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Killer/Medium 2014-10-09 unsolvable?
http://www.calcudoku.org/forum/viewtopic.php?f=16&t=628
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Author:  eyot  [ Thu Oct 09, 2014 9:58 am ]
Post subject:  Killer/Medium 2014-10-09 unsolvable?

Is it just me or is http://www.calcudoku.org/killersudoku/en/2014-10-09/9/2 unsolvable?

I'm having a huge problem filling in the 8 cage in the top-right corner. I can't find any solution that doesn't break other constraints/numbers in other rows/columns.

I don't need specific answers (yet), and I'll avoid spoilers if you haven't gotten to it yet, but I'm happy to walk through my thinking if needed. Just hoping someone will tell me I'm not crazy.

Author:  sheldolina  [ Thu Oct 09, 2014 10:35 am ]
Post subject:  Re: Killer/Medium 2014-10-09 unsolvable?

I didn't have any problem with the 8 cage. This is how I solved it (you might want to put a checkmark next to "show coordinates". also, contains major spoilers if you haven't solved the puzzle yet, of course):

1. The top-right block contains the numbers 1 through 9 once each, so the sum is 45.
2. The i3 cell is the only cell of the block not contained in the 22, 8 and 13 cages. The sum of these cages is 43. Therefore, the number in the i3 cell must be a 2 to complete the sum of 45.
3. There are only two ways to make a sum of 8 without using duplicate numbers: 1+3+4 and 1+2+5. The second way is not possible, since there is already a 2 in i3, so there can't be one in the 8 cage. You now know the combination of the 8 cage is 1,3,4.
4. Since there is a 2 in i3, there must be an 8 in i4.
5. You can now use the same trick of step 2 for the block on the right containing the 22 cage: 45 - 22 - 8 - 3 = 12, so the sum of the cells h6 and i6 is 12.
6. h6 and i6 form a 16 cage together with h7 and i7. This means the sum of h7 and i7 is 4, so these cells will have a 1 and a 3.
7. Now if you put the 4 in the 8 cage in h2, the 1 and 3 will be in i1 and i2, leaving no way to fill in the 1 and 3 in h7 and i7. Therefore, the 4 must be in i1 or i2.
8. If you put the 4 in the 8 cage in i1, the 1 and 3 will be in h2 and i2. Combined with the 1 and 3 in h7 and i7, this would mean the puzzle had no unique solution. Therefore, the 4 must be in i2.
9. You can't fill in the 1 and 3 yet, but their positions will become clear later when you've solved most of the rest of the puzzle.

Author:  eyot  [ Thu Oct 09, 2014 10:47 am ]
Post subject:  Re: Killer/Medium 2014-10-09 unsolvable?

Thanks, sheldolina... I followed everything you said, but I approached it a slightly different way and ran into the problem.

Keeping your steps 1-6 intact, I followed a slightly different path:

7. Since i7/h7 sum to 4, then g7 must be 7 (so the box sums to 45)
8. If g7 = 7 there can be no other 7 in g. Since we know i3 = 2 and there can be no 7 in the 8-cage, either h1 or h3 must be a 7, with no other 7s in h
9. g5/g6 must be a 1 + 2 combination
10. Since there is a 2 in g (5 or 6) and i (i3) there must be a 2 in h9 (the 21 cage cannot contain a 2 and step 6 proves there is no 2 in h7/i7
11. Step 7 places a 7 in col g. Step 8 places a 7 in col h, therefore the 13 cage in the lower right must contain a 7.
12. For the 13 cage in the lower right we have solved two of the values - 7 and 2 - so the remaining value must be 4 in either i8 or i9

Now we have a 4 in col i which breaks your solution that puts a 4 in i2

So if the 8 cage is right, the 13 cage cannot be, and vice versa.

Am I wrong?

Author:  sheldolina  [ Thu Oct 09, 2014 10:59 am ]
Post subject:  Re: Killer/Medium 2014-10-09 unsolvable?

eyot wrote:
Thanks, sheldolina... I followed everything you said, but I approached it a slightly different way and ran into the problem.

Keeping your steps 1-6 intact, I followed a slightly different path:

7. Since i7/h7 sum to 4, then g7 must be 7 (so the box sums to 45)
8. If g7 = 7 there can be no other 7 in g. Since we know i3 = 2 and there can be no 7 in the 8-cage, either h1 or h3 must be a 7, with no other 7s in h
9. g5/g6 must be a 1 + 2 combination
10. Since there is a 2 in g (5 or 6) and i (i3) there must be a 2 in h9 (the 21 cage cannot contain a 2 and step 6 proves there is no 2 in h7/i7
11. Step 7 places a 7 in col g. Step 8 places a 7 in col h, therefore the 13 cage in the lower right must contain a 7.
12. For the 13 cage in the lower right we have solved two of the values - 7 and 2 - so the remaining value must be 4 in either i8 or i9

Now we have a 4 in col i which breaks your solution that puts a 4 in i2

So if the 8 cage is right, the 13 cage cannot be, and vice versa.

Am I wrong?


The bolded part is where it goes wrong. In your step 7, you placed a 7 in g7. Therefore, the lower right block already contains a 7, and the 13 cage in the lower right cannot contain a 7. This means it also does not contain a 4, so there is no conflict with the 4 in i2.

Author:  eyot  [ Thu Oct 09, 2014 7:05 pm ]
Post subject:  Re: Killer/Medium 2014-10-09 unsolvable?

Thanks! that's the bit I was missing - dueling 7's. That's what I get for running the killers in the middle of the night.

:)

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