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| Problem Kakuro Puzzle http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=12&t=1245 |
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| Author: | nj3h [ Sun Feb 16, 2014 7:53 pm ] |
| Post subject: | Problem Kakuro Puzzle |
I cannot find any Kakuro forums that are active. I am hoping someone here can offer a suggestion to help me solve the puzzle displayed below. I have place an "8", a "4", and eliminated some combination possibilities. But that is it. Anyone see a path forward? Regards, George Attachment: Kakuro Puzzle 16 Feb 2014.jpg [ 228.17 KiB | Viewed 16127 times ] |
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| Author: | HATMAN [ Mon Feb 24, 2014 3:29 pm ] |
| Post subject: | Re: Problem Kakuro Puzzle |
George There is an innie/outie ring: i/o TR r3c6 = r5c7 + 3 i/o BL r5c3 = r7c4 + 1 i/o TL r3c6 = r5c3 i/o BR r7c4 = r5c7 + 2 r3c6 = 4/5/6 r5c7 = 1/2/3 r5c3 = 4/5/6 r7c4 = 3/4/5 Maurice |
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| Author: | Smythe Dakota [ Tue Feb 25, 2014 2:00 am ] |
| Post subject: | Re: Problem Kakuro Puzzle |
I prefer to think of these as a ring of doubularities. If you draw boxes around the two cells r3c6 and r5c7, you can see that these two cells isolate a portion of the puzzle, in the upper right corner, from the rest of the puzzle. The horizontal sum in this corner is 7+25+7+4+11, or 54. The vertical sum is 9+12+23+13, or 57. Of the two cells in this doubularity, r3c6 contributes only to the vertical sum, while r5c7 contributes only to the horizontal sum, while all the other cells contribute to both. It follows that the extra vertical cell, r3c6, must exceed the extra horizontal cell, r5c7, by 3. Thus r3c6 cannot be any of 1,2,3. And it already can't be any of 7,8,9. Thus it must be one of 4,5,6, and r7c7 must be one of 1,2,3. Right away this solves one of the other cells in that corner of the puzzle. Since r3c6 must be at least 4, r2c6 must be at most 3. This, in turn, means that the three cells r2c7, r2c8, r2c9 must add up to at least 22, which means one of them must be 9. The only one that can be 9 is r2c8, so there you go. Then, cells r5c7 and r7c4 isolate a puzzle portion in the lower right corner, and by comparing horizontal and vertical sums in that corner, one sees that r7c4 must exceed r5c7 by 2. Thus r7c4 must be one of 3,4,5. Similarly, moving around to cell r5c3 and looking at the lower left corner, one sees that r5c3 must be one of 4,5,6. This is the same as the original, r3c6, which is heartening because the horizontal and vertical sums in the upper left corner differ by 0. Looking at doubularities in this manner eliminates many candidates for various neighboring cells. Bill Smythe |
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| Author: | nj3h [ Wed Feb 26, 2014 9:14 pm ] |
| Post subject: | Re: Problem Kakuro Puzzle |
]Maurice and Bill, Thank you both so much for you replies with techniques I hadn't even considered. I am really impressed!!! I have finally solved the puzzle. What is interesting is, solving the upper right, the upper left, bottom right, and the bottom left are independent of each other after using the technique that you gentlemen provided. Sorry for the delay in getting back, but been away with father-in-law in hospital 225 miles away who is very near the end. Just got back for a few days yesterday. Regards, George |
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| Author: | Smythe Dakota [ Fri Feb 28, 2014 4:15 am ] |
| Post subject: | Re: Problem Kakuro Puzzle |
That's what I like about Kakuro. You almost invent a new solving technique every time you solve a puzzle. It doesn't get old, like Sudoku does. Incidentally, Maurice and I were telling you essentially the same thing, but I was wordier. Whether that makes it easier or harder to understand, you can decide for yourself.  Bill Smythe |
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