Problem: What strategies can you find for solving KenKen puzzles? What makes a KenKen difficult?
Fill in "gimme" squares. These are regions consisting of a single cell.
Use arithmetic to find the remaining possibilities. For example, if we have a 7+ region, with 2 and 3 placed, and one cell remaining, then the last cell must be 2. If we had a 3+ with two empty cells, then the cells could be filled as (1, 2), or (2, 1). It might not be obvious which regions have few possibilities. For example, an 8+ region consisting of four cells, in a 4x4 grid, has only 3 possibilities. Namely, (1, 3, 4), (2, 3, 3), and (2, 2, 4).
Use the shape and position of the region to prune combinations. For example, a 4+ region, consisting of two cells along a single row, cannot be filled by (2, 2), and therefore must be filled by (1, 3) or (3, 1).
The aforementioned Sudoku, and KenKen strategies, are sufficient for solving this particular puzzle:
Use the row sum, or column sum, to determine the final remaining number, or possibilities. For example, in this puzzle:
We know the third column must add up to 1 + 2 + 3 + 4 = 10. But the top region in the third column reads 8+. Thus, the bottom cell must be 10 - 8 = 2. It's helpful to know the formula for the triangular numbers, so you can compute the row or column sum mentally for larger grids. That is, if Tn is the nth triangular number, then Tn = 1 + 2 + ... + n, but also, Tn = n(n + 1)/2. The latter is much easier to compute mentally than the former!
Here's another way to use the row or column sum. Refer to the following puzzle state:
Have a look at the third row. We know where the 1, 3, 4, and 5 go, so all that's left are the 2 and 6. If we place the 2 in the 15+ region, then the two remaining cells in the 15+ region must add to 13, which is impossible, since the largest number available in a 6x6 puzzle is 6. Thus, the 2 must go in the 6x region, and the 6 in the 15+ region.
Solve a system of equations. Refer to the rightmost column in the following puzzle:
The 4 and 8+ region add to 12, but the entire column adds to 21. Thus, the numbers in the 3- region must add to 21 - 12 = 9. But also, their difference is 3, and so we have x + y = 9, and x - y = 3. Adding these equations together, we get 2x = 12, and so x = 6. But x + y = 9, and so y = 3.
Save large addition regions for last. It might be easy to determine which numbers must go in those squares, but the permutation of those numbers might be very difficult to find.
Identify how many times each number can be placed, based on the shape of the region. For example, take a look at the smaller 20+ region in the following puzzle: