3D Medusa extends Simple Colouring (or 'Single's chains') into a third dimension. Simple Colouring looked for pairs of X in rows, columns and boxes. Wherever the chains led they stuck to the same candidate number. This is good for tracking an elimination when you have made notes on a paper Sudoku for a particular number but it limits the scope of the strategy. The way we extend the search is up through the bi-value cells which contain two different numbers. You can think of the different candidate numbers as existing in a third dimension lifting up from the page with 1 at the bottom and 9 at the top.

The devastating effect of colouring is that we are showing that ALL of one colour will be the solution. We don't know which set yet - but if any one of those cells becomes the solution we can know for certain ALL the cells of the same colour

The devastating effect of colouring is that we are showing that ALL of one colour will be the solution. We don't know which set yet - but if any one of those cells becomes the solution we can know for certain ALL the cells of the same colour

There are six different ways we eliminate - six contradictions. The first is in the example to the right. It doesn't matter where you start on the grid. In this example I've started with the 4s in row B. By colouring one green and the other yellow we mentally draw a line between them, done graphically on the diagram. Going into the third dimension in B7 when we colour the 4 yellow we can colour the 9 green - since there are only two values left in the cell.

Continue to look for bi-value and bi-location candidates and you soon build up a web of connections. This is where the image of Medusa was perhaps attached to this strategy - her head being a tangle of snakes.

Continue to look for bi-value and bi-location candidates and you soon build up a web of connections. This is where the image of Medusa was perhaps attached to this strategy - her head being a tangle of snakes.

When you have built up a web of connections, alternating between two colours you might find a cell with the same colour set twice. This has been ringed in H2. Since we know that if yellow candidates have the potential to be ALL true we can't have a situation where two yellow numbers are competing for the same cell. This is a contradiction and therefore we can state that no yellow numbers can be the solution!

Rule 1 is:**If two candidates in a cell have the same colour - all of that colour can be removed - and the opposite colour are all solutions**

** Update October 2015 **: My old definition of Rule 1 only made a negative assertion about the yellow candidates - they can be removed. But **Steve Jacobs**, also programming a solver, alerted me to the fact that all green candidates MUST be the solutions to their cells - a positive assertion. This is self-evident for bi-value cells (where only green and yellow exist and green becomes a Naked Single) but also for cells like H1 in the example. The 1 in H1 becomes a Hidden Single. This is true because of the binary either/or connections in the Medusa web. My solver will continue to only remove the yellow candidates - hidden singles will be eliminated in the next step - but for pen and paper solvers - go fill in the 'green' solved cells. Jacobs' corollary also applies to Rule 2 and Rule 6.

Note: this rule does not exist in Simple Colouring since the same number does not appear twice in the same cell.

As an exercise, try colouring any of the highlighted cells starting from a different position. You may end up swapping the colours around and you may find some new connections. But eventually - in this example - you will get two of the same colour on H2. This is a very powerful yet simple strategy.

Rule 1 is:

Note: this rule does not exist in Simple Colouring since the same number does not appear twice in the same cell.

As an exercise, try colouring any of the highlighted cells starting from a different position. You may end up swapping the colours around and you may find some new connections. But eventually - in this example - you will get two of the same colour on H2. This is a very powerful yet simple strategy.

This rule **is** shared with Simple Colouring. Its the same principle as the first rule but we are looking for two coloured occurrences of X in the same unit (row, column or box) as opposed the two of the colour in the same cell.

The example shows most of the links between bi-value and bi-location candidates, coloured between green and yellow. Ringed in red are two 7s in column 7. Since both cannot be true neither can be true and all yellow coloured candidates can be removed.

(Example requires three Medusa Rules 6 before Rules 2 comes into play)

The example shows most of the links between bi-value and bi-location candidates, coloured between green and yellow. Ringed in red are two 7s in column 7. Since both cannot be true neither can be true and all yellow coloured candidates can be removed.

(Example requires three Medusa Rules 6 before Rules 2 comes into play)

If you had unticked 3D Medusa in the solver this example would have been found by a number of later strategies, particularly Alternating Inference Chains as the pattern is a classic Nice Loop. 3 and 7 alternate. It doesn't matter where you start in a Nice Loop but you can trace the on / off or green/blue round the loop. 3s and 7s neatly occur twice in units and cells.

But 3D Medusa is not about loops, its about the network of links. This example just happens to be the same formation. We know that either ALL the blue candidates will be true, or ALL the green ones. If there are any another candidates in any cell with two colours, they cannot be solutions. Hence the 8 can be removed from C2. In Nice Loop terms, this is an*off-chain* elimination.

But 3D Medusa is not about loops, its about the network of links. This example just happens to be the same formation. We know that either ALL the blue candidates will be true, or ALL the green ones. If there are any another candidates in any cell with two colours, they cannot be solutions. Hence the 8 can be removed from C2. In Nice Loop terms, this is an

Simple Colouring cannot produce this elimination since it is restricted to a single candidate number.

If we can eliminate "off chain" in a cell we can certainly do so off-chain in a unit. In this example there are two sets of eliminations (blue and red lines) that point to 6s. 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 the same number as members of the blue/green links.

The 6 in B1 is removed because of the coloured 6s along the row in B9 and down the column in H1. In a similar way the blue 6 in C2 and the green 6 in B9 point to 6 in C8

This rule is shared with Simple Colouring.

The 6 in B1 is removed because of the coloured 6s along the row in B9 and down the column in H1. In a similar way the blue 6 in C2 and the green 6 in B9 point to 6 in C8

This rule is shared with Simple Colouring.

A few steps later in the same puzzle we get a cluster of 4s, 6s and an 8 using the same observation.

Since February 2015 I have combined Rule 4 with the old Rule 5. I want to thank a reader going by the name

This type of elimination looks to be the most complex - but inconveniently it is the most common. It's well worth looking out for. The rule says

**If an uncoloured candidate can see a coloured candidate elsewhere (it shares a unit) and an opposite coloured candidate in its own cell, it can be removed.**.

So its a combination of unit and cell - the colours green and blue are found looking along a unit and within the same cell. The example to the right demonstrates this with four eliminations.

The logic is very appealing. Consider 1 in E5. If 1 were the solution to the cell it would remove a green 1 from E6 AND a blue 7 from its own cell in E5. Since we know ALL blue or ALL green must be solutions we have a contradiction.

So its a combination of unit and cell - the colours green and blue are found looking along a unit and within the same cell. The example to the right demonstrates this with four eliminations.

The logic is very appealing. Consider 1 in E5. If 1 were the solution to the cell it would remove a green 1 from E6 AND a blue 7 from its own cell in E5. Since we know ALL blue or ALL green must be solutions we have a contradiction.

This type of elimination looks to be the most complex - but inconveniently it is the most common. It's well worth looking out for. The rule says

**If an uncoloured candidate can see a coloured candidate elsewhere (it shares a unit) and an opposite coloured candidate in its own cell, it can be removed.**.

So its a combination of unit and cell - the colours green and blue are found looking along a unit and within the same cell. The example to the right demonstrates this with four eliminations.

The logic is very appealing. Consider 1 in E5. If 1 were the solution to the cell it would remove a green 1 from E6 AND a blue 7 from its own cell in E5. Since we know ALL blue or ALL green must be solutions we have a contradiction.

So its a combination of unit and cell - the colours green and blue are found looking along a unit and within the same cell. The example to the right demonstrates this with four eliminations.

The logic is very appealing. Consider 1 in E5. If 1 were the solution to the cell it would remove a green 1 from E6 AND a blue 7 from its own cell in E5. Since we know ALL blue or ALL green must be solutions we have a contradiction.

Anton Delprado in the comments below has discovered another way we can use 3D Medusa and I'm pleased to include it in the solver. It's almost a reflection of Rule 5. Take any cell that doesn't have any colors from the coloring and see if all the candidates can see the same color. If that color were the solution (and remember, all of one or all of the other will be) then all the candidates in that cell would be removed - leaving an empty cell!

Rule 6 highlights in cyan the cell that catches this. You can see the 2 and 5 in cell C1 can see the yellow 2 in C9 and 5 in H1. (Yellow is used to show eliminated cells). All green coloured candidates are solutions to their cells.

Rule 6 highlights in cyan the cell that catches this. You can see the 2 and 5 in cell C1 can see the yellow 2 in C9 and 5 in H1. (Yellow is used to show eliminated cells). All green coloured candidates are solutions to their cells.

This puzzle has an amazing series of Medusa calls, using many different rules. It ends with Rule 7. I wanted to show a second puzzle to emphasise that the cell we are comparing the Medusa net to can have any number of candidates. These are 4,6 and 9 in C6. Green candidates have been turned yellow because they are eliminated, but you can see that the 4, 6 and 9 can all see the same color somewhere along the column or row.

You can be certain that it will be one color or the other, never equally both. Because this strategy is easier to spot and somewhat follows on from Rule 5, the solver looks for it before Rule 6. But too late to re-number them now. Well spotted Anton!

To end this article I want to show you some special puzzles discovered by Klaus Brenner starting with this 37 elimination Rule 1 Medusa that completely solves the puzzle from that point. We go from 35 known numbers to 70 and the rest is trivial!

There are two candidates in A1 with the same color, 5 and 7. So All of those of that color can be removed.

However, the initial puzzle is not trivial and a very large number of steps are required before this mega medusa. Certainly an extreme grade.

A slightly different puzzle, 22 clues, gives us these 10 eliminations using Rule 6. I don't know another with more in one step.

These puzzles require the 3D Medusa strategy at some point but are otherwise trivial.

They make good practice puzzles.

They make good practice puzzles.

## Comments

Comments Talk## Tuesday 22-Dec-2015

## ... by: rafirafi

As a cell is just a kind of unit you can also apply rule 6 to other units.P.ex. all candidates for a value in a row are not colored but can see the same color, because a given value must appear one time in a row, we know that every candidate with this color can be removed.

an example :

163...987794..3256258769..4..6...72.87.6...4.4....76..341..687.68..7...2.273.8461

the solver gives a cell forcing chain after a few basic steps

but we can use medusa rule 6 instead :

the candidates of values 9 in the central box are all removed by a net of 9 (check Show strong (bi-location) links in boxes, rows, colums, there is only one net with the value 9).

another example :

...4.........15....5.3.7..8135.8..9...2..9.53..9.53.......76.34..4..1...7..5..92.

after 2 locked candidates

There is a net with the chain +A6(2)-A6(8)+J6(8)-J6(4)+J5(4)-J5(3)+H5(3)

If you check all the candidates 2 in column 5: A5(2), C5(2), H5(2)

A5(2) and C5(2) can see +A6(2) - same box

H5(2) can see +H5(3) - same cell

## Monday 31-Aug-2015

## ... by: Nini Software

You may also extend two virtual networks, each one extended with the assumption that one color of the 3D Medusa strong chain is true.Then you get the same rules for candidates eliminations or color wrap, plus the extra rule : if both virtual networks are coloring a same candidate, then this candidate is true.

This very powerful method, named the virtual coloring, is published in french by Bernard Borrelly.

I recently implemented it with Microsoft Excel2003:

http://nini-software.fr/site/uploads/Sudoku/Sudoku%20V46.15%20by%20Nini%20Software.zip

The virtual coloring has successfully and very shortly resolved all your extreme puzzles of august 2015

## Friday 7-Nov-2014

## ... by: Rainer Ernst

4. Here is an example:This is a Set with normal 3D-Medusa coloring. The bivalue coloring of this chain is exhausted but no contradiction has been found.

Now you can shift to the second level of coloring and chose any colored candidate that if it was true would color other candidates.

Look at the yellow 5 candidate in R2C6. If yellow WAS true, then the 5 in R2C6 would clean R3C5 and R5C6 and thus both cells would be 2. You thus can color these in your 2nd level color that is related to the the 1st level color with the strength of "would be true if yellow was true" i here chose orange as related to yellow.

Though you must not color them yellow as the omission of an unambiguous bivalue link would corrupt the strong link bivalue information "it sure is either / or" and thus lead you into traps like corrupt a possible contradiction at a later stage or simply be false. Without further information these candidates are not proven to be part of the yellow set. They will be the solution if yellow IS true, but if the other color (green) turns out to be true, the implications that came solely from the untrue color are not valid. This is because something that is not true cannot have any implications - unless it is within the full and strict bivalue rule, that leaves only one possibility, if the other is untrue.)

Now if R3C5 and R5C6 were 2 this would leave a Naked Pair of 5s and 9s in C8 (R3C8 and R5C8) thus cleaning 5 and 9 from R6C8 leaving 4 and 2 here while 4 is already green.

Since this would be the situation if yellow was true and since there is only one other possibility for a 2 in C8 that is already taken by a yellow 4, you now might think of coloring the candidate 2 in R6C8 orange. This would not be false but here comes a special feature of second level coloring: We here have a situation where the logical flipside information of the strict bivalue first level coloring that has been cut in the shift to the second level (omission of strict bevalue link and thus only presumed second level colors) is retrieved from another location:

(1) Since candidate 4 is already first level green and the allegedly orange (yellow if yellow was true) is naturally different from the 4, both flipsides of the original first level Medusa information reunite within the cell/unit and thus this 2 in R6C8 is convicted to be first level yellow - with all power for further expansion of 1st level coloring (or 2nd level half information).

(2) This reunion of both flipsides of bivalue information would be valid even if an oppositely colored other candidate would be of the second level opposite color (here related to 1st level green), validly convicting both oppositely colored candidates into original 1st level colors (featuring that, It does not say anything about the truth of one color though).

Other possibilities might be:

(3). The already colored other candidate is of the same 1st or 2nd level color -> this is valid contradiction of all the first level color referents (only the first level color is contradicted!)

(4) The already colored candidate is of opposite 2ND level color but the same candidate (oh no!) -> in this cell this candidate is the solution but without further implications for all colors (if not by secondary means).

(5) The already colored candidate is of opposite 1ST level color and the same candidate -> the candidate and with him his 1st level color are the solution.

Back to our example set:

By omission of a strict bivalue link and obliged shift to second level coloring but retrieval of the flipside information from the green 4 in R6C8 could be validated to be (green 4 and yellow 2). Thus classicaly, 5 and 9 can be eliminated.

The now yellow 2 in R6C8 would clean all but one candidate in R6C7 thus with another shift to 2nd level coloring giving us an orange 9 in R6C7 -> followed up by an orange 5 in R6C3. Since Unit C3 has already an oppositely colored other 5 (green 5 in R4C3) bivalue flipside informations again are reunited -> candidate 5 in R6C3 convicted first level yellow -> 5 in R5C3 classically eliminated (R6 now could be classically colored and eliminated in first level green/yellow).

Now look at Row 5:

The now first level yellow 9 together with the second level orange 2 clean for a second level orange 5 in R5C8 and all of these together with the yellow 7 in R1C9 clean for an orange 3 in R5C9.

R1C9 and R5C9 clean for a second level orange 5 in R2C9. But Row 2 already contains a same-flipside colored candidate 5 in R2C6 -> Therefor all yellow candidates are contradicted and green is the solution.

## Thursday 23-Oct-2014

## ... by: Rainer

3.If color x candidates and/or color x2 (assumed x-candidates) would clean all but one candidate of a cell, this candidate can also be colored x2. And ever so on, if new x2 colored candidates would help to leave another cell with only one possible candidate. Within that second layer of coloring, do not use y2 to clean for other y2-candidates because the validity of this information it not transported. Color y2 must be used only in bivalue-dependence with color x2 but every resulting x2 then can validly be used to clean cells for other x2 in cooperation with every other x and x2.

The resulting x2 and y2 candidates will leave a complex web of what would be a third, fourth etc. layer of coloring, thus if you use only two second-layer colors they cannot be intertwined within themselves, BUT they still carry the information to validly contradict the original x-candidates: as soon as (two color x2) or (one color x and one color x2) occur within one unit.

thx

## Thursday 23-Oct-2014

## ... by: Rainer

Dear Readers,sometimes i use a second level of Medusa coloring which shirks from the bivalue necessity. It harvests a second layer of coloring which is not definite to the original (true/certain) chain however its contradictions are true:

Within unites you often find candidates which - in a second next step - would be the only possible candidate for a color. Yet they cannot decisively be colored because their presumed link to the original strict bivalue and thus certain chain is open. This might be for example due to several candidates in a linking unit one of which would have to be of the opposite color, or maybe because of a color cleaning all but this candidate from a cell.

The candidate so (in the second next step, i.e: under omission of an unambiguous link) might be the only possibility for color x of the ground chain but still be ambiguous for color y (in fact it has to be for otherwise the chain would simply continue full-value at this point).

Now if you start to color the alleged candidate under question with a (different but) associated color x2 to the color x ( the reference in the original chain color x2 is derive from) the information this carries is "IF color x of the original chain WAS true - then the alleged candidate would also be true". Because of that feature, with associated bivalue coloring beginning from that "IF x then x2" candidate results a second level of coloring that is capable of contradicting color x (but not color y!) from the original chain.

Attention:

As soon as color x and color x2 contradict each other, then all color x, but NOT color x2 candidates are contradicted. This is because of the original chain not being fully closed, the transportation of truth is unidirectional.

If you start the second level from only one derived candidate x2 then the following also is true:

If color x2 OR color y2 contradict themselves the contradicting color in layer 2 of course can be eliminated, If the contradicting color is x2 (that is derived from the ground chain color x shirking the assumed ambiguous color y link) then x2 AND x are contradicted! If y2 contradicts itself, then only y2 but neither x nor y are contradicted - if not by newly found inferences after the eliminations.

The second level of colors can validly be intertwined only to contradict color x (original chain) beginning from every candidate that *would* unambiguously be color x2, IF color x WAS true. However if starting the second layer from several presumed links x2, then only the strictly bivalue connected parts of the second layer are capable of contradicting within themselves. This is because each ambiguous link might individually be true (one might use more associated colors here but that would make it more demanding for the working memory).

A second layer cannot be validly intertwined starting with mixed x2 and y2.

## Friday 9-May-2014

## ... by: Brett Yarberry

Rules 1-6 can easily be combined. If a possibility is weakly connected to both colors, then it can be removed. This is because of the fact that if either one of these colors is true, than that possibility must be false.## Thursday 19-Sep-2013

## ... by: Mr Turner

Rule 7 can be extended. If all candidates of a single value in a row, column, or box can see a single color, then that color is invalid.Also, if all candidates in a cell (or a value in a row, column, or box) but one can see a single color, then that remaining candidate must be that color. If this candidate is part of another chain, then the two chains are now intertwined.

There may be some directionality induced here. It's not clear to me that if that candidate is colored that it implies the original chain coloring must be true.

## Saturday 7-Sep-2013

## ... by: Daniel

For Rule 6 - Two colours Unit + Cell, it says: If an uncoloured candidate can see a coloured candidate elsewhere (it shares a unit) and an opposite coloured candidate in its own cell, it can be removed... please advise if the 'coloured candidate' found elsewhere MUST be the same number ( in this case both the uncoloured and the coloured number are 1).Will it work if they are not the same number?

## Monday 2-Sep-2013

## ... by: jack

In rule 1 example (first one at top of page) should there be a link between J5 & J2 and/or H4 and H2? Otherwise the link between H2 & J2 seems arbitrary?## Thursday 17-Jan-2013

## ... by: voidsky

The sample Sudoku Solver provides for 3D Medusa Rule2 seems to me as if it is about Rule6.I am enjoying this site almost everyday. Thank you for your great work.

## Tuesday 31-Jan-2012

## ... by: Anton Delprado

I have been using this strategy for a while now and I will propose a different type which I have used a few times in solving grids.Namely if a cell has only two values and it sees those values in other cells with the same colour. For example if the cell has a 2 and 5 and the cell sees a blue 2 and a blue 5 in different rows, columns or boxes. If blue is the "correct" colour then that cell cannot be 2 or 5 leaving no values left. Therefore blue cannot be the "correct" colour and all cells take their green values.

Works for cells with more values of course but that will turn up less frequently.

## Monday 30-Jan-2012

## ... by: Ehsan

AbbasThe Type 4 example works fine. What maybe confusing you is the line Andrew draws between those two colored 5's in column 7, since as you said it implies a conjugate pair (or strong link) between those two 5's. Indeed it's a weak link between those two fives, but that does not ruin the example for 3D Medusa. Focus instead on the other numbers for the cells in question (D7 and F7). Both these cells are bivalue cells, so if one of them numbers is the answer, the other cannot be. In D7, there is clearly a strong link between the 3 in C7 and the 3 in D7. So if 3 in C7 is purple, then 3 in D7 must be the opposite color (green). Now since D7 is a bivalue cell, if one of the candidates becomes colored, then the other candidate can take on the opposite color. You can think of this as a strong link between the 3 and the 5 in D7. Same exact concept holds for the 5 in F7, as there is a clear strong link between the 2 in F3 and the 2 in F7. Therefore, since the 2 in F7 is green, the 5 in F7 has to be the opposite color (purple) because it's a bivalue cell.

So now you have two opposite colored candidates that share the same unit (Column 7 and Box 6). Therefore, any other 5's that share a unit (column, row, or box) with these oppositely colored 5's can now be eliminated.

## Wednesday 4-Jan-2012

## ... by: Abbas

Dear SirI am really confused with Type 4 , for your example there is weak link between two 5s and you colored only that weak links ! whats going on there ???

beacuse of that weak link you eleiminate other 5s

I would be grateful if you ould explain

## Friday 31-Dec-2010

## ... by: Jmf

In response to Oliver: I agree that unique "rectangles" as such can turn corners. In this example it is with 1 and 7, but can even have a 3rd number in the mix. You can remove 1 from B1 and from B8. That will avoid the unique corner rectangle. and yes leaves 1 in B6## Monday 8-Nov-2010

## ... by: Jen G

You stated above under rule 1 that:"you can't just set every cell with a green number to be the solution. When you eliminate a yellow number it *might* leave only a green number - in which case green is the solution. But look at H1. 1 is marked as green but there are other un-coloured candidates there. 1 *could* be the solution but it might not be. Always remove the yellow numbers and see what's left - don't just set green as the solution automatically."

This is incorrect reasoning. When one color is proven false, the other color *must* be true for ALL the cells it's in. Even ones with uncolored candidates alongside the 'true' color. This is the nature of conjugate (bilocation) pairs... and because Medusa coloring strictly sticks to coloring based on conjugate pairs..when you find out one color is false, then the other must be true everywhere it's located.

In example 1 above, the candidates labeled green represent the conjugate pairs of the candidates labeled yellow in a given box/row/column. We learn that the yellow candidates cannot be valid. Therefore, ALL of the green labeled candidates are true and valid. Cell H1 MUST be 1 (the green candidate) because we know the yellow 1 in H2 is invalid. Because we colored only conjugate pairs and any of their bivalue cell mates... we can assume that when one color is false the other is true.

Another way to look at it is if you enter in anything other than the color you learn to be true in those cells that have uncolored cohorts you will be in error... in essence you will be missing that candiate for the area it's conjugate pair occured in.

If we don't put the green labeled 1 in cell H1, then there will be no 1 in that box and row,,,,an invalid result. Cell H1's 1 was colored green because it formed a conjugate pair with the 1 colored yellow in H2... we know yellow everywhere is false and so H1 must be green candidate 1.

Likewise, in B2, if you don't enter the green labeled 4, then you will not have a 4 in that box since we know the 4 in cell C2 is invalid. In H4, you must enter the green labeled 3 or else there isn't one in that box and row. You can find more occurances in the examples on this page and whenever you implement medussa. Try putting anything in the cell other than the color we learn to be true and you will end up not having a valid puzzle because entire candidates will be missing from the box/row/column the colored conjugate pair occrured in.

Just remember: When the "color X" half of all the conjugate pairs we label in Medussa is proven false, then the "color Y" half of the pairs are ALL proven true. Even though Medusa can sometimes look messy and links jump all around the board, because these links are based strictly on occurances of pairs of candidates only, they allow us this basic assumption.

Jen G

Seattle

## Monday 11-Oct-2010

## ... by: wbngai

Suggestion: For "color consistency", we better use blue and green in Rules 1 and 2; yellow for digits to be removed## Saturday 25-Sep-2010

## ... by: Ian Binnie

I have been trying to wrap my mind around this strategy, but cannot reconcile the bi-value and bi-location candidates.If I have a row with 15 56 156

I colour the 1 in 15 green which sets the 1 in 156 yellow (bi-location)

and 5 in 15 yellow (bi-value) which sets the 6 in 56 yellow

This sets the 6 in 156 green(bi-location)

156 now contains 1 yellow 6 green

Rule 3 - Two colours in a cell implies that this can not be 5, but this is obviously nonsense

## Monday 7-Jun-2010

## ... by: suneet

Hello Andrew,In the latest version you have removed multi-coloring and Multi-Value X-wing. What is the reason.

I know at least some problem which can be solved by multi-coloring but not by the 3D Medusa and simple coloring type 4 and 5.

Please explain.

Strange, but these strategy Multi coloring etc are present in kendoku 6 by 6 solver.

Regards suneet.

## Saturday 3-Apr-2010

## ... by: Cliff

I have been using this strategy for quite some time. It’s easy and mindless. Sort of therapeutic. Just start with the largest set of single digits tied together with hard links and branch out to another digit at a bi-value cell. If that gets you know where find the next largest set and use two different colors. Now comes the headache. When these two sets tie together with a soft link at different places there should be some rules as to which colors in the two sets belong together. I have to draw lines. If I get nowhere with colored sets they sometimes help find an alternation inference chain. If the headache becomes too severe I admit failure and guess which color is correct. Usually a quick solution but no satisfaction.## Thursday 25-Mar-2010

## ... by: Raj

The addendum to Rule 1 should be unnecessary. Since only exclusively conjugate pairs are used in coloring, if one color is false, the other must be true within the given group (row, column or box).## Wednesday 24-Mar-2010

## ... by: Oliver Paulsen

Hello Andrew,my request do not concern Medusa but the top sudoku on this page. I want to know if there is some kind of unique rectangles (I call them forbidden bubbles) but in a larger form. There are 2 floors with 1 and 7 in line 1 and 5 and there is a strong link of them in row 9. The roof is in line 2 - cell B1 (147) and cell B8 (137). To avoid 17 for 6 times in this circle I would argue that the 7 in the roof-cells of line 2 have a strong link. There is no other place to stay in line 2. But the 1 can find a place in cell B6. So I would eleminate both 1 in the roof-cells and in B1 has to be the last 1 of this row. Is this argumentation which refers to unique rectangles type 4B correct in this case? Please let me hear

Kind regards Oliver