Full Analytical Solution (FAS) for the 9x9 on 8/18/2015:
The empty puzzle:
1. Pre-filled cells.
2. The 9: at 7d must be [9,1].
3. The 45x at 6b must be [9,5].
4. The 4: at 8b must be [2,8].
Through step 4:
5. Subtracting the known sums from rows 1-3 leaves 17 for the 2- at 2d, the 12x at 2f, and the two leftover cells at 3b and 3h. The leftovers must sum as least 3 and the 12x sums either 7 or 8, leaving a max value of 7 for the 2-. With the 1 blocked in column d, that means the 2- is [2,4].
6. The 12x cannot be [2,6] because this would block the 2s in rows 2-3, while forcing the leftovers at 3b and 3h to be [1,2], which is impossible. So the 12x at 2f must be [3,4] and the leftovers must be [1,3].
7. Since both leftovers are part of 3- cages, the corresponding 4b and 4h cells must be [4,6].
8. The 14+ at 4c must therefore be [9,5] due to the blocked 6s in row 4.
9. The 15+ at 2b must be [7,8] due to blocked 9s in columns b-c.
10. The 15+ at 2g must therefore be [6,9]
Through step 10:
11. Subtracting known sums from columns a-c leaves 15 for the 3- at 3b and the leftover cell at 6c. The only values that work are a [3,6] for the 3- and a 6 for 6c.
12. The 3- at 3h is therefore [1,4].
13. The 3- at 6c is also therefore [6,3] due to the 9 at 7d.
14. The 10+ at 6h must be [2,8] due to blocked 1s and 4s in column h as well as blocked 3s in rows 6-7.
15. Subtracting known sums from rows 4-6 leaves 18 for the 6- in 4f and the leftover cell at 6h. The only values that work are a [2,8] for the 6- and an 8 for 6h.
Through step 15:
16. Subtracting known sums from columns d-f leaves 22 for the 1- at 7f and the leftover cells at 3f and 6f. The only values that work are an [8,7] for the 1-, a 2 for 4f, and a 5 for 6f.
17. The 7 in row 7 must be in 7a, which also makes 7i a 4.
18. Subtracting known sums from rows 7-9 leaves 36 for the 3240x at 7i and the 1- at 8g. The only values for the 3240x that include a 4 sum at 27 and 29, which mean that the 1- must be 7 or 9, which means that 8g must be a 4.
Through step 18:
19. 3a and 3i must both be [6,9].
20. With the 9s, 8s, 6s, 5s, and 3s blocked in columns b-c, the only choices for 9b-c are [7,4,2,1].
21. 9b-c must contain a 7 because no other combination allows for usable values in 8-9a. ([1,2] leads to sum 18 which is impossible, [1,4] leads to sum 16, which is blocked by 7a, and [2,4] leads to sum 15, which is blocked by 7a and 3a.)
22. 8a cannot be a 6 because it allows for no usable values for 9a. (With two 7s and a 6 in the 28+ cage, 9a and the non-7 in 9b-c would sum 8: [1,7] is blocked by both 7s, [2,6] is blocked by the 6, and [4,4] is impossible.)
23. The only place for a 6 in row 8 is 8i.
24. The only option for the 3240x cage is now [5,4,9,6,3] which makes 8h 5 to balance of the sum of rows 7-9.
25. This forces a resolution of the cells in the 3240x at 7i, the 15+ at 2g, as well as 3a and 3i.
Through step 25:
26. 8a cannot be a 3 because it allows for no usable values for 9a. (As in step 23, the remainders in the 28+ cage would sum 11: [1,10] is impossible, [2,9] is blocked by the 9 in row 9, and [4,7] is blocked by both 7s.)
27. 8a must therefore be 9 and 8e must be 3.
28. 9a-c must be [1,4,7] due to the blocked 3 in row 9.
29. The remainder of [2,6,8] are easily resolved in 9d-f.
Through step 29:
30. 1g must be a 2.
31. The only place for a 2 in column c is 8c, resolving the 4: at 8b, the 15+ at 2b, and 9c.
32. The only place for a 2 in column b is now 5b.
33. 1c must be a 4, which means 1b must be 1, further resolving 9a-b.
34. 4-6a must be [3,4,2] and 1-2a must be [8,5].
Through step 34:
35. The rest of the puzzle is fairly simple.
Final solution:![Image](http://oi64.tinypic.com/se3zbc.jpg)