by **Squirrel** » Sun Oct 15, 2006 5:54 pm

Yes, both puzzles only have 1 solution and are solvable by logic.

Heres the solution to the second double killer puzzle. As you enter in the numbers, please place them in both grids 1 (left) and 2 (right). As I dont know all the shortforms for all the techniques, I just wrote it out in full. Hope you can follow it.

rule of 45 gives you

r6c3 = 5

r4c6 = 4

r2c7 = 4 (using regions from both grids)

r7c2 = 4 (using regions from both grids)

r6c9 = 3 (bottom four rows of second grid)

r4c1 = 7 (top 4 rows of 2nd grid)

unique sums in regions at r4c1 and r5c9:

r5c1 = 3

r5c9 = 6

This is where you get into the fun logic. The region at r6c4g1 and r6c5g2 share 3 cells in common (r6c5, r7c4, r8c4). The highest those cells can total is 17 (which would yeild a 1 in r6c4 (look at grid 1)), and the lowest they can equal is 15 (which would yeild a 9 in r7c5 (look at grid 2)). That means that r6c4 is either a 1 (the sum of the three square is 17), or a 2 (the sum ofthe three squares is 16). The sum of the three squares cant be 16 because of 3 in r6c9. (These possible sums also tell us that r7c5 = 7 or 9)

Directly to the left of the two grids we have been working with are two more which share 3 cells (located at r6c3g12). You can use the same logic using the possible sums of the three cells in common, or else you can say that the difference in the sums between the two regions is 3, so r6c4 (look at the specified region in grid 2) is three less than r7c2 (look at the specified region in grid 1). That means that because r6c4 = 1 or 2, r7c2 = 4 or 5. However, r7c2 cant be a 5 because r6c3 = 5, and is in the same region.

Therefore,

r7c2 = 4

r6c4 = 1

r7c5 = 7

You should be able to solve the rest of it fairly easily from here on. The rest of the sloution is still there if you need it.

looking at grid 2, we can see:

the region at r6c1 has a remainder of 8 which has to be a 2 and a 6.

r6c1 = 2

r6c2 = 6

the region at r6c3 has a remainder of 16 which has to be 7 and 9

r7c3 = 9

r8c3 = 7

using rule of 45 in bottom left nonet, r9c23 = 9, and has to contain a 3 (because of the 3 at r5c1).

r9c2 = 3

r9c3 = 6

looking at the grid 1:

rule of 45 reveals

r4c4 = 3

looking at grid 2:

the region at r3c4 has a remainder of 15

r4c3 = 8

r3c4 = 7

PENCILMARKS: looking at the region at r9c2, the remainder is 12 which has to be an 8 and a 5

looking at grid 1:

the region at r5c4 = 20, which has to be either 974 or 875 and therefore must contain a 7.

r5c6 = 7

the only place for a 4 in the left center nonet is r5c3

r5c3 = 4

the only spot for a 4 in the center center nonet is r6c6 (if r6c5 = 4 the the remainder in the region at r6c4 = 13, which has no possible sloutions)

r6c6 = 4

rule of 45 reveals

r4c9 = 9

you can now complete the center left nonet:

r4c2 = 1

r5c2 = 9

the only spot for a 9 in center center nonet is r6c5

r6c5 = 9

looking at both grids:

the regions at r2c6g1 and r2c6g2 share 3 cells and have a difference of 5. Therefore r4c6 is 5 greater than r3c5.

r4c6 = 6

r3c5 = 1

the regions at r2c7g1 and r2c7g2 share 3 cells and have a difference of 0. Therefore r3c7 = r4c6

r3c7 = 6

looking at grid 2:

r6 is missing a 7 and an 8, r6c7 <> 8 because of the region it is in

r6c7 = 7

r6c8 = 8

unique sum in region at r3c8

r4c8 = 5

unique sum in region at r6c6

r7c6 = 1

unique sum in region at r6c5: r78c4 = 8 and has to be 2 + 6

r7c4 = 2

r8c4 = 6

1 number left in r4

r4c5 = 2

looking at grid 1:

unique sum in region at r3c2: r3c23 = 11, and has to be 3 + 8

r3c2 = 8

r3c3 = 3

rule of 45 gives us

r1c2 = 5

1 number left in c2:

r2c2 = 7

looking at grid 2:

unique sum in region at r3c1

r3c1 = 4

only place for a 2 in top center nonet is r1c6 (cant be r23c6 because of region)

r1c6 = 2

unique sum in region at r2c6: r23c6 = 13, and has to be 8 + 5

r2c6 = 8

r3c6 = 5

2 numbers left in r3:

r3c7 = 9

r3c9 = 2

unique sum in region at r2c7

r2c7 = 5

only 1 place for a 6 in bottom right nonet:

r7c7 = 6

unique sum in region at r6c8: r7c89 = 8 and has to be 3 + 5

r7c8 = 3

r7c9 = 5

rule of 45 on r8 reveals r8c1 + r8c9 = 5 and has to be 1 + 4

r8c1 = 1

r8c9 = 4

only two numbers left in bottom left nonet:

r7c1 = 8

r9c1 = 5

unique sum in region at r1c9: r12c9 = 9 and has to be 1 + 8

r1c9 = 8

r2c9 = 1

two numbers left in top right box:

r1c7 = 3

r1c8 = 7

one number left in r9:

r9c9 = 7

unique sum in region at r1c5:

r1c5 = 6

two numbers remain in c1:

r1c1 = 9

r2c1 = 6

two numbers remain in r1:

r1c3 = 1

r1c4 = 4

one number remains in top left nonet:

r2c3 = 2

two numbers remain in top center nonet:

r2c4 = 9

r2c5 = 3

get rid of penclmarks:

r9c4 = 8

r9c5 = 4

two numbers remain in center center nonet:

r5c4 = 5

r5c5 = 8

one number remains in c5:

r8c5 = 5

looking at grid 1:

rule of 45 reveals:

r9c8 =2

two numbers remain in right center nonet:

r5c7 = 2

r5c8 = 1

two numbers remain in bottom center nonet:

r8c6 = 3

r9c6 = 9

three numbers remain in bottom right nonet:

r8c7 = 8

r8c8 = 9

r9c7 = 1

Hope that helps,

~Squirrel