This is an excellent candidate eliminator. The name derives from the fact that it looks like an X-Wing - but with three corners, not four. The forth corner is where the candidate can be removed but it leads us to much more as we'll see in a minute.

Lets look at Figure 1 for the theory.

A, B and C are three different candidate numbers in a rectangular formation. Three of the corners have two candidates AC, AB and BC. The cell marked AB is the key. If the solution to that cell turns out to be A then C will definitely occur in the lower left corner.If AB turns out to be B then C is certain to occur in the top right corner. C is a*complementary* pair.

Lets look at Figure 1 for the theory.

A, B and C are three different candidate numbers in a rectangular formation. Three of the corners have two candidates AC, AB and BC. The cell marked AB is the key. If the solution to that cell turns out to be A then C will definitely occur in the lower left corner.If AB turns out to be B then C is certain to occur in the top right corner. C is a

So whatever happens, C is certain in one of those two cells marked C. The red C can be 'seen' by both Cs - the cell is a confluence of both BC and AC.

It's impossible for a C to live there and it can be removed.

In Figure 2 I'm demonstrating the sphere of influence two example cells have, marked red and blue. X can 'see' all the red cells, Z can 'see' all the blue ones. In this case there are two cells which overlap and these are 'seen' by both.

It's impossible for a C to live there and it can be removed.

In Figure 2 I'm demonstrating the sphere of influence two example cells have, marked red and blue. X can 'see' all the red cells, Z can 'see' all the blue ones. In this case there are two cells which overlap and these are 'seen' by both.

If our A, B and C are aligned more closely they can 'see' a great deal more cells than just the corner of the rectangle they make. In Figure 3 BC can see AB because they share the same box. AC can see AB because they share the same row. BC and AC can see all the cells marked with a red C where this Y-Wing can eliminate whatever number C is.

I have found an superb 'tough' Sudoku puzzle with a sequence of five Y-Wings and this illustrates teh full range of this strategy. The first three are pictured here but you can load the puzzle into the solver to see the remaining examples.

The first Y-Wing finds the AB cell in A2 which links 8 with the pair on B3 and the 3 in J2. Common to both the*pincer* cells is 4 which must go in either B3 or J2 so 4 in H3 can be removed.

The first Y-Wing finds the AB cell in A2 which links 8 with the pair on B3 and the 3 in J2. Common to both the

The second Y-Wing gets two candidates because of the alignment in column 1. The 8s in B1 and C1 can both see the cells A2 and G1 which also contain 8. These *pincer* cells are linked to the pair [4,3] in J2.

The third step is included because it shows a very neat rectangular alignment which almost mirrors the theory diagram. You couldn't ask for a clearer example. 4 must go in B1 or D6 otherwise 7 and 8 would be used up and there'd be nothing to go in B6. The 4 in D1 is the lone candidate that can see the yellow cells and should be removed.

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