# Stuck on (a couple of) diagramless kakuros

I was working through OAPC 6 (http://oapc.wpc2009.org/archive.php?id=46). I was having trouble with the diagramless kakuros. The rules for this puzzle are as follows: Place the numbers $$1-n$$ (7 and 9 respectively) into the grid to form a valid kakuro. The black squares should have symmetry with respect to 180 degree rotation about the centre. The clues represent sums formed in the grid. They are in order of where the first number in the sum appears, going across the first row, then the second row, etc. Every number in the puzzle must be in a sum of at least 2 numbers in each direction (so no singletons, neither horizontally nor vertically). All white squares should be connected.

I have worked through the one on the left a bunch of times and keep running into a problem in the top right corner. I am not convinced the puzzle is possible and was hoping someone can give it a try and either confirm my suspicion or give me a hint/solution.

The second one I only worked through twice, so I am less convinced it is necessarily broken. However, I have not managed to solve it and ran into the same problem both times. So I guess for that second one, once I am asking about the first one, I am just more interested in a "yes or no" with regards to whether it is possible or not.

• The site shows a 6.2% success rate for that section, so it appears that at least a few people managed to solve it. AFAIK, it's not required that the top left corner be occupied. Have you tried setting the 3-sum squares somewhere other than top left? – shoover Nov 28 '19 at 21:15
• @shoover Your point about the top left is true. I am pretty sure I was allowing for that, although I did happen to find that the top left is forced to be occupied. As for the 6.2% success rate, these are just 2 out of 4 problems of this type and the success rate includes the other 2 which are a bit easier. – Jobo Nov 29 '19 at 3:29

Looking at the 7x7 grid, my answer is

Impossible

First I worked out the length range for each solution.

Across
3(2), 7(2-3), 6(2-3), 17(3-5), 25(5-6), 15(3-5), 18(3-5), 11(2-4), 14(3-4), 7(2-3), 9(2-3)

Down
4(2), 6(2-3), 22(4-6), 13(2-4), 5(2), 24(5-6), 14(3-4), 12(2-4), 16(3-5), 8(2-3), 8(2-3)

After some unsuccessful (even wrong) attempts to work it by logic, I resorted to making a C program to figure out the possible grids. The sequence of the Across solutions was fairly easy, but not so for the Down solutions, as they vary accoring to the starting row.

I found one grid fitting the symmetry and possible length and sequence of the lights:

The top left corner is obvious. The top right corner has these possibilities below. I showed the green cell as 1-7 although there can't be a 6 (22 is 1 2 3 4 5 7).

This allows a range for the blue cell to be found from the 17 Across clue.

Then the minimum for the green cell is 17 - 7 - 4 = 6.

Which resolves to this

Now there is a problem with the 25 Across:
25 - 3 - 5 = 17 but the only possibility for a 17 is (4 6 7).
But none of those can go in the yellow cell.

• I ran into the exact same problem. This definitely supports my suspicion that it is a real problem. Thank you so much! I will accept your answer soon if no one addresses the second part of my question! – Jobo Nov 29 '19 at 21:25
• Perhaps a typo in the puzzle. I don't have time to look at the 9x9 yet. – Weather Vane Nov 29 '19 at 21:30
• I got to the same point. The sum of the horizontal clues is equal to the sum of the vertical ones, so a typo is not likely. The grid can be filled if you allow the digits 1 to 9 according to an online solver I tried, but I don't know if that is a unique solution. – Jaap Scherphuis Nov 29 '19 at 23:08
• @Jobo I explored the 9x9 puzzle and failed to find any grid that fits the clues given, let alone try to solve them. I can't easily prove a negative though. – Weather Vane Dec 1 '19 at 19:56
• Thank you so much! I also got stuck on the grid phase. That is dissapointing; the other puzzles of this type were quite neat! – Jobo Dec 4 '19 at 15:49