## Can this one be solved?

For fans of Kakuro

### Can this one be solved? Jeff

Posts: 708
Joined: 01 August 2005

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...
udosuk

Posts: 2698
Joined: 17 July 2005

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.
nj3h

Posts: 47
Joined: 07 July 2005

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

udosuk wrote: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'.
Jeff

Posts: 708
Joined: 01 August 2005

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...
udosuk

Posts: 2698
Joined: 17 July 2005