Lesson 35

Sue de Coq

The rule

When cells at a box-line crossing collectively have two more distinct candidates than cells, a bivalue helper outside the box in the line and another outside the line in the box can draw disjoint pairs from that pool. Every pool digit must then be placed exactly once among the crossing cells and helpers. The line helper's digits can be removed from the line outside the crossing cells and that helper; the box helper's digits can be removed from the box outside the crossing cells and that helper. Any pool digits in neither helper are confined to the crossing cells and can be removed from both the line and the box outside those cells.

Step by step

1. These 2 cells sit where the outlined box crosses a column. Together they hold 4 different candidates, two more than the number of cells, so every digit that lands in them must come from that small pool.
2. Two more helper cells take their candidates from the same pool, and each of them has only two candidates left. One helper is in the same column and its two candidates are 2, 6. The other is in the same box and its two candidates are 3, 8. The two pairs have no candidate in common. So there are now 4 cells and 4 digits: the crossing cells and the two helpers each take a different digit from the pool, and every pool digit is used exactly once.
3. However the digits fall, 2, 6 can only be in the crossing cells or in the column helper, and 3, 8 can only be in the crossing cells or in the box helper. So we can remove 2, 6 from all the remaining cells of the column, and 3, 8 from all the remaining cells of the box.