The New Sudoku Players' Forum

[Jeff]

[udosuk]

This is indeed a challenging kakuro puzzle!

Here is how to solve it:

Using basic techniques (sums/combinations), we could reach the following state:



Here Red+Yellow cells=30, and the Red cells cannot be 4 or 5 otherwise we'll have a naked subset of {1245}=12 forcing the remaining 2 cells to sum to 18, impossible...



Since Blue+Green cells must sum to 25-2-7-3=13, the Blue cell cannot be 9 as the Green cells sum to at least 1+4=5, so it must be 8...



Here it gets quite tricky... We can use the "difference" technique (a popular technique for killers) to work it out:

Across (Violet+Orange):
21+28+24+22=95

Down (Violet):
24+16+30+13=83

Orange=95-83=12=6+6

(Alternatively, you can work on the right hand side of the grid, so the 2 cells above the Orange cells sums to:
(8+15+13+23)-(14+18+15)=59-47=12=3+9)

Solution:


I think we probably could apply some Unique Rectangles on the bottom left area to work around the "difference" technique, but I'm not familiar in that department...

[nj3h]

Kakuro Works says it is a valid puzzle. Therefore there must be a logical way to solve it. I wish KW had a solver log, though, to see steps in some of the puzzles. KW runs out of hints and clues real fast. LOL.

[Jeff]

Thank you very much Udosuk, and sorry for being so late to reply.

Here it gets quite tricky... We can use the "difference" technique (a popular technique for killers) to work it out:

Across (Violet+Orange):
21+28+24+22=95

Down (Violet):
24+16+30+13=83

Orange=95-83=12=6+6

(Alternatively, you can work on the right hand side of the grid, so the 2 cells above the Orange cells sums to:
(8+15+13+23)-(14+18+15)=59-47=12=3+9)

The Sum-Difference technique is very powerful indeed. Thank you for sharing that.



BTW, I just found that a '7' can be fixed in r4c7 as follows:

The universal set for 27 with 4 digits consists:
27=3+7+8+9
27=4+6+8+9
27=5+6+7+9

Since r3c7 is 3 or 5, the possible combinations reduce to:
27=3+7+8+9
27=5+6+7+9

Since both combinations have 7, r4c7 must be a '7'.

[udosuk]

Yes, combination checking is very powerful too!

Alternatively, you can see you have to make 18 with r346c7...

If r4c7=8, you have to get 10 from r36c7, which is impossible, so r4c7=7...

Another method is the "subtraction combo" I used a lot in killer puzzles:

r346c7=18 comes from {35678}, with a total sum of 29...
So a pair totalling 11 has to be eliminated, which must be {38} or {56}...
Therefore 7 must be there and it could only be in r4c7...