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A kakuro puzzle

This kakuro puzzle comes from the newest edition of Topple, an indie puzzle magazine I run. For those who are unfamiliar with kakuro, here are some instructions:

Kakuro consists of a blank grid with sum-clues in various places. The object is to fill all empty squares using numbers 1 to 9 so the sum of each horizontal block equals the clue on its left, and the sum of each vertical block equals the clue on its top. In addition, no number may be used in the same block more than once.

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  • $\begingroup$ Is there something special about this kakuro? $\endgroup$ – Oray Jun 3 '17 at 19:26
  • $\begingroup$ It depends on what you mean by special. It's most likely not particularly hard for people who enjoy this type of puzzle, if that's what you mean. $\endgroup$ – Topple Jun 3 '17 at 19:46
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    $\begingroup$ there are thousands of kakuro on internet. if this is just a regular kakuro, it is not interesting at all... $\endgroup$ – Oray Jun 3 '17 at 19:49
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    $\begingroup$ You don't "lose the rights" to them. You still have the freedom to do whatever you want - it's just that other people may use them as long as credit is given. $\endgroup$ – Deusovi Jun 3 '17 at 20:23
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    $\begingroup$ I meant it when I said it comes down to how you define "special" when asking if this Kakuro is special. I think it's very special because I paid someone $50 to make it and get it tested. Now it might not be particularly hard or innovative, but I love that I can do my part to help the puzzle industry. Yes you can play tons of puzzles for free, but why not enjoy THIS free puzzle that actually supported someone trying to make these puzzles for a living? $\endgroup$ – Topple Jun 3 '17 at 20:57
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The final solution is:

enter image description here

Here are the steps:

First of all, focus on the top-left corner:

enter image description here
Here the only possibility for the vertical 23-run is 9 8 6. But 9 or 8 in d would render the 12-run unsolvable, so d=6. Then a+b+c=6, so they are 1,2,3 in some order. But there's no possible combination for 27 with 1 or 2, so a=3. Then the 27-run must be 3 7 8 9 in some order; but g can't be 9 or 8, so g=7, f=1; so b$\ne$1$\implies $b=2,c=1,e=2. Since the horizontal 23-run is 9 8 6, and we need 12 more for the 15-run, the only possibilities for the rest of the 15-run are 4 8 or 3 9. But in the first case, we'd need another 16 for the 22-run, forcing a 7 between a and g, which is bad since g=7. So the part is something like:
enter image description here

Now we'll solve four very similar regions. Focus on this part:

enter image description here
Clearly the only possibilities for the 12-run are 7 5 or 8 4. But the first case would imply d=1, b=7=a, which is bad. So (a,b,c,d)=(8,6,2,4). Similar strategy can be applied to the other 2x2 parts, giving:
enter image description here
Now that we know the 23-run considered in the first paragraph is 6 8 9, we can do some usual sum-chasing to arrive at the following: enter image description here

Now,

we apply some reasoning along the line "S has the only combination a1 a2 a3, but that cell can't have a1 or a2, so it's a3", to arrive at the following: enter image description here

Next, consider the top-right corner.

The only possibility for the vertical 24-run is 9 8 7 in some order, and for 30, it's 9 8 7 6 in some order. So their intersection must be 9 or 8. If it's 9, then the 16-run must be 7 9, and the 24-run would be 9 8, which means we need to complete the 19 run from 8, 9 and 2 more numbers, which is impossible. So the only possibility is: 8 9 * * and 9 7 * *. Also, since the last two asterisks are 1 2 in some order, and there's no combination for 4-cell 26-run with a 1, we get that these two rows are 8 9 7 6 and 9 7 2 1. Now some sum-chasing leads to this:
enter image description here

Now let's try the bottom-left corner.

The horizontal 23-run has to be 9 6 8, or 6 9 8; but the first possibility would mean we need to get a sum of 4 from 3 cells in the 13-run, so it has to be 6 9 8. The other three cells in the 13-run have to be 1 2 4 in some order. How 1 or 2 in that 12 run isn't possible, so the next row is 4 8.
enter image description here
Now focus on the remaining 2x4 grid. By summing by rows and then by column, we conclude that the yellow cells must add up to 13; but the only possible combination for the 11-run is 1 2 3 5 in some order, so these yellow cells must be 5 8. Now the remaining cells in the bottom row are 1 2 4 in some order, and 4 fits only in the second cell. Now sum chasing yields the following:
enter image description here

To finish things off, look at the bottom right:

it's simple to get that the 23-run is 6 8 9 and the 7-run begins with 4. Now look at the 22-run. It's first element is 1 or 2 and the third element is among 1 2 3. So the other two cells must sum up to at least 22-(2+3)=17; but that's the maximum for two cells, so equality holds, and the 22-run is precisely 2 9 3 8. Now some quick casework yields the following:
enter image description here

Now it's solved completely!

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  • $\begingroup$ That is exactly correct! $\endgroup$ – Topple Jun 4 '17 at 21:43

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