### ... by: S. Lee

I love this site, but I want to point out some point.Besides the chaining strategies, there is also a family of strategies that make use of the concept of 'wing'. Indeed, the following strategies

XY-Wing

XYZ-Wing

XYZW-Wing

Aligned Pair Exclusion (possibly partially)

Sue-De-Coq (possibly partially)

can be understand as special cases of the strategy named 'Bent-Naked subset' as is illustrated in Sudoku Assistant (http://www.stolaf.edu/people/hansonr/sudoku/explain.htm). This strategy admits two rules as follows:

Setting) Let A and B be a subset of two overlapping units (such as row-box). Let

Wa = [portion of A that does not overlap with B] = A-B

Wb = [portion of B that does not overlap with A] = B-A

U = [union of A and B] = A∪B

so that both Wa and Wb form two wings of U. Assume that U is 'bent-locked', in the sense that the number of candidates on the cells in U is equal to the number of cells in U.

Rule 1) If there is no common candidate between two wings Wa and Wb, then A behaves like a genuine naked subset with candidates of Wa in each unit, and likewise for B.

Rule 2) If there is a unique candidate k between two wings Wa and Wb, then for every cell that can see the candidate k in U, we can eliminate the candidate k from the cell.

The power of this strategy is that we do not care what happens in the intersection part A∩B, except that it affects the total number of the candidate in U = A∪B. Conceptually it involves neither the concept of bi-valued cells nor that of tri-valued cell, though they appear in many examples. Even this does not involve the notion of ALS. All the matter reduces to the problem of candidate counting in U, Wa and Wb.

Of course, the disadvantage of this strategy is that there is no guide for the choice of A and B. Thus it is of no harm to understand that both APE and Sue-De-Coq are special guidelines for choosing the pair (A, B).

In the 1st APE example, we actually have

Wa = {1,2}[F5]

Wb = {2,5,6}[E7|E8]

U = {1,2,5,6}[F5|E7|E8|F9]

Clearly U forms a bent-locked subset. Since there is a unique candidate, namely 2, in both wings Wa and Wb, we can eliminate the candidate 2 in F8 as F8 can see all the candidates 2 in the subset U by the Rule 2.

Other two examples can also be analyzed by exactly the same manner:

Extended APE example 1:

Wa = {5,6,9}[D9|F9]

Wb = {1,9}[G8]

U = {1,5,6,9}[D9|F9|G8|H9]

=> 9 in J9 is eliminated by Rule 2

Wa = {5,6,9}[D9|F9]

Wb = {3,6}[J7]

U = {3,5,6,9}[D9|F9|J7|J9]

=> 6 in H9 is eliminated by Rule 2

Extended APE example 2:

Wa = {1,5,9}[A2|A9]

Wb = {8,9}[B4]

U = {1,5,6,8,9}[A2|A9|B4|A4|A5]

=> 9 in A6 is eliminated by Rule 2

Sue-De-Coq example 1:

Wa = {2,8}[B2]

Wb = {3,5}[F3]

U = {2,3,5,8}[B2|F3|D2|E2]

=> By the Rule 1,

The triple A = [B2|F3|D2] behaves like a naked pair with candidates {2, 8} in column 2, eliminating 8 in C2, 2 in G2 and 8 in J2.

The triple B = [F3|D2|E2] behaves like a naked pair with candidates {3, 5} in box B1, eliminating 3 in E3.

Sue-De-Coq example 2:

Wa = {6,8}[E2]

Wb = {1,3,7}[D9|F9]

U = {1,3,6,8,9}[E2|D9|F9|E7|E8]

=> By the Rule 1,

The triple [E2|E7|E8] behaves like a naked pair with candidates {6,8} in the row E, eliminating 6 and 8 in the other cells in the row E.

The quadraple [D9|F9|E7|E8] behaves like a naked triple with candidates {1,3,7} in the box B3, eliminating 1, 3, 7 in the other cells in the box B3.

In fact, we may extend the bent-naked subset strategy to a more general setting, though it becomes slightly harder to identify, as the naked triple is harder to apply than the naked pair.

Andrew Stuart writes:

Duplicates a posting found elsewhere an left on more relevant pages.