Phistomefel's Theorem (sometimes also called

Phistomefel's Ring) stipulates that if we divide any solved sudoku grid like this:

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`AA.....AA`

AA.....AA

..CBBBC..

..B...B..

..B...B..

..B...B..

..CBBBC..

AA.....AA

AA.....AA

then the contents on A equal the contents on B+C. In other words, if we know which digits go exactly on region A, those exact same digits will appear on B+C, and vice versa.

Further explanation on this can be found here:

Neuer (?) Fund über Sudoku-GeometrieNote: Probably this isn't news to you, but I couldn't find another topic about this hereThis can probably be used on classic sudoku as a way to spot fishes like

keith's technique, but where this property really shines is on figure-based sudoku variants like Killer, Arrow, etc.

I've seen some killer sudokus where JSudoku needs recursion with +30 guesses to find the solution, and a human can find it with a pretty straight-forward path using this technique (more on that later).

Furthermore, computers could use this technique in a more generalized way: columns, rows and stacks can be swapped, giving other patterns that are still perfectly valid since permutations don't affect the theorem, they simply move the regions around. The only distinction between B and C appart from the distinction used to prove the equality is that C shares a box with A; that difference is often omitted to simplify the possible patterns. A human can easily spot the symmetric ones, but there are exactly 729 (3^6) unique patterns.

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`AA.....AA`

AA.....AA

..CBBBC..

..B...B..

..B...B.. Default

..B...B..

..CBBBC..

AA.....AA

AA.....AA

AA.....AA

..CBBBC..

AA.....AA

..B...B..

..B...B.. Swapping r23

..B...B..

..CBBBC..

AA.....AA

AA.....AA

..CBBBC..

AA.....AA

AA.....AA

..B...B..

..B...B.. Swapping r13

..B...B..

..CBBBC..

AA.....AA

AA.....AA

(...)

....B..B.

....B..B.

....B..B.

...A.AA.A

BBB.C..C. Swapping c23-c78-r23-r78-c{123|456}-r{123|456}

...A.AA.A

...A.AA.A

BBB.C..C.

...A.AA.A

You can download the full list here:

phistomefel.txtThis is an

example I made some days ago:

Anti-King Little Rose (the Anti-King restriction must be included manually on JSudoku)

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`3x3::k:17:18:5645:19:20:21:5646:22:23:24:5645:25:2049:2049:2049:26:5646:27:5645:28:4870:4870:4357:3586:3586:29:5646:30:2055:4870:4357:4357:4100:3586:2051:31:32:2055:3850:3850:33:4100:4100:2051:34:35:2055:2825:3850:3851:3851:4620:2051:36:5648:37:2825:2825:3851:4620:4620:38:5647:39:5648:40:2056:2056:2056:41:5647:42:43:44:5648:45:46:47:5647:48:49:`

JSudoku can only find 4 digits using "Deduce All Moves", and requires 13 guesses with "Recursively Solve", 20 if we use recursion from start.

But here's how it can be solved using this trick:

**Solution path (first steps): **ShowWe use this pattern, which is symmetrical, therefore not hard to spot by a human:

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`A.A...A.A`

.C.BBB.C.

A.A...A.A

.B.....B.

.B.....B. Swapping r23-c23-r78-c78

.B.....B.

A.A...A.A

.C.BBB.C.

A.A...A.A

The crucial point here is the 8-cages are {125} or {134} so they can't have a 6 or higher, and the 22-arrows are {679} or {589} so they can't have a 4 or lower.

Now it shouldn't be hard to see that {679} is impossible in an arrow because it uses two "non-fillable" cells using the BC region at once, that would mean another of the arrows must also be {679} making the disbalance even higher. In other words, all 22-arrows are {589}.

Not only that, but if 5 is in the middle (C) of an arrow then we are defining on the A region two "non-fillable" cells at once again. Also, there will be four 5s in the A region which can only come from the 8-cages. All the 8-cages must be {125}.

That means we have this.

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` 12 . 589 . . . 589 . 12`

. 89 . 125 125 125 . 89 .

589 . 12 . . . 12 . 589

. 125 . . . . . 125 .

. 125 . . . . . 125 .

. 125 . . . . . 125 .

589 . 12 . . . 12 . 589

. 89 . 125 125 125 . 89 .

12 . 589 . . . 589 . 12

and from there there are various ways to proceed, the more obvious one is the 19-cage.

We still don't have any digit, but as a comparison, if we use that as a starting point on JSudoku, it can solve it without recursion and without many hard steps:

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`79 Naked Single`

2 Hidden Single

3 Unique Pair

5 Naked Pair

3 Hidden Pair

5 Unique Triplet

3 Intersection

3 Naked Triplet

1 Hidden Triplet

1 Odd Pairs

3 Odd Triplets

1 Conflicting Pair

4 Quadruple Innies & Outies

1 Pointing Pair

1 Generalized Naked Subset

5 Turbot Fish

1 Y-Wing