BUG+1: Why Must You Assume a Unique Solution?

Is finding one cell with three candidates enough?

BUG+1 uses a special state in a puzzle with a unique solution, where almost all empty cells have two candidates. But the appearance of “one cell with three candidates and two in all the others” is not enough. You must also check how often each number appears in each unit and, above all, you need the assumption of a unique solution.

Uniqueness is separate from the rule that each row, column and box contains 1 through 9 once. There may be two or more completed boards that follow all those rules. Check uniqueness first for a board you enter yourself or one with givens removed.

Conditions — from the assumption to cell counts and occurrences in each unit

1. The current board must have a unique solution, and any numbers previously entered or candidates removed must also be correct.
2. Exactly one empty cell must have exactly three candidates, and every other empty cell must have exactly two.
3. If you hypothetically remove one number from the cell with three candidates, each remaining candidate number must appear exactly twice in every row, column and box where it occurs. Numbers already filled in have zero candidate occurrences in their unit.
4. Before removal, that number therefore appears three times in each of the special cell's row, column and box. Do not stop after counting three occurrences in just one row or column.

A state where every empty cell has two candidates and every candidate occurs twice in each unit where it appears is called a BUG. If that state has a solution, flipping the choices gives another solution, so it cannot have exactly one. It may also have no solution. In a board with a unique solution, that state cannot be left intact, so fill the special cell with the one extra candidate that breaks the BUG.

Example — what is the extra candidate in r7c6?

c1c2c3c4c5c6c7c8c9

r1r2r3r4r5r6r7r8r9

68

7

36

89

5

1

39

4

2

9

1

5

2

4

3

7

8

6

28

4

23

7

6

89

39

5

1

4

2

8

69

3

69

5

1

7

5

6

9

4

1

7

2

3

8

1

3

7

58

2

58

4

6

9

3

8

26

56

7

256

1

9

4

7

9

1

3

8

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6

2

5

26

5

4

1

9

26

8

7

3

★

★ r7c6 has candidates 2, 5 and 6. Every other empty cell has only two candidates. Count the positions for 6.

• Row 7: three cells, r7c3, r7c4, r7c6.
• Column 6: three cells, r4c6, r7c6, r9c6.
• Box 8: three cells, r7c4, r7c6, r9c6.

Removing 6 from r7c6 leaves two positions for 6 in each of these three units, and every other candidate's count in each unit is also 0 or 2. Backtracking has confirmed that this board has one solution. Therefore, r7c6=6. Placing 2 or 5 would be incompatible with the uniqueness assumption. After placing the number, remove candidates for 6 from the same row, column and box, then return to basic techniques.

When you cannot use it, and common mistakes

• You do not know how many solutions there are. Use this assumption only on boards whose uniqueness is guaranteed by the generator or has been checked separately. Enabling BUG+1 in a logical solver is not itself a uniqueness check.
• Two or more cells have three candidates. This does not meet this article's BUG+1 conditions.
• Cell candidate counts match, but the counts in each unit do not. This is not a BUG, so you cannot place the extra candidate.
• You have left out a candidate. A BUG shape created by incorrect notes is not valid evidence.
• You think a BUG must have exactly two solutions. What matters here is that it cannot have a unique solution. If some parts can vary independently, there may be more solutions.

A counterexample with multiple solutions — the same candidate conditions do not force 8

c1c2c3c4c5c6c7c8c9

r1r2r3r4r5r6r7r8r9

8

7

6

9

5

1

3

4

2

9

1

5

2

4

3

78

78

6

2

4

3

7

6

8

9

5

1

4

2

8

6

3

9

5

1

7

5

6

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1

7

2

38

38

1

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7

8

2

5

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6

9

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8

2

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7

6

1

9

4

7

9

1

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6

5

4

1

9

2

78

378

38

★

This counterexample was found by removing givens from a completed board produced by the generator and simplifying it with logical solving. r9c8 is the only cell with three candidates, and the number that looks like the extra candidate is 8. In row 9, 8 remains at r9c7·r9c8·r9c9; in column 8, at r2c8·r5c8·r9c8; and in box 9, at r9c7·r9c8·r9c9: three occurrences in each. Other candidates also occur 0 or 2 times, so all BUG+1 candidate conditions hold, except for uniqueness.

However, the solution count is at least 2 (the checker stops after finding two). If BUG+1 is enabled without checking uniqueness, the solver proposes r9c8=8. The following two completions also obey every row, column and box rule, so that placement is not forced. Keep all numbers in the diagram unchanged except for the seven cells below.

Empty cell	Completion A	Completion B
r2c7	7	8
r2c8	8	7
r5c8	3	8
r5c9	8	3
r9c7	8	7
r9c8	7	3
r9c9	3	8

r9c8 is 7 in A and 3 in B. The conclusion that “r9c8 must be 8” is therefore wrong. A solution with 8 also exists, but choosing one possible solution is different from a placement that holds in every solution. This counterexample is a real failure caused by multiple solutions, not a false pattern caused by missing candidates.

Practice puzzle

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r1r2r3r4r5r6r7r8r9

8

24

39

49

7

6

23

5

1

1

24

39

5

49

8

7

6

23

5

6

7

3

1

2

4

9

8

6

9

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2

3

1

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7

7

3

1

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6

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9

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5

2

8

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9

1

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39

7

2

49

8

5

6

1

34

39

5

6

1

49

7

23

8

234

4

1

8

6

2

3

5

7

9

★

★ r8c9 has candidates 2, 3 and 4. This board also has a unique solution. Which candidate should you place? Write the coordinates in the three units where it appears three times.

Show answer

r8c9=3. In row 8, 3 appears at r8c1, r8c7, r8c9; in column 9, at r2c9, r7c9, r8c9; and in box 9, at r7c9, r8c7, r8c9. Removing 3 from r8c9 leaves every candidate, including those in other empty cells, occurring twice in each unit where it appears. The uniqueness assumption means you must avoid that BUG state, so place the extra candidate, 3.

Further reading

The prerequisites are How to Use Sudoku Candidate Notes and Do You Need Trial and Error to Solve Sudoku?. Next, read Introduction to AIC: What Can Alternating Strong and Weak Links Remove?.

The example boards are original boards found using the sudoku-core generator and logical solver, with seeds, difficulty and step numbers recorded in the metadata. Definition reference: HoDoKu Uniqueness.