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This puzzle has the condition that half cells cannot contribute the same amount to a cage as a full cell.
1. 4(3) cage at R4C2 can only be 2 + 0.5 + 1.5 (cannot be 1 + 0.5 + 2.5 because 1 would be repeated in the cage) -> R4C3 = 2, R4C2 + R5C3 = {13}
Each of the four corner nonets has two half outies. Each of these outie cells “sees” all except two cells inside that nonet. Thus R1C4 “sees” all cells in N1 except for R2C1 and R3C3 and its complementary outie R4C1 “sees” all cells in N1 except for R1C2 and R3C3.
Each of the three-cell cages contains two half-cells and one full cell. Therefore to make an integer total the two half-cells must contain two even numbers or two odd numbers.
2. 45 rule on N7 2 half-outies R6C1 + R9C4 = 7 -> R6C1 + R9C4 = 14 = {59/68}
2a. R6C1 + R9C4 “see” all cells in N7 except for R7C3, R8C1 (for R9C4) and R9C2 (for R6C1)
2b. R8C1/R9C2 cannot be 8,9 to make 5(3) cage in R8C1
2c. 5(3) cage in R8C1 can only be 2.5 + 0.5 + 2 or 2.5 + 1 + 1.5 (cannot be 6 in R8C1/R9C2 contributing 3 because the remaining cells would each contribute 1, which is specifically forbidden) -> R8C1/R9C2 = 5, R9C1 = {12} and the other of R8C1/R9C2 = {13}, 5(3) cage contains both of 1,5 and one of 2,3
2d. R8C1/R9C2 = 5 -> R6C1 + R9C4 = {59} -> R7C3 = 9 (because 9 cannot be in R8C1/R9C2)
2e. 5 must be in R68C1, locked for C1
2f. 5 must be in R9C24, locked for C9
3. R7C3 = 9 -> half-cells R7C2 + R8C3 = 5
3a. For 21(7) cage at R6C1, half-cells R6C1 + R9C4 = 7, half-cells R7C2 + R8C3 = 5 -> remaining half-cells R7C1 + R8C2 + R9C3 = 9
3b. For 12(5) cage at R7C1, half-cells R7C1 + R8C2 + R9C3 = 9 -> remaining half-cells R8C1 + R9C2 = 3 -> R9C1 = 2, R8C1 + R9C2 = 6 = {15}
4. For 7(3) cage at R9C4, R9C4 is odd -> R9C6 must be odd
45 rule on N9 2 half-outies R6C9 + R9C6 = 5 -> R6C9 + R9C4 = 10 = {19/37}
R9C4 = {59} -> cage contribution = 2.5 or 4.5 -> cage = 4.5 + 1 + 1.5 (cannot be 2.5 + 1 + 3.5 because R9C45 = [51] clashes with R9C2, cannot be 2.5 + 3 + 1.5 because 3 would be repeated in the cage, cannot be {2.5 + 4 + 0.5 because R9C46 = [51] clashes with R9C2) -> R9C456 = [913], R6C9 = 7, R6C1 = 5, R8C1 = 1, R9C2 = 5
5. For 9(3) cage at R4C9, R6C9 is odd -> R4C9 must be odd
5a. 45 rule on N3 2 half-outies R1C6 + R4C9 = 3 -> R1C6 + R4C9 = 6 = {15}
5b. 9(3) cage = 0.5 + 5 + 3.5 or 2.5 + 3 + 3.5 = [157/537], 5 locked for C9 and N6
6. For 9(3) cage at R1C4, R1C6 is odd -> R1C4 must be odd
6a. 45 rule on N1 2 half-outies R1C4 + R4C1 = 6 -> R1C4 + R4C1 = {39/57} = [39/57]
6b. 9(3) cage at R1C4 = 1.5 + 7 + 0.5 or 2.5 + 6 + 0.5 (cannot be 1.5 + 5 + 2.5 because 5 would be repeated in the cage) = [371/561] -> R1C6 = 1, R4C9 = 5, R5C9 = 3, R5C3 = 1, R4C2 = 3
6c. 14(3) cage at R4C1 = 3.5 + 8 + 2.5 or 4.5 + 7 + 2.5 -> R456C1 = [785/975], 7 locked for C1 and N4
7. 3 in R6 must be in 5(3) cage at R5C4, cannot be 0.5 + 3 + 1.5 or 1.5 + 3 + 0.5 (because 3 would be repeated in the cage) -> R6C5 = 3, R5C4 must be odd -> 5(3) cage = 2.5 + 1 + 1.5 -> R56C4 = [51], R1C4 = 3, R1C5 = 7 (step 6b), R4C1 = 9 (step 6a), R5C1 = 7 (step 6c)
8. R4C1 = 9 “sees” all cells in N1 except for R1C2 and R3C3, R3C3 cannot be 9 -> R1C2 = 9
8a. R1C2 is odd -> R2C1 must be odd = 3, 12(3) cage at R1C1 = 6 + 4.5 + 1.5 -> R1C1 = 6
9. R1C2 + R2C1 = [93] = 12 -> half-cells R1C2 + R2C1 = 6 -> half-cells R1C3 + R2C2 + R3C1 = 7
9a. R1C4 + R4C1 = [39] = 12 -> half-cells R1C4 + R4C1 = 6, half-cells R1C3 + R2C2 + R3C1 = 7 -> half-cells R2C3 + R3C2 = 4 -> R3C3 = 5
9b. Half-cells R2C3 + R3C2 = 4 -> R2C3 + R3C2 = 8 = {17} -> R2C3 = 7, R3C2 = 1
9c. R2C2 = 2 (hidden single in N1)
10. 1 in R4 only in 8(3) cage at R4C7, R4C8 and R5C7 must both be even or both odd; cannot be odd = {19} because 0.5 + 3 + 4.5 clashes with R5C9 -> R4C7 = 1, half-cells R4C8 + R5C7 = 7 -> R4C8 + R5C7 = 14 = {68}
11. R9C6 = 3 “sees” all cells in N9 except for R7C7 and R9C8, 3 blocked from R9C8 -> R7C7 = 3, R8C3 = 3 (hidden single in N7)
11a. 14(3) cage at R7C2 = 3.5 + 9 + 1.5 -> R7C2 = 7
12. R6C9 = 7 “sees” all vacant cells in N9 except for R9C8 -> R9C8 = 7, R3C7 = 7 (hidden single in C7)
13. R1C6 = 1 “sees” all vacant cells in N3 except for R2C9 -> R2C9 = 1
13a. R4C9 = 5 “sees” all vacant cells in N3 except for R1C8 -> R1C8 = 5
13b. 7(3) cage at R1C8 = 2.5 + 4 + 0.5 -> R1C9 = 4
14. R3C1 = 4 (hidden single in N1), R1C3 = 8 (final cell in N1), R1C7 = 2 (final cell in R1), R7C1 = 8 (final cell in C1)
15. R3C7 = 7 -> half-cells R2C7 + R3C8 = 6 -> R2C7 + R3C8 = 12 = {39} -> R2C7 = 9, R3C8 = 3
16. R9C8 is odd -> R8C9 must be odd -> R8C9 = 9
16a. 16(3) cage at R8C9 = 4.5 + 3.5 + 8 -> R9C9 = 8
16b. R2C8 = 8 (hidden single in N3), R3C9 = 6 (final cell in N3), R7C9 = 2 (final cell in C9)
16c. R4C8 + R5C7 (step 10) = {68} -> R4C8 = 6, R5C7 = 8
17. R7C7 = 3 -> half-cells R7C8 + R8C7 = 3 -> R7C8 + R8C7 = 6 = {15} -> R7C8 = 1, R8C7 = 5, R9C7 = 6 (hidden single in N9), R8C8 = 4 (final cell in N9), R9C3 = 4 (final cell in R9), R8C2 = 6 (final cell in N7), R6C2 = 8 (hidden single in C2), R5C2 = 4 (final cell in C2), R6C3 = 6 (final cell in N4), R6C7 = 4 (final cell in C7)
18. 9 in R3 must be in 15(3) cage at R2C6, R2C6 and R3C5 must both be even or both odd -> 15(3) cage = 2 + 4 + 9 or 2.5 + 4.5 + 8 -> R2C6 + R3C56 = [489/598]
18a. R3C4 = 2 (hidden single in R3)
19. For 10(3) cage at R7C4, R7C4 and R8C5 must both be even (because there are no remaining odd candidates) -> R8C4 = 7 (cannot be 8 because R7C4 + R8C5 would require odd candidates), half-cells R7C4 + R8C5 = 3 -> R7C4 + R8C5 = 6 = {24} -> R7C4 = 4, R8C5 = 2, R8C6 = 8 (final cell in R8)
20. R8C6 = 8 -> R2C6 + R3C56 (step 18) = [489] -> R2C6 = 4, R3C56 = [89]
[And the rest are hidden singles and final cells …]
21. R2C5 = 5, R2C4 = 6, R4C4 = 8, R4C5 = 4, R4C6 = 7, R5C5 = 9, R6C8 = 9, R5C8 = 2, R5C6 = 6, R6C6 = 2, R7C5 = 6, R7C6 = 5
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