Comments  Talk
... 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.
Andrew Stuart writes: A very good observation. Of course, if this applies to a cell, why not a unit. I will try and add this to the detection very soon. On your second point, I'll try and find an example. Interesting. It sounds like one of the other rules bent slightly. There are quite a few now!
... 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?
... 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?
Andrew Stuart writes: Yes you can put those links in. The links I chose to draw in the example were done as if I was tracing the network myself and showing enough to get the first contradiction. I started in the top left. I haven't shown every possible pairing since this would overwhelm the picture. You can find a contradiction in a partial network  additional links will merely be confirmatory (and isolated links that can't be connected to you main network are irrelevant and can't help you).
... 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.
Andrew Stuart writes: It does indeed find a Rule 5 and several Rule 6 before the Rule 2. I shall try and find an example that starts with Rule 2.
... 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.
Andrew Stuart writes: You are very correct. I've checked this and it is sufficiently distinct to warrant it's own rule number. Very pleased you posted about this. Well done.
... by: Ehsan
Abbas
The 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.
... by: Abbas
Dear Sir
I 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
... 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
... 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 uncoloured 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
... by: wbngai
Suggestion: For "color consistency", we better use blue and green in Rules 1 and 2; yellow for digits to be removed
... by: Ian Binnie
I have been trying to wrap my mind around this strategy, but cannot reconcile the bivalue and bilocation candidates.
If I have a row with 15 56 156
I colour the 1 in 15 green which sets the 1 in 156 yellow (bilocation) and 5 in 15 yellow (bivalue) which sets the 6 in 56 yellow This sets the 6 in 156 green(bilocation)
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
... by: suneet
Hello Andrew,
In the latest version you have removed multicoloring and MultiValue Xwing. What is the reason. I know at least some problem which can be solved by multicoloring 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.
... 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 bivalue 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.
... 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).
... 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 roofcells 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 roofcells 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
Andrew Stuart writes: If you untick 3D Medusa it does find a Hidden Unique Rectangle, but not the one you identified. A deadly pattern is must be four cells sharing exactly 2 rows, 2 columns and 2 boxes, but you refer to many more than that so I don't immediatey get your example. You might have a loop there. In the next version of the solver I hope so present all alternatives at any one stage so if there is a pattern there is should pop out. 
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