Sudoku Written by Brian Hamilton and Karan Hamilton.
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Ten pencil marks decide the next move in Evil puzzle 85, one of the boards our Evil sudoku page deals. It is 18 moves in: 40 cells empty, no naked or hidden single anywhere, and our Sudoku Helper, with eighteen techniques to try, still finds a move with just one of them, an Almost Locked Set. Follow the digits 1, 5 and 9 through the places where each is forced into a pair, colouring as you go, and ten candidates join into one two-colour web. Two more, the 1 in R3C1 and the 9 in R3C7, lose whichever colour turns out to be true. Strike them, and singles finish the grid.
That is 3D Medusa: colouring that crosses from one digit to another inside two-candidate cells. Below are its links, all six rules in their standard numbering, and what we found running it over 500 of our Evil puzzles.
What is 3D Medusa in sudoku?
3D Medusa is Simple Colouring carried across digits. Simple Colouring follows one digit through the rows, columns and boxes where it has only two places left, colouring those places alternately. Medusa colours candidates instead of cells: each node is one digit in one cell, such as the 5 in R2C2, so the web can pass from a 5 to a 9 inside a cell that holds only those two. Rows and columns are two dimensions, the digits a third, hence the name. It also goes by Medusa colouring or multi-digit colouring.
It uses two kinds of link and nothing else:
- Inside a cell: a cell with exactly two candidates, a bivalue cell. One of the two is the answer.
- Inside a house: a conjugate pair, where only two cells of a row, column or box can still take the digit. One of the two holds it.
Give the two ends of each link different colours and every candidate you reach gets a colour forced by the first. The finished cluster can be filled in only two ways: every green candidate is true, or every blue one is. You never need to know which. The rules look for a candidate that loses under both colours, or a colour that cannot be true.
Chain guides usually define a strong link as “at least one of these two is true”. Medusa’s links are stricter: a cell holds one digit and a digit fills one place in a house, so every link has exactly one true end. That is why the colours work in both directions. A cell with three candidates gives no link, because “not this one” no longer means “that one”.
How do you colour a Medusa cluster?
Start in a two-candidate cell and give its candidates one colour each. For every newly coloured candidate, check its cell (two candidates?) and its row, column and box (two places for its digit?), and give the other end of each link the opposite colour. Stop when nothing new gets coloured. On Evil puzzle 85, with colour A drawn green and circled and colour B blue:
- R2C8 holds {1,5}: its 1 is A, its 5 is B.
- Row 2 has 1 only in R2C4 and R2C8, so R2C4’s 1 is B. Column 4 has 1 only in R2C4 and R4C4, and box 2 only in R2C4 and R3C6, so those two 1s are A.
- R3C6 holds {1,9}, so its 9 is B.
- Back to R2C8’s blue 5. Row 2’s other 5 is R2C2 and box 3’s is R3C7: both A. R2C2 holds {5,9}, so its 9 is B, and box 1’s other 5, in R3C1, is B too.
- Row 3 has 5 only in R3C1 and R3C7, already B and A. That fits.
Ten candidates, five of each colour, and no more links lead out. If a link ever joins two candidates of one colour while you build, recheck your pencil marks: on a correct grid it cannot happen.
Now look along row 3. R3C1 holds a blue 5 and an uncoloured 1, and that 1 shares the row with R3C6’s green 1. If green is true, R3C6 is 1, so R3C1 isn’t. If blue is true, R3C1 is 5, so it isn’t 1 either. The 1 in R3C1 loses under both colours, so it goes. The 9 in R3C7 is the mirror image: its cell holds a green 5, and it sees R3C6’s blue 9. Both are rule 5 below.
Two candidates is a small haul, but they were the only rivals in two columns: R1C1 becomes the last place for 1 in column 1 and R1C7 the last 9 in column 7. Green turns out to be the true colour. The board also has a second cluster, of 16 candidates, with a move of its own; our solver plays the smaller one.
What are the six 3D Medusa rules?
We number them as SudokuWiki does. Andrew Stuart’s page there is the usual reference, and his solver reports Medusa moves by these numbers. Other guides order the rules differently (one popular guide lists “twice in a unit” first), so go by what a rule says, not its number. SudokuWiki added a seventh in July 2026, for a colour that would leave a digit no place in a row, column or box; our detector has the six below, and every count on this page is for those six.
| Rule | SudokuWiki’s name | What you see | What goes |
|---|---|---|---|
| 1 | Twice in a cell | Two candidates of one colour in the same cell | Every candidate of that colour, on every digit |
| 2 | Twice in a unit | One digit, in one colour, twice in a row, column or box | Every candidate of that colour |
| 3 | Two colours in a cell | A cell holding one candidate of each colour | The other candidates in that cell |
| 4 | Two colours elsewhere | An uncoloured candidate that sees its own digit in both colours | That candidate |
| 5 | Two colours unit + cell | An uncoloured candidate that sees its digit in one colour, in a cell that holds the opposite colour | That candidate |
| 6 | Cell emptied by colour | An uncoloured cell whose every candidate sees its digit in the same colour | Every candidate of that colour |
“Sees” means shares a row, column or box. Rules 1, 2 and 6 prove a colour false, which makes every candidate of the other colour true. Rules 3, 4 and 5 remove uncoloured candidates and leave open which colour wins.
Rules 3 to 5 each remove a candidate that is false under either colour. In rule 3 the cell’s answer is one of its two coloured candidates. In rule 4 the digit lands in a cell the candidate sees under green, and in another under blue. Rule 5 mixes the two: under one colour the candidate’s own cell takes a different digit, and under the other its digit lands in a cell it sees.
The coloured candidate in the cell and the one the candidate sees must be different colours. If both were green, blue being true would make both false, and nothing would stop the uncoloured candidate from being the answer. Written with “same colour”, rule 5 removes candidates it has no right to.
When does a colour contradict itself?
Rules 1, 2 and 6 catch a colour that cannot be true. Evil puzzle 335, 16 moves in, has a cluster of just seven candidates on 3, 4 and 5. R8C3 is not part of it, and its only candidates are 3 and 4. Blue’s 3 is in R9C3, just below it in the same column and box; blue’s 4 is in R2C3, higher up the same column. If blue were true, R8C3 would be left with nothing. So blue is false: rule 6.
A false colour goes everywhere at once, not just where it was caught: the blue 4s in R2C3 and R6C1, the blue 5 in R9C1 and the blue 3 in R9C3. And because one colour must be true, every green candidate is now an answer: R6C1 and R9C3 are 5, R2C1 is 4. Our helper’s eighteen techniques finish from there.
That is why contradictions pay so well. In our runs they were 96 of the 277 Medusa moves and removed a median of 19.5 candidates each; the other rules typically removed one. Checking rules 1, 2 and 6 in that order, rule 1 caught 54 of the 96, rule 6 25 and rule 2 17.
How often is 3D Medusa needed?
On Easy, Medium, Hard and Expert, never. We deal those levels only when singles, locked candidates and naked pairs can finish a board, and across 200 boards per level Medusa was not once the simplest move. A Medusa deduction existed at every Hard position without a single and at 92% of the Expert ones, but something easier always came first.
We ran the first 500 puzzles of the Evil bank through our helper’s eighteen techniques plus five more, Medusa among them, always taking the simplest move. Medusa was the move 277 times, on 183 boards, and at 70.5% of the positions with no single some cluster had a deduction. The catch: Medusa comes after every technique our helper has except Almost Locked Sets, and at 268 of its 277 moves the helper had an ALS-XZ move ready instead. Only 9 times did Medusa move a position where the helper had nothing. On our boards Medusa is mostly a second way through, not the only one, and a good one if ALS-XZ is not in your toolkit. Our Extreme puzzles never bring it into play: on 499 of the first 500, none of the 23 techniques finds even a first move.
| Rule | Could fire (all positions) | Part of the move (277 Medusa moves) |
|---|---|---|
| 1 · twice in a cell | 878 | 54 |
| 2 · twice in a unit | 1,950 | 17 |
| 3 · two colours in a cell | 4,243 | 20 |
| 4 · two colours elsewhere | 31,545 | 79 |
| 5 · two colours unit + cell | 3,722 | 134 |
| 6 · cell emptied by colour | 8,559 | 25 |
The first 500 Evil puzzles, tougher than the 5,000 taken as a whole. Left: every rule each cluster could apply at the 26,429 positions our solver passed through, 50,897 in all; a cluster that could apply two rules counts for both. Right: the 277 positions where Medusa was the simplest move; one move can use several of rules 3–5, which are only checked when a cluster has no contradiction.
Counted wherever it could fire, rule 4 dominates, with 31,545 of the 50,897. Counted where Medusa was actually the move, rule 5 leads: it was part of 134 moves, rule 4 of 79. SudokuWiki’s count over 50,000 puzzles points the same way, rule 5 at 57% of its Medusa moves and rule 4 at 22%. Order explains part of it: a rule-4 elimination whose chain never leaves one digit is plain Simple Colouring, which comes first, while rule 5 always involves two digits. If you learn one Medusa rule first, make it rule 5.
Of the clusters behind our 277 Evil Medusa moves, 179 had more than 20 candidates (the median was 26), and only 2 used just two digits. The chains that actually prove each move were short: a median of 8 candidates, and 84 moves needed only 6. Across every position our measurement walked, on 11,300 boards plus random test positions, all 550,196 Medusa eliminations agreed with the puzzle’s solution.
How do you find a Medusa move by hand?
Colouring a whole cluster on paper is slow. Evil puzzle 357, 43 moves in, is typical: one cluster of 29 candidates over six digits, which removes three.
No single elimination needs all 29. For every candidate it removes, our detector records the shortest run of links that proves it, and for the 1 in R8C4 that is six candidates in the four shaded cells:
- Suppose R8C4 is not 6. Column 4 has 6 only in R8C4 and R2C4, so R2C4 is 6.
- R2C4 holds {6,9}, so it is not 9. Row 2 has 9 only in R2C4 and R2C8, so R2C8 is 9.
- R2C8 holds {1,9}, so it is not 1. Column 8 has 1 only in R2C8 and R8C8, so R8C8 is 1.
So R8C4 is 6 or R8C8 is 1, and either way R8C4 is not 1: rule 5 again, between a green 6 and a blue 1. The same cluster removes the 2 from R8C4 and the 7 from R8C5. R8C4 is left with 6, R9C4 becomes the last 1 in column 4 and R9C5 the last 7 in column 5, and 30 naked singles in a row finish the puzzle. Our helper would have played an ALS-XZ here instead.
Short proofs suggest hunting from the candidate you want gone:
- Make the pencil marks complete and correct. One wrong mark makes the colours lie, and earlier eliminations count: R1C2 in Evil 357 has two candidates only because earlier moves removed its 8 and 9. Our pencil marks guide shows how to keep them that way.
- Pick a target: a candidate X in a cell of three or more candidates, where another candidate there, Y, has only two places in one of the cell’s houses.
- Colour Y green and walk outward for four or five links, alternating colours.
- Stop at a blue X that the target sees: rule 5 removes it.
- Watch for contradictions on the way: two greens in a cell, two green copies of one digit in a house, or a cell whose every candidate sees green. Any of these makes blue true everywhere, which beats any single target.
On paper, mark one colour with a small dot beside the candidate and the other with a ring round it. It survives a photocopier, and it is what our diagrams do: the green candidates are also circled.
Stuck after Simple Colouring? Where 3D Medusa fits
Medusa contains Simple Colouring: limit it to one digit and its rules 2 and 4 become the two rules our helper’s Simple Colouring hint uses, which the helper numbers 1 and 2. So when one-digit colouring has run dry, colour across digits next. Medusa itself is not among the eighteen techniques in our Sudoku Helper, and our Sudoku Trainer doesn’t drill it. When the helper offers only an Almost Locked Set, or nothing at all, that is the moment to start colouring.
| Technique | Follows | Links it may use | Our helper |
|---|---|---|---|
| Simple Colouring | One digit | Conjugate pairs | Has it |
| 3D Medusa | Candidates of any digit | Conjugate pairs and two-candidate cells | Doesn’t |
| X-cycles and X-chains | One digit | Conjugate pairs, plus any two cells that see each other | Has the Skyscraper, one four-cell shape |
| XY-chains | Two-candidate cells | The cell itself, plus any two cells that see each other | Has the XY-Wing, a three-cell chain |
“Any two cells that see each other” is a weak link, which Medusa never uses. In turn, neither chain finds the Evil 357 proof: it runs through the three-candidate cell R8C4 and changes digit twice.
Two candidates that are each other’s only alternative also cannot both be true, so every Medusa link can serve as a weak link too, and any Medusa proof can be written as an alternating inference chain (= strong, − weak). The Evil 357 proof reads (6)r8c4 = (6)r2c4 − (9)r2c4 = (9)r2c8 − (1)r2c8 = (1)r8c8. The contradictions are forcing chains with the bookkeeping done for you: assume one colour and watch the grid break.
When no cluster has a move either, turn to chains that use weak links, X-cycles and XY-chains before forcing chains, and check for a BUG, a pattern that relies on the puzzle having one solution. Our sudoku techniques guide shows where each of them sits.