The New Sudoku Players' Forum

[4ivonne]

Hi,

I'm stucked on this kakuro puzzle.
I'm already solved it like below, but now I don't know the next step.
Are there any advanced techniques to solve this like in Sudoku?
I can't see any hidden triple etc...
Or is the next step just "try and error"?
Please help...

Thanx,
Ive

Code: Select all

    \    | 13 \    | 30 \    |    \    |  9 \    | 18 \    |    \    |    \    | 13 \    | 10 \    |
    \ 14 |    5    |    9    |    \  7 |    2    |    5    |    \    | 12 \ 13 |    6    |    7    |
    \ 10 |    8    |    2    | 16 \ 16 |    7    |    9    | 18 \  8 |    1    |    4    |    3    |
    \    |    \ 13 |    4    |    9    | 19 \ 10 |    4    |    1    |    2    |    3    |    \    |
    \    |  7 \ 22 |    6    |    7    |    9    |  5 \ 14 |    5    |    9    | 24 \    | 12 \    |
    \ 12 |    4    |    8    |    \ 11 |    1    |    2    |    8    | 22 \  6 |    5    |    1    |
    \  4 |    3    |    1    | 17 \ 26 |    2    |    3    |    4    |    1    |    7    |    9    |
    \    |    \    | 23 \ 16 |    9    |    7    |    \    |    \ 10 |    5    |    3    |    2    |
    \    |  9 \  5 |    4    |    1    |    \    |    \    | 25 \ 16 |    7    |    9    |    \    |
    \ 15 |_23_56___|______7_9|_2345____| 20 \    |  7 \ 15 |    6    |    9    | 31 \    |  9 \    |
    \ 32 |_23_56___|______7_9|_2345____|__3_56___|123_5____|_______89|    \  6 |    4    |    2    |
    \  4 |    1    |    3    | 17 \ 19 |____56_8_|_2_456___|_______89| 11 \  8 |    1    |    7    |
    \    |    \    | 15 \ 12 |    5    |    7    | 23 \  9 |    2    |    4    |    3    |    \    |
    \    | 11 \ 15 |    1    |    3    |    2    |    9    |  7 \ 16 |    7    |    9    |  6 \    |
    \ 24 |    7    |    8    |    9    |    \  8 |    6    |    2    |    \  9 |    8    |    1    |
    \ 10 |    4    |    6    |    \    |    \ 13 |    8    |    5    |    \ 11 |    6    |    5    |

sums
KakuroSum COL   7   r9c3,r10c3,
   25,52,34,43,
KakuroSum COL   16   r9c2,r10c2,
   79,97,
KakuroSum COL   8   r9c1,r10c1,
   26,62,35,53,
KakuroSum COL   17   r10c6,r11c6,
   89,98,
KakuroSum COL   11   r10c4,r11c4,
   38,56,65,
KakuroSum COL   7   r10c5,r11c5,
   16,25,52,34,
KakuroSum ROW   15   r9c1,r9c2,r9c3,
   294,672,375,573,
KakuroSum ROW   16   r10c1,r10c2,r10c3,
   295,592,394,673,574,
KakuroSum ROW   19   r11c4,r11c5,r11c6,
   829,649,568,658,
KakuroSum ROW   16   r10c4,r10c5,r10c6,
   619,529,628,358,538,


[udosuk]

There're 2 advanced Kakuro/Killer Sudoku techniques you can apply to solve this puzzle,
namely the "odd/even parity analysis" and the "partial sum analysis"...

Code: Select all

|#####|13\##|30\##|#####| 9\##|18\##|#####|#####|13\##|10\##|
|##\14|  5  |  9  |##\ 7|  2  |  5  |#####|12\13|  6  |  7  |
|##\10|  8  |  2  |16\16|  7  |  9  |18\ 8|  1  |  4  |  3  |
|#####|##\13|  4  |  9  |19\10|  4  |  1  |  2  |  3  |#####|
|#####| 7\22|  6  |  7  |  9  | 5\14|  5  |  9  |24\##|12\##|
|##\12|  4  |  8  |##\11|  1  |  2  |  8  |22\ 6|  5  |  1  |
|##\ 4|  3  |  1  |17\26|  2  |  3  |  4  |  1  |  7  |  9  |
|#####|#####|23\16|  9  |  7  |#####|##\10|  5  |  3  |  2  |
|#####| 9\ 5|  4  |  1  |#####|#####|25\16|  7  |  9  |#####|
|##\15|.2356|..79.|.2345|20\##| 7\15|  6  |  9  |31\##| 9\##|
|##\32|.2356|..79.|.2345|.356.|.1235|..89.|##\ 6|  4  |  2  |
|##\ 4|  1  |  3  |17\19|.568.|.2456|..89.|11\ 8|  1  |  7  |
|#####|#####|15\12|  5  |  7  |23\ 9|  2  |  4  |  3  |#####|
|#####|11\15|  1  |  3  |  2  |  9  | 7\16|  7  |  9  | 6\##|
|##\24|  7  |  8  |  9  |##\ 8|  6  |  2  |##\ 9|  8  |  1  |
|##\10|  4  |  6  |#####|##\13|  8  |  5  |##\11|  6  |  5  |

These are the sums of the 4 triplets of cells:

Pink=15
Yellow=16
Green=16
Blue=19

And there are 2 methods to determine the value of the middle yellow cell:

Method 1 (odd/even parity analysis):
1. The green cells cannot be [628], otherwise we'll have two 5s in the blue cells
2. The green cells cannot be [358] or [538], otherwise the yellow cells with 1 odd and 2 even numbers can never sum to 16
From (1) & (2) we can conclude the rightmost green cell cannot be 8, so it must be 9, and the middle yellow cell must be 7

Method 2 (partial sum analysis):
1. The green cells cannot be [628], otherwise we'll have two 5s in the blue cells
2. The sum of the middle yellow cell and the rightmost green cell is at most 8+9=17
=> The sum of the 4 remaining yellow & green cells must be at least 32-17=15
=> One of the leftmost yellow/green cells must be 6 (otherwise those 4 cells will not exceed 2+3+4+5=14)
=> Either the yellow cells=[672], or the green cells=[619] (they cannot be [628] from (1))
=> In both cases the middle yellow cell must be 7

Hence, the pink cells must be [294], the yellow & green cells [673529] and the blue cells [658].

Here is the solution:

[4ivonne]

I just found another way to solve this puzzle.
I don't know how this strategy is called, but I helps me to solve this by just looking at the row with the yellow and green cells.

The yellow cells can be
[295], [592], [394], [673], [574]

The green cells can be
[619], [529], [628], [358], [538]

To fill the total sum of 32, I need to combine the 2 sum-pairs,
but the 2 parts should not contain the same digits.
So the only valid combinations are

yellow+green
[295]+[628]
[394]+[628]
[673]+[529]
[574]+[619]
[574]+[628]

So some digit canidates for this row can be removed.
The remaining valid digits are:

|2356|.79.|.345|.56.|.12.|.89.|

The 2 in the rightmost yellow cell can be removed
In the leftmost green cell the 3 is removed.
In the middle green cell the 3 and the 5 ist removed.

So we can remove the 8 from the leftmost blue cell
and the 2 and 4 from the middlle blue cell.

The possible digits for the blue cells are now
|.56.|.56.|.89.|

So, a pair of 5 and 6 = 11.
The rightmost blue cell needs to be 19 -11 = 8.

=> the rightmost green cell is 9.
=> the middle yellow cell is 7...

[udosuk]

... to solve this by just looking at the row with the yellow and green cells.

The yellow cells can be
[295], [592], [394], [673], [574]

The green cells can be
[619], [529], [628], [358], [538]

To fill the total sum of 32, I need to combine the 2 sum-pairs,
but the 2 parts should not contain the same digits.
So the only valid combinations are

yellow+green
[295]+[628]
[394]+[628]
[673]+[529]
[574]+[619]
[574]+[628]

So some digit canidates for this row can be removed.
The remaining valid digits are:

|2356|.79.|.345|.56.|.12.|.89.|
...

This looks like trial and error to me... You're writing down all possiblities, and work out which are valid and which are not... "Just looking at the row with the yellow and green cells" doesn't mean it's not T&E... It's still T&E on a single row...

Anyway, from your 5 "valid combinations" of yellow+green, it's easy to see a [574] in yellow will lead to two 3s in the pink cells and a [628] in green will lead to two 5s in the blue cells... That leaves us with only one valid combination, and you can fill all 6 yellow+green cells at once...

Alternatively, when you just look at the 5 possibilities for the yellow cells ([295], [592], [394], [673], [574]), you notice all of them either contain a 3 or a 5... So the green cells cannot be [358] or [538]... And subsequently you can eliminate all 3s in the green cells... Which soon will let you work out the rightmost blue cell similar to what you did...