In Simple Coloring Rule 2: C1=8 and C4=3?

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

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.

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.

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.

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

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?

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.

I really don't understand rule 2, what does it mean if two of the same colors see each other?

I only see rules 2 and 4. Am I missing 1 and 3 or is that intentional?

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.

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?

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.

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)?

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

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......

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?

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

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.

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

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. "

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?
Thanks for your help.

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.

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?

How do you know which candidate to link? In your example above for candidate 5, could you have used candidate 2 instead?

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)?

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

I think rule 2 should include the condition that they be on the one chain.

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?

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

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?

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.

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.

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.

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.

How do you choose which colour to start a particular chain with?

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 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

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.

For Glisten and Donald Waddell

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

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?

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"?

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

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?

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?

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.

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.

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?

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.

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.

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!

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 ?

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

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.

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?

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

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.

"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.

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.

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?

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.