The goal is to fill in the blanks on the board to create a sudoku, a 9×9 grid of numbers subject to the constraints that the digits 1-9 can only appear once in each row, column and 3×3 box, for example:
| 7 | 1 | 9 | 3 | 4 | 2 | 6 | 8 | 5 |
| 3 | 4 | 8 | 6 | 1 | 5 | 7 | 9 | 2 |
| 6 | 2 | 5 | 7 | 8 | 9 | 1 | 3 | 4 |
| 1 | 6 | 3 | 4 | 2 | 8 | 9 | 5 | 7 |
| 9 | 7 | 2 | 1 | 5 | 6 | 8 | 4 | 3 |
| 5 | 8 | 4 | 9 | 7 | 3 | 2 | 6 | 1 |
| 8 | 3 | 7 | 5 | 9 | 1 | 4 | 2 | 6 |
| 4 | 9 | 6 | 2 | 3 | 7 | 5 | 1 | 8 |
| 2 | 5 | 1 | 8 | 6 | 4 | 3 | 7 | 9 |
Using the mouse, trackpad or your finger, drag the tiles onto the board to fill in the gaps. When you release the tile, it will snap into place. If you try to drop a tile on top of a hint (a number that's been given to you by the game), the tile will be returned to its starting position.
The pencil marks in the blank cells show which numbers can go where; if a cell shows 0 possibilities, you've made a mistake somewhere along the way. If you get stuck, click on the "Hint" button to receive part of the solution.
When you've solved the puzzle, your performance will be rated according to how many additional hints you needed: 5 stars for zero additional hints, 0 stars for five or more. If the game is in a state where there are only naked singles remaining on the board, the game will autocomplete to save you the hassle of moving the final few tiles into position.
To start a new game, either dismiss the "Congratulations!" popup or click the "New Game" button. If you're a Sudoku wizard, you can select a harder difficulty from the dropdown menu and your choice is remembered as a preference.
If you want to share a game with others, click the "Share" button and copy the
link from the popup. You can use a similar mechanism to import games from
elsewhere by appending ?s=... to the URL in your browser bar, for example:
sudoku.html?s=.....4......7..16.....3.7.2.4...7..9..8.2....7..6..2.1.34..1..6.6.........73.29.5
When a puzzle is generated or imported, the engine attempts to solve it using logical elimination and assigns a grade depending on what sort of strategies are required.
Very easy: the puzzle may require you to locate a few hidden singles but otherwise can be completed simply by filling in the grid. |
|
Easy-medium: the puzzle will require some slightly more advanced strategies to solve, such as using pairs and triples to eliminate possibilities from the grid. |
|
Medium-tough: the puzzle will typically require a wider variety of basic strategies to solve and may even require use of an advanced strategy, such as simple colouring, to eliminate possibilities. You will likely need pen and paper to solve this without requiring additional hints. |
|
Tough: the puzzle will certainly require a number of advanced strategies to solve and these will certainly need pen and paper to apply without requiring additional hints. On the harder difficulty, tough puzzles may require you to use a couple of diabolical strategies as well. |
|
Diabolical: the puzzle will require you to use a wide variety of advanced and diabolical strategies to solve. At the very least, you will need to understand the X-Cycle and XY-Chain strategies. |
|
Diabolical-extreme: a puzzle that the game engine was unable to solve using logical strategies. This will likely be hard from the start and will require application of the most advanced strategies to solve. |
The basic strategies outlined here will serve to complete most "Low / Medium" difficulty puzzles without requiring additional hints. The tougher puzzles may require one or two of the advanced strategies outlined below.
The key to any strategy is carefully studying the pencil marks, the small numbers that show the possible values of blank cells. In case you think it's cheating to have the computer calculate these for you, it's not really: it just saves you from a tedious and error-prone exercise.
The simplest strategy is to scan the pencil marks for singles,
values that appear as unique possibilities for a given unit (a unit
being a row, column or 3×3 box).
5
7
2
The red-coloured 1 in A2 is a naked single, i.e. the only possibility for a given cell and allows elimination of all other 1s in the row, column and box visible to the cell. The green-coloured 8 in C3 is a hidden single: it initially looks like there are multiple possibilities but since it's the only cell in the 3×3 box that has an 8 as a possibility, any other pencil marks can be ignored.
If you find two cells in a unit that have the same two numbers as unique possibilities, you can eliminate those two values as possibilities from all other cells in the same unit. If the pair intersects with another unit, you can eliminate those possibilities from all other cells in the other unit as well.
| 2 |
1
3
5
6
7
|
4 |
5
7
|
1
3
7
9
|
1
3
5
|
8 |
5
6
7
9
|
6
7
9
|
| 8 |
3
5
7
|
3
5
9
|
4 |
3
7
9
|
6 | 1 |
5
7
9
|
2 |
|
5
9
|
1
5
6
7
|
5
9
|
5
7
8
|
1
7
8
9
|
2 | 3 |
5
6
7
9
|
4 |
The pair of 5s and 9s coloured red are the only two possibilities for the two cells. This allows us to eliminate 5s and 9s in the same 3×3 box and in the same row as well, since the pair lies along a horizontal line. The eliminated possibilities are coloured blue. Eliminating the 5 and 9 in cell B3 reveals 3, coloured green, as part of the solution.
Identifying naked doubles may be extended to triples as well. This does not
require 3 unique possibilities in 3 cells, for example, {1,4,7},{1,4,7},{1,4,7}.
This happens but is rare. Instead, look for 3 unique possibilities across 3 cells in a unit so that
{1,4}{4,7},{1,7} would be a naked triple along with, for example,
{1,4,7}{1,4},{1,7}. Naked triples are much rarer than
naked pairs. Actually, that's not quite true: you'll see them all the time when
you've solved five or six cells in a unit, by which time they're useless.
| 9 |
6
8
|
3 | 4 | 1 | 2 | 7 |
6
8
|
5 |
| 7 |
4
5
6
8
|
1 |
5
6
|
5
8
|
3 |
4
6
8
|
9 | 2 |
|
4
6
|
4
5
6
8
|
2 |
5
6
7
|
9 |
5
7
8
|
1
4
6
|
1
4
6
|
3 |
The {4.6},{1,4,6},{1,4,6} triple, coloured red, in C1, C7 and C8
allows removal of the 4 and 6, coloured blue, from C2 and C4. Removing the 6 from
C4 reveals the 6 in B4, coloured green, as a hidden single for box 2.
Naked quads (i.e. four unique possibilities spread across four cells in a unit) are also a very remote possibility. Here's an example where the quad is pretty obvious.
| 7 | 4 | 6 |
1
2
5
|
1
2
8
|
3 |
1
8
|
1
8
9
|
1
2
5
8
9
|
|
1
2
5
|
1
2
3
5
|
1
2
3
5
|
1
2
5
|
9 |
2
4
5
8
|
1
3
4
6
7
8
|
1
3
4
6
8
|
1
2
5
7
8
|
| 8 |
1
2
3
5
|
1
2
3
5
9
|
6 |
1
2
4
|
7 |
1
3
4
|
1
3
4
9
|
1
2
5
9
|
Note that there's only a quad on row B in cells B1-B4. If you look at cells
B1, B2, B3 and C2, which also have {1,2,3,5} as unique possibilities,
you'll see one of the reasons why quads are so rare: what you
usually have is something much simpler, in this case a naked single.
You won't find higher order sets in a 9×9 sudoku. If you somehow stumbled upon what looked like a quin, you would also have a quad which would eliminate it since 5 + 4 = 9.
Two sets of unique possibilities may be obscured by superfluous ones. If you manage to locate the pair you can remove other possibilities from the cells in which the pair appears.
|
1
7
9
|
1
6
|
6
7
9
|
| 2 |
1
6
8
|
4 |
| 3 |
1
8
|
5 |
| 6 |
1
2
5
|
2
7
|
|
5
7
|
9 | 3 |
|
1
7
|
4 | 8 |
|
5
9
|
2
5
|
1 |
| 8 | 3 |
6
9
|
| 4 | 7 |
2
6
|
Hidden triples are a possibility but extremely rare, which is fortunate because they're very hard to spot. As with a naked triple, you're looking for three possibilities in a unit that can only go in three cells. However, they're obscured by superfluous possibilities.
|
1
2
|
1
2
|
6 |
|
4
5
|
4
5
8
9
|
4
9
|
| 3 |
4
7
8
|
4
7
|
|
1
7
|
1
3
7
|
5 |
|
4
6
|
4
6
7
9
|
2
4
7
9
|
| 8 |
3
4
7
|
2
3
4
7
|
|
2
7
|
2
3
7
|
1 |
|
5
6
|
5
6
|
8 |
| 9 |
3
4
7
|
3
4
7
|
The triple is {5,6,9} located in B2, B5 and B9. This allows
the elimination of superfluous possibilities in B2 and B5 revealing the 8 in B3
as a hidden single.
In a 9×9 sudoku, a hidden quad will typically only be found when there are no solved cells in a unit. Here's an example I stumbled across.
|
1
2
5
|
3 |
1
4
5
6
|
4
5
|
9 | 7 |
1
2
6
8
|
2
5
|
1
2
4
5
6
8
|
| 9 |
5
6
7
|
1
4
5
6
|
2 | 3 | 8 |
1
6
|
5
7 |
1
4
5
6
|
| 8 |
2
7
|
4
5
|
4
5
|
1 | 6 |
2
3
9
|
2
3
7
9
|
2
3
9
|
This elimination strategy requires you to notice that the only possible locations for a value in one unit (for example, a row or column) all intersect with another unit (a box) or vice versa. When you find this arrangement, you can eliminate possibilities from the second unit.
|
1
4
5
9
|
1
4
5
9
|
2
5
9
|
1
3
5
|
1
2
3
4
5
6
|
1
5
6
|
7 | 8 |
1
2
3
5
|
| 7 | 6 | 3 |
1
5
8
|
1
2
5
8
|
1
5
|
1
2
|
4 | 9 |
|
1
4
5
8
|
1
4
5
8
|
2
5
|
7 |
1
2
3
4
5
6
8
|
9 |
1
2
3
6
|
2
3
6
|
1
2
3
5
|
The only locations for 6 in row A are the red coloured values in A5 and A6. Both cells are also in box 2 which means that 6 cannot be elsewhere in box 2, permitting the elimination of the blue 6 in C5. If you're keen-eyed, you will also notice that there's a pointing pair in box 3 at C7 and C8 that would have allowed you to make the same elimination.
There are a couple of others to find in these three rows.
|
1
4
5
9
|
1
4
5
9
|
2
5
9
|
1
3
5
|
1
2
3
4
5
6
|
1
5
6
|
7 | 8 |
1
2
3
5
|
| 7 | 6 | 3 |
1
5
8
|
1
2
5
8
|
1
5
|
1
2
|
4 | 9 |
|
1
4
5
8
|
1
4
5
8
|
2
5
|
7 |
1
2
3
4
5
8
|
9 |
1
2
3
6
|
2
3
6
|
1
2
3
5
|
The pointing-triple in B4, B5 and B6 allows elimination of the four blue-coloured 5s from A4, A5, A6 and C6.
|
1
4
5
9
|
1
4
5
9
|
2
5
9
|
1
3
|
1
2
3
4
6
|
1
6
|
7 | 8 |
1
2
3
5
|
| 7 | 6 | 3 |
1
5
8
|
1
2
5
8
|
1
5
|
1
2
|
4 | 9 |
|
1
4
5
8
|
1
4
5
8
|
2
5
|
7 |
1
2
3
4
8
|
9 |
1
2
3
6
|
2
3
6
|
1
2
3
5
|
Either B4 or B5 must be the location of 8 in row B. This excludes the blue-coloured 8 from C6 in the intersecting box.
The strategies outlined here will probably require you to use pen and paper to locate the opportunities. This is because the game only updates the pencil marks for you when you place a tile on the board. That means you'll have to keep track of possibilities eliminated by naked pairs, pointing pairs and so forth. The strategies in this section will allow you to solve any puzzle graded tough or medium-tough
Simple colouring is the simplest of many alternate inference chain strategies. Here's how it works:
If you're lucky, you'll end up with one or more logical contradictions that you can use to eliminate one or more possibilities. Note that this is not a guessing strategy since there is no real significance to TRUE/FALSE here. You could start with FALSE or call the possibilities RED/GREEN and it wouldn't make a difference.
There are two rules that can be applied after colouring in possibilities. The first is the "two colours elsewhere" rule which allows elimination of a possibility if it can see two other possibilities with opposing colours. This is shown in the example below where none of the simple strategies from earlier will serve to progress the puzzle.
|
4
6
|
5 | 7 | 2 |
1
3
|
8 | 9 |
1
3
|
4
6
|
| 9 | 1 | 8 | 7 |
3
4
6
|
4
6
|
5 |
2
3
4
|
2
4
6
|
| 2 |
3
4
6
|
3
4
|
5 | 9 |
1
4
6
|
7 |
1
4
|
8 |
| 1 | 8 | 5 | 6 | 2 | 9 | 4 | 7 | 3 |
| 3 |
4
7
|
9 | 8 | 5 |
4
7
|
2
6
|
2
6
|
1 |
|
4
7
|
2 | 6 | 1 |
4
7
|
3 | 8 | 5 | 9 |
|
7
8
|
9 | 1 |
3
4
|
7
8
|
2 |
3
6
|
4
6
|
5 |
|
4
6
8
|
3
4
6
|
2 |
3
4
|
6
8
|
5 | 1 | 9 | 7 |
| 5 |
6
7
|
3
4
|
9 |
1
6
|
1
6
7
|
2
3
|
8 |
2
4
|
The chain is fairly easy to follow: C3-J3-J9-G8-G4-H3. Now look at the blue-coloured 4 in C8. It can see both the TRUE 4 in C3 and the FALSE 4 in G8. C3 and G8 are linked: one or the other of them must be a 4 meaning that C8 cannot be 4. Eliminating 4 from C8 reveals 1 as a naked single.
The second rule that applies to simple colouring is "two colours in the same unit". This allows us to eliminate all possibilities of a given colour if that colour appears twice in any row, column or box. This is shown in the example below which is the same puzzle as the example above but a later stage.
|
4
6
|
5 | 7 | 2 | 1 | 8 | 9 | 3 |
4
6
|
| 9 | 1 | 8 | 7 | 3 |
4
6
|
5 |
2
4
|
2
4
6
|
| 2 |
3
6
|
3
4
|
5 | 9 |
4
6
|
7 | 1 | 8 |
| 1 | 8 | 5 | 6 | 2 | 9 | 4 | 7 | 3 |
| 3 | 4 | 9 | 8 | 5 | 7 |
2
6
|
2
6
|
1 |
| 7 | 2 | 6 | 1 | 4 | 3 | 8 | 5 | 9 |
| 8 | 9 | 1 |
3
4
|
7 | 2 |
3
6
|
4
6
|
5 |
|
4
6
|
3
6
|
2 |
3
4
|
8 | 5 | 1 | 9 | 7 |
| 5 | 7 |
3
4
|
9 | 6 | 1 |
2
3
|
8 |
2
4
|
There are two chains of interest here: A1-G1-H4-G4-G8-B8 and A1-G1-J3-C3-C6-B6. You can also trace A1-A9 and J3-J9 to complete the colouring. Note that there are two FALSE cells in row B (and column 9 for that matter). This allows all FALSE 4s (coloured blue) to be eliminated: since they're linked and it's impossible for some of them to be 4 (because 4 can only appear once in a row or column), logically none can be 4. There is an additional logical implication when this rule applies: since one colour or the other is actually true, if you are able to eliminate the possibilities of one colour, the differently coloured possibilities MUST be the solutions to the cells in which they appear.
The X-Wing strategy involves finding two rows with only two possibilities for a given number x both in the same column OR two columns with only two possibilities for a given number x both in the same row. This pattern is called an X-Wing. The two possibilities across the diagonals are interlocked: if x is part of the solution in one of the corners it must be part of the solution in the opposite corner.
|
4
6
7
|
5 |
3
4
7
|
2 |
1
3
4
6
|
8 | 9 |
1
3
4
|
4
6
|
| 9 | 1 | 8 | 7 |
3
4
6
|
4
6
|
5 |
2
3
4
|
2
4
6
|
| 2 |
3
4
6
|
3
4
|
5 | 9 |
1
4
6
|
7 |
1
3
4
|
8 |
| 1 | 8 | 5 | 6 | 2 | 9 | 4 | 7 | 3 |
| 3 |
4
7
|
9 | 8 | 5 |
4
7
|
2
6
|
2
6
|
1 |
|
4
7
|
2 | 6 | 1 |
4
7
|
3 | 8 | 5 | 9 |
|
7
8
|
9 | 1 |
3
4
|
7
8
|
2 |
3
6
|
4
6
|
5 |
|
4
6
8
|
3
4
6
|
2 |
3
4
|
6
8
|
5 | 1 | 9 | 7 |
| 5 |
3
4
6
7
|
3
4
7
|
9 |
1
6
7
|
1
6
7
|
2
3
|
8 |
2
4
|
In both rows F and G, 7 can only go in columns 1 and 5. Since either {F1,G5} or {F5,G1} are both 7, 7 can be eliminated as a possibility from other cells in columns 1 and 5. Eliminating 7 in A1 reveals 7 as a hidden single in A3.
Here's an example of an X-Wing on two columns.
| 7 | 3 |
1
4
9
|
2 |
4
5
6
|
5
6
|
8 |
1
5
9
|
1
5
|
| 5 | 2 | 6 | 1 | 9 | 8 | 7 | 4 | 3 |
|
1
8
|
1
9
|
1
4
8
9
|
4
5
|
7 | 3 |
1
5
9
|
6 | 2 |
| 9 | 5 | 2 | 3 |
1
6
|
7 |
1
6
|
8 | 4 |
| 4 | 7 |
1
3
8
|
5
6
8
|
1
5
6
|
2 |
1
3
5
6
|
1
5
|
9 |
|
1
8
|
6 |
1
3
8
|
4
5
8
|
1
4
5
|
9 |
1
3
5
|
2 | 7 |
| 3 | 4 |
1
9
|
5
6
9
|
8 |
5
6
|
2 | 7 |
1
5
|
| 2 | 8 | 7 |
5
9
|
3 | 1 | 4 |
5
9
|
6 |
| 6 |
1
9
|
5 | 7 | 2 | 4 |
1
9
|
3 | 8 |
Columns 6 and 9 can only accept a 5 in rows A and G. This means that either {A6,G9} must be a 5 or else {A9,G6} must be a 5, allowing the elimination of the blue-coloured 5s in rows A and G. This doesn't immediately reveal another part of the solution, but as we'll see in the next section, it does open up a move which will.
The XY-wing strategy involves finding a {2,2,2} triple spread
across two intersecting units - either a row and a column, a row and a box or
a column and a box (if the three cells are in just one unit, it's a naked triple,
covered previously). The three cells will contain values {x,y},{x,z},{y,z}
The cell containing {x,y} is called the hinge while the
other two cells, visible to the hinge but not to each other, are called pincers.
This pattern is called an XY-Wing and allows z to be eliminated as
possibilities from all cells that can see both pincers since either pincer
(or both) must be z.
The first example follows immediately from the previous X-Wing example.
| 7 | 3 |
1
4
9
|
2 |
4
6
|
5
6
|
8 |
1
9
|
1
5
|
| 5 | 2 | 6 | 1 | 9 | 8 | 7 | 4 | 3 |
|
1
8
|
1
9
|
1
4
8
|
4
5
|
7 | 3 |
1
5
9
|
6 | 2 |
| 9 | 5 | 2 | 3 |
1
6
|
7 |
1
6
|
8 | 4 |
| 4 | 7 |
1
3
8
|
6
8
|
1
5
6
|
2 |
1
3
5
6
|
1
5
|
9 |
|
1
8
|
6 |
1
3
8
|
5
8
|
1
4
5
|
9 |
1
3
5
|
2 | 7 |
| 3 | 4 |
1
9
|
5
9
|
8 |
5
6
|
2 | 7 |
1
5
|
| 2 | 8 | 7 |
5
9
|
3 | 1 | 4 |
5
9
|
6 |
| 6 |
1
9
|
5 | 7 | 2 | 4 |
1
9
|
3 | 8 |
The hinge at A8 forces a 5 into either A9 or H8. With this pattern, there are 5 cells that can see both pincers: B8, C8, G9, H9 and J9. G9 contains a 5 that can be eliminated, revealing 1 as part of the solution. To show that the XY-Wing pattern eliminates the candidate in G9, prove to yourself that the hinge forces a 5 into at least one of the pincers: if the value in A8 is a 1, A9 must be a 5; if it is 9, H8 must be a 5. This means that G9 cannot be a 5.
The second example is almost picture-book perfect.
| 7 | 8 | 4 | 6 | 9 | 1 | 2 | 3 | 5 |
|
1
2
|
2
5
|
1
2
5
|
4 | 8 | 3 | 9 | 6 | 7 |
| 3 | 6 | 9 | 5 | 7 | 2 | 8 | 1 | 4 |
| 4 |
2
7
|
1
2
7
|
8 | 6 | 9 |
1
5
|
2
5
|
3 |
|
1
2
|
9 | 6 | 3 | 4 | 5 |
1
7
|
2
7
|
8 |
| 5 | 3 | 8 | 2 | 1 | 7 | 4 | 9 | 6 |
| 6 | 1 |
5
7
|
9 | 3 | 8 |
5
7
|
4 | 2 |
| 8 |
2
5
7
|
3 | 1 |
2
5
|
4 | 6 |
5
7
|
9 |
| 9 | 4 |
2
5
|
7 |
2
5
|
6 | 3 | 8 | 1 |
This is a nice, rectangular XY-Wing across an intersecting row and column. The hinge is located at D8 while the pincers are at D2 and H8. The possibility that the pincers have in common is 7. There is only one cell that can see both pincers, located at H2. We can eliminate the blue-coloured 7 as a possibility from that cell revealing 7 as a hidden single in both pincer cells.
This is similar to the XY-Wing strategy. The pattern that you're searching for
is a hinge containing three possibilities {x,y,z} and two pincers
containing {x,z},{y,z}. The three cells need to be located across
two different units, either a row and a box or a column and a box. This pattern
is called an XYZ-Wing and when you find such an arrangement you can eliminate
z from the two cells that can see all three members of the wing.
| 3 |
6
8
|
5
7
8
|
2
5
7
|
5
6
7
|
1 | 9 |
2
7
|
4 |
| 2 |
5
7
|
4 | 9 | 3 |
5
7
|
8 | 1 | 6 |
| 9 |
1
6
|
1
7
|
8 | 4 |
2
6
7
|
2
7
|
5 | 3 |
| 6 |
1
3
|
1
2
7
|
4
5
7
|
5
7
|
8 |
2
3
4
|
2
7
|
9 |
| 4 |
3
8
|
7
8
|
1 | 2 | 9 |
3
7
|
6 | 5 |
| 5 |
2
7
|
9 | 3 |
6
7
|
4
6
7
|
2
4
7
|
8 | 1 |
| 7 | 4 |
2
5
|
2
5
|
1 | 3 | 6 | 9 | 8 |
| 1 | 9 | 6 |
4
7
|
8 |
4
7
|
5 | 3 | 2 |
| 8 |
2
5
|
3 | 6 | 9 |
2
5
|
1 | 4 | 7 |
The hinge is at A3 with pincers at B2 and E3. The possibility that all three cells have in common is 7 (coloured red) and one of them must contain this value. To see for yourself that this is the case, consider all three possibilities in A3 to see how the pincers are affected: if it's 7, it's 7; if it's 5, B2 will be 7; if it's 8, E3 will be 7. This means that 7 cannot be a value in the two cells that can see all of the wing members, B2 and C2. Eliminating the blue-coloured 7 from C2 leaves 1 (coloured green) as a naked single.
If an X-Wing involves identifying the same two unique columns as possibilities
for a given number on two rows (or the same two unique rows on two columns), a
swordfish is when you find the same three unique column possibilities on
three rows (or row possibilities on three columns). What makes the pattern much
harder to spot, however, is that a swordfish is almost never perfect, that is
you generally won't see the same three column possibilities {x,y,z} on three
rows {a,b,c}. Instead the three unique column possibilities are spread
across the three rows so that n may be restricted to {x,y,z}
on row a, {x,z} on row b and {x,y}
on row c or any other combination of 3s and 2s. It's easier to
visualise with an example.
7
9
6
1
4
3
1
6
6
8
3
6
1
5
3
7
9
3
2
6
1
5
4
7
1
4
5
6
2
4
The second example is a column-based swordfish and a good example of the
{2,2,2} variety.
|
4
6
|
7
9
|
5
7
9
|
4
7
|
2 | 8 |
3
5
6
7
|
3
5
|
1 |
| 1 | 3 | 8 | 6 |
5
7
|
9 |
2
5
7
|
4 |
2
7
|
|
4
6
|
2 |
5
7
|
1 |
4
5
7
|
3 |
5
6
7
|
8 | 9 |
| 9 | 5 | 6 |
4
7
|
3 |
4
7
|
1 | 2 | 8 |
| 7 | 8 | 4 | 5 | 1 | 2 | 9 | 6 | 3 |
| 3 | 1 | 2 | 8 | 9 | 6 | 4 | 7 | 5 |
| 2 | 4 |
1
7
|
9 | 8 | 5 |
3
7
|
1
3
|
6 |
| 5 | 6 | 3 | 2 |
4
7
|
1 | 8 | 9 |
4
7
|
| 8 |
7
9
|
1
7
9
|
3 | 6 |
4
7
|
2
5
7
|
1
5
|
2
4
7
|
Columns 2, 4 and 6 only allow 7s on rows A, D and J. This allows us to eliminate other possibilities from the three rows. Once again I've coloured in the "holes" in the swordfish to make the pattern more visible. The five eliminations don't reveal another part of the puzzle but they open up a slightly tricky-to-spot XY-Wing which does.
Note: because the swordfish pattern is extremely tricky to spot, you won't find them in puzzles generated when you've selected "Low / Medium" difficulty.
Several of the strategies in this section are generalisations of those from the previous sections. Often, they're a lot of work to eliminate a single possibility. In principle, you can use these along with the strategies outlined earlier to solve any puzzle graded Diabolical. X-Cycles and XY-Chains are by far the most common patterns so if you can get the hang of these, you can tackle the majority of diabolical puzzles.
X-Chains are the second alternating inference chain strategy that you might apply (Simple Colouring above is the first). Before you can start to apply it, however, you need to understand the concept of weak and strong links.
Pick a number and look at the places where it can go in a unit. Assuming it's not a naked or hidden single, there's a logical association between any two possibilities: if possibility A is part of the solution, B isn't and if B is part of the solution, A isn't. If it is possible for neither A nor B to be part of the solution, this is a weak link: both assertions can't be true but they can both be false. If, however, you are able to make the assertion that if A isn't part of the solution, B must be or vice versa, you have a strong link: both assertions can't be true but they can't both be false either. If you look back at the Simple Colouring examples above, you can see that the strategy involves following strong links.
X-Chains involve a slightly different strategy: instead of following strong links, you're following alternating strong and weak links. In addition, the goal is to loop back to the starting cell. These are the rules:
When you have an even length chain, you can eliminate x from cells that can see both ends of the chain. This is the most common result of the X-Chain / X-Cycle strategy.
|
5
6
8
|
1 | 4 | 7 |
5
8
|
2 | 3 | 9 |
5
6
8
|
|
5
6
8
|
2
3
8
|
2
5
8
|
9 | 4 |
1
3
5
8
|
1
5
6
|
5
6
8
|
7 |
| 7 |
3
8
|
9 |
3
5
8
|
6 |
1
3
5
|
4 | 2 |
1
5
8
|
| 3 | 4 |
5
8
|
2 | 1 |
5
6
8
|
5
6
9
|
7 |
5
6
9
|
| 2 | 6 | 7 |
4
5
|
3 | 9 | 8 | 1 |
4
5
|
|
5
8
|
9 | 1 |
4
6
|
5
8
|
7 |
2
6
|
3 |
2
4
6
|
|
1
4
|
2
8
|
2
8
|
3
5
6
|
9 |
3
5
6
|
7 |
4
5
6
|
1
3
5
6
|
|
4
9
|
5 | 6 | 1 | 7 |
3
8
|
2
9
|
4
8
|
2
3
8
9
|
|
1
9
|
7 | 3 |
5
6
8
|
2 | 4 |
1
5
6
9
|
5
6
8
|
1
5
6
8
9
|
A word about chain notation, since there's a compact and readable way to
describe them. If a value is turned ON, this indicated by a '+'; a value switched
OFF is indicated by '-'. This is followed by the value under consideration and then
the cell in square brackets. Following this notation, the chain in the example
is: -5[A5]+5[F5]-5[E4]+5[E9].
The strong links mean that the value of E9 depends on value of A5. Either A5 is 5 or E9 is 5. Without being able to say which, we can state that A9 cannot be 5 because it is able to see two differently coloured possibilities.
If the loop is OFF-ON all the way round, x can be eliminated from any cell outside the loop which is able to see two loop members.
|
5
6
8
|
1 | 4 | 7 |
5
8
|
2 | 3 | 9 |
6
8
|
|
6
8
|
2
3
|
2
5
|
9 | 4 |
1
3
5
|
1
5
6
|
5
6
8
|
7 |
| 7 |
3
8
|
9 |
3
8
|
6 |
1
5
|
4 | 2 |
1
5
|
| 3 | 4 |
5
8
|
2 | 1 |
5
6
8
|
5
6
9
|
7 |
5
6
9
|
| 2 | 6 | 7 |
4
5
|
3 | 9 | 8 | 1 |
4
5
|
|
5
8
|
9 | 1 |
4
6
|
5
8
|
7 |
2
6
|
3 |
2
4
6
|
|
1
4
|
2
8
|
2
8
|
3
5
6
|
9 |
3
5
6
|
7 |
4
5
6
|
1
3
5
6
|
|
4
9
|
5 | 6 | 1 | 7 |
3
8
|
2
9
|
4
8
|
2
3
8
9
|
|
1
9
|
7 | 3 |
5
6
8
|
2 | 4 |
1
5
6
9
|
5
6
8
|
1
5
6
8
9
|
This example follows a few steps after the previous one. The loop traced
is -5[A5]+5[F5]-5[E4]+5[E9]-5[C9]+5[C6]-5[A5]
When you have a continuous loop like this, the eliminations are possible
because you can assert that alternate members of the chain (although you don't
know which) MUST be part of the solution. Therefore any cell outside the loop
that can see two members of the loop must be seeing part of the solution and
therefore cannot be part of the solution.
Note that you could create a continuous loop from the same starting location that goes
-5[A5]+5[F5]-5[F1]+5[A1]-5[A5]. This does not lead to any eliminations
but you may notice that it makes an X-Wing. I mentioned that more advanced
strategies are generalisations of less advanced ones. Well, X-Cycles incorporate
the X-Wing and the {2,2,2} Swordfish.
If the last link in the chain is OFF and able to see the first link in the chain, the possibility in the first cell is part of the solution.
|
3
4
6
|
2 |
4
5
|
1
3
5
6
|
1
3
|
9 | 8 |
3
6
|
7 |
|
1
3
6
8
|
1
3
|
1
8
|
4 | 7 |
3
6
|
9 | 2 | 5 |
| 7 |
3
5
|
9 |
2
3
5
6
8
|
2
3
8
|
3
5
6
8
|
1 |
3
6
|
4 |
|
1
4
8
9
|
1
4
9
|
1
8
|
1
3
6
8
|
5 |
3
4
6
8
|
2 | 7 |
3
6
|
|
1
4
8
|
7 | 2 |
1
3
6
8
|
1
3
8
|
3
4
6
8
|
5 | 9 |
3
6
|
| 5 | 6 | 3 |
2
9
|
2
9
|
7 | 4 | 8 | 1 |
|
1
3
9
|
1
3
5
9
|
7 |
3
5
9
|
3
9
|
2 | 4 | 6 | 8 |
| 2 | 8 | 6 | 7 | 4 | 1 | 3 | 5 | 9 |
|
3
4
9
|
3
4
5
9
|
4
5
|
3
5
8
9
|
6 |
3
5
8
|
7 | 1 | 2 |
The loop traced in the example is -5[A3]+5[A4]-5[G4]+5[G2]-5[C2]+5[A3].
The different signs at each end reveal a logical contradiction: if A3 is not part
of the solution, A3 is part of the solution or, equivalently, if A3 is not part
of the solution, neither is C2. Since that is impossible, this is proof that
the 5 in A3 MUST be part of the solution.
Strong links aren't restricted to units. The two possibilities in bi-value cells are strongly linked too which is obvious when you think about it. XY-Chains leap from bi-value cell to bi-value cell, turning off one possibility and turning on the other until you complete the chain or fail to. If you manage to complete a chain, one end of the chain or the other MUST be part of the solution which allows you to eliminate possibilities able to see both ends. Here's how you construct them:
The following example shows the shortest possible chain (we'll see why the shortest XY-chain is four links shortly).
| 4 |
1
2
5
8
9
|
5
6
8
9
|
1
5
8
9
|
1
2
5
8
|
2
5
6
8
9
|
7 | 3 |
5
9
|
|
2
8
|
2
5
7
8
9
|
5
8
9
|
5
8
9
|
2
5
7
8
|
3 | 4 | 1 | 6 |
|
1
6
|
1
3
5
7
9
|
3
5
6
9
|
4 |
1
5
7
|
5
6
9
|
2 | 8 |
5
9
|
| 9 | 6 | 2 |
3
8
|
3
8
|
1 | 5 | 7 | 4 |
|
1
8
|
1
4
5
8
|
4
5
8
|
7 | 9 |
2
4
5
|
3
6
|
2
6
|
2
3
8
|
| 3 |
4
5
8
|
7 | 6 |
2
5
|
2
4
5
|
1 | 9 |
2
8
|
| 7 |
2
4
|
4
6
|
1
3
8
9
|
1
3
8
|
8
9
|
3
6
9
|
5 |
2
3
|
| 5 |
3
9
|
3
9
|
2 | 6 | 7 | 8 | 4 | 1 |
|
2
6
8
|
2
8
|
1 |
3
5
9
|
4 |
5
9
|
3
9
|
2
6
|
7 |
XY-Chains can be described using the same notation as earlier. This length 4
chain is -8[F9]+2[F9]-2[E8]+6[E8]-6[J8]+2[J8]-2[J2]+8[J2]. We have
logical proof that either F9 is 8 or J2 is 8, which means that F2 cannot be,
allowing 8 to be eliminated from that cell. This elimination leaves the 8 in F9 as a hidden
single for row F.
As to why 4 is the minimum length, imagine a chain that went
-A[X]+B[X]-B[Y]+C[Y]-C[Z]+A[Z]. This is certainly possible but
if you call cells X and Z "pincers" and cell Y
"the hinge", you'll see that we already have a name for this pattern: XY-Wing,
showing once again how an advanced strategy is a generalisation of a simpler
one.
Chains can become arbitrarily long. Here's an example further along in the same puzzle.
| 4 |
1
2
5
8
9
|
5
6
8
9
|
1
5
8
9
|
1
2
5
8
|
2
6
8
9
|
7 | 3 |
5
9
|
|
2
8
|
2
5
7
8
9
|
5
8
9
|
5
8
9
|
2
5
7
8
|
3 | 4 | 1 | 6 |
|
1
6
|
1
3
5
7
9
|
3
5
6
9
|
4 |
1
5
7
|
6
9
|
2 | 8 |
5
9
|
| 9 | 6 | 2 |
3
8
|
3
8
|
1 | 5 | 7 | 4 |
|
1
8
|
1
5
8
|
4
5
8
|
7 | 9 |
4
5
|
3
6
|
2
6
|
2
3
|
| 3 |
4
5
|
7 | 6 |
2
5
|
2
4
|
1 | 9 | 8 |
| 7 |
2
4
|
4
6
|
1
3
9
|
1
3
8
|
8
9
|
6
9
|
5 |
2
3
|
| 5 |
3
9
|
3
9
|
2 | 6 | 7 | 8 | 4 | 1 |
|
2
6
8
|
2
8
|
1 |
3
5
|
4 |
5
9
|
3
9
|
2
6
|
7 |
The chain is -2[B1]+8[B1]-8[E1]+1[E1]-1[C1]+6[C1]-6[C6]+9[C6]-9[J6]+5[J6]-5[E6]+4[E6]-4[F6]+2[F6]-2[F5]+5[F5]-5[F2]+4[F2]-4[G2]+2[G2]
which is perhaps at the limits of human solvability. Note that a complete chain
is found after seven links but since {B1,F6} doesn't create
any eliminations, the chain is followed for a few more links until
we find endpoints that do. Note also that three eliminations is a
pretty decent outcome for this strategy.
|
4
6
|
5 | 7 | 2 |
1
3
|
8 | 9 |
1
3
|
4
6
|
| 9 | 1 | 8 | 7 |
3
4
6
|
4
6
|
5 |
2
3
4
|
2
4
6
|
| 2 |
3
4
6
|
3
4
|
5 | 9 |
1
4
6
|
7 |
1
4
|
8 |
| 1 | 8 | 5 | 6 | 2 | 9 | 4 | 7 | 3 |
| 3 |
4
7
|
9 | 8 | 5 |
4
7
|
2
6
|
2
6
|
1 |
|
4
7
|
2 | 6 | 1 |
4
7
|
3 | 8 | 5 | 9 |
|
7
8
|
9 | 1 |
3
4
|
7
8
|
2 |
3
6
|
4
6
|
5 |
|
4
6
8
|
3
4
6
|
2 |
3
4
|
6
8
|
5 | 1 | 9 | 7 |
| 5 |
6
7
|
3
4
|
9 |
1
6
|
1
6
7
|
2
3
|
8 |
2
4
|
Next, locate the bi-value cells with coloured possibilities and apply the opposite colour to the cell partners:
|
4
6
|
5 | 7 | 2 |
1
3
|
8 | 9 |
1
3
|
4
6
|
| 9 | 1 | 8 | 7 |
3
4
6
|
4
6
|
5 |
2
3
4
|
2
4
6
|
| 2 |
3
4
6
|
3
4
|
5 | 9 |
1
4
6
|
7 |
1
4
|
8 |
| 1 | 8 | 5 | 6 | 2 | 9 | 4 | 7 | 3 |
| 3 |
4
7
|
9 | 8 | 5 |
4
7
|
2
6
|
2
6
|
1 |
|
4
7
|
2 | 6 | 1 |
4
7
|
3 | 8 | 5 | 9 |
|
7
8
|
9 | 1 |
3
4
|
7
8
|
2 |
3
6
|
4
6
|
5 |
|
4
6
8
|
3
4
6
|
2 |
3
4
|
6
8
|
5 | 1 | 9 | 7 |
| 5 |
6
7
|
3
4
|
9 |
1
6
|
1
6
7
|
2
3
|
8 |
2
4
|
The newly coloured possibilities (2, 3 and 6) can be used as the basis of their own Simple Colouring networks (this is where the "3D" in the strategy name comes from - the layered stack of colour networks). Colour these possibilities, taking care to alternate colours correctly as you follow the links.
|
4
6
|
5 | 7 | 2 |
1
3
|
8 | 9 |
1
3
|
4
6
|
| 9 | 1 | 8 | 7 |
3
4
6
|
4
6
|
5 |
2
3
4
|
2
4
6
|
| 2 |
3
4
6
|
3
4
|
5 | 9 |
1
4
6
|
7 |
1
4
|
8 |
| 1 | 8 | 5 | 6 | 2 | 9 | 4 | 7 | 3 |
| 3 |
4
7
|
9 | 8 | 5 |
4
7
|
2
6
|
2
6
|
1 |
|
4
7
|
2 | 6 | 1 |
4
7
|
3 | 8 | 5 | 9 |
|
7
8
|
9 | 1 |
3
4
|
7
8
|
2 |
3
6
|
4
6
|
5 |
|
4
6
8
|
3
4
6
|
2 |
3
4
|
6
8
|
5 | 1 | 9 | 7 |
| 5 |
6
7
|
3
4
|
9 |
1
6
|
1
6
7
|
2
3
|
8 |
2
4
|
It just so happens that this third round of colouring didn't open up any new bi-value cells but if it did, you would continue colouring them as well. It didn't reveal any eliminations beyond the one that we already knew about but it wasn't a complete waste of time either since the logical assertion that all possibilities of one colour are either all part of the solution or none of them are still holds. If any subsequent move eliminates just one coloured possibility, you can solve a dozen cells in one go.
The advantage of the 3D Medusa strategy over Simple Colouring is both the increase in the number of elimination candidates and the addition of a bunch more elimination rules including two that permit elimination of an entire colour in one go.
If colouring results in two possibilities of the same colour in a single cell, all candidates of that colour may be eliminated. Equivalently, all possibilities of the other colour are part of the solution.
|
4
6
8
|
3 | 5 |
7
8
|
4
7
|
1 | 9 | 2 |
4
6
|
|
4
6
|
1 |
4
7
9
|
3
5
9
|
2 |
4
6
9
|
3
5
7
|
5
6
|
8 |
|
2
4
6
8
|
2
7
9
|
2
4
7
9
|
3
5
8
9
|
3
4
5
|
4
5
6
9
|
3
5
7
|
1 |
3
4
6
|
|
2
4
|
5 | 3 |
2
9
|
8 |
4
9
|
6 | 7 | 1 |
| 1 | 8 |
4
7
|
6 |
4
5
7
|
4
5
|
2
3
|
9 |
2
3
|
| 9 |
2
7
|
6 |
2
7
|
1 | 3 | 4 | 8 | 5 |
| 3 | 4 | 8 | 1 | 9 | 7 |
2
5
|
5
6
|
2
6
|
| 7 | 6 | 1 |
3
5
|
3
5
|
2 | 8 | 4 | 9 |
| 5 |
2
9
|
2
9
|
4 | 6 | 8 | 1 | 3 | 7 |
The colouring starts with the 8 in A1 which colours the 8 in C1 FALSE:
+8[A1]-8[C1]. There are a number of paths that colour the 2 in
C1, for example, +8[A1]-8[A4]+7[A4]-7[F4]+2[F4]-2[F2]+2[D1]-2[C1].
It's worthwhile to fill in the remainder
of the colour network to increase the number of eliminations. Since having two
parts of the solution in a single cell is clearly impossible, we can conclude
that no FALSE-coloured candidate can be part of the solution. Furthermore, we
can assert that all TRUE-coloured candidates are the solutions to the cells
in which they appear.
If any unit contains two possibilities with same value and same colour, all candidates of that colour can be eliminated (and candidates of the other colour placed). This rule is shared with the Simple Colouring strategy, but 3D Medusa creates more ways for this pattern to emerge.
|
3
4
5
|
2 |
5
7
9
|
3
5
|
4
9
|
6 |
1
5
7
8
|
1
3
5
8
9
|
1
5
7
8
|
|
3
5
6
|
6
7
|
1 |
3
5
|
2
9
|
8 | 4 |
2
3
5
6
9
|
2
5
6
7
|
|
3
4
5
6
|
8 |
5
9
|
1 | 7 |
2
4
|
2
5
6
|
2
3
5
6
9
|
2
5
6
|
| 9 | 1 | 6 |
7
8
|
5 | 3 |
2
7
8
|
4 |
2
7
8
|
| 7 | 5 | 2 | 8 |
4
6
8
|
1
4
|
3 |
1
6
8
|
1
6
8
|
| 8 | 4 | 3 |
6
7
|
2
6
|
1
2
|
9 |
1
5
6
|
1
5
6
7
|
|
1
2
|
3
6
|
8 | 4 |
3
6
|
5 |
1
2
|
7 | 9 |
|
1
2
|
3
7
|
5
7
|
6
8
|
3
8
|
9 |
5
6
|
1
2
|
4 |
|
5
6
|
9 | 4 | 2 | 1 | 7 |
5
6
8
|
5
6
8
|
3 |
The critical path starts with the 7 in A3 and goes +7[A3]-7[B2]+7[B9], leaving the 7
in B9 coloured TRUE. We then start colouring from the 6 in B2 and follow the
links to F4: +6[B2]-6[G2]+6[G5]-6[H4]+6[F4]. Finally, we switch
back to 7s, colouring the 7 in F4 FALSE and linking to the 7 in F9 which is coloured TRUE.
The presence of two TRUE-coloured 7s in column 9 means that all TRUE-coloured
candidates may be eliminated and all FALSE-coloured candidates placed. Once
again, it's worth completing the network to maximise the number of solved cells.
If a cell contains possibilities with two different colours, uncoloured possibilities can be eliminated.
|
5
6
|
5
6
|
7
9
|
3
9
|
1 | 2 |
3
7
|
8 | 4 |
| 3 | 2 | 1 | 8 | 7 | 4 | 6 | 9 | 5 |
| 8 |
4
7
9
|
4
7
9
|
5 | 6 |
3
9
|
1
2
|
1
2
|
3
7
|
|
2
7
|
8 | 6 | 4 | 3 | 5 | 9 |
2
7
|
1 |
| 9 | 1 | 3 | 2 | 8 | 7 | 5 | 4 | 6 |
|
2
4
5
7
|
4
5
7
|
2
4
7
|
6 | 9 | 1 |
2
7
8
|
3 |
7
8
|
|
1
2
4
6
7
|
3 |
2
4
7
8
|
1
9
|
2
4
|
6
8
|
1
7
8
|
5 |
7
8
9
|
|
1
2
4
|
4
9
|
5 | 7 |
2
4
|
8
9
|
1
3
8
|
6 |
3
9
|
|
1
6
7
|
6
7
9
|
7
8
9
|
1
3
9
|
5 |
3
6
8
9
|
4 |
1
7
|
2 |
Note that I've not bothered showing the complete network here, just the
critical paths. The first chain is +3[A4]-3[J4] leaving
a FALSE-coloured 3 in J4. Returning to the 3 in A4 gives
+3[A4]-3[A7]+3[C9]-3[H9]+9[H9]-9[G9]+9[G4]-1[G4]+1[J4] colouring
another possibility in J4 and creating the elimination. Remember that the colour networks are either-or
logical statements. Therefore J4 is either 1 or it is 3. We don't know which
but we do know that it isn't 9.
As with Simple Colouring, an uncoloured possibility that can see two differently coloured possibilities elsewhere can be eliminated.
|
5
6
|
5
6
|
7
9
|
3
9
|
1 | 2 |
3
7
|
8 | 4 |
| 3 | 2 | 1 | 8 | 7 | 4 | 6 | 9 | 5 |
| 8 |
4
7
9
|
4
7
9
|
5 | 6 |
3
9
|
1
2
|
1
2
|
3
7
|
|
2
7
|
8 | 6 | 4 | 3 | 5 | 9 |
2
7
|
1 |
| 9 | 1 | 3 | 2 | 8 | 7 | 5 | 4 | 6 |
|
2
4
5
7
|
4
5
7
|
2
4
7
|
6 | 9 | 1 |
2
7
8
|
3 |
7
8
|
|
1
2
4
6
7
|
3 |
2
4
7
8
9
|
1
9
|
2
4
|
6
8
9
|
1
7
8
|
5 |
7
8
9
|
|
1
2
4
|
4
9
|
5 | 7 |
2
4
|
8
9
|
1
3
8
|
6 |
3
9
|
|
1
6
7
|
6
7
9
|
7
8
9
|
1
3
9
|
5 |
3
6
8
9
|
4 |
1
7
|
2 |
If the layout looks familiar, it's the move immediately before the rule 3
elimination above. Once again, I've not bothered showing the entire network
since there's nothing interesting. Start at the 3 in A4 and follow the chain around clockwise
to the 9 in G9: +3[A4]-3[A7]+3[C9]-3[H9]+9[H9]-9[G9]. Next colour the
9 in A4 and link to the 9s in A3 and C6, colouring them TRUE. G4 and G6 can see
differently coloured 9s, revealing the eliminations.
If an uncoloured possibility can see a coloured possibility in the same unit and there is a different possibility in the same cell with the opposite colour, the uncoloured possibility can be eliminated.
|
1
2
3
8
|
3
5
8
|
9 |
1
3
5
7
8
|
3
5
|
1
4
|
2
6
|
1
5
6
7
8
|
1
4
5
6
|
|
1
2
3
8
|
6 |
2
3
8
|
1
3
5
7
8
|
3
5
9
|
1
4
9
|
2
5
9
|
1
5
7
8
|
1
4
5
|
| 4 |
5
8
|
7 |
1
5
8
|
2 | 6 |
5
9
|
1
5
8
|
3 |
| 9 | 2 | 4 |
1
5
6
|
7 | 3 | 8 |
1
5
6
|
1
5
6
|
| 5 | 1 | 6 | 4 | 8 | 2 | 3 | 9 | 7 |
| 7 |
3
8
|
3
8
|
1
5
6
|
5
6
9
|
1
9
|
4 |
1
5
6
|
2 |
|
3
6
8
|
4 |
3
5
8
|
9 | 1 |
5
8
|
7 | 2 |
5
6
8
|
|
2
6
8
|
9 |
2
5
8
|
2
6
|
4 | 7 | 1 | 3 |
5
6
8
|
|
2
6
8
|
7 | 1 |
2
3
6
|
3
6
|
5
8
|
5
6
|
4 | 9 |
This is a very pretty colouring chain: +2[A1]-2[A7]+6[A7]-6[J7]+5[J7]-5[J6]+8[J6]-8[J1]
which creates eliminations at both ends of the chain. The logic works because the
null hypothesis (that the eliminated possibility is part of the solution) results
in an impossibility. If the 8 in A1 were part of the solution, it would result
in the elimination of both the FALSE-coloured 8 in J1 and the TRUE-coloured 2
in A1. Since either TRUE possibilities or FALSE possibilities are part of the
solution, any choice that results in the elimination of two differently-coloured
possibilities must be wrong. The same argument eliminates the 2 in J1.
If all the possibilities of an uncoloured cell can see coloured possibilities of just one colour, all candidates of that colour may be eliminated and candidates of the other colour placed.
| 3 | 6 | 9 |
4
5
8
|
1
4
8
|
2 |
4
8
|
1
4
5
8
|
7 |
| 8 | 5 | 2 | 3 |
1
4
7
|
1
4
7
|
9 |
1
4
|
6 |
| 7 | 1 | 4 |
5
6
8
9
|
6
8
9
|
5
6
8
|
2 | 3 |
5
8
|
| 6 | 3 | 5 | 2 |
4
7
8
|
4
7
8
|
1 |
4
7
|
9 |
| 2 | 4 | 7 | 1 | 5 | 9 |
3
8
|
6 |
3
8
|
| 9 | 8 | 1 |
4
6
7
|
3 |
4
6
7
|
4
7
|
2
5
|
2
5
|
| 4 | 7 | 8 |
5
6
|
1
2
6
|
1
5
6
|
3
6
|
9 |
2
3
|
| 1 | 2 | 6 |
7
8
9
|
7
8
9
|
3 | 5 |
7
8
|
4 |
| 5 | 9 | 3 |
4
6
7
8
|
2
4
6
7
8
|
4
6
7
8
|
6
7
8
|
2
7
8
|
1 |
The colouring starts at A5 and the critical path is
+5[A5]-5[A8]+5[C9]-8[C9]+8[E9]-3[E9]+3[G9]-3[G7]+6[G7]. The 5
in G4 can see the TRUE-coloured 5 in A5 while the 6 can see the TRUE-coloured 6
in G7. If TRUE were the correct colour, this would result in the elimination
of all possibilities in G4. We can conclude, therefore, that TRUE candidates
cannot be part of the solution. Since this rule results in the elimination of
an entire colour, I've shown the complete network in the example.
The Unique Rectangle strategy is based upon the assumption that the puzzle has a unique solution. If that assumption is invalid, so is the strategy. If the assumption is applicable, a certain type of pattern should be impossible:
|
1
3
6
|
2
8
|
7 |
1
4
|
5 |
3
8
|
1
2
3
6
8
|
9 |
1
2
4
8
|
|
1
3
5
|
8
9
|
4 | 6 |
3
9
|
2 |
1
3
8
|
7 |
1
5
8
9
|
|
1
3
5
6
|
2
8
9
|
5
6
|
1
4
7
9
|
3
4
9
|
3
7
8
|
1
2
3
6
8
9
|
5
6
|
1
2
4
5
8
|
| 2 |
1
3
|
5
6
|
4
7
9
|
3
4
6
9
|
3
5
7
|
1
6
7
|
5
6
|
8 |
|
5
6
7
|
1
3
|
8 |
7
9
|
2
3
6
9
|
3
5
7
|
1
2
6
7
|
4 |
1
2
5
9
|
|
5
6
7
|
4 | 9 | 8 |
2
6
|
1 |
2
6
7
|
3 |
2
5
|
| 4 | 7 | 3 | 2 | 1 | 9 | 5 | 8 | 6 |
| 9 | 5 |
1
2
|
3 | 8 | 6 | 4 |
1
2
|
7 |
| 8 | 6 |
1
2
|
5 | 7 | 4 | 9 |
1
2
|
3 |
Look at the rectangle H3:H8:J3:J8. You can solve these four cells by putting either a 1 or a 2 in H3 without eliminating possibilities elsewhere in the puzzle. This means that there are at least two solutions to the puzzle. This pattern, 2 identical pairs in a rectangle that occupies 2 boxes, is known as The Deadly Pattern and, you can use the fact that the solution to a uniquely solvable puzzle will not include this pattern to help solve it.
Note that the Deadly Pattern has to occupy two boxes. The other shaded cells, C3:C8:D3:D8, do not form the pattern since the corners of the rectangle occupy four boxes.
When trying to apply the strategy, look for a cell with a pair of possibilities
{x,y}. Now look for three other cells with {x,y} as
possibilities (they do not have to be cells with only two possibilities and,
indeed, at least one of them must have at least three). If the cells form a
rectangle with the corners occupying two boxes, you can see whether they form
one of the six Unique Rectangle patterns and make eliminations accordingly.
When you locate the pattern, the cells that have just {x,y} as
possibilities are known as the floor while those that have more than
two possibilities are known as the roof.
If there are three floor cells, the Deadly Pair may be eliminated from the roof cell.
|
3
4
5
|
2 |
5
7
9
|
3
5
|
4
9
|
6 |
1
5
7
8
|
1
3
5
8
9
|
1
5
7
8
|
|
3
5
6
|
6
7
|
1 |
3
5
|
2
9
|
8 | 4 |
2
3
5
6
9
|
2
5
6
7
|
|
3
4
5
6
|
8 |
5
9
|
1 | 7 |
2
4
|
2
5
6
|
2
3
5
6
9
|
2
5
6
|
| 9 | 1 | 6 |
7
8
|
5 | 3 |
2
7
8
|
4 |
2
7
8
|
| 7 | 5 | 2 | 8 |
4
6
8
|
1
4
|
3 |
1
6
8
|
1
6
8
|
| 8 | 4 | 3 |
6
7
|
2
6
|
1
2
|
9 |
1
5
6
|
1
5
6
7
|
|
1
2
|
3
6
|
8 | 4 |
3
6
|
5 |
1
2
|
7 | 9 |
|
1
2
|
3
6
7
|
5
7
|
6
8
|
3
6
8
|
9 |
1
2
5
6
|
1
2
5
6
|
4 |
|
5
6
|
9 | 4 | 2 | 1 | 7 |
5
6
8
|
5
6
8
|
3 |
If H7 were 1, H1 would be 2, G1 would be 1 and G7 would be 2, creating the Deadly Pattern which we assume is impossible: swapping 1s and 2s between the four cells wouldn't invalidate the rules of Sudoku. The same thing would happen if H7 were 2. Therefore, we can state that neither 1 nor 2 can be the solution to H7 and eliminate them.
If the roof cells (there may be two or three) have the same third possibility, z, we can conclude that z must be the solution to one of those cells and eliminate it from other cells that are not part of the rectangle but which can see all roof cells.
| 5 |
2
7
|
2
7
|
6 | 3 | 8 |
1
4
9
|
4
9
|
1
4
|
| 8 | 6 | 3 | 4 | 9 | 1 |
2
5
|
2
7
|
5
7
|
|
1
4
|
1
4
|
9 | 5 | 7 | 2 | 8 | 3 | 6 |
|
2
4
7
|
2
4
7
|
6 | 9 |
5
8
|
3 |
4
5
|
1 |
5
7
8
|
|
4
7
9
|
5 | 8 | 1 | 6 |
4
7
|
3 |
4
7
9
|
2 |
|
1
3
4
7
9
|
1
3
4
7
9
|
1
4
7
|
2 |
5
8
|
4
7
|
4
5
9
|
6 |
5
7
8
|
|
2
3
4
7
|
2
3
4
7
|
2
4
7
|
8 |
1
2
|
9 | 6 | 5 |
1
4
|
| 6 |
1
2
8
|
1
2
|
3 | 4 | 5 | 7 |
2
8
|
9 |
|
2
4
9
|
2
4
8
9
|
5 | 7 |
1
2
|
6 |
1
2
4
|
2
4
8
|
3 |
Both roof cells, D9 and F9, have 7 as an additional possibility. One of them must be 7 to avoid the Deadly Pattern. The subsequent eliminations in B9 and E8 leave a naked single in B9 and a hidden single in B8.
A Type 3 Unique Rectangle arises when there are two floor cells and the two
roof cells don't have possibilities in common. If there are two surplus
possibilities, {a,b}, we can see that
at least one of the roof cells must be a or b which doesn't
produce an elimination. If, however, the roof cells pair up with another cell
that has {a,b} as its only possibilities in the same unit,
{a,b} can be eliminated from other cells in the unit.
| 6 |
5
7
|
5
7
|
1 | 4 | 9 | 3 | 2 | 8 |
| 4 | 2 | 1 | 6 | 8 | 3 | 7 | 5 | 9 |
| 9 | 8 | 3 |
5
7
|
5
7
|
2 | 1 |
4
6
|
4
6
|
| 1 |
3
5
7
|
2
5
6
7
8
|
9 |
2
5
6
7
|
5
7
8
|
4 |
3
8
|
3
6
7
|
|
2
5
7
8
|
3
4
5
7
9
|
2
4
5
6
7
8
|
4
5
7
|
2
5
7
|
4
7
8
|
6
9
|
1 |
3
6
7
|
|
7
8
|
7
9
|
6
7
8
|
3 |
6
7
|
1 | 5 |
8
9
|
2 |
|
5
7
|
1 | 9 | 8 | 3 |
4
5
6
7
|
2 |
4
6
7
|
4
5
6
|
|
2
8
|
4
5
7
|
2
8
|
4
5
7
|
1 |
4
5
6
7
|
6
9
|
3
9
|
3
4
5
6
|
| 3 | 6 |
4
5
7
|
2 | 9 |
4
5
7
|
8 |
4
7
|
1 |
The 5 in G9 and 7 and G8 pair up with the {5,7} in G1 meaning that in row G, 5 and 7 can only be in two of those three cells. This allows us to eliminate 5 and 7 from G6.
Sometimes, it is not possible to make a pair using the roof cells, either because there are more than two surplus possibilities in the roof cells or because there is no cell outside the rectangle to pair with. However, it may still be possible to make a triple. Here's a very nice example where the roof cells form a pair in one unit and a triple in another.
|
2
7
|
2
3
7
9
|
5 |
4
9
|
6 |
2
3
|
4
7
9
|
8 | 1 |
|
6
7
|
3
4
6
7
9
|
4
6
7
9
|
5 | 8 | 1 | 2 |
7
9
|
3
6
7
|
| 8 | 1 |
2
6
9
|
4
9
|
2
3
|
7 |
4
6
9
|
5 |
3
6
|
| 4 |
2
6
7
|
1
2
6
7
|
8 |
1
2
3
|
2
3
6
|
6
7
9
|
2
3
6
7
9
|
5 |
| 3 | 8 |
1
2
6
7
|
1
2
|
5 | 9 |
6
7
|
4 |
2
6
7
|
| 9 | 5 |
2
6
|
7 | 4 |
2
3
6
|
1 |
2
3
6
|
8 |
| 5 |
2
4
6
9
|
2
4
6
9
|
1
2
6
|
7 | 8 | 3 |
1
2
6
|
2
4
6
|
|
1
2
6
7
|
2
4
6
7
|
3 |
1
2
6
|
9 | 5 | 8 |
1
2
6
7
|
2
4
6
7
|
|
1
2
6
7
|
2
6
7
|
8 | 3 |
1
2
|
4 | 5 |
1
2
6
7
|
9 |
The roof cells in A7 and C7 form a reasonably easy-to-spot {6,7} pair with D7. However, there's also a rather less easy-to-spot {3,6,7} triple in box 3 with B9 and C9. The eliminations leave two naked singles.
When it is not possible to make eliminations outside the rectangle, it may still be possible to make eliminations from the roof cells themselves. If the roof cells share a unit and one of the Deadly Pair candidates can only go in a roof cell, the other candidate may be eliminated from the roof.
|
3
4
5
|
2 |
5
7
9
|
3
5
|
4
9
|
6 |
1
5
7
8
|
1
3
5
8
9
|
1
5
7
8
|
|
3
5
6
|
6
7
|
1 |
3
5
|
2
9
|
8 | 4 |
2
3
5
6
9
|
2
5
6
7
|
|
3
4
5
6
|
8 |
5
9
|
1 | 7 |
2
4
|
2
5
6
|
2
3
5
6
9
|
2
5
6
|
| 9 | 1 | 6 |
7
8
|
5 | 3 |
2
7
8
|
4 |
2
7
8
|
| 7 | 5 | 2 | 8 |
4
6
8
|
1
4
|
3 |
1
6
8
|
1
6
8
|
| 8 | 4 | 3 |
6
7
|
2
6
|
1
2
|
9 |
1
5
6
|
1
5
6
7
|
|
1
2
|
3
6
|
8 | 4 |
3
6
|
5 |
1
2
|
7 | 9 |
|
1
2
|
3
6
7
|
5
7
|
6
8
|
3
6
8
|
9 |
5
6
|
1
2
|
4 |
|
5
6
|
9 | 4 | 2 | 1 | 7 |
5
6
8
|
5
6
8
|
3 |
One of H2 or H4 must be 3. However, the other cannot be 6 or you'd end up with the Deadly Pattern. Therefore, 6 can be eliminated from both cells.
When the roof cells do not share a unit (i.e. they're opposite corners of the rectangle), it may still be possible to remove Deadly Pair candidates from the roof cells. If one or other of the Deadly Pair only appear in their rows / columns in cells forming corners of the rectangle, that possibility can be eliminated from the roof cells.
|
4
6
8
|
3 | 5 |
7
8
|
4
7
|
1 | 9 | 2 |
4
6
|
|
4
6
|
1 |
4
7
9
|
3
5
9
|
2 |
4
5
6
9
|
3
5
7
|
5
6
|
8 |
|
2
4
6
8
|
2
7
9
|
2
4
7
9
|
3
5
8
9
|
3
4
5
|
4
5
6
9
|
3
5
7
|
1 |
3
4
6
|
|
2
4
5
|
2
5
|
3 |
2
9
|
8 |
4
9
|
6 | 7 | 1 |
| 1 | 8 |
4
7
|
6 |
4
5
7
|
4
5
|
2
3
|
9 |
2
3
|
| 9 |
2
7
|
6 |
2
7
|
1 | 3 | 4 | 8 | 5 |
| 3 | 4 | 8 | 1 | 9 | 7 |
2
5
|
5
6
|
2
6
|
| 7 | 6 | 1 |
3
5
|
3
5
|
2 | 8 | 4 | 9 |
|
2
5
|
2
5
9
|
2
9
|
4 | 6 | 8 | 1 | 3 | 7 |
5 only occurs as a possibility in row D in D1 and D2. Likewise in row J, the only locations that can take a 5 are J1 and J2. This is also true for column 1 (D1 and J1) and column 2 (D2 and J2). Assigning 5 to either D1 or J2 would create the Deadly Pattern while 2 would not. This allows us to eliminate 5 from the two roof cells, leaving 5s as hidden singles in D2 and J1.
If none of the previous patterns work, find (one of) the bi-value cells in the rectangle and look at the cell forming the opposite corner of the rectangle. If one of the Deadly Candidates in that cell forms strong links in its row and column, the other candidate may be eliminated from that cell.
| 5 |
1
6
7
|
1
6
|
2 | 4 |
1
3
|
8
9
|
8
9
|
3
7
|
|
1
4
|
1
4
7
8
|
1
4
8
|
6 |
1
3
|
9 | 5 | 2 |
3
7
|
| 3 | 2 | 9 | 7 | 5 | 8 | 6 | 4 | 1 |
|
1
2
6
|
1
4
6
|
7 |
1
4
8
|
6
8
|
5 |
2
4
|
3 | 9 |
| 8 | 9 |
3
4
6
|
3
4
|
2 |
3
4
6
|
1 | 7 | 5 |
|
1
2
|
5 |
1
3
4
|
1
3
4
|
9 | 7 |
2
4
|
6 | 8 |
|
6
9
|
6
8
|
5 |
3
4
8
|
7 |
3
4
|
3
8
9
|
1 | 2 |
|
1
4
9
|
3 |
1
4
8
|
5 |
1
8
|
2 | 7 |
8
9
|
6 |
| 7 |
1
8
|
2 | 9 |
1
3
6
8
|
1
3
6
|
3
8
|
5 | 4 |
Rather pleasingly, there are two type 1 hidden unique rectangles here, centred on E4. The first is E4:E3:F3:F4 while the second is E4:E6:G4:G6. In the first, F3 is the opposite corner cell and there are two strong links to the 3 from E3 and F4. To prove that this forces the 4 out of F3, disprove the null hypothesis. Setting F3 to 4 forces 3 into E3 and F4 because of the strong links. In turn, this leaves 4 as the only candidate in E4, creating the Deadly Pattern. Since this is impossible, we can conclude that 4 cannot be the solution in F3.
The other rectangle is slightly anomalous because there are two bi-value cells (in most hidden unique rectangles, there is just one). The strong links to the 4 in G6 from E6 and G4 force the 3 out of G6 leaving a naked single behind.
The type 2 hidden rectangle has two floor cells and two roof cells with floor cells and roof cells paired up and sharing either rows or columns. If one of the Deadly Candidates form a strong link between one pair, but not the other, the other candidate may be eliminated from the other roof cell. An example will make this clearer.
| 5 |
1
6
7
|
1
6
|
2 | 4 |
1
3
|
8
9
|
3
8
9
|
3
7
9
|
|
1
4
|
1
4
7
8
|
1
4
8
|
6 |
1
3
|
9 | 5 | 2 |
3
7
|
| 3 | 2 | 9 | 7 | 5 | 8 | 6 | 4 | 1 |
|
1
2
6
|
1
4
6
|
7 |
1
4
8
|
6
8
|
5 |
2
4
|
3
9
|
3
9
|
| 8 | 9 |
3
4
6
|
3
4
|
2 |
3
4
6
|
1 | 7 | 5 |
|
1
2
|
5 |
1
3
4
|
1
3
4
|
9 | 7 |
2
3
4
|
6 | 8 |
|
6
9
|
6
8
|
5 |
3
4
8
|
7 |
3
4
|
3
8
9
|
1 | 2 |
|
1
4
9
|
3 |
1
4
8
|
5 |
1
8
|
2 | 7 |
8
9
|
6 |
| 7 |
1
8
|
2 | 9 |
1
3
6
8
|
1
3
6
|
3
8
|
5 | 4 |
If the layout looks familiar, it's the move just before the previous example. The floor and roof cells are aligned on columns 8 and 9. The Deadly Candidates are 3 and 9. In column 8, 3 can only go into D8 and A8 so the two possibilities are strongly linked. As for the logic of this eliminating the 9 from A9, it is easy to disprove the null hypothesis. Setting A9 to 9 forces a 3 into D9 which in turn forces a 9 into D8. A8 is now the only cell in column 8 which can take a 3, creating the Deadly Pattern. Therefore, we can assume that 9 cannot be the solution to A9 and eliminate it. We could also use the strong links between the 9s in A9 and D9 to eliminate the 3 in A8, but as it happens, the naked single that is left in D9 will remove it anyway.
This is an extension of the previous bent sets we've seen (XY-Wings and XYZ-Wings). The pattern is four cells in two units with one cell able to see the other three, four possibilities across the four cells with one non-restricted common digit. What this means is one possibility where not all the cells containing that possibility are visible to each other. If there are two such digits, the pattern is invalid and you can look elsewhere. An important thing to note is that the hinge cell need not (and, indeed, is unlikely to) contain all four possibilities. The cells containing the non-restricted common digit form the pincers and you can eliminate that possibility from any cell able to see all of them.
|
2
6
9
|
7 |
2
3
4
6
9
|
3
4
8
|
3
4
8
9
|
1 | 5 |
2
6
9
|
3
6
9
|
| 1 | 8 | 5 |
2
3
|
2
3
6
9
|
2
6
9
|
3
9
|
7 | 4 |
|
2
6
9
|
4
6
9
|
2
3
4
6
9
|
3
4
7
|
3
4
5
9
|
4
5
7
9
|
8 |
1
2
6
9
|
1
3
6
9
|
| 8 |
6
9
|
7 |
1
3
|
3
5
6
|
5
6
|
2 | 4 |
1
3
5
9
|
| 4 | 2 |
6
9
|
1
3
8
|
7 |
5
6
8
|
3
9
|
1
9
|
1
3
5
9
|
| 3 | 5 | 1 | 9 |
2
4
|
2
4
|
6 | 8 | 7 |
|
2
6
7
|
1 |
2
4
6
|
5 |
2
4
8
9
|
2
4
8
9
|
4
7
|
3 |
6
8
9
|
|
7
9
|
4
9
|
8 | 6 | 1 | 3 |
4
7
|
5 | 2 |
| 5 | 3 |
2
4
6
|
2
4
7
8
|
2
4
8
9
|
2
4
7
8
9
|
1 |
6
9
|
6
8
9
|
What do you know? Two at once! Both sets have 9 as the non-restricted common digit. It is present in all four cells of the A9:B7:G9:J9 set but in only two of the G3:J3:H2:E3 one. The latter set provides the more useful elimination, however, since the naked single left behind makes the rest of the puzzle trivial. As to how the hinge forces a 9 into one or both of the pincers, consider G3. if it's 4 or 6, a 9 will be forced into one of the pincers directly. If it's 2, only 4 and 6 are left in J3 and this will force a 9 into one or other of the pincers.
If you're so inclined, you can search for a set with five possibilities in five cells - the VWXYZ-Wing.
The basic strategies are pretty obvious once you you've applied them a couple of times, but the more advanced strategies might require some reinforcement. The following sample games all use examples from this page at some point.
Medium: requires identification of two pointing pairs,
a naked triple and the hidden triple from the example.
Medium-Tough: contains the second X-Wing example and the first
sample XY-Wing. There's also another X-Wing, but you don't need to spot it to solve
the puzzle.
Tough: contains a naked triple, the first sample X-Wing
and the two simple colouring examples.
Tough: contains the second swordfish example and requires
you to spot the subsequent XY-Wing to solve.
Tough: a gentle introduction to diabolical strategies. This
requires at least one X-chain / X-cycle to solve (and you can use the OFF-OFF
discontinuity example if you want, although there are others) but is otherwise
straightforward.
Tough: the only tough bit is the WXYZ-Wings and since they're
both in the example, this puzzle is actually quite easy.
Diabolical: XY-chains everywhere! This puzzle uses
both XY-chain examples with a bunch more between the two and one after.
Diabolical: this puzzle contains the type 1 unique rectangle
example and the first type 4 example together with the rule 2 3D medusa.
Diabolical: this puzzle contains both hidden unique rectangle
examples, one after the other.