intresting grid:

as daj put it the

xy-wing

you are trying to build doesnt remove any candidate from the marked cell.

because it is not where the xy wing eliminates the candiate:

as the red cell doesnt directly see the same dgit elimination from all views of the marked cells.

think of it this way

i find its the easiest:(when looking for them as well)

if you have a pattern of 3 sperate cells the

middle connection

when active as

a or b,

the cells it see = the same digit expressed

the elimination is in a cell that see both the wings and is the expressed digit.

example

- Code: Select all
`B(12) -> X`

| |

| v

A(23) - (C)31

when

A=2 => B=1

A=3 => C=1

thus X <>1

the same thing applies to both C & D

in all cases x <>1

am example of this in work is shown above by daj as well.

there is however a complex move:

als-xy

that does accomplish what you are asking though.

{there is many varations that could be used here is an example}

A=r1c17 - {259}, B=r7c1,r9c3 - {579}, C=r8c17 - {279}, X,Y=2,7, Z=9 => r1c3<>9

another way to do it is an

xy-chain:

(as noted by daj) here is the easiest one.

(both operate off the same cells)

9 r9c3 -5- r9c7 -9- r8c7 -2- r1c7 -5- r1c1 => r1c3,r8c1<>9

5 r9c7 -9- r8c7 -2- r1c7 -5- r1c1 -9- r8c1 -7- r7c1 => r7c9,r9c3<>5

or: easier approaches..

any one of the following solves the puzzle as well.

Skyscrapers:

5 in r1c1,r2c9 (connected by r7c19) => r1c7,r2c3<>5

5 in r2c9,r9c7 (connected by r29c3) => r1c7,r7c9<>5

5 in r2c3,r7c1 (connected by r27c9) => r1c1,r9c3<>5

5 in r7c1,r9c7 (connected by r1c17) => r7c9,r9c3<>5

or

Empty rectangles:

5 in b1 (r9c37) => r1c7<>5

5 in b1 (r19c7) => r9c3<>5

or:

2-string kytes

5 in r1c7,r7c1 (connected by r7c9,r9c7) => r1c1<>5

5 in r1c1,r9c7 (connected by r7c1,r9c3) => r1c7<>5

5 in r2c9,r9c3 (connected by r7c9,r9c7) => r2c3<>5

5 in r2c9,r7c1 (connected by r1c1,r2c3) => r7c9<>5

5 in r2c3,r9c7 (connected by r1c7,r2c9) => r9c3<>5

Some do, some teach, the rest look it up.