### Re: German Puzzle championship

Posted:

**Tue 30 Apr, 2013 11:28 pm**Step 5:

Let's look at the 3rd column. We are missing the 2,3 so the remaining cells

are 1568. The 6 and 8 can't be directly under the 7. If the 5 were there,

there would be no way to complete the 21 in the middle row (we'd need two

things that add up to 7, we already have a 5, a 34 is a conflict, and 16 couldn't

be placed without conflict with either the 6 in the row above or that 5 we just tried

to place.

So the 1 is under the 7. The 8 must be in the middle of the remaining 3 cells buffering the 5 and 6.

The 5 can't be above the 8, since we've got 5's blocked out for both the 5th and 6th rows.

Thus the column goes 1 6 8 5 at the bottom.

This also resolves our 58 and 35 pairs at right.

Meanwhile, we can fill in the middle row, since the remaining two cells

must add to 11. 29 and 56 are clearly out. the spot next to the 9 couldn't

contain either a 3 or an 8. So it's 74 (in that order to avoid 76 conflict).

So now we're at:

3 1 4 _ 3 1 7

5 7 9 3 8 6 4

8 2 7 1 6 4 9

_ _ 1 7 4 9

+ + 6 + + 5 8

+ + 8 + + 3 5

+ + + + _ 8 1

Step 6: Finishing up

The 9 column at bottom left can't be 234 (period).

If it's a 612, the 6 is in the middle.

If it's a 513, there's a problem since given the already filled in

numbers at right, you'd need 5 on the bottom and 3 on top, but

the 3 would conflict with the 6 next to it. Thus it is 162 or 261.

Now the rest of the 6th row. We are missing 2 things that add up

to 8. We have a 3, 5, and 6 already, so missing 17. The other 3

numbers are 249. Meanwhile, the only possible numbers for the last

couple slots in the 5th column are 19, so putting that together gives

1 in row 4, 9 in row 5. The 2 in row 6 is next to the 8 (since the 4 can't be)

and the 4 is at the beginning of row 6.

We have also resolved the order of the 1 and 2 in column 2 now, 2 on top.

The rest of column 4 must be an 8 and 9, with the 8 on the bottom.

The rest of column 1 must be a 6 and 7, with the 6 on the bottom.

Put all of that together and you have:

3 1 4 _ 3 1 7

5 7 9 3 8 6 4

8 2 7 1 6 4 9

_ _ 1 7 4 9 _

7 2 6 9 1 5 8

4 6 8 2 9 3 5

6 1 5 8 _ 8 1

Let's look at the 3rd column. We are missing the 2,3 so the remaining cells

are 1568. The 6 and 8 can't be directly under the 7. If the 5 were there,

there would be no way to complete the 21 in the middle row (we'd need two

things that add up to 7, we already have a 5, a 34 is a conflict, and 16 couldn't

be placed without conflict with either the 6 in the row above or that 5 we just tried

to place.

So the 1 is under the 7. The 8 must be in the middle of the remaining 3 cells buffering the 5 and 6.

The 5 can't be above the 8, since we've got 5's blocked out for both the 5th and 6th rows.

Thus the column goes 1 6 8 5 at the bottom.

This also resolves our 58 and 35 pairs at right.

Meanwhile, we can fill in the middle row, since the remaining two cells

must add to 11. 29 and 56 are clearly out. the spot next to the 9 couldn't

contain either a 3 or an 8. So it's 74 (in that order to avoid 76 conflict).

So now we're at:

3 1 4 _ 3 1 7

5 7 9 3 8 6 4

8 2 7 1 6 4 9

_ _ 1 7 4 9

+ + 6 + + 5 8

+ + 8 + + 3 5

+ + + + _ 8 1

Step 6: Finishing up

The 9 column at bottom left can't be 234 (period).

If it's a 612, the 6 is in the middle.

If it's a 513, there's a problem since given the already filled in

numbers at right, you'd need 5 on the bottom and 3 on top, but

the 3 would conflict with the 6 next to it. Thus it is 162 or 261.

Now the rest of the 6th row. We are missing 2 things that add up

to 8. We have a 3, 5, and 6 already, so missing 17. The other 3

numbers are 249. Meanwhile, the only possible numbers for the last

couple slots in the 5th column are 19, so putting that together gives

1 in row 4, 9 in row 5. The 2 in row 6 is next to the 8 (since the 4 can't be)

and the 4 is at the beginning of row 6.

We have also resolved the order of the 1 and 2 in column 2 now, 2 on top.

The rest of column 4 must be an 8 and 9, with the 8 on the bottom.

The rest of column 1 must be a 6 and 7, with the 6 on the bottom.

Put all of that together and you have:

3 1 4 _ 3 1 7

5 7 9 3 8 6 4

8 2 7 1 6 4 9

_ _ 1 7 4 9 _

7 2 6 9 1 5 8

4 6 8 2 9 3 5

6 1 5 8 _ 8 1