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Platon Sudoku · Sudoku Study

W-Wing · Step-by-step example

Two nonseeing cells hold the same pair {X,Y}; a strong link on X has one end seen by each wing. Remove the other pair digit from a candidate seeing both wings.

This course is adapted from Platon Sudoku App. Read the complete example here, then continue learning and practising in the app.

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Recognize the pattern

Two nonseeing cells hold the same pair {X,Y}; a strong link on X has one end seen by each wing.

How to find it

Find matching pairs in cells that do not see each other, then trace an exact two-place link for one pair digit through a row, column, or box.

What it allows

Remove the other pair digit from a candidate seeing both wings.

Avoid the mix-up

The matching pair alone is not enough. Check both link ends, its exact two-place condition, and the target’s sight of both wings.

Step-by-step example · Strong row

This example starts from a saved intermediate position. Its candidates include earlier logical eliminations; they are not fresh automatic notes.

Blue: evidence. Green: conclusion. Crossed-out digits: eliminated candidates. R1C1 means row 1, column 1.

Step 1 · This is a W-Wing

Start with R4C8 and R6C3: both contain only 9 and 6, so they are the two wings.

Start with R4C8 and R6C3: both contain only 9 and 6, so they are the two wings.
Start with R4C8 and R6C3: both contain only 9 and 6, so they are the two wings.
Step 2 · See the complete W-Wing

Each wing sees one end of the 6 strong link. In row 5, 6 can go only in R5C8 and R5C3, so one must be 6. Every target (R6C8=9) sees both wings.

Each wing sees one end of the 6 strong link. In row 5, 6 can go only in R5C8 and R5C3, so one must be 6. Every target (R6C8=9) sees both wings.
Each wing sees one end of the 6 strong link. In row 5, 6 can go only in R5C8 and R5C3, so one must be 6. Every target (R6C8=9) sees both wings.
Step 3 · Case 1: R5C8 is 6

If R5C8=6, it rules out 6 from R4C8. That wing must be 9. Since R6C8=9 sees R4C8, it cannot be 9.

If R5C8=6, it rules out 6 from R4C8. That wing must be 9. Since R6C8=9 sees R4C8, it cannot be 9.
If R5C8=6, it rules out 6 from R4C8. That wing must be 9. Since R6C8=9 sees R4C8, it cannot be 9.
Step 4 · Case 2: R5C3 is 6

If R5C3=6, it rules out 6 from R6C3. That wing must be 9. Since R6C8=9 sees R6C3, it cannot be 9.

If R5C3=6, it rules out 6 from R6C3. That wing must be 9. Since R6C8=9 sees R6C3, it cannot be 9.
If R5C3=6, it rules out 6 from R6C3. That wing must be 9. Since R6C8=9 sees R6C3, it cannot be 9.
Step 5 · Combine the two cases

The two cases cover the strong link, so at least one wing is 9. Every target sees both wings; remove R6C8=9.

The two cases cover the strong link, so at least one wing is 9. Every target sees both wings; remove R6C8=9.
The two cases cover the strong link, so at least one wing is 9. Every target sees both wings; remove R6C8=9.

What it allows

The two cases cover the strong link, so at least one wing is 9. Every target sees both wings; remove R6C8=9.

W-Wing · Remove the other pair digit from a candidate seeing both wings.
W-Wing · Remove the other pair digit from a candidate seeing both wings.

Step-by-step example · Strong column

This example starts from a saved intermediate position. Its candidates include earlier logical eliminations; they are not fresh automatic notes.

Blue: evidence. Green: conclusion. Crossed-out digits: eliminated candidates. R1C1 means row 1, column 1.

Step 1 · This is a W-Wing

Start with R1C7 and R5C8: both contain only 6 and 7, so they are the two wings.

Start with R1C7 and R5C8: both contain only 6 and 7, so they are the two wings.
Start with R1C7 and R5C8: both contain only 6 and 7, so they are the two wings.
Step 2 · See the complete W-Wing

Each wing sees one end of the 7 strong link. In column 9, 7 can go only in R1C9 and R5C9, so one must be 7. Every target (R4C7=6) sees both wings.

Each wing sees one end of the 7 strong link. In column 9, 7 can go only in R1C9 and R5C9, so one must be 7. Every target (R4C7=6) sees both wings.
Each wing sees one end of the 7 strong link. In column 9, 7 can go only in R1C9 and R5C9, so one must be 7. Every target (R4C7=6) sees both wings.
Step 3 · Case 1: R1C9 is 7

If R1C9=7, it rules out 7 from R1C7. That wing must be 6. Since R4C7=6 sees R1C7, it cannot be 6.

If R1C9=7, it rules out 7 from R1C7. That wing must be 6. Since R4C7=6 sees R1C7, it cannot be 6.
If R1C9=7, it rules out 7 from R1C7. That wing must be 6. Since R4C7=6 sees R1C7, it cannot be 6.
Step 4 · Case 2: R5C9 is 7

If R5C9=7, it rules out 7 from R5C8. That wing must be 6. Since R4C7=6 sees R5C8, it cannot be 6.

If R5C9=7, it rules out 7 from R5C8. That wing must be 6. Since R4C7=6 sees R5C8, it cannot be 6.
If R5C9=7, it rules out 7 from R5C8. That wing must be 6. Since R4C7=6 sees R5C8, it cannot be 6.
Step 5 · Combine the two cases

The two cases cover the strong link, so at least one wing is 6. Every target sees both wings; remove R4C7=6.

The two cases cover the strong link, so at least one wing is 6. Every target sees both wings; remove R4C7=6.
The two cases cover the strong link, so at least one wing is 6. Every target sees both wings; remove R4C7=6.

What it allows

The two cases cover the strong link, so at least one wing is 6. Every target sees both wings; remove R4C7=6.

W-Wing · Remove the other pair digit from a candidate seeing both wings.
W-Wing · Remove the other pair digit from a candidate seeing both wings.

Step-by-step example · Strong box

This example starts from a saved intermediate position. Its candidates include earlier logical eliminations; they are not fresh automatic notes.

Blue: evidence. Green: conclusion. Crossed-out digits: eliminated candidates. R1C1 means row 1, column 1.

Step 1 · This is a W-Wing

Start with R2C3 and R5C7: both contain only 8 and 1, so they are the two wings.

Start with R2C3 and R5C7: both contain only 8 and 1, so they are the two wings.
Start with R2C3 and R5C7: both contain only 8 and 1, so they are the two wings.
Step 2 · See the complete W-Wing

Each wing sees one end of the 1 strong link. In box 3, 1 can go only in R2C9 and R3C7, so one must be 1. Every target (R5C3=8) sees both wings.

Each wing sees one end of the 1 strong link. In box 3, 1 can go only in R2C9 and R3C7, so one must be 1. Every target (R5C3=8) sees both wings.
Each wing sees one end of the 1 strong link. In box 3, 1 can go only in R2C9 and R3C7, so one must be 1. Every target (R5C3=8) sees both wings.
Step 3 · Case 1: R2C9 is 1

If R2C9=1, it rules out 1 from R2C3. That wing must be 8. Since R5C3=8 sees R2C3, it cannot be 8.

If R2C9=1, it rules out 1 from R2C3. That wing must be 8. Since R5C3=8 sees R2C3, it cannot be 8.
If R2C9=1, it rules out 1 from R2C3. That wing must be 8. Since R5C3=8 sees R2C3, it cannot be 8.
Step 4 · Case 2: R3C7 is 1

If R3C7=1, it rules out 1 from R5C7. That wing must be 8. Since R5C3=8 sees R5C7, it cannot be 8.

If R3C7=1, it rules out 1 from R5C7. That wing must be 8. Since R5C3=8 sees R5C7, it cannot be 8.
If R3C7=1, it rules out 1 from R5C7. That wing must be 8. Since R5C3=8 sees R5C7, it cannot be 8.
Step 5 · Combine the two cases

The two cases cover the strong link, so at least one wing is 8. Every target sees both wings; remove R5C3=8.

The two cases cover the strong link, so at least one wing is 8. Every target sees both wings; remove R5C3=8.
The two cases cover the strong link, so at least one wing is 8. Every target sees both wings; remove R5C3=8.

What it allows

The two cases cover the strong link, so at least one wing is 8. Every target sees both wings; remove R5C3=8.

W-Wing · Remove the other pair digit from a candidate seeing both wings.
W-Wing · Remove the other pair digit from a candidate seeing both wings.

Practice

This course is adapted from Platon Sudoku App. Read the complete example here, then continue learning and practising in the app.

View on Google Play