BUG+1 (bivalue universal grave plus one) is a late-game rule. When every empty cell has two candidates except one cell with three, the digit seen three times in that cell's row, column and box goes there. It is valid only in a puzzle with exactly one solution.
How to spot it in play
BUG+1 turns up near the end of a hard puzzle, once every empty cell has pencil marks and no single is left. The sign is a grid of pairs: nearly every empty cell has two candidates.
Scan for a cell with three. If there is exactly one, count its candidates in its row, column and box. If one digit appears three times in all three houses and the others twice, that digit goes in.
The count is only as good as your marks, so check them first.
Find the cell with three
Seventeen cells are empty. Sixteen hold two candidates. Row seven, column three holds three: a 2, a 5 and a 9.
Count its row
Row seven has four empty cells. The 2 and the 5 each appear in two. The 9 appears in three: columns two, three and eight.
Count its column
The 2 is in rows three and seven, the 5 in rows seven and nine, and the 9 in rows four, seven and nine.
Count its box
Box seven has three empty cells. The 2 is in two of them, the 5 in two, and the 9 in all three.
Write the 9
The 9 is seen three times in all three houses, so it goes in. Strike the 2 and the 5.
Why it has to be the 9
Suppose the cell were not a 9, and strike that 9. Every empty cell has two candidates, and every digit has exactly two places in each row, column and box that still needs it. That state is a bivalue universal grave, or BUG.
A BUG never has exactly one solution. It has none, or two or more, a proven property of the pattern. A puzzle with one solution can't reach that state, so the 9 can't be struck.
Striking the 2 instead leaves row seven with one 2 and three 9s. That is not a BUG, so nothing follows. Only the digit seen three times leaves every house with pairs.
Every cell has two candidates, and each house holds each digit twice.
Row seven keeps one 2 and three 9s. This is not a BUG, so it proves nothing.
BUG+1 is valid only because the puzzle has exactly one solution. If you are solving a puzzle that might have several, such as one you made up or one from a source that doesn't check, don't use it. The 9 could be wrong.
How the in-game hint counts
The in-game hint calls this step BUG+1 and strikes the 2 and the 5. It counts each candidate of the cell with three across every empty cell: seven 9s, eight 2s and six 5s here. The digit with an odd count goes in. Both counts name the same digit, since every row but that cell's holds each digit twice or not at all.
After the 9, singles finish this grid.
Where this sits on the path
BUG+1 opens the last chapter of the path, and it is the only technique on the path that relies on the puzzle having one solution. The restricted common candidate comes next. Next come almost locked sets: groups of cells that are one cell short of a naked subset.
On a different grid, every empty cell has two candidates except row five, column nine, which holds a 2, a 4 and a 5. Which digit goes there?
Count each one in row five, column nine and box six.
Итог
- Ищите
- Pencil marks in every empty cell, no single left, and every cell holding two candidates except one with three.
- Область
- One cell: the only one with three candidates.
- Результат
- The digit seen three times in that cell's houses goes in. The other two go.
- Признак
- A grid of pairs late in a hard puzzle.
- Если не сработало
- If two cells have three candidates, or one has four, this rule doesn't apply. On a puzzle that might have several solutions, don't use it.