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**

Now I stated that if the solution is not one colour then it must be the other. There is a catch - 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.

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:

Now I stated that if the solution is not one colour then it must be the other. There is a catch - 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

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 quite a few links between 3s, 5s and 6s. Most have been drawn on the diagram. 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 5s in column 7 are these. The 5s in ACG7 are removed because of the coloured 5s in column 7. The 5s in box 6 can see blue and green 5s in D7 and F7.

This rule is shared with Simple Colouring.

The 5s in column 7 are these. The 5s in ACG7 are removed because of the coloured 5s in column 7. The 5s in box 6 can see blue and green 5s in D7 and F7.

This rule is shared with Simple Colouring.

Given the above possibilities, its tempting to generalise. The example to the right we have some 6s in row F. Both can see a green 6 in E9 and a blue 6 F5. This is not exactly like Rule 4 but very close. The 6s can be removed because they can see **two different coloured candidates of 6 elsewhere**.

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 7 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).

Rule 7 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).

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.

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.

## Comments

Comments Talk## 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