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In addition to standard Kakuro rules, the grey shaded cells must form a valid Sudoku solution.

Corrected Puzzle

One sum in this puzzle has been corrected from 28 to 23. Fortunately, it was in a location that nobody has resolved yet, so everybody's progress up to this point is still valid. My apologies.

If you are building on a previous solution, please verify that you are working with the correct grid. See this grid for the circled correction.

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  • $\begingroup$ Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :) $\endgroup$ – Rubio Apr 18 '17 at 6:28
  • $\begingroup$ @Rubio Yes, I was just giving it some time in case someone posted a more comprehensive answer. $\endgroup$ – paramesis Apr 18 '17 at 9:06
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3rd partial answer.

Looking at the right column of the top middle and bottom middle sudoku sqaures, we can find that there are only possibilities for 6 unique numbers. That is, 123456. 6 unique numbers over 6 squares means that those numbers must be in those squares (in some order). This gives enough information to place 789 in the remaining space in the middle square.

Furthermore, 1 and 5 must occur in the right column of the top middle square. That only leaves a feasible remainder of 6 and 2.

We have determined the column of the 9 in the middle square (right), and the kakuro clue now contains a 9 that overlaps the middle column in the top middle square, the the 9 must be in the left column there to complete the sudoko. By elimination, the bottom middle square, centre column must contain a 9.


More results. This was filled in by searching for where 1's are valid in the columns. I think it's right but I need to check the working again.


The final solution is

final

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  • $\begingroup$ Sorry, I don't understand where all of this comes from. "There is enough information to place 789 in the middle square." - why? "Also, the bottom middle square, centre column must contain a 9." - why? Please could you add some more explanations to this answer? $\endgroup$ – Rand al'Thor Mar 26 '17 at 19:23
  • $\begingroup$ I've updated the answer to cover where the first image comes from. $\endgroup$ – Jay Mar 26 '17 at 23:09
  • $\begingroup$ Oh, very nice! +1 :-) $\endgroup$ – Rand al'Thor Mar 26 '17 at 23:10
  • 1
    $\begingroup$ Nice one Jay! Could you provide some sort of description on how you got the final solution? $\endgroup$ – Mike Limburg Apr 6 '17 at 7:01
9
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Kakuro is a game about sums, so to progress we need to check the sum of areas which are not necessarily just rows and columns. We also know that each sudoku row, column and block has a sum of 45. With some double counting we can deduce the sum of several 3-block areas:

The areas with sum 4 and 28 can only be filled with 1/2/1 and 9/8/9 respectively.

Next we look at a huge area. One time counting the sum of all the red areas and one time counting the sum of all the blue areas.

proof that grid is unsolvable

This gives us

sadly a contradiction. Even though the areas cover the same squares, red gives a sum of 545 while blue gives a sum of 550. This makes the puzzle unsolvable. Hopefully this is just a typo from the asker.

The typo was fixed and we can go on. My pictures still use the old grid including the typo. Filling in all the 3 block areas that have a sum of 4 or 26 gives us

Now we can solve some of the kakuro hints

Continuing to the left

Next we check the 6th sudoku row. The remaining tiles outside the sudoku have to have a sum of 6 which means the four squares have to be filled with 1/2/1/2.

That's all for now. Feel free to continue from this state, but remember that the typo from 28 to 23 is not fixed in my images.

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  • $\begingroup$ You are correct. One of the sums was entered incorrectly. I have verified all sums and can assure they are now correct. Thank you. $\endgroup$ – paramesis Mar 29 '17 at 23:25
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Partial answer, just to get things started.

In this answer, set-bracket notation {a,b,c} always denotes an unordered n-tuple, and round-bracket notation (a,b,c) always denotes an ordered n-tuple. Each image is thumbnail-size but you can click on it to get the full version.

  • Consider the bottom left chunk. 7 in three must be {1,2,4}, so the very bottom left cell must be 4 (1 or 2 can't be part of 21 in three) and the two above it are {1,2}. Also, the two to the right of it must be {8,9}.

    If the middle one is 9, then the two above that are {1,2}, so the second-to-bottom row starts with {1,2}, contradiction. So the bottom row is (4,8,9), and we can swiftly fill in the rest of this little rectangle to get:

    Bottom left chunk filled in

  • A couple more random deductions. Firstly, consider the bottom right chunk. 16 in five must be {1,2,3,4,6}, and nothing smaller than 5 can be part of 22 in three, so the bottom left cell of this chunk must be 6.

    Secondly, consider the bottom second-to-left chunk. 35 in five must be {9,8,7,6,5}, and 9 or 8 can't be part of 13 in four. So the second cell of that 35 in five must be 7, 6, or 5; then the rest of the 13 in four is respectively {1,2,3}, {1,2,4}, or {1,3,4}. But nothing smaller than 4 can be part of 28 in four, so the second cell of that 13 in four must be 6 or 5 and the third must be 4.

    Couple more random bits filled in

  • Now let's add in a load of little red numbers for possibilities: where we don't know a cell's value for definite, but we can get it down to five or fewer possibilities. Most of these are quite trivially deduced: 6 in three is {1,2,3}, 7 in three is {1,2,4}, 8 in three is {1,2,5} or {1,3,4}, that kind of thing. (Leave a comment if there are any deductions you don't get in this bit, and I'll explain them better.)

    A couple more random cells and plenty of possibles

  • As excellently deduced by Karan Atree (see their answer for the details), the top second-to-right chunk can be entirely filled. After getting as far as Karan's answer did in that chunk, we can fill in a few more possibilities in red close by:

    Karan's deductions and a few more possibles

  • Consider the upper right grey block. Clearly 1 can't be in either of its top two rows, so there must be a 1 somewhere in its bottom row. This means 1 can't appear in the bottom row of either of the other two upper grey blocks.

    In the upper middle grey block, the right-hand column must be {1,2,3} or {1,2,4} or {1,2,5} or {1,3,4}. Either way, we must have a 1 somewhere in this column (and therefore in one of the upper two cells). This means 1 can't appear in the right-hand column of the centre grey block or of the lower middle one.

    Similarly, in the lower left grey block, the left-hand column must be {1,2,3} or {1,2,4} or {1,2,5} or {1,3,4}. Either way, we must have a 1 somewhere in this column, so 1 can't appear in the left-hand column of either of the other two left grey blocks.

    In the lower middle grey block, the 1 can't be in the left-hand column (1 can't be part of 32 in five) or in the right-hand column (by above), so it must be in the middle column, and that 30 in five is {9,8,7,5,1}. After all this Sudoku-ing, we have:

    More reds after Sudoku deductions

  • As argued by Jay in their answer, in the sixth column of the Sudoku block, the top three cells must be among 1,2,3,4,5 and the bottom three must be among 2,3,4,6, so the middle three must be {7,8,9}. Thus 1 and 5 must be somewhere among the top three, which means they are {1,2,5} and the cell above them is 6, with the one to the left of it being 9.

    In the fifth column of the Sudoku block, the top three cells cannot be 9 (since there's a 9 just above them) and neither can the middle three (since 9 is on the right of the centre grey block) or the seventh (since it's part of 18 in five). So one of the bottom two cells must be 9, and whichever it is can't have 6 next to it.

    After some more Sudoku deductions

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6
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Another Partial Answer,

enter image description here

Consider the cells filled above

enter image description here

Pic 1 -
Here, the 12 show at E1 requires to be made in 4. Which means, either a 1236 or a 1245. Also the 8 on D1 can be made in either 125, 134. However the 27 on B3 requires to be made in 4(C3,D3,E3,F3). D3 and E3 need to together make atleast, a 10, so that the total left to be made reduces to 17, which is possible in two (C3,F3). If E3 is a 5 (highest of 1245, making the vertical 12), D3 must be atleast 6 (cannot be a 5, since repeating numbers aren't allowed). However 8 cannot be made in 3 moves using a 6. Therefore to make the vertical 12 (E1), E3 must be 6 (by the 1236 configuration).

Pics 2 and 3 -
Now D3 can either be a 4 or a 5. It cannot be a 4, by seeing the 3rd pic. Say D3 is 4, therefore D2 is a 1 or 3. But looking at the 13 in C2, if D2 is 1, E2 has to be 3 or above (for F2 to be a 9 or lower). And if D2 is a 3, E2 must be a 1 minimum. And since the 12 in E1 requires 6 to be made now, in three moves, it requires a 1-2-3. So, now the pair D2-E2 can be a 3-1,1-3 or 3-2. Thereby making F2 a 9or8. However, the pair C3-F3 also requires a 17, therefore an 8-9 pair. So in total F2-F3 would be a 9-8 or an 8-9 and F4-F6 would need 4 to be made in 3 moves, which isn't possible. Hence going back to statement 1, D3 cannot be a 4. Phew.

Now to fill up the rest

enter image description here

GRID 1 and 2 and 3 -
D3 is a 5!! So as shown in Grid 2, D2 can be a 1 or a 2. But E2 has to be a 3 in both cases, to ensure F2 is 9 or below (to make up the 13 at C2). If D2 were 1, the F2 would be 9. Now C3-F3 together are 7-9,9-7. If F2 is 9, F3 is 7 then to reach the total 21 as shown in F1, the remaining F4-F6 need to make a 5, which again isn't possible. Therefore D2 has to be 2, E3 a 3, F2 an 8, F3 a 7 (cannot be an 9, then F4-F6 would need to make 4, again not possible), C3 a 9.

Grid 4 -
D4 is obviously a 1. Therefore E4 has to be a 2, and E5 a 1.

Grid 5 -
F4-F6 are 1-2-3 in any order. Since D4 is 1 and E4 is 2, F4 has to be a 3. Since E5 is 1, F5 has to be a 2 and F6 has to be a 1. Phew!

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Some additional progress. I failed to keep track of every specific step I made, so I have outlined only the major steps.

edit: I see Mike also posted an update while I was typing mine. I think mine contains a few additional steps w.r.t. his.

I've tried to fill out every single entry where at least one number can be removed. Adding up to e.g. 32 using 5 entries can never contain a 1, while adding up to 16 using 5 entries has to be done with {1,2,3,4,6} and so on.

The sudoku has helped with some additional elimination:
Consider e.g. the middle row of the sudoku blocks... Here we know that the rightmost row will contain a 9 in the middle block, thus the 9 in the top block must sit in the leftmost row, leaving the bottom block with a guaranteed 9 in the middle row. The Kakuro's 32 eliminates "1" for the leftmost row in the bottom middle block, so that also means there is a guaranteed "1" in the middle row of the bottom middle block.

Back to the leftmost row we now know it has 3 out of {2,5,7,8}. We have 4 entries left there to make 32-6=26 for the Kakuro. There is no way this can be done by including the 5, so this one can be eliminated. That leaves us with a guaranteed 17 total from the sudoku entries, along with the 6 we already found before. Leaves us with 9 for the bottom entry for the kakuro sum to 32.

In the middle row (bottom middle sudoku block) we now thus know that we're left with {1,5,9} allowing us to do get rid of the "5" in {5,7,8} in the kakuro entries.

If we now look at the center sudoku block, we know that the 2 empty rows will need to be filled with numbers out of {1,2,3,4,5,6}. Since we know everything in the bottom middle block, we can reduce the possibilities here. "1" and "5" must go in the left row, "2" must go in the middle row.

For the top row, middle sudoku block we can do something similar: "7" and "8" must be in the middle row over there. Along with the 9 we already had, we know that "6" cannot be in that same row due to the kakuro constraint, leaving us with either "3" or "4" as the final entry. Because of this, we can eliminate some more for the kakuro sum to 15.

Because of this, we also have enough information in the bottom row middle sudoku to reduce each entry to at most 3 possibilities.

After some additional elimination of single entries, I am now left with the following:
Current status

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  • $\begingroup$ You did nicely found that 5 just above the mid mid sudoku block. There was no possibility left on this one, totally overlooked it... (7, 8 and 9 are too large and can't fit in the 10 that's left, and a 3, 3 and 4 combination isn't allowed...) $\endgroup$ – Mike Limburg Mar 28 '17 at 12:35
  • $\begingroup$ Yeah, but now it feels like I've reached a point where I get the feeling there's just one single possibility somewhere which has to be found in order to continue eliminating. And I've been stuck at this point so long that I thought now is the time to share my progress. And apparently you thought the same at approximately the same amount of progress :) $\endgroup$ – slvrbld Mar 28 '17 at 12:41
  • $\begingroup$ Hehe.. just got some spare minutes left while working. But yes - there are I think a few options left, easily 'overlooked'. We might need to use different techniques or look very closely to each one of the kakuro numbers.. $\endgroup$ – Mike Limburg Mar 28 '17 at 12:44
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In addition to rand al'thor his/her last answer, we can update the sudoku slightly.

- In the sudoku block mid bottom: we do know there's a 2 forced on the left column of this block. When we fill that in the series of the 32 kakuro, the only possibility of this 32 is 2, 7, 8, 6, 9. When that's filled, we can almost complete the complete middle sector of the sudoku.

- The 29 kakuro, found on the top bottom of this sudoku, consists at least of a 7 and 8 now, leaving 29-7-8-9=5. 5 consists of 1/4 or 2/3, thus the only characters available in the sudoku are 3 and 4 (1 and 2 already in this block).

I removed some possibilities in the existing solution as well. Updated puzzle can be found here:

Updated

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