example for this Rule for Simple Colouring, and your example is very good, so I've poached it. Rule 4 now added to the page above. Thanks for sharing and pointing it out. - [Del]

## ... by: Guy Renauldon

Friday 3-Jun-2011

The 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

## ... by: glisten

Monday 23-May-2011
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
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"?
Andrew Stuart writes:
I believe the example in Rule 5 happened before the examples above it. I have to search quite hard to find examples, and they are not always in the order of the documentation. Only that really :) - [Del]

## ... by: Zophuko

Thursday 12-May-2011
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
Andrew Stuart writes:
In the example for Rule 2, you will notice that A1 contains a 5 as well. If the eliminations of all one colour left no number X in some rows, columns or boxes, something very wrong would be going on. Infact, your chains would be very different in the first place. Rule 2 works because there are 3 or more candidates of X in such units. I suggest tracing the chain network from any starting point - you can only make chains along a unit with 2 candidates. You should get the same result every time. - [Del]

## ... by: John

Friday 8-Apr-2011
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?

## ... by: Zophuko

Wednesday 30-Mar-2011
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?
Andrew Stuart writes:
Several chains might co-exist at the same time but with no link between them. In this example the solver has found one chain and highlighted those numbers which form that chain, but it is not required to highlight all links in all chains. I don’t believe any of the links you found can be connected to that chain. The solver only returns the first chain that actually does something - there might be other chains and there might be other eliminations on other chains at this point. I am planning a version of the solver which will allow one to cycle through every elimination at a given point, which might show shorter solve routes over all - [Del]

## ... by: Zophuko

Tuesday 29-Mar-2011
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.

## ... by: Zophuko

Sunday 27-Mar-2011
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.
Andrew Stuart writes:
The connections you list are between cells that do not see each other, that is they share no common row, column or box. Not every cell can 'see' every other cell. A link is always along one common 'unit', sometimes two (same row and box for example). - [Del]

## ... by: Ed

Saturday 5-Feb-2011
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?

## ... by: John

Saturday 29-Jan-2011
I believe the easiest way to state this rule is

We 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.

## ... by: Lenni Nero

Thursday 27-Jan-2011
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.

## ... by: ozzy

Tuesday 25-Jan-2011
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!

## ... by: Robin

Tuesday 4-May-2010
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

Friday 26-Feb-2010
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

Thursday 25-Feb-2010
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: Lea Hayes

Wednesday 10-Feb-2010
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

Tuesday 5-Jan-2010
The 11-link chain in the last example may be shortened to a 5-link chain by linking B5 to A6.

## ... by: Steve

Tuesday 1-Dec-2009
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: Mike Wallis

Thursday 22-Oct-2009
"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

Sunday 28-Jun-2009
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

Thursday 18-Jun-2009
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?