Cage combinations

From Sudopedia
Jump to navigationJump to search

A Killer Sudoku contains cages and cage sums. Knowing the number of cells in a cage and its sum, we can determine the possible cage combinations, i.e., the possible combinations of digits that can go into the cage. For example, if we have a cage of 2 cells whose sum is 5, written 15[2], then either it contains the digits 1 and 4, or the digits 2 and 3. The cage combinations are written in set notation as {1,4} or {2,3}, or simply {14} or {23}.

Manually enumerating the cage combinations for arbitrary cages can be very tedious, but some helper programs such as SumoCue can automatically compute the cage combinations for the player. However, for some cages, the cage combinations are easy to enumerate and can lead to candidate eliminations, either simply because the candidate is not part of any combination, or in conjunction with some other technique. A suggested way of enumerating the combinations is to start from the largest possible digit and then working systematically downwards, so a cage 11[3] would be enumerated as {8,2,1} or {7,3,1} or {6,4,1} or {6,3,2} or {5,4,2}. One should be careful to list all combinations.

Example 1

For a cage 24[3], its only combination is {7,8,9}. Then we can eliminate the digits 1 to 6 from all three cells in the cage. Further, if all the three cells all belong to the same line or nonet, we can then apply Naked Triple to that line or nonet.

Example 2

Given a cage 28[4], there are two possible cage combinations: {4,7,8,9} or {5,6,8,9}. Either way, the cage must contain the digits 8 and 9 in 2 of the 4 cells. Knowing this, we might be able to apply the Locked Candidates technique. Further, if a cell whose candidates are 8 and 9 sees all 4 cells, we may be able to apply the Killer Subset technique.

Example 3

Suppose we have a cage 15[3] such that

  • Cell A has candidates 3, 4, 5.
  • Cell B has candidates 7, 8.
  • Cell C has candidates 1, 2, 3, 4, 5, 6.

We could enumerate the permutations (3+7+5 or 3+8+4 or 4+8+3 or 5+7+3 or 5+8+2) and conclude that 1 and 6 can be eliminated from the cell C.

Alternatively, we can also consider the minimum and maximum possible values of the sum A + B. Since 3 <= A <= 5 and 7 <= B <= 8, we have 10 <= A + B <= 13. Due to the constraint A + B + C = 15, we have 2 <= C <= 5. This means that we can eliminate 1 and 6 from cell C.