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Kakuro Solver

The Kakuro Solver is coming soon and will allow you to step through a solution for all of our puzzles.

Sudoku Solver
Kakuro Helper
In order to help you with solving Kakuro Puzzles we are providing this helper which works out all the possibles for you in a given cell/cage combination. In order to open the helper in a new window please click on the button below.
Kakuro Fig 1
Figure 1
Figure 2
Figure 2
Figure 3
Figure 3
Figure 4
Figure 4
Figure 5
Figure 5
Figure 6
Figure 6
Kakuro   Kakuro Solved

The basics:

  • The clue represents the sum of the numbers in the adjacent block of empty cells
  • The numbers 1 to 9 are used to fill the grid
  • Each block may contain only one occurrence of a digit

So how is it solved? If you've come from sudoku, then you have a head start. You will be using logic to seek out individual numbers and you will be keeping track of "candidate" numbers in each cell, just as you would in sudoku.

Candidates are the possible numbers that can go in cells. For example, in a block of two empty cells, where the clue is 3 both cells will have candidates of 1 and 2. With these small clues you may see the solution straight away, but if not, you should note the candidates in the cells.

It will have already become apparent that there is only one possible combination for the solution to the clue 3: 1 and 2. The same goes for 4 in two cells, which is 1 and 3. The clue 10 for a four-cell block will produce 1, 2, 3 and 4 - no other combination will work - and there are many more examples like these. Combinations where there is only one outcome are very important to kakuro solving and I have included a table of them.

Making a Start
Let's look at a real kakuro. Where do you start? Well, before we commence entering lots of candidate numbers there is an important rule to learn. We have already learned that there are some fixed combinations of numbers, such as 3 in two = 1 and 2, 4 in two = 1 and 3. If two of these blocks intersect and share a unique number then the point where they intersect must be that common number. If we examine the puzzle, there are quite a few of these fixed combinations:

  • 3 in two = 1, 2
  • 4 in two = 1, 3
  • 7 in three = 1, 2, 4
  • 16 in five = 1, 2, 3, 4, 6

Are there any combinations that intersect and share a unique digit? Yes, look at the bottom row of Figure 1. Here the 4 intersects with the 3 in the central black square. They both share the number 1 uniquely, so that must be the number that goes in the cell that they both share. Before we move on to find some more of these, it is important to remember that they must share the number uniquely. For example, the 16 intersects with the top 4, but they both contain 1 and 3, so the cell at the intersection could be either 1 or 3 (but that fact may in itself be a help later on).

Back at our puzzle there are two more places where 3 and 4 intersect (Figure 2), so both of the intersecting cells resolve to 1. Having solved those three squares directly, there are now some holes that can be filled as a result (Figure 3).





Reducing Combinations
Now we are into a phase of reducing combinations. Look at the last cell of the block for the 7 in Figure 4. It can only be 1 or 4, but we also know that we need a total of 10 to complete the 15 block that intersects with the last cell of the 7 block in two. Making 10 with either a 1 or 4 in it would give us either 1 and 9 or 4 and 6. The 1 and 9 combination doesn't work, because the only other square available in that 15 block intersects with the 16 block where only 1, 4 or 6 are available as candidates. So, it must be 4 and 6 and the 4 must be the last digit in the 7 block. The remaining 6 can nosw complete the 15 block. We can also complete the 7 block with the remaining 1 to complete the sum.

Locked Values
Tackling the 16 block in Figure 5, where just two cells remain, we must make 5 from a 1 and a 4 - the only digits remaining in that known combination. We can see that the second cell cannot be a 1, because there is already a 1 in the 15 block - it's a locked value - so it must be 4. And we have solved the 16 block.

All that remains in Figure 6 is the last number in the 15 block, which must be a 7 to make up the sum, and then the 5 in the 12 block, which is proved by the sum of the 6 block. Job done!



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