Rule of Parity in Killer Sudoku
The rule of parity in Killer Sudoku ignores the digits in a cage and looks only at whether they are odd or even. That sounds like throwing information away, but it is often the one thing a stubborn cage will tell you, and it can wipe out half of a cell's candidates in a single line of arithmetic.
What the pattern is
Adding two even numbers gives an even number. Adding two odd numbers also gives an even number. Only an odd count of odd digits produces an odd total. So the parity of a sum is fixed by how many odd digits go into it, and nothing else.
Now take any group of cells with a known total: a cage, or a group produced by the 45 rule, since every row, column and box totals 45. Classify each cell as odd-only, even-only or mixed, according to its remaining candidates. If exactly one cell is mixed, its parity is forced: the others already decide the parity of everything else, and the total decides the rest. Every candidate of the wrong parity in that cell can go.
When to look for it
- A cage or a pseudo-cage where most cells have already been reduced to
{2, 4},{1, 3, 5}or similar single-parity sets. - Larger groups than you would normally enumerate. Parity scales to eight or ten cells where a combination list would not.
- Any total from the 45 rule. Because 45 is odd, a box whose complete inner cages total an even number leaves cells that must total an odd number.
- Two-cell cages, where an odd sum means one odd and one even digit, and an even sum means both the same.
How to apply it, step by step
- Take a group of cells whose total you know, and note whether that total is odd or even.
- Look at each cell's candidates. Mark it odd, even or mixed.
- Stop unless exactly one cell is mixed. Two mixed cells give you no conclusion on their own.
- Count the odd-only cells. That count, plus the unknown cell's parity, must match the total's parity.
- Remove every candidate of the wrong parity from the mixed cell.
A worked example
This position uses the 45 rule on a group instead of a single cage. Rows D and E add up to 2 × 45 = 90. The nine cages touching them add up to 222, so their cells outside rows D and E add up to 132. The solved cells among them give 75, which leaves 57 for the 11 empty ones: F1, F2, F3, F5, F8, G1, G3, H3, J1, B4 and B5. Watch how those cells split by parity.
G1 and J1 are , F2 is and F3 is , so all four are odd. F1 , F8 , B4 , B5 , G3 and H3 hold only even digits. F5 is the only mixed cell. Four odd digits make an even sum, and even digits keep it even, so the other ten cells contribute an even total. The group sums to the odd 57, so F5 must be odd. Remove 4 and 8 from F5, which keeps 3 and 5.
The same steps work on a single cage. Once every cell but one has a settled parity, the cage total fixes the parity of the last cell. Combination logic usually reaches that cage first, which is why parity earns its place on large groups.
What the sum tells you
| Sum of the group | Number of odd digits in it | Two-cell case |
|---|---|---|
| Even | Even count: 0, 2, 4, … | Both odd or both even |
| Odd | Odd count: 1, 3, 5, … | One odd and one even |
Common mistakes
- Two mixed cells. With two unknown parities the equation has two solutions. You need exactly one mixed cell.
- Miscounting the odd cells. Only cells whose every candidate is odd count as odd. A cell holding
{3, 4}is mixed. - Forgetting placed digits. A solved cell inside the group is odd or even, and it counts.
- Expecting a placement. Parity halves a candidate set. It almost never names a digit by itself, so pair it with cage combinations to finish the job.
Where you meet it
Parity belongs to the tough tier and turns up from hard Killer Sudoku upwards, usually late in a solve when several cells have been narrowed but none is settled. It is cheap to test, so it is worth a sweep whenever a puzzle stalls with lots of two-candidate and three-candidate cells and no obvious cage to enumerate.