Rules

Killer Sudoku

Killer Sudoku (also known as Sumdoku or Samunamupure) is a puzzle that's similar in many ways to Sudoku, but with some added challenges. For more information about Sudoku, see this wikipedia article.

As with Sudoku, the objective of Killer Sudoku is to fill the grid with the numbers 1 to 9 such that each row, column and nonet (3x3 group of cells) contains each number only once. As well as this, a Killer Sudoku grid is divided into a number of cages, shown with dashed lines. The sum of the numbers in a cage must equal the small number in its top-left corner. The same number cannot appear more than once in a cage.

Here's an example Killer Sudoku puzzle:

Greater-Than Killer Sudoku

Greater-Than, or Comparison, Killer Sudoku has exactly the same rules as Killer Sudoku, except that not every cage has a sum in its top-left corner. Instead, some cages are linked together with symbols describing whether their sums are less than (<), equal to (=) or greater than (>) each other.

Here's an example Greater-Than Killer Sudoku puzzle:


Hints & Tips

Single Combinations

A good way to start solving a Killer Sudoku puzzle is to look for single combinations - cages that can only contain a single combination of numbers. For example, look for the 2-cell cage that sums to 3 on the right-hand side of the puzzle above. The only two numbers that can possibly appear in the cage are 1 and 2, as they are the only two numbers that sum to 3.

Although we don't know which cell contains the 1 and which cell contains the 2, we can deduce (using the rules of Sudoku) that neither 1 nor 2 can appear anywhere else in the middle row or the middle-right nonet. Finding single combinations like this can often help to narrow down the possibilities and "find a way in" to the puzzle.

The 45 Rule

This is probably the most widely used strategy. As we know, the rules of Sudoku state that each row, column and nonet must contain the numbers 1 to 9 exactly once. The numbers 1 to 9, when added together, make 45, so the numbers in each row, column and nonet, when added together, must also equal 45.

Look at the middle column in the puzzle above. It contains four 2-cell cages, with sums of 7, 9, 9 and 11. Only one cell, in the middle of the puzzle, is not in one of these cages. We can calculate the sum of all of the numbers in these cages by summing the cages themselves: 7 + 9 + 9 + 11 = 36. The 45 rule says that the numbers in the column must sum to 45, so the middle number must be 45 minus 36, which is 9.

The 45 rule doesn't only apply to single rows, columns or nonets. The sum of all numbers in two columns must be 45 + 45, which is 90. The sum of all numbers in three nonets must be 45 + 45 + 45, which is 135, and so on. However, whatever the size of the section, the 45 rule will only work if you can find a block of cages that you can sum and compare with a set of rows, columns or nonets.

Impossible Combinations

Sometimes having certain combinations of cages grouped together can give you a clue as to their contents. For example, look at the two cages below. The possible combinations of numbers in the 6-cage are 1 & 5 and 2 & 4. The possible combinations in the 5-cage are 1 & 4 and 2 & 3.

However, if the 6-cage were to contain 2 & 4, in whichever order, then neither of the combinations in the 5-cage would fit. Because of this, the 6-cage must contain the numbers 1 & 5. Likewise, the 5-cage must contain the numbers 2 & 3.

These are just a few solving strategies to help you get started. To see how they're put into use, take a look at the Solving An Example Puzzle section below. There are plenty more strategies to discover (and invent!) as you go along. Good luck!


Solving An Example Puzzle

This example should give you a good idea of how these and other techniques are used in practice. A variety of strategies are used to solve the puzzle.

Let's solve the example puzzle above. To start with we use the 45 rule. We can calculate the sum of the cells highlighted in green by summing their cages: 7 + 9 + 9 + 11 = 36. The 45 rule tells us that the cells in the middle column sum to 45, so the middle cell must be 45 - 36 = 9.

Next, we look at the highlighted 2-cell cage that sums to 16. This cage only has a single combination, 7 & 9. The rules of Sudoku tell us that the 9 cannot be in the middle nonet, as it already contains a 9, so the 7 must be on the right and the 9 on the left.

Let's use the 45 rule again. We calculate the sum of the cells highlighted in green by summing their cages: 9 + 13 + 13 + 16 = 51. The 45 rule tells us that the cells in the bottom-right nonet sum to 45, so the 'extra' cell that isn't in the nonet must be 51 - 45 = 6.

Likewise, we can calculate the value of the symmetrically opposite cell (at the top of the puzzle) by summing the cages highlighted in purple and subtracting 45: 17 + 12 + 10 + 9 - 45 = 3.

Now look at the left-hand nonet. Here we can calculate the sum of the highlighted cells by summing the cages, 12 + 15 + 7, and the completed cell, 9, giving 43. Using the 45 rule again (I told you it was probably the most widely-used strategy!), we can deduce that the remaining cell is 45 - 43 = 2.

This time we look at a larger section. The highlighted cells show a block of cages that sum to 175. The 45 rule tells us that each column sums to 45, from which we can deduce that four columns must sum to 4 x 45, which is 180. There is only one cell in these four columns that isn't in the highlighted section, so we can calculate its value by subtracting 175 from 180, giving 5.

We now have only one empty cell in the 20-cage in the middle. Its value is the cage sum, 20, minus the other cells in the cage: 20 - 9 - 5 = 6.

Now let's look at the combinations that can fit into the cage highlighted green. Its sum is 14, which means its possible combinations are 5 & 9 and 6 & 8. However, there's already a 6 in the same column as the cage, so the rules of Sudoku tell us that it cannot contain a 6. Given this, it must contain 5 & 9. We don't yet know in which order, so we enter both values using 'pencil marks'.

As this cage now contains a 5, the rules of Sudoku tell us that no other cells in the column can contain a 5. This, and the 5 that's already in the middle nonet, tell us that none of the cells highlighted yellow can contain a 5. So, what about the 11-cage in the middle at the bottom. Can that contain a 5? The answer is no, because if it did, it would also have to contain a 6 (as its sum is 11) and there's already a 6 in the same nonet.

Given this, the only cell in the nonet that can contain a 5 is the one highlighted purple. This cell is in a 2-cell cage that sums to 9, so we can deduce that the other cell in the cage must be 4.

We can use a similar technique to find the next number. The 2-cell cage highlighted green has only a single combination, 7 & 9. Again we don't yet know the order of the values in the cage, so we enter them using pencil marks.

As this cage contains a 7, the cells highlighted yellow cannot contain a 7. What's more, the 2-cell cage at the top cannot contain a 7, as its total sum is 7.

That only leaves one cell in the nonet that can contain 7, and that's the one highlighted purple. This cell is in a 2-cell cage that sums to 9, so we can deduce that the other cell in the cage must be 2.

The highlighted 2-cell cage sums to 10, so its possible combinations are 1 & 9, 2 & 8, 3 & 7 and 4 & 6. However, none of 1 & 9, 3 & 7 or 4 & 6 will fit because of the 6, 7 and 9 that are already in the column, so the cage must contain 2 & 8. The rules of Sudoku tell us that the 2 cannot be at the top, so the 8 must be at the top and the 2 at the bottom.

The 2-cell cage highlighted green sums to 7, so its possible combinations are 1 & 6, 2 & 5 and 3 & 4. However, neither 2 & 5 nor 3 & 4 will fit because of the 2 and 4 that are already in the row, so the cage must contain 1 & 6. We don't yet know the order, so we enter them using pencil marks.

Now look at the 2-cell cage highlighted purple. It sums to 8, so its possible combinations are 1 & 7, 2 & 6 and 3 & 5. However, neither 1 & 7 nor 2 & 6 will fit because of the 1 and 2 that are already in the row, so the cage must contain 3 & 5. The rules of Sudoku tell us that the 5 must be on the right, so the 3 must be on the left.

This leaves just one cell remaining in the nonet, which is highlighted yellow. It must contain a 1, as that is the only number that doesn't yet appear in the nonet. This cell is in a 2-cell cage that sums to 5, so we can deduce that the other cell in the cage must be 4.

Now look at the right-hand nonet. We can calculate the sum of the highlighted cells by summing the cages, 10 + 3 + 19, and the completed cell, 5, giving 37. Using the 45 rule, we can deduce that the remaining cell is 45 - 37 = 8.

The cells highlighted yellow contain the values 1, 2, 3, 4, 5, 6 and 8, so the final two cells in the row, highlighted green, must contain the values 7 and 9, although we don't yet know in which order.

However, although we don't know the order, we do know that they sum to 16. We also know that these two cells, along with the cell highlighted purple, make up a 3-cell cage that sums to 19. For this reason, the cell highlighted purple must be 19 - 16 = 3.

We can calculate the sum of the highlighted cells in the middle row by summing the cages, 15 + 20 + 3, and the completed cell, 3, giving 41. Using the 45 rule, we can deduce that the remaining cell is 45 - 41 = 4.

The 2-cell cage highlighted green only has a single combination, 1 & 2. We don't yet know in which order the numbers appear, so we enter pencil marks.

That leaves the 2-cell cage highlighted purple. Its possible values are 1 & 9, 2 & 8, 3 & 7 and 4 & 6. However, none of 1 & 9, 2 & 8 or 3 & 7 will fit because of the 1, 2, 3, 7, 8 and 9 that are already elsewhere in the nonet, so it must contain 4 & 6. Again we don't yet know in which order the numbers appear, so we enter pencil marks.

Now we can use a variant of the 45 rule. We can calculate the sum of the cells highlighted green and yellow by summing the cages contained by them: 11 + 10 + 19 + 16 = 56. The 45 rule tells us that the cells highlighted green and purple sum to 45. So, the green cells plus the yellow cells equal 56, but the green cells plus the purple cell equals 45. From this we can deduce that the sum of the yellow cells is 11 (i.e. 56 - 45) more than the value of the purple cell.

Let's now look at the possible combinations within the yellow cells. They are 4 & 7, 4 & 9, 6 & 7 and 6 & 9. If they contained 4 & 7, the purple cell would have to contain 0 (as the value of the purple cell is 11 less than the sum of the yellow cells), meaning that the yellow cells cannot contain 4 & 7.

If the yellow cells contained 6 & 9, which sum to 15, the purple cell would have to contain 4. There is already a 4 in the same row as the purple cell, which means that the yellow cells cannot contain 6 & 9.

So, the yellow cells can contain either 6 & 7 or 4 & 9. Both of these combinations sum to 13, so irrespective of which combination is in the yellow cells, the purple cell must contain a 2.

The cells highlighted green can contain either 1 & 9, 3 & 7 or 4 & 6. However, neither 3 & 7 nor 4 & 6 will fit because of the 4 and the 7 that are already in the same row, so they must contain 1 & 9. We don't yet know the order, so we enter pencil marks.

Now we know that the cells highlighted green must contain a 9, we can deduce that the cell highlighted purple cannot contain a 9, meaning that it must contain a 5. This means the other cell in its cage must contain a 9.

The cells highlighted green cannot contain a 9 because of the 9s that are already elsewhere in the 2nd and 3rd rows. The cells highlighted yellow cannot contain a 9 either, given the 3 that's already in the cage.

So, that leaves the cell highlighted purple as the only one in the top-left nonet that can contain a 9.

The 2-cell cage highlighted blue can contain the combinations 1 & 6, 2 & 5 or 3 & 4. However, neither 2 & 5 nor 3 & 4 will fit because of the 3, 4 and 5 that are already elsewhere in the nonet, so it must contain 1 & 6. We don't yet know the order, so we enter pencil marks.

We can calculate the sum of the cells highlighted green by summing the cages 12 + 17 and the value 9 to make 38. The 45 rule tells us that the cells highlighted yellow and purple must sum to 7 (i.e. 45 - 38).

We already know that the yellow cell must contain either 1 or 6. As the yellow cell and the purple cell sum to 7, this means the purple cell must also contain either 1 or 6.

The purple cell is part of a 2-cell cage that sums to 11. If it contained a 1, the other cell in the cage would have to contain a 10, which it obviously can't do. So, the purple cell must contain a 6. This means the other cell in its cage must contain a 5.

The 6 we entered in the previous step means that neither the cell highlighted green nor the cell highlighted purple can contain a 6.

The cell highlighted purple is in a cage containing the values 1 & 6, meaning that it must contain a 1 and the other cell in the cage must contain a 6.

The cell highlighted green is in a cage containing the values 4 & 6, meaning that it must contain a 4 and the other cell in the cage must contain a 6.

The possible combinations in the 3-cell cage containing the green, purple and yellow cells are 1 & 7 & 9, 2 & 6 & 9, 2 & 7 & 8, 3 & 5 & 9, 3 & 6 & 8, 4 & 5 & 8 and 4 & 6 & 7. However, the only combination that will fit is 2 & 7 & 8 because of the 1, 3, 6 and 9 that are elsewhere in the row and the 5s that are already in the top-middle and top-right nonets.

Neither the cell highlighted purple nor the cell highlighted yellow can contain the 2, because of the 2 that's already in the top-right nonet, so the cell highlighted green must contain the 2.

The cell highlighted purple cannot contain the 8, because of the 8 that's already in the column. So, the cell highlighted purple must contain the 7 and the cell highlighted yellow must contain the 8.

This leaves only one empty cell in the top-middle nonet, which is highlighted blue. It must contain an 8, because that's the only number that isn't already in the top-middle nonet.

The possible combinations in the 2-cell cage highlighted green are 1 & 8, 2 & 7, 3 & 6 and 4 & 5. However, none of 1 & 8, 2 & 7 or 4 & 5 will fit because of the 5, 7 and 8 that are already elsewhere in the column. So, the cage must contain 3 & 6. We don't yet know the order, so we enter pencil marks.

The two remaining cells in the top-right nonet, highlighted purple and yellow, must contain the values 3 and 4, as they are the only two numbers that don't already appear in the nonet.

As one of the cells highlighted green must contain 3, this means the cell highlighted purple cannot contain 3, so it must contain 4. This means the cell highlighted yellow must contain 3.

The cell highlighted green can contain the values 1, 2 or 9, because the values 3, 4, 5, 6, 7 and 8 are already elsewhere in the column.

If it contained a 1, the cell highlighted blue would have to contain a 6, which it can't do because of the 6 that's already in the cage. If it contained a 9, the cell highlighted blue would have to contain -2, which it obviously can't do! Given this, the cell highlighted green must contain a 2, meaning that the cell highlighted blue must contain a 5.

We already know that the cell highlighted purple must contain either 1 or 2. Because of the 2 we just entered in the cell highlighted green, we now know that it must contain a 1. This means the other cell in its cage must contain a 2.

Likewise, we know the cell highlighted yellow must contain either 1 or 9. Because of the 1 we just entered in the cell highlighted purple, we now know that it must contain a 9. This means the other cell in its cage must contain a 1.

The possible combinations in the 2-cell cage highlighted green are 4 & 9, 5 & 8 and 6 & 7. However, neither 5 & 8 nor 6 & 7 will fit because of the 5, 6 and 8 that are already elsewhere in the column, so it must contain 4 & 9. We don't yet know the order, so we enter pencil marks.

We already know that the cell highlighted purple must contain either 7 or 9. Because we now know that the cage highlighted green must contain a 9, we know that the cell highlighted purple must contain a 7. This means the cell highlighted yellow must contain a 9.

The 3-cell cage containing the cells highlighted green and purple has only a single combination, 1, 2 & 4. Neither of the purple cells can contain the 2 because of the 2s that are already elsewhere in their columns, so the green cell must contain the 2.

That means the purple cells must contain the 1 and the 4. We don't yet know the order, so we enter pencil marks.

Because the purple cells must contain a 4, we can now determine the order of the values in the cage highlighted yellow. In turn, because we now know the position of the 9 in the yellow cage, we can determine the order of the values in the cage highlighted blue.

The 2-cell cage highlighted purple can contain either 3 & 9, 4 & 8 or 5 & 7. However, neither 3 & 9 nor 4 & 8 will fit because of the 4 and the 9 that are already elsewhere in the column, so it must contain 5 & 7. We know that the 5 must be at the top and the 7 at the bottom because of the 5 and the 7 that are already elsewhere in the bottom two rows.

The possible combinations in the green cells are 1 & 7, 2 & 6 and 3 & 5. However, neither 1 & 7 nor 3 & 5 will fit because of the 5 and the 7 that we've just entered, so they must contain 2 & 6. We know that the 2 must be at the top and the 6 at the bottom because of the 2 and the 6 that are already elsewhere in the second and third rows.

Because the green cells must contain a 6, we can now determine the order of the values in the cage highlighted yellow.

We know that the cells highlighted yellow cannot contain a 9 because of 9s that are already elsewhere in the puzzle. So, that leaves the cell highlighted green as the only cell in the nonet that can contain a 9.

The cage highlighted purple can contain the combinations 1 & 9, 2 & 8, 3 & 7 and 4 & 6. However, none of 1 & 9, 2 & 8 or 4 & 6 will fit because of the 2, 6 and 9 that are already elsewhere in the nonet, so it must contain 3 & 7.

We know that the 7 must be at the top and the 3 at the bottom because of the 3 and the 7 that are already elsewhere in the second and third rows.

The cells highlighted green can contain 1 & 6, 2 & 5 or 3 & 4. However, neither 1 & 6 nor 2 & 5 will fit because of the 2, 5 and 6 that are already elsewhere in the row, so they must contain 3 & 4. We know that the 3 must be on the left because of the 3 that's already in the 4th column, meaning that the 4 must be on the right.

The 4 we just entered means that we can now determine the order of the values in the cells highlighted purple.

The cells highlighted green must contain 3 and 8, as these are the only two numbers that don't already appear in the nonet. We can determine that the 3 must be at the top and the 8 at the bottom because of the 3 we recently entered in the bottom row.

The 3 we just entered means that we can now determine the order of the values in the cells highlighted purple.

The cell highlighted green must contain 1, as that's the only number that isn't already in its row.

Likewise, the cell highlighted purple must contain 8, as that's the only number that isn't already in its row.

This, in turn, means that the cell highlighted yellow must contain 7.

The cells highlighted yellow, purple and green must contain the values 1, 6 and 8, as they are the only values that don't already appear in the nonet.

The cells highlighted yellow and purple cannot contain the 6 because of the 6s that are already elsewhere in the 1st and 2nd columns, so the cell highlighted green must contain the 6.

Likewise, the cell highlighted yellow cannot contain the 1 because of the 1 that's already elsewhere in the 1st column, so the cell highlighted purple must contain the 1. That means the cell highlighted yellow must contain the 8.

The cells highlighted blue must contain the values 3 and 5, as they are the only two numbers that don't already appear in the row. We can determine that the 3 must be on the left and the 5 on the right because of the 3 and 5 that are already elsewhere in the 1st and 2nd columns.

The cells highlighted green must contain the values 7 and 8. We can determine that the 8 must be on the left and the 7 on the right because of the 7 that's already in the 2nd column.

The cells highlighted purple must contain the values 4 and 5. We can determine that the 4 must be on the left and the 5 on the right because of the 4 and 5 that are already in the 2nd and 3rd columns.

The cells highlighted yellow must contain the values 1 and 8. We can determine that the 1 must be at the top and the 8 at the bottom because of the 1 and 8 that are already in the 2nd and 3rd rows.

That's it, we've solved the puzzle!