SudokuSolver Forum

0. Prelims
16(2)n6 = {79}

1. Outies c1 -> r5c23 + r8c2 = +5
-> They must be [12][2] or [31][1]
-> (since 8(3)n6 has a 1) -> r6c8 = 1

2. Outies - Innies r1234 -> r5c6 = r4c1 + 6
-> max r4c1 = 3 and since 1 already in r5c23 -> min r4c1 = 2
-> r4c1, r5c23, r5c6, r5c89 = [2][31][8]{25} or [3][12][9]{34}

3. 12(3)n1? Either:
a) r5c23, r8c2 = [12][2] -> Only place for (12) in c1 is in n1 -> 12(3)n1 = {129}
b) r5c23, r8c2 = [31][1] -> Only place for (13) in c1 is in n1 -> 12(3)n1 = {138}

4. Since r5c6 from (89) -> 15(4)r3c6 = {124}[8] or {123}[9]

5. Outies c6789 -> r189c5 = +7 = {124}

6. Max r567c5 = +16 -> Min r234c5 = +22
-> 9 in r23c5
Also -> 3 in r67c5

7. Max r234c5 = +24 -> Min r567c5 = +14 -> Max r7c4 = 3.
Since 3 already in r67c5 -> Max r7c4 = 2
-> in n8 r7c4, r89c5 = {124} with 4 in r89c5
-> r1c5 from (12)

8. Innies n69 -> r4c9 + r9c7 = +8.
Since (17) already in n6 -> the innies cannot be {17}
Since r89c5 = {14} or {24} -> r9c7 cannot be from (124) (17(4)r8c5)
-> r4c9 from (235) and r9c7 from (653)
-> 3 in n6 placed in r4c9 or r5c89

9.28(4)r2c3 contains a 9.
Since 9 in r23c5 -> 9 in the 28(4) must be in r2c3 or r2c5.
-> 9 nowhere else in r2.

10. (68) in n6 placed in r56c7 or r6c9
-> At least one of (68) in r56c7
-> Only way 9 can go in 19(4)r5c7 is if it is {1369}
which is impossible since none of (139) can go in r56c7
-> HS 9 in c7 -> r4c7 = 9
-> r4c8 = 7

11. Given r4c8 = 7 -> Innies c89 = r1c89,r2c8 = +15
-> r2c67 = +10.
Also Outies r1 -> r2c128 + r3c1 = +11
Since 12(3)n1 from {129} or {138} -> r1c1 from (89)

12. 8 in n2 and r2?
If r1c1 = 8 -> Max r1c4 = 7
If r1c1 = 9 -> Min r1c23+r2c2 = {345} = +12 -> Max r1c4 = 7
-> 8 not in r1c4
28(4)r2c3 must contain an 8.
8 in r2c6 puts 8 in 28(4) in r3c3 leaves no place for 8 in r1
-> 8 in n2 in r23c45
-> 8 in r2 in r2c345
-> 8 nowhere else in r2
Also - since (89) placed in r2c345 -> neither of (89) in r3c3

13. -> 8 in c7 in the 19(4)!
-> 19(4) cannot contain a 7! (Would have to be {1378} and none of (137) can go in r56c7)
-> 7 in c7 in r12c7

14. H10(2)r2c67 = {46} or {37}
But making it {46} would put 10(3)r1c5 = {12}[7] and (46) in r1 in r1c234
Which since 1 is in r23c1 and 8 is in r2c345 leaves no solution for 19(4)r1c2
-> H10(2)r2c67 = {37}

15. (Step 11) Outies r1 = r2c128 + r3c1 = +11
Since 3 placed in r2c67 and 1 already in c2 and c8
-> r2c128, r3c1 can only be [162][2] or [1{25}][3]
-> r2c1 = 1 - nowhere else in r2
-> 12(3)n1 = [813] or [912]

16. Min r4c23 = {45} = +9 -> Max r3c2 = 8
But 8 in r3c2 puts 14(2)r6c2 = [95] and r4c23 = [45] leaves no place for 5 in c1
-> 8 not in r3c2

17. (89) in n3 can only go in r13c89.
Since r1c1 from (89) -> At least one of (89) in r3c89
But since 1 cannot go in either r2c9 or r4c9 -> r3c89 cannot be {89}
-> One of (89) in each of r1c89 and r3c89
-> Since neither of (89) can go in r3c123 -> One of (89) in r3c45
-> One of (89) in r2c3

18. Breakthrough!
(Step 15) r2c28 = [62] or {25}
But the former puts 12(3)n1 = [912] and 28(4)n12 with [87] in r23c3
Leaves no solution for 19(4)r1c2. (r1c23 with 2 values from (345)).
-> r2c28 = {25}

19. -> 12(3)n1 = [813]
-> r2c3 = 9
-> r3c5 = 9
Also r4c1 = 2
-> r5c23, r5c6, r5c89 = [31][8]{25}
-> r8c2 = 1
Also r4c9 = 3 -> r9c7 = 5
Also 9 in r1c89 and 8 in r3c89
-> 21(4)n3 = {468}[3]
-> H15(3)n3 (r1c89,r2c8) = [915]
-> r2c2 = 2
Also -> 7 not in r1c7
-> r123c7 = [372]
-> r2c6 = 3
Also -> r1c56 = [25]
-> r89c5 = {14} and r7c4 = 2
-> r9c6 = 7
Also r567c5 = [7]{35}
Also r234c5 = [896]
Also r34c6 = {14}
-> 17(3)r6c6 = [2]{69}

20. Continuing...
-> r89c4 = {38} or [58]
But {38} leaves no solution for 25(4)r8c4
-> r89c4 = [58]
-> r9c23 = [93]
-> 14(2)r6c2 = {68}
Also r67c5 = [53]
Also r34c4 = [71]
-> r34c6 = [14]
-> r4c23 = [58] and 14(2)r6c2 = [68]
-> r3c2 = 4
-> r2c9 = 4 and r3c89 = {68}
etc.

1. R456+C12
a) 16(2) = {79} locked for R4+N6
b) Outies C1 = 5(2+1) = [122/311] -> 1 locked in R5C23 for R5+N4+24(6) and 1 locked in R58C2 for C2
c) 8(3) = 1{25/34} -> R6C8 = 1
d) 14(3): R89C1 <> 1,2 since R8C2 = (12)
e) Hidden Killer triple (789) in 12(3) for C1 since each of 24(6) and 14(3) can only have one of them -> 12(3) <> 5,6
f) 1 locked in 12(3) @ C1 for N1 -> 12(3) = 1{29/38} since {147} blocked by Killer pair (47) of 24(6)

2. C456+R89
a) Outies C6789 = 7(3) = {124} locked for C5
b) 17(4) must have one of (12) -> R7C4 = (12); 17(4) = 3{158/167/257} <> 9 -> 3 locked for C5
c) 9 locked in R23C5 @ C5 for N2; CPE: R3C3 <> 9
d) 28(4) = 89{47/56} -> 9 locked for R2; CPE: R2C12 <> 8
e) Naked triple (124) locked in R7C4+R89C5 for N8
f) 4 locked in R89C5 @ N8 for C5+17(4)
g) 17(4) @ R8 = 4{139/157/238/256}: R9C7 <> 1,2 since R89C5 = (124)

3. R456+N9
a) Innies N69 = 8(1+1) = [26/35/53]
b) Hidden Killer pair (68) in 19(4) @ N6 -> 19(4) = {1369/1378/1468/1567/2368/2458/2467}
c) Innies+Outies R1234: 6 = R5C5 - R4C1 -> R4C1 = (23), R5C5 = (89)
d) 15(4) = 12{39/48} since R5C6 = (89); R3C67+R4C6 <> 8,9
e) 6,8 locked in R5C7+R6C79 @ N6
f) Innies+Outies N9: 13 = R5C7+R6C79 - R9C7 -> R5C7+R6C79 = 68{2/4/5} <> 3 since R9C7 = (356)
g) 19(4) <> 13{69/78} since R56C7 <> 1,3,7,9 -> 19(4) <> 9

4. R12+C789
a) Hidden Single: R4C7 = 9 @ C7, R4C8 = 7
b) Outies C89 = 10(2) <> 1,5
c) Using Outies C89 = 10(2): Innies+Outies R1: -1 = R2C28 - R1C1 -> R2C28 <> 7,8 since R2C28 <> 1; R1C1 = (89)
d) Using Outies C89 = 10(2): Outies R1 = 11(3+1) -> R2C8 <> 4,6 because 4 of R2C12+R3C1 = {124} sees R2C8

5. C4567 !
a) Innies C6789 = 20(2+2): R1C7 <> 1 because R1C5 = (12) and [71]+[93] blocked by Killer triple (789) in each of R5C6 + 17(3)
b) 17(4) @ R8: R9C6 <> 3 since 8,9 only possible there
c) 19(4) + R9C7 must have exactly two of (1234) @ C67 since each of 17(3) + R1C56 @ 10(3) + Outies C89 must have exactly one of (1234) and 3 of (1234) are in 15(4)
d) ! 5,6 locked in 19(4) + R9C7 for C7 since either 19(4) = {1567} or Killer pair (56) locked in 19(4) + R9C7 for C7
e) ! Innies C6789 = 20(2+2) <> 9 because [17]+[93] forces 19(4) = {1567} which is blocked by R1C7 = 7
f) ! Consider placement of 1 in C5 -> R1C6 <> 1
- i) R1C5 = 1
- ii) 1 in 17(4) @ R8 = {14}[75] and R1C5 = 2 -> Innie N69 = R4C9 = 3 -> R4C1 = 2 -> R3C7 = 2 (HS @ 15(4)) -> 1 locked in 15(4) for C6

6. C6789
a) 1 locked in 15(4) @ C6 for 15(4)
b) 1 locked in 19(4) @ C7 = 16{48/57} for N9; 6 locked for C7
c) Innies N69 = 8(1+1) = {35} -> CPE: R56C7 <> 5
d) 19(4) = {1468} locked for C7 since R56C7 = (468)
e) Hidden Single: R9C7 = 5 @ C7
f) Innie N69 = R4C9 = 3
g) 8(3) = {125} -> 2,5 locked for R5+N6

7. C123
a) Outies C1 = 5(2+1) = [311] -> R5C3 = 1, R5C2 = 1, R8C2 = 1
b) R4C1 = 2
c) 12(3) = {138} locked for C1+N1 -> R1C1 = 8
d) Hidden Single: R9C5 = 1 @ R9, R7C4 = 2, R8C5 = 4 -> R8C6 = 7

8. C1456+R4
a) 10(3) = {235} -> R1C5 = 2, R1C7 = 3, R1C6 = 5
b) 17(4) @ R5 = {2357} -> R5C5 = 7; 3,5 locked for C5
c) R3C7 = 2, R2C7 = 7
d) Outie C89 = R2C6 = 3
e) 17(3) = {269} -> R6C6 = 2; 6,9 locked for C6+N8
f) 25(4) = {3589} since R89C5 = (358) -> R8C4 = 5; 3,8,9 locked for R9; 9 also locked for N7
g) 14(3) = [167] -> R9C1 = 6, R8C1 = 7
h) 5 locked in 17(3) @ R4 = {458}

9. Rest is singles.

1.75. I used two forcing chains where one uses interesting Killer quad properties.

Prelims

a) R4C78 = {79}
b) R67C2 = {59/68}
c) 10(3) cage at R1C5 = {127/136/145/235}, no 8,9
d) 8(3) cage at R5C8 = {125/134}
e) 28(4) cage at R2C3 = {4789/5689}, no 1,2,3

Steps resulting from Prelims
1a. Naked pair {79} in R4C78, locked for R4 and N6
1b. 8(3) cage at R5C8 = {125/134}, 1 locked for N6
1c. 28(4) cage at R2C3 = {4789/5689}, CPE no 8,9 in R2C12

2. 45 rule on C1 3(2+1) outies R5C23 + R8C2 = 5
2a. Min R5C23 = 3 -> max R8C2 = 2
2b. Min R58C2 = 3 -> max R5C3 = 2
2c. Min R5C3 + R8C2 = 2 -> max R5C2 = 3
2d. Max R8C2 = 2 -> min R89C1 = 12, no 1,2 in R89C1
[There’s more, but I didn’t spot it until later.]

3. 45 rule on C6789 3 outies R189C5 = 7 = {124}, locked for C5
3a. 17(4) cage at R8C5 can only contain two of 1,2,4 -> no 1,2,4 in R9C67

4. 17(4) cage at R5C5 = {1358/1367/2357} (cannot be {1259/1268} because 1,2 only in R7C4), no 9, 3 locked for C5
4a. 1,2 only in R7C4 -> R7C4 = {12}
4b. Naked triple {124} in R7C4 + R89C5, locked for N8, 4 also locked for C5
4c. 9 in C5 only in R23C5, locked for N2
[I missed 9 in R23C5, CPE no 9 in R3C3, 28(4) cage at R2C3 = {4789/5689}, 9 locked for R2, which would have simplified later step(s) and led to more opportunities, which Afmob and wellbeback used.]

5. 45 rule on N69 2 innies R4C9 + R9C7 = 8 = [26/35/53]
5a. R4C9 + 8(3) cage at R5C8 must contain 3, locked for N6

6. 17(4) cage at R8C5 contains 4 = {1349/1457/2348/2456}
6a. 8,9 of {1349/2348} must be in R9C6 -> no 3 in R9C6

7. 45 rule on R1234 1 outie R5C6 = 1 innie R4C1 + 6, R4C1 = {123}, R5C6 = {789}

8. Naked triple {123} in R4C1 + R5C12, locked for N4 and 24(6) cage at R4C1, no 1,2,3 in R7C1

9. 17(3) cage at R3C2 = {368/458/467} (cannot be {179/269/278/359} because R4C23 must contain two of 4,5,6,8), no 1,2,9
9a. 3,7 of {368/467} must be in R3C2 -> no 6 in R3C2

10. 15(4) cage at R3C6 = {1239/1248/1257/1347} (cannot be {1356/2346} because R5C6 only contains 7,8,9), no 6
10a. R5C6 = {789} -> no 7,8,9 in R3C67 + R4C6

11. Hidden killer triple 7,8,9 in 12(3) cage at R1C1, 24(6) cage at R4C1 and 14(3) cage at R8C1 for C1, none of these cages can contain more than one of 7,8,9 -> each must contain one of 7,8,9
11a. 12(3) cage = {129/138/147/237} (other combinations don’t contain one of 7,8,9), no 5,6
11b. 24(6) cage = {123459/123468/123567}
11c. 12(3) cage = {129/138/237} (cannot be {147} which clashes with 24(6) cage), no 4
11d. Killer triple 1,2,3 in 12(3) cage and R4C1, locked for C1

12. 8(3) cage at R5C8 = {125/134}
12a. 5 of {125} must be in R5C89 (R5C89 cannot be {12} which clashes with R5C3), no 5 in R6C8
12b. 4 of {134} must be in R5C89 (R5C89 cannot be {13} which clashes with R5C23, ALS block), no 4 in R6C8
12c. 8(3) cage = {125} => naked triple {125} in R5C389, locked for R5 => R5C2 = 3 or 8(3) cage = {134}, naked quad {1234} in R5C2389, locked for R5 -> 1,2,3 in R5C2389, locked for R5

13. 45 rule on R89 4 innies R8C3678 = {1369/1378/1567/2359/2368/3457} (cannot be {1279/1459/1468/2458/2467} which clash with R8C25, ALS block)
13a. Killer triple 1,2,4 in R8C2, R8C5 and R8C3678, locked for R8
13b. Min R8C12 = 6 -> max R9C1 = 8

14. 45 rule on C89 3(2+1) outies R2C67 + R4C7 = 19
14a. Max R24C7 = 17 -> min R2C6 = 2
14b. R4C7 = {79} -> R2C67 = 10,12, but cannot be {19} (because no 1,9 in R2C6) -> no 1 in R2C7
14c. R2C67 = 10 doesn’t contain 5, R2C67 + R4C7 cannot be [57]7 -> no 5 in R2C6

[I was a bit slow to spot …]
15. R5C23 + R8C2 = 5 (step 2)
15a. R5C2 cannot be 2 (because R5C3 + R8C2 = 3 = {12} “see” R5C2) -> R5C2 = {13}
15b. R5C23 + R8C2 = [122/311], 1 locked for R5 and C2
15c. 8(3) cage at R5C8 = {125/134} -> R6C8 = 1
15d. 1 in R5 only in R5C23, locked for N4, clean-up: no 7 in R5C6 (step 7)
15e. 3 in R6 only in R6C456, locked for N5

16. 1 in C1 only in 12(3) cage at R1C1, locked for N1
16a. 12(3) cage (step 11c) = {129/138}, no 7

17. 15(4) cage at R3C6 (step 10) = {1239/1248}, no 5

18. R4C9 + R9C7 (step 5) = [26/35/53]
18a. 19(4) cage at R5C7 = {1468/1567/2359/2368/2458/2467} (cannot be {1279/1369/1378} because 1,3,7,9 only in R78C7, cannot be {1459/3457} which clash with 8(3) cage at R5C8)
18b. 19(4) cage = {1468/2359/2368/2458/2467} (cannot be {1567} = {56}{17} which clashes with R4C9 + R9C7, killer combo clash)
18c. 19(4) cage = {1468/2368/2458/2467} (cannot be {2359} = [52]{39} which clashes with R4C9 + R9C7, killer combo clash), no 9 in R78C7

19. R4C9 + R9C7 (step 5) = [26/35/53]
19a. Consider combinations for 19(4) cage at R5C7 (step 18c) = {1468/2368/2458/2467}
19(4) cage = {1468/2368/2467}, no 5
or 19(4) cage = {2458}, locked for C7 => R9C7 = {36} => R4C9 = {25}, killer pair 2,5 in R4C9 and 8(3) cage at R5C8
-> no 5 in R56C7
19b. 19(4) cage = {1468/2368/2467}, 6 locked for C7 => R9C7 = {35} => R4C9 = {35}
or 19(4) cage = {2458} with 2 in R6C7 => R4C9 = {35}
or 19(4) cage = {2458} with R56C7 = {48} => 8(3) cage at R5C8 = {125}, locked for N6 => R4C9 = 3
-> R4C9 = {35}, R9C7 = {35}
19c. R4C9 + R9C7 = {35}, CPE no 3,5 in R789C9
19d. Killer pair 3,5 in R4C9 and 8(3) cage at R5C8, locked for N6
19e. Min R8C9 = 6 -> max R9C89 = 9, no 9 in R9C8, no 8,9 in R9C9

20. 17(4) cage at R8C5 (step 6) = {1349/1457/2348/2456}
20a. R9C7 = {35} -> no 5 in R9C6

21. 17(3) cage at R3C2 (step 9) = {368/458/467}
21a. Consider placements for 3 in N4
R4C1 = 3 => R4C9 = 5 => R4C5 = {68} => 17(3) cage = {458/467} (cannot be {368} which clashes with R4C5)
or R5C2 = 3 => 17(3) cage = {458/467}
-> 17(3) cage = {458/467}, no 3

22. Consider placements for 3 in N4
R4C1 = 3 => 12(3) cage at R1C1 = {129}, locked for N1 => R2C5 = 9 (only remaining place for 9 in 28(4) cage at R2C3), R4C46 = {12} (hidden pair in R4) => 23(4) cage at R3C4 = {2678} (only remaining combination with one of 1,2 and no 9) => R4C4 = 2
or R5C2 = 3 => R4C1 = 2
-> 2 in R4C14, locked for R4
22a. 15(4) cage at R3C6 (step 17) = {1239/1248}, 2 locked for R3

[Taking the forcing chain in step 22 further, omitting some of the explanation]
23. R4C1 = 3 => 12(3) cage at R1C1 = {129}, locked for N1 => R2C5 = 9 => R4C4 = 2, 23(4) cage at R3C4 = {2678}, R7C4 = 1 => R567C5 = 16 = {358} (cannot be {367} which clashes with 23(4) cage), locked for C5 => R34C5 = [76], R3C4 = 8 => R2C3 = 8 (only remaining place for 8 in 28(4) cage at R2C3)
or R5C2 = 3 => R4C1 = 2 => 12(3) cage at R1C1 = {138}
-> 8 in 12(3) cage at R1C1 + R2C3, locked for N1
23a. 28(4) cage at R2C3 = {4789/5689}, 8 locked for R2

24. 9 in C7 only in R24C7
24a. R2C67 + R4C7 = 19 (step 14) = [397]/{37}9/{46}9, no 2 in R2C6, no 2,5 in R2C7

25. 8 in C7 only in 19(4) cage at R5C7 (step 18c) = {1468/2368/2458}, no 7
25a. 15(4) cage at R3C6 (step 17) = {1239/1248}
25b. Consider placement for 3 in R4
R4C1 = 3 => R5C6 = 9 (step 7), R4C6 = 1, R4C9 = 5 => R9C7 = 3 (step 5), R3C67 = [32] => 19(4) cage = {1468}
or R4C9 = 3 => R9C7 = 5 (step 5) => 19(4) cage = {1468/2368}
-> 19(4) cage = {1468/2368}, no 5, 6 locked for C7
25c. R4C1 = 3 => R5C6 = 9 , R4C6 = 1, R4C9 = 5 => R9C7 = 3, R3C67 = [32] (as in step 25b)
or R4C9 = 3 => R4C1 = 2, R5C6 = 8 (step 7) => R3C67 + R4C6 = {124}
-> no 3 in R3C7
25d. R2C67 + R4C7 (step 24a) = [397]/{37}9/[64]9, no 4 in R2C6

26. Consider placement for 3 in R4
R4C1 = 3 => R5C6 = 9 (step 7), R4C9 = 5 => R9C7 = 3 (step 5) => R89C5 + R9C6 = 14 = {24}8
or R4C9 = 3, R4C1 = 2 => R5C6 = 8 (step 7), R9C7 = 5 (step 5) => R89C5 + R9C6 = 12 = {14}7/{24}6
-> 8 must be in R59C6, locked for C6, R9C6 = {678}, no 9

27. 17(3) cage at R6C6 = {269/359/467}
27a. 2,4 of {269/467} must be in R6C6 -> no 6,7 in R6C6

[At last more placements …]
28. Consider combinations for 17(3) cage at R6C6 (step 27) = {269/359/467}
17(3) cage = {269}, locked for C6 => R5C6 = 8 => R9C6 = 7 => R2C6 = 3
or 17(3) cage = {359} contains 3, locked for C6
or 17(3) cage = {467}, locked for C6 => R2C6 = 3
-> 3 must be in R2C6 + 17(3) cage, locked for C6
28a. R3C67 + R4C6 = {124} = 7 -> R5C6 = 8 (cage sum), R4C1 = 2 (step 7), R5C23 = [31], R8C2 = 1 (hidden single in N7, or from step 15b)
28b. 12(3) cage at R1C1 = {138} (only remaining combination), locked for C1 and N1
28c. 8(3) cage at R5C8 = {125} (only remaining combination), 2,5 locked for R5 and N6 -> R4C9 = 3, R9C7 = 5 (step 5)
28d. 28(4) cage at R2C3 = {4789/5689}, 8 locked for N2

29. 8 in R4 only in R4C23, locked for N4, clean-up: no 6 in R7C2
29a. 17(3) cage at R3C2 (step 21a) = {458} (only remaining combination), CPE no 5 in R6C2, clean-up: no 9 in R7C2
29b. Naked triple {458} in R347C2, locked for C2, 4 also locked for 17(3) cage, no 4 in R4C3

30. 6 in R4 only in R4C45, locked for N5 and 23(4) cage at R3C4, no 6 in R3C45
30a. R5C5 = 7 -> 17(4) cage at R5C5 (step 4a) = {1367/2357}, no 8
30b. Naked triple {356} in R467C5, locked for C5 -> R23C5 = [89]
30c. 28(4) cage at R2C3 = {4789/5689} -> R2C3 = 9

31. R4C7 = 9 (hidden single in C7), R4C8 = 7

32. 23(4) cage at R3C4 contains 6 and 9 = {1679/3569}, no 4
32a. 3,7 only in R3C4 -> no 1,5 in R3C4

33. Caged X-Wing for 4 in 17(3) cage at R3C2 = {458} and R3C67 + R4C6 = {124} for R34, no other 4 in R34

34. 21(4) cage at R2C9 contains 3 = {3468/3567}, no 1,2, 6 locked for N3
34a. R4C7 = 9 -> R2C67 = 10 (step 14) -> R1C89 + R2C8 = 15
34b. 9 in N3 only in R1C89 = 15 -> R1C89 + R2C8 = {159/249}, no 3,7,8
34c. 8 in N3 only in 21(4) cage = {3468} -> R2C9 = 4, R3C89 = {68}, locked for R3
34c. R1C89 + R2C8 = {159} -> R1C9 = 1, R12C8 = [95], R3C7 = 2
34d. R1C5 = 2 -> R1C67 = 8 = [53], R2C7 = 7, R2C6 = 3 (cage sum), R123C4 = [467], R3C23 = [45], R4C23 = [58], R7C2 = 8 -> R6C2 = 6, R6C79 = [48], R6C13 = [97], R5C1 = 4, R7C1 = 5 (cage sum)

35. R89C5 = [41], R9C7 = 5 -> R9C6 = 7 (cage sum)

and the rest is naked singles.

I'll rate my walkthrough for Messy One 14 at least Hard 1.75. I used a lot of forcing chains.