Kakuro is great for anybody who enjoys sudoku but is looking for something different, something a bit more challenging. And challenging it is, indeed! These puzzles can be quite daunting for the first-timer, what with every white cell being empty. Those converting from sudoku to kakuro are likely accustomed to starter clues. Well, the black squares hold all the clues we need.
Around the edges of the white squares are black squares with diagonal lines running from the top left to the bottom right. A number to the right of the line is a row number, while one to the left is a column number. Using any or all digits from 1 to 9 with no repeats, the cells in a row must have a sum equal to the row number and those in a column to the column number. For instance, the number 8 split over two cells (henceforth denoted as 8|2) can be 7 and 1, 6 and 2, or 5 and 3, but not 4 and 4. Get it? The answers are all there, sitting in those row and column numbers. It’s all just a matter of how we use them, and getting the most out of them is two parts logic, one part clever trickery. The best place to start is with unique sums
Every kakuro book has a list of unique sums. A unique sum is a row or column number that has only one possible combination of cell numbers. These include the highest, lowest, second highest and second lowest possible sums for each size. With practice, these can be memorized, their recollection becoming second nature. But here’s a list for reference in the meantime:
(3|2 = 1,2 4|2 = 1,3 17|2 = 9,8 16|2 = 9,7) (6|3 = 1,2,3 7|3 = 1,2,4 24|3 = 9,8,7 23|3 = 9,8,6) (10|4 = 1,2,3,4 11|4 = 1,2,3,5 30|4 = 9,8,7,6 29|4 = 9,8,7,5) (15|5 = 1,2,3,4,5 16|5 = 1,2,3,4,6 35|5 = 9,8,7,6,5 34|5 = 9,8,7,6,4) (21|6 = 1,2,3,4,5,6 22|6 = 1,2,3,4,5,7 39|6 = 9,8,7,6,5,4 38|6 = 9,8,7,6,5,3) (28|7 = 1,2,3,4,5,6,7 29|7 = 1,2,3,4,5,6,8 42|7 = 9,8,7,6,5,4,3 41|7 = 9,8,7,6,5,4,2)
Any possible sum over 8 cells (ranging from 36 to 44) is unique since only one number is left out. Just subtract the sum from 45 to figure out what’s missing. Similarly, a row or column with 9 cells will invariably add up to 45 and thus include every one-digit number once. If you see a puzzle that breaks this rule, you have a dud.
To get started, skim around for unique sums, focusing your attention on shorter sets. When you find one, look at what it intersects. This can tell you which entry goes where. The easiest ones are those that intersect other unique sums. For example, suppose you have a 17|2 intersecting a 16|2. Since the only common entry these two share is the number 9, that must go in their intersect cell. By simple process of elimination, you can place the 8 and the 7 in their respective positions as well.
Suppose we don’t have that convenience though. Well, outside of unique sums, many sums have numbers that absolutely must and absolutely cannot be included in them. One such example (there are far too many to list here) is that 8|3 must have a 1 in it. This is because the lowest possible 3-cell sum without a 1 in it is the sum of 2, 3 and 4, which is 9. 14|2 cannot have a 7 since no repeats are allowed, nor can it have any number below 5 since only one-digit entries are allowed.
So, let’s say we find a 4|2, and it intersects an 11|2. Since we can’t put the number 10 in a cell, the 3 would have to go where the 4|2 and 11|2 intersect. In more advanced puzzles, our best start may be that 4|2 intersecting a 6|3. The intersect point can be 1 or 3, which gives us a clue for the 6|3 but not for the 4|2. Sometimes, where we start may not be where we determine our first entry, though.
If we look at the other cells in the 6|3, we may find a more telling clue. We may find another 4|2, meaning that this one is also either 1 or 3.. This would mean that the one remaining must be 2. It’s not where we were aiming to start, exactly, but it’s a starting point nonetheless. That brings us to another point.
Keep track of remaining possibilities for each cell, especially in a long row or column, that you narrow down to two or three possibilities. If two cells have two common possibilities (or three cells have three) and they’re the only possibilities for each one, those numbers can be eliminated from all other cells in that row or column. As a group, they’ve been assigned a place; now it’s just a matter of sorting them among those cells.
For example, suppose you have 37|6 with a 30|4 running through the middle of it. One cell in the 30|4 has already been determined to be a 6. 37|6 can be 9,8,7,6,4,3 or 9,8,7,6,5,2. So, that intersect point can be 9, 8, or 7. Well, if a pair of 17|2s also intersects the 37|6, then 9 and 8 have each been designated to either of those two cells, leaving only 7 at the intersect cell of 30|4 and 37|6. It often helps to look at possibilities in multiple sets that intersect the same set. This brings us to a slightly more advanced technique.
Let’s look back at the earlier scenario of a 4|2 intersecting a 6|3. To recap, we know that the intersect point must be a 1 or a 3, but we need more information to narrow it down further. So, let’s look at what else intersects the 6|3. We could find it also crosses an 11|2, where that intersect point must be 2 or 3. It might also intersect a 10|2, fully adjacent to the 11|2. If the 11|2 intersect is a 2, the 10|2 must be a 3. Otherwise, the other entry in both would be 9. If the 11|2 intersect is a 3, the 10|2 must be a 1. This cannot be, since it eliminates both possibilities for the 4|2 intersect. Therefore, the 11|2 intersect must be a 2, and now we know the remaining entries for the 11|2, 10|2, 6|3 and 4|2. Had the 11|2 and 10|2 intersected another 2-cell set, in this case a 16|2, we could have employed another technique.
In any quadrilateral block of cells, the row sums must add up to the same as the column sums. This is another way you can check for a dud, as many sites use handmade puzzles rather than computer-generated ones. Anyway, find a block of cells with one extra sticking out from it. If this cell is off to the side, the row sums will include it, but not the column sums. Vice versa goes for if it sits just above or just below the block. The difference between the sum of the row sums and the sum or the column sums is what goes in this stray cell.
For example, please refer to this crude diagram. Seem familiar? If we add 4, 11 and 10, we get 25. This covers every cell in the picture. If we add 6 and 16, we get 22. This includes every cell except that one at the top of 4|2. Since 25 minus 22 is 3, that cell must be 3. Granted, we could also have looked at the 16|2 and determined that the 7 can’t be in the 11|2, because that would put a 4 in the 6|3. Actually, that would be the better way to do it, since it involves keeping track of fewer numbers in your head. But it’ll do for explaining the technique.
A more advanced twist on this technique, which may be necessitated in more advanced puzzles, involves a block of cells with one extra to the side and one extra to the top or bottom. The difference between the row sums and the column sums is the difference between these two cells. This can be a dead end, or it can tell you a lot. If the difference is 3, the smaller one can only range from 1 to 6, and the larger from 4 to 9. See, depending on what else we know, this may or may not tell us anything. If the larger cell, in this scenario, is in a 10|4 or 11|4, we know that it must be 4 or 5, making the other 1 or 2 respectively. If the smaller one is in a 30|4 or 29|4, then that cell would have to be 6 or 5, making the other 9 or 8 respectively.
The best way to learn any of these techniques is to put them to use. With practice, you can come to understand how and why they work, allowing you to apply them to scenarios beyond the ones covered here. Perhaps, you may even devise your own techniques. Additionally, recalling unique sums will become second nature, as will knowing what absolutely cannot or absolutely must be in a given set. The road is long and convoluted, but with a sharp eye and a sharper mind, you could become a kakuro master.