The Tridagon (a.k.a."Thor's Hammer") is a pattern first described by Denis Berthier and much developed on New Sudoku Player's Forum between 2022 and 2025. I have been reluctant at first to include it as it seemed contrived - relying on a spread of candidates that are very unlikely to appear in randomly created puzzles, let alone published ones. But people have generated long lists of puzzles where Tridagon is necessary and as this site is (partially) devoted to exploring the very hardest puzzles, I rolled up my sleeves and got stuck in. I am going to follow the outline given on Phil's Folly as it compresses the pattern down to it's essence and by-passes a lot of speculative development.

While it has a distinctive pattern it is a difficult strategy to prove logically. I've found Ryokousha Cyclical Parity theory the most convincing (good links to further reading on the forum) and Rangsk's video on this very watchable.

This pattern is interested in boxes with triples that span all three columns and rows in a box. There are six ways to do this. Three of these ways are said to be "rising" (the orange cells) since looking at them from left to right the cells migrate up and to the right as you read them. This is to do with their relative adjacency not their numbers, although we need the numbers to show which are which and show they fit without conflict. The yellow cells are "falling".

All six patterns are either rising or falling so we can think of their "parity". If rising they go 1 to 2 and 2 to 3 and 3 to 1 since it is "cyclical". Hence the theory talks of these boxes having "cyclical parity".

Next, for every set of three numbers we are interested in how to fill each stack (three vertical boxes) and band (three horizonal boxes). It is a property of Sudoku that

• All rising patterns have a different parity for the band and the stack
• All falling patterns have the same parity for the band and the stack

So how is this useful? If you have a rising pattern in box 4 (doesn't matter which, so drawing it in a line) and you have a rising pattern in box 1 – the diagonal – we are forced to consider a falling pattern for box 2. Which I've coloured in yellow. If we try another rising pattern there is no way to fit {1,2,3} into box 1 that doesn't cause a conflict.

It turns out that if you have four boxes in a square (or rectangle) that all possess three cells with the same triple in them - and three of the patterns are rising and one is falling, or three are falling and one is rising - then this form a deadly pattern.

To save the puzzle from having multiple solutions some additional candidates are required to be present on one of the cells in the pattern. If there is one cell with any number of additional candidates that cell can have the triple removed from it. These extra candidates are called Guardians.

The pattern that stood out when first identified looked a little like "Thor's Hammer" so that is another name of this strategy.

Here is the first and classic example from Phil's Folly docs. I am breaking my rule about finding my own examples. I don't believe I can do that in a reasonable amount of time so I will pinch all the examples and provide a link instead.

There is a single 5 guardian in F6 which must be the solution to this cell.

It is not a difficult pattern to spot so the solver tries at the start of the 'diabolicals'. But I've bundled it under "various static patterns' as I don't think it will be discovered much outside the test lists.

Just to emphasis that the patterns within each box can be any of the three suitable for the parity. Not just straight diagonals.

It is important to appreciate the the triples are similar to Naked and Hidden Triples. In that it matters there are three numbers IN TOTAL across three cells, not that there are the same three numbers IN EACH cell. So while most examples will be {123} + {123} + {123} it could be as sparse as {12} + {13} + {23}.

Here is an example: 4 found to be the solution in C4.