Arrow · Medium

Arrow Combinations in Arrow Sudoku

List every way an arrow can add up from its candidates. A digit in none of them goes.

The idea

When the bounds are not enough, list every way the arrow can add up, using only the candidates in its cells. A candidate that is in none of the ways cannot be right.

Two cells on an arrow that share a row, column or box cannot hold the same digit, so some ways drop out. Cells that share none of them may repeat a digit, so 1 + 1 can count.

Worked examples

2 positions from real Arrow puzzles on this site, the familiar case first. Step through each one: the board lights what each step talks about, and the red lines show what rules what out.

Two cells that share a house

Medium puzzle · Arrow Combinations

Two cells on the lit arrow are in the same row, column or box, so they cannot hold the same digit.

Step 1 of 5Step 2 of 5Step 3 of 5Step 4 of 5Step 5 of 5

A digit can repeat

Medium puzzle · Arrow Combinations

The two cells on the lit arrow share no row, column or box, so they may hold the same digit.

Step 1 of 5Step 2 of 5Step 3 of 5Step 4 of 5Step 5 of 5

Where it fits

MediumArrow Combinations is a Medium technique: Medium Arrow puzzles are the first to need it.

The Arrow overview explains the rule and the levels.

Try it on a Medium Arrow puzzle. Next Step points it out when it is the move.

The worked examples, every step

Two cells that share a house

Medium Arrow puzzle · Arrow Combinations

  1. Two cells on the lit arrow are in the same row, column or box, so they cannot hold the same digit.
  2. List every way the arrow from R6C8 can add up, using only the candidates in its cells. R5C9 and R4C8 share a row, column or box, so they cannot repeat a digit.
  3. Reading R5C9 + R4C8 = R6C8, there are 2: 4 + 2 = 6 and 7 + 2 = 9.
  4. None of them puts 7 in R6C8. Eliminate 7 from R6C8.
  5. Done: 7 is eliminated from R6C8.

A digit can repeat

Medium Arrow puzzle · Arrow Combinations

  1. The two cells on the lit arrow share no row, column or box, so they may hold the same digit.
  2. List every way the arrow from R6C5 can add up, using only the candidates in its cells.
  3. Reading R6C6 + R7C7 = R6C5, there are 5: 1 + 1 = 2, 1 + 5 = 6, 1 + 8 = 9, 2 + 1 = 3 and 2 + 7 = 9.
  4. None of them puts 5 in R6C5 or 3 or 6 in R6C6. Eliminate 5 from R6C5 and 3 and 6 from R6C6.
  5. Done: 5 is eliminated from R6C5; 3 and 6 are eliminated from R6C6.

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