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Here is a picture of the puzzle
http://i22.tinypic.com/347v2wk.gif

Can anyone tell me the secret to cracking open this puzzle? It is not like most other kakuro puzzles, which can be solved without guessing or trial and error? Is there some special strategy to solving this without trial and error or guessing?

NOTE that I am not interested in the final solution. I could get that if I want to. What I want is tips and strategies.
Thank you.

2007-11-05 10:12:39 · 4 answers · asked by Dr D 7 in Science & Mathematics Mathematics

I know that much Alex, usually I am able to solve even the so called advanced problems. But show me where in this puzzle you see any of the standard combinations you describe.

2007-11-05 10:29:06 · update #1

There are many square where you can narrow down the possibilites, but I don't see one single square that you can fill in with 100% certainty.

2007-11-05 11:39:40 · update #2

4 answers

Boring, but the top and bottom of the vertical 28-bar in the middle can be determined. Sum the acrosses of the left side, subtract the verticals of the left side, and you get 6 for that top spot. The 14,5 box in the lower middle can no longer be 8641 (because the 6's would collide) so it must be 9532.
The 2x2 block just above it then falls very easily.

(shortly later) I guess I saw the same start as the guy before me.

2007-11-06 04:54:44 · answer #1 · answered by xwdguy 6 · 3 0

I can see a way to start:

Edit: realised my instructions weren't all that clear.

Look at the last three columns of the grid. By comparing horizontal and vertical totals you can deduce, for instance, that the first four cells of the vertical 41 must add to 23 (and hence the bottom four must add to 18). Similarly, if you take all of the last three columns plus the vertical 11 in the lower section of the fourth column from right, you can show that the bottom cell in the fifth column from right must have a 4 in it (this cell has 22 to the left and 13 beneath it). Using a similar process you can establish that the cell at the top of the same column must be 6.

Then work on the 2x2 area near the bottom of that column with the 14 and the 5 going across and the 12 going down; there's now only one way to fill in these cells, and the remainder of the middle bit can then be filled in.

I'll leave the rest up to you.

2007-11-05 16:23:58 · answer #2 · answered by Scarlet Manuka 7 · 3 0

The secret is not to bother!
The puzzles from this particular site "kakuropuzzle" are very, very often practically impossible, and I'm not using the phrase in a euphomistic way. The only way to get anywhere with them, as far as I can see, is do an awful lot of guessing, so much so as to be totally impracticable.
I liked what you did with that 137/2007 problem!
Oh, by the way, I think I got 1 square definitely. Second row from the bottom, 5th column in from the right (next to the "22") is a 4. If you're interested on how I got it, write to me!
I didn't bother with the left side of the puzzle since I saw I couldn't do the right side.

2007-11-05 13:29:44 · answer #3 · answered by rrsvvc 4 · 3 1

First work out which numbers can only be made by one combination
eg 17 in two squares can only be 9 and 8
and 24 in three squares can only be 9, 8 and 6

Find the combinations and see which numbers fit in which squares until you find one which only fits in one square. Continue this as much as possible then it should be easy to work it from there.

2007-11-05 10:20:04 · answer #4 · answered by Alex P 2 · 1 1

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