Box-line reduction: when a line claims a digit for one box

What is box-line reduction?

Choose a row or a column and a digit that is still missing from it. Find the empty cells of that line where the digit is a candidate. If all of them lie inside the same 3×3 box, you have a box-line reduction. There can be two such cells or three, since a line crosses a box in exactly three cells.

The consequence is that the digit is locked into the part of the box that the line passes through. Every other cell of that box, off the line, can lose the digit as a candidate.

Why the elimination is safe

The line needs the digit exactly once, and all its possible places are inside one box. So the box’s copy of the digit must be on that line; there is no other way to satisfy the line. Because the box can contain the digit only once, the box cells off the line cannot hold it. As with every locked candidates move, you do not need to know which cell of the line is correct.

How to find it

With complete pencil marks, scan each row and each column one digit at a time. For each digit, ask whether its candidates in that line fit inside one box. Rows and columns that already have five or six digits placed are the best places to start, because they leave fewer candidates to check.

A quicker check comes from the way lines cross boxes. A row passes through three boxes. If a digit is already placed, or blocked, in two of those three segments, the remaining segment claims it. Look for rows or columns where two of the three box segments are full or nearly full.

Once you find a claim, look at the other six cells of the box. If none of them has the digit as a candidate, nothing changes. If some do, erase it, and then check those cells for a naked single and the box for a hidden single.

Worked example

In the example board, look at row 7. Its empty cells are R7C5 (row 7, column 5) with candidates 1 and 7, R7C7 with 1, 3 and 5, R7C8 with 1 and 7, and R7C9 with 3 and 5. Digit 5 is missing from the row, and R7C5 cannot hold it because of the 5 in R3C5. So 5 can only go in the two highlighted cells, R7C7 and R7C9.

Both of them are in box 9, the bottom-right box. Row 7 therefore claims 5 for box 9: the box’s 5 must be in row 7, so the other cells of box 9 cannot be 5. Box 9 has two other empty cells, R8C8 with 1, 7 and 9, and R9C7 with 5 and 9. Removing 5 from R9C7 leaves only 9.

R9C7 is a naked single and must be 9, shown in gold. Before the reduction, 9 could also go in R9C4 in the same row and in R2C7 or R3C7 in the same column, so nothing pointed to this cell. The claim from row 7 did.

Common mistakes

Getting the direction backwards. In box-line reduction you look at a line and eliminate inside the box. In a pointing pair you look at a box and eliminate along the line. If you erase from the wrong place, you can remove the correct digit and make the puzzle unsolvable.

Overlooking one candidate on the line. The claim only holds if every candidate for the digit in that line is inside the box. One stray candidate in another box breaks it.

Removing the digit from the claiming cells themselves. The two or three cells on the line keep the candidate; only the box cells off the line lose it.

When to use it and what comes next

Use box-line reduction together with pointing pairs whenever singles run out. Quietdoku’s Medium puzzles need exactly these two forms of locked candidates on top of singles, and a hint will name locked candidates when that is the next step.

After this, the natural next steps are naked pairs in Hard puzzles and, later, the X-Wing, which applies the same “the digit must be on one of these lines” reasoning across two rows at once.