Cage Splitting in Killer Sudoku

Cage splitting in Killer Sudoku takes a cage that hangs over the edge of a region and cuts it in two, giving each half its own sum. One awkward cage becomes two small ones, and small cages are where Killer Sudoku gives up its digits.

What the pattern is

Every row, column and box holds the digits 1 to 9 exactly once, so each of them totals 45. That is the 45 rule, and it is the engine behind this technique.

Pick a unit, usually a box. Add up the cages that sit completely inside it. Whatever is missing from 45 belongs to the cells of the cage that straddles the boundary. You now know that cage's inside total, and subtracting it from the cage's printed sum gives you the outside total. One cage has become two virtual cages, each with a sum you can enumerate.

When to look for it

• A box or line has exactly one cage poking over its edge, and whole cages cover the rest of its cells.
• The straddling cage has two or three cells on the inside, so the split leaves a small sum rather than a vague one.
• Single-cell innies and outies gave you nothing, because more than one cell was left over.
• A locked set sits inside the unit. You do not need the individual digits of a naked pair, only its total, and that is enough to finish the arithmetic.

How to apply it, step by step

1. Choose a unit you can account for. Its total is 45.
2. Add the sums of every cage lying entirely inside it, plus any placed digit no such cage covers.
3. Subtract that from 45. The remainder is the total of the straddling cage's cells inside the unit.
4. Subtract the inside total from the cage's own sum. The remainder is the total of its cells outside the unit.
5. Treat each half as a cage in its own right. List the digit sets that reach its sum, discard any that repeat a digit used elsewhere in the parent cage, and strike the surviving candidates from the cells.

A worked example

Box 1 (r1c1..r3c3) totals 45

  cage A  20   r1c1 r1c2 r2c1 r2c2       (inside)
  cage B  15   r1c3 r2c3 r3c3            (inside)
  cage C  17   r3c1 r3c2 | r3c4          (crosses into box 2)

  inside part of C  = 45 - 20 - 15 = 10   ->  r3c1 + r3c2 = 10
  outside part of C = 17 - 10      =  7   ->  r3c4 = 7

  pairs making 10:  1+9  2+8  3+7  4+6   (5+5 repeats)
  3+7 puts a second 7 in cage C          -> also out

  r3c1, r3c2 -> {1,2,4,6,8,9}   (3, 5 and 7 removed)

Cages A and B account for seven of the box's nine cells and 35 of its 45 points, so the two remaining cells, both belonging to cage C, must total 10. Cage C is worth 17, so its third cell is 17 - 10 = 7 and r3c4 is solved outright.

That placement then feeds back. Because cage C already contains a 7, the pair {3, 7} is dead for the inside half, on top of the impossible {5, 5}. Three candidates leave each of the two cells from a single piece of arithmetic.

Cage splitting versus innies and outies

Common mistakes

• Counting a cage twice. Only cages with every cell inside the unit count towards the inner total. A second straddler breaks the method.
• Forgetting the parent cage. The two halves still share one cage, so a digit used on one side cannot reappear on the other.
• Stopping at the inside half. The outside total is usually the more valuable of the two, especially when it covers one or two cells.
• Ignoring placed digits. A solved cell not covered by a complete inside cage still contributes to the inner total.

Where you meet it

Cage splitting is a tough-tier technique. Easy and medium Killer puzzles fall to plain combinations and single-cell innies, but from hard Killer Sudoku upwards you will meet boxes that only open once you cut a straddling cage in half. It is worth practising until the arithmetic is automatic, because the diabolical techniques above it assume you already see these sums.