This is a record of the original Tag solution. Some typos have been corrected. There is also some annotation, while other red comments and steps were in the original Tag.
Andrew1. R12C1 = {69/78}
2. R12C9 = {17/26/35}, no 4,8,9
3. R56C1 = {49/58} (cannot be {67} which would clash with R12C1)
3a. Killer pair 8/9 in R1256C1 for C1
4. R56C2 = {69/78}
4a. Killer pair 8/9 in R56C12 for N4
5. R56C8 = {18/27/36/45}, no 9
6. R56C9 = {15/24}
7. 20(3) cage in R234C5 = {389/479/569/578}, no 1,2
8. 7(3) cage in N45 = {124}
9. 24(3) cage in R7C234 = {789}, locked for R7
10. 13(4) cage in N14 = {1237/1246/1345}, no 8,9
11. 26(4) cage in N36, no 1
12. 14(4) cage in N7, no 9
13. 45 rule on N7 2 outies R79C4 = 16 = {79}, locked for C4 and N8
14. 8 in R7 locked in R7C23, locked for N7
15. 45 rule on N9 2 outies R79C6 = 7 = {16/25/34}
15a. 8 in N8 locked in 22(5) cage
16. 3 in N4 locked in R4C123, locked for R4
17. 45 rule on N1 2 outies R2C4 + R4C1 = 7 = {16/25/34}
18. 45 rule on N3 2 outies R2C6 + R4C9 = 13 = {49/58/67}
19. 45 rule on C6789 3 innies R158C6 = 20, no 1,2
20. 45 rule on C9 2 innies R34C9 – 9 = 1 outie R9C8, max R34C9 = 17 -> max R9C8 = 8
Time to start thinking harder! Probably need to work the combinations more. It looks like there are still too many candidates to be able to do any useful contradiction moves yet.
I also noticed the following which I haven't included above because they don't (yet) lead directly to any eliminations
Valid combinations for 23(4) cage in N78 are {1679/2579/3479/3569}
45 rule on R7 3 innies R7C159 = 11, R7C678 = {136/145/235}, R7C159 = {146/245/236}
45 rule on C1234 3 innies R158C4 = 12 = {138/156/246/345} -> R2346C4 = {1268/1358/2348/2456}
45 rule on C1 3 outies R239C2 = 10
45 rule on C1 5 innies R34789C1 = 17 = 123{47/56}
45 rule on C1 2 innies R34C1 – 3 = 1 outie R9C2
45 rule on N4 3 innies R4C123 – 10 = 1 outie R6C4
45 rule on N6 3 innies R4C789 – 12 = 1 outie R6C6
EdThanks for getting us started Andrew. Had to use my favourite combining steps and then doing hypotheticals to make any impression. Probably too many steps for a tag solution.
21. 7(3)n4 = {124} -> no 4 r6c1
21a. no 9 r5c1
22. 8(2)c9 = {17/26/35}
22a. 6(2)c9 = {15/24}
22b. -> 2 locked for c9 in these 2 cages
23. 26(4) n3 = 9{278/368/458/467} ({5678} blocked by 8(2)n3)
24. 1 locked for r6 in r6c349. Here's how.
24a. r6c34 = {24} -> r6c9 = 1 ([15] in r56c9 blocked by 1 in r5c3)
24b. only other options for r6c34 = {12/14} include 1
24c. -> 1 locked for r6
24d. no 8 r5c8
25. 13(4)n1 = 1{237/246/345}(no 8,9)
25a. 15(2)n1 must have [8/9], not both
25b. -> 24(5)n1 must have [8/9] for n1 -> {24567} combo not possible
25c. 24(5)n1 cannot have both 8 and 9 because of 15(2) -> {12489} combo not possible
25d. 24(5)n1 cannot have both 7 and 9 because of 15(2) -> {12579/13479} blocked
26. 9 in c2 locked in 15(2) or r7c2
27. Now need some hypotheticals to make progress
27a. "45" n4 -> r6c4 + 10 = r4c123 = 11, 12 or 14 and must have 3
27b. r6c4 = 1 -> r4c123 = 11 = {137} ({236} blocked by 2 in r56c3)
27c. r6c4 = 2 -> r4c123 = 12 = {237} ({345} blocked by 4 in r56c3)
27d. r6c4 = 4 -> r4c123 = 14 = {347}
27e. r6c4 = 4 -> r4c123 = 14 = {356}
28. However, r4c123 = 14 = {356} is blocked. Here's how.
28a. 20(4)n2 now = {1478/1568/2378/2468/2567/3458/3467}
28b. "45" n1 -> 2 outies = 7.
28c. Since r6c4 = 4 (step 27e) in this hypothetical -> r4c1 != 3
28c. -> r4c23 = 3{5/6}
28d. the only combo's in 20(4) that allow 3{5/6} are {3458/3467} with {48/47} in r34c4
28e. but this means 2 4's in c4
28f. -> r4c123 cannot be {356}
29. r4c123 = {137/237/347} = 37{1/2/4}(no 5 or 6)
29a. no 12 r2c4
30. 7 Locked in r4c23 for n4, r4 and must be in 20(4)n2 = 7{148/238/256/346}
30a. no 6 r2c6 (step 18)
31. 13(2)n4 = {58}(hidden 5 n4): Locked n4, c1
32. 15(2)n1 and n4 = {69}:locked for c12, n14
33. 14(4)n7 = {1247/2345} = 24{17/35}: 2 and 4 locked n7
33a. 5 only in r9c2 -> no 3 r9c2
34. 23(4)n7 = {1679/3569} = 69{17/35}
35. 13(4)n1 = {1237/1345} = 13{27/45}
36. 24(5) n1 must have 8 for n1
36a. 13(4) must have both 1 and 3, only 1 of which can 'hide' in r4c1 -> {13578} blocked from 24(5) (note: 3 can't hide in r2c4 when 1 in r4c1 since 2 outies n1 = 7)
36aa. 24(5)n1 now = {12678/14568/23478/23568}
36b. 24(5), only combo's with 5 also have 6 {14568/23568} which is only in r2c4 -> no 5 r2c4
36c. no 2 r4c1
Andrew37. Any 6 in the 24(5) cage in N12 must be in R2C4. If R2C4 = 6, R4C1 = 1 (step 17) -> 1 in N1 must be in 24(5) cage
37a. No {23568} in 24(5) cage
38. 45 rule on C123 4 outies R2679C4 – 13 = 2 innies R4C23; R79C4 = 16 (step 13) -> R26C4 + 3 = R4C23
Try hypotheticals
38a. R6C4 = 1 -> R4C123 = {137}, R4C23 = 7{1/3} -> R2C4 = {46}
38b. R6C4 = 2 -> R4C123 = {237}, R4C23 = {27} -> R2C4 = 4
38c. R6C4 = 4 -> R4C123 = {347}, R4C23 = 7{3/4} -> R2C4 = {34} but cannot be 4 so R4C23 = {37}, R2C4 = 3
39. Now try looking at those hypotheticals starting from R2C4 and the effect on the 20(4) cage in N254 which contains 7{148/238/256/346} (step 30)
39a. R2C4 = 3 -> R6C4 = 4, R4C1 = 4, R4C23 = {37}, 20(4) = {2378} ({3467} clashes with R6C4)
39b. R2C4 = 4 and R6C4 = 1 -> R4C1 = 3, R4C23 = {17} but 20(4) = {1478} clashes with R2C4 so cannot have R2C4 = 4 and R6C4 = 1
39c. R2C4 = 4 and R6C4 = 2 -> R4C1 = 3, R4C23 = {27}, 20(4) = 7{238/256}
39d. R2C4 = 6 -> R6C4 = 1, R4C1 = 1, R4C23 = {37}, 20(4) = {2378} ({3467} clashed with R2C4)
40. To summarise steps 38 and 39
40a. There is a one-one relationship between R2C4 and R6C4, R2C4 = 3,4,6 -> R6C4 = 4,2,1
40b. 20(4) cage in N254 = 7{238/256} = 27{38/56}, no 1,4
41. To continue further and look at the effect of these hypotheticals on the 24(5) cage in N12
41a. R2C4 = 3, R6C4 = 4, R56C3 = {12}, 24(5) cage = {23478} -> R1C2 = 2, R123C3 = {478}
41b. R2C4 = 4, R6C4 = 2, R56C3 = {14}, 24(5) cage = {23478} -> R1C2 = {2378}, R123C3 = {2378}
41c. R2C4 = 6, R6C4 = 1, R56C3 = {24}, 24(5) cage = {12678} -> R1C2 = 2, R123C3 = {178}
41d. R2C4 = 6, R6C4 = 1, R56C3 = {24}, 24(5) cage = {14568} -> R1C2 = 4, R123C3 = {158}
41e. In summary, combining all these hypotheticals, R1C2 = {23478}, no 1,5, R123C3 = {1234578}
EdLook good - only just missed the first placement. I see my hypothetical disease is contagious. Great work Andrew.
We are probably going to have to move to other areas now - one of Para's far-flung relationship between two digits in a cage combination, or one of Richard's wild "45"s.
Quote:"Andrew"
"41b. R2C4 = 4, R6C4 = 2, R56C3 = {14}, 24(5) cage = {23478} -> R1C2 = {2378}, R123C3 = {2378}"
This one is not quite right: because r4c3 = {237} -> 1 of 2,3 or 7 have to be in r1c2
41f. no 8 in r1c2
42. r7c2 = 8 (hidden single c2)
43. an extention of Andrews step 39
43a. r2346c4 = [3{28}4/4{38}2/4{56}2/6{28}1]
43b. 2 locked for c4
ParaJust a two quick steps. Very basic.
44. 9(2) in R5C8 can't be {45} because of 6(2) in R5C9.
45. 45-test on C6789: R158C6 = 20: no options with 4 in R15C6
Two more steps
46. 1 locked in R12356C3.
46a. 24(5) = {12678/14568}: 1 locked in R123C3.
46b. 24(5) = {23478}: R2C4 = 3 -->> R4C1 = 4 -->> R56C3 = {12}
46c. 24(5) = {23478}: R2C4 = 4 -->> R4C1 = 3 -->> R4C3 = {27} -->> one of {27} in R123C3 : Killer Pair {27} in R1234C7 -->> R56C3 = {14}
46d. R89C3: no 1
46e. 23(4)in R8C2: only 1 in R8C2. So R8C2: no 7.
47. Building on 46-->> R1C2: no 3
47a. 24(5) = {12678/14568}: R1C2: no 3
47b. {23478}: R2C4 = 3: R1C2: no 3
47c. 24(5) = {23478}: R2C4 = 4 -->> R4C1 = 3 -->> R4C3 = {27} -->> one of {27} in R123C3: can't have both {27} in R123C3 so R1C2 = {27}: R1C2: no 3
RichardOk
just trying to catch up and see a couple of things. Hope I've picked up the marks pick correctly !!
48. 13(3)n2 - no 6,8 in r1c5
48a. only options with 8 are {148} -r1c45 must be {14} or {238} r1c5 must be 2
48b. only options with 6 is {256} - r1c5 must be 2
49. 45 on n5 - 5 innies total 23. no placement with 4 or 8 at r6c6
49a. {12479} - 7 must be at r6c6
49b. {14567} - 7 must be at r6c6
49c. {23459} - 3 must be at r6c6
49d {23468} - ditto
49e. {13469} - ditto
49f. {12569} - no 4 or 8
49g. {12578} - 7 must be at r6c6
49h {13568} - 3 must be at r6c6
49i. {12389} - ditto
50. 45 on n6. r4c789 minus r6c6 is 12 -> no 5 at r6c6 because:
50a. when r6c6 is 5, r4c789=17(3)={269} or {458}
50b. {269} would eliminate all possibilities for 6(2) and 9(2) n6 since -> 6(2)={15}, 9(2)=no options)
50c. {458} would eliminate all possibilities from 6(2)n6
That brings up the half century, which is generally more than England's batsmen can do.
ParaOK some building on Richards steps.
51. 45 on N6. R4C789 - R6C6 = 12
51a. R6C6 = 9 -->> R4C789 = {489}
51b. R6C6 = 7 -->> R4C789 = {289/469}: {568} clashes with 9(2) and 6(2) cage in N6
51c. R6C6 = 6 -->> R4C789 = {189/468}: {459} clashes with 6(2) cage in N6
51d. R6C6 = 3 -->> R4C789 = {159/249}: {258/456} clashes with 6(2) cage in N6; {168} clashes with 6(2) and 9(2) cage in N6
51e. R6C6 = 2 -->> R4C789 = {158/248}: {149} clashes with 6(2) cage in N6
52. 45 on N6. R4C789 - R6C6 = 12
52a. R6C6 = 9 -->> R4C789 = {489}: 18(3) in R5C7 can't be {189/459}
52b. R6C6 = 7 -->> R4C789 = {289/469}
52c. R6C6 = 6 -->> R4C789 = {189/468}: 18(3) can't be {468}
52d. R6C6 = 3 -->> R4C789 = {159/249}
52e. R6C6 = 2 -->> R4C789 = {158/248}
52f. Concluded 18(3) in R5C7 = {279/369/378/567}: no 1, 4
53. 45 on N6. R4C789 - R6C6 = 12
53a. R6C6 = 9 -->> R4C789 = {489}: 9(2) no [18].
53b. R6C6 = 7 -->> R4C789 = {289}: 9(2) no [18]; {469} -->> 6(2) = {15}: 9(2) no [18]
53c. R6C6 = 6 -->> R4C789 = {189/468}: 9(2) no [18]
53d. R6C6 = 3 -->> R4C789 = {159}: 9(2) no [18]; {249} -->> 6(2) = {15}: 9(2) no [18]
53e. R6C6 = 2 -->> R4C789 = {158/248}: 9(2) no [18]
53f. Concluded 9(2) in N6: no [18]
AndrewNice moves Para and Richard.
A small extension to part of Para's moves.
54. R6C6 = 7 -->> R4C789 = {289} (part of step 52b), R56C9 = {15}, R56C8 = {36}, R56C7 = {47} (4 hadn't been eliminated at step 52b) but 18(3) cage cannot be {477} -> R4C789 cannot be {289}
Then an extension to one of Richard’s moves
55. “49. 45 on n5 - 5 innies total 23. no placement with 4 or 8 at r6c6
49a. {12479} – 7 must be at r6c6
49b. {14567} - 7 must be at r6c6”
In both cases from step 52b as amended by step 54, R6C6 = 7 -->> R4C789 = {469} which clashes with {1249/1456} in R4C456 -> 5 innies in N5 cannot be {12479/14567}
“49c. {23459} - 3 must be at r6c6
49e. {13469} – ditto
49i. {12389} – ditto”
In each of these cases from step 52d, R6C6 = 3 -->> R4C789 = {159/249} which clashes with the 9 in the N5 innies -> 5 innies in N5 cannot be {12389/13469/23459}
55a. The remaining combinations for 5 innies in N5 are {12569/12578/13568/23468}
ParaOne more elimination
56. 45 on C9: R34C9 – R9C8 = 9
56a. R9C8 = 1 -->> R34C9 = {46} -->> R56C9 = {15}: 22(4) can't be {1489/1579/1678}
56b. R9C8: no 1
I posted that Para’s 56a also works as
56a. R9C8 = 1 -->> R34C9 = {46} -->> R56C9 = {15}, no valid combinations for R12C9
RichardContinuing on from the good work around n6 . . .
somebody check step my step 58. Can I use UR in this instance? not one of my stronger techniques but it looks right.
[edited to recognise the fact that 57 duplicated one of Para's moves and also to add a cleanup on step 59]
57. 45 on n6. innies=30
57a. {35679} not possible because it clashes with 9(2)
57b. {25689} ditto
57c. {45678} ditto
57d. {24789} - if r6c7 was 4, r5c6 would need to be 7 which doesn't fit the combos for 18(3)n56
57e. {34689} - if r6c6 was 4, r5c6 would need to be 3 which doesn't fit the combos for 18(3)
57f. {15789} - no 4
57g. conclusion -> no 4 in r6c7
58. unique rectangle at r7c34 r9c34 with {79} in 2 row, 2 columns, 2 nonets and 2 cages
58a. can't have 7 or 9 at r9c3 otherwise puzzle has non-unique solution
59. Can't have 1 at r1c9 as it eliminates all places for 1 in n6
59a. r1c9=1 -> no 1 at r56c9
59b. r1c9=1 -> no 1 at r1c45 -> r3c6=1 -> no 1 at r4c78
no other places in n6 for 1
59c. clean-up - no 7 at r2c9
EdOK Andrew. They all look good. And you've opened up a little teeny crack. My final step isn't as productive as first thought: found a mistake. So, don't get a headache over them: they make you really concentrate.
Marks pic (edited out): didn't include the 7,9 in r9c3 since there is some disquiet about: we must not be desperate enough yet.
First, one extra part of step 52c. R6C6 = 6 -->> R4C789 = {189/468}
60. The {189} is blocked.
60a. since r6c6 = 6 -> r5c2 = 6 -> 6 for n6 must be in r4c789 = {468} only
60b. R6C6 = 9 -->> R4C789 = {489}
60c R6C6 = 7 -->> R4C789 = {469}
60d. R6C6 = 6 -->> R4C789 = {468}
60e. R6C6 = 3 -->> R4C789 = {159/249}
60f. R6C6 = 2 -->> R4C789 = {158/248}
61. Now combining this with 5 innies n5 = {12569/12578/13568/23468}
61a missing but not worth changing now unless the reference to 61h is changed in step 6261b. R6C6 = 9 -->> R4C789 = {489} and rest of innies n5 = {1256}
61c R6C6 = 7 -->> R4C789 = {469} and rest of innies n5 = {1258}
61d. R6C6 = 6 -->> R4C789 = {468} and rest of innies n5 = {1259} ({1358/2348} both blocked by 8 in r4c789)
61e. R6C6 = 3 -->> R4C789 = {159} and rest of innies n5 = {2468}
({1568} blocked by 5 in r4c789)61f. R6C6 = 3 -->> R4C789 = {249} and rest of innies n5 = {2468}
Why wasn’t {1568} also included? It doesn’t seem to be immediately blocked by r4c789. Hopefully this doesn’t affect later moves61g. R6C6 = 2 -->> R4C789 = {158} and rest of innies n5 = all blocked by 5 and 8 in r4c789
61h. R6C6 = 2 -->> R4C789 = {248} and rest of innies n5 = {1569} ({1578/3468} blocked by 8 in r4c789
62. However,61h is also blocked
62a. 2 in r6c6 and other 4 innies = {1569}
62b. -> 1 in r6c4 -> r4c123 = 11 = [1]{37} (step 27b)
62c. -> rest of 20(4) = {28} only (not 5,6 or 9!):
62d. -> r6c6 !=2
63. In summary
63a. no 2 r6c6
63b. r4c789 = {489/469/468/159/249}
63c. 5 innies n5 = {12569/12578/23468}
63d. 2 locked n5
64. 22(4)n5 = {1579/3469/3478} ({1489/1678/3568/4567} blocked by 5 innies)
65. No 2 in r9c8
65a. "45"c9 -> r9c8 + 9 = r34c9
65b. r9c8 = 2 -> r34c9 = 11 = [38] -> r239c8 = {69}[2} -> 9(2)n6 Clash
..............................{47} -> r239 = {69}[2} -> 9(2)n6 Clash
…………………..{56} -> r239 = {78}[2] not possible, no 9 in 26(4) clashes with step 2365c -> no 2 r9c8
66. Building more on steps 43a, 41 and 47
66a. r2346c4 = [3{28}4] and r1c2 = 2 -> 13(3)n2 = [148]/{157}
66b. r2346c4 = [4382] and r1c2 = 2 -> 13(3)n2 = {157}
66c. r2346c4 = [4382] and r1c2 = 7 -> 13(3)n2 = {56[2]56}
66d. r2346c4 = [4{56}2] and r1c2 = 2 -> 13(3)n2 = {139/157}
66e. r2346c4 = [4{56}2] and r1c2 = 7 -> 13(3)n2 = {139/38[2]38} ({256}blocked by r3c4)
66f. r2346c4 = [6{28}1] and r1c2 = 2 -> 13(3)n2 = [319/418/517]
66g. r2346c4 = [6{28}1] and r1c2 = 4 -> 13(3)n2 = [319/517] ({238} blocked by r3c4)
For completeness r2346c4 = [6{28}1] and r1c2 = 7 is blocked by r56c3 = {24}66h. In summary 13(3)n2 = {139/148/157/238/256} ({247/346} eliminated)
Andrew67. R4C789 = {489/469/468/159/249} (step 63b) [4/5], killer pair 4/5 in R4C789 and R56C9 for N6, no 5 in 18(3) cage in N56 = {279/369/378}
13(4) in N14 13{27/45}
24(5) in N12 {12678/14568/23478}
13(3) in N2 {139/148/157/238/256}
26(4) in N36 9{278/368/458/467}
20(4) in N254 27{38/56}
18(3) in N56 {279/369/378}
14(4) in N7 24{17/35}
23(4) in N78 69{17/35}
R4C123 37{1/2/4}
R4C789 {159/249/468/469/489}
5 innies in N5 {12569/12578/23468}
5 innies in N6 89{157/247/346}
RichardAnother couple building from Ed's position.
68. no 2 at r4c7 or r4c8 as it removes all placements for 2 in r6
68a. r4c78=2 -> no 2 in n6r6
68b. r4c78=2 -> no 2 in n6r5 -> r5c3=2 -> r6c34<>2
68c. no other 2's in r6 - so can't have a 2 at r4c7 or r4c8
ParaOK a bit more. We'll get there.
69. 45 on C9: R9C8 + 9 = R34C9
69a.R9C8 = 3 -->> R34C9 = {39}: {48} clashes with 22(4) in N9; 8 then locked in R789C9, now R34C9 = [75] clashes with 22(4) in N9
69b.R9C8 = 4 -->> R34C9 = {49}: {58/67} blocked because R23C8 would be {49} and then 2 4's in C8.
69c.R9C8 = 5 -->> R34C9 = {59}: {68} blocked: R34C9 = {68} -->> R23C8 = {39} -->> R56C8 = {27} -->> R56C9 = {15} -->> no options left for 8(2)N1.
69d.R9C8 = 6 -->> R34C9 = {69}: [78] blocked: R34C9 = [78] -->> R23C8 = {29}: no options left for 9(2) in N6
69e.R9C8 = 7 -->> R34C9 = {79}
69f.R9C8 = 8 -->> R34C9 = {89}
69g. Conclusion 9 locked in r34C9 for C9 and 26(4) in R2C8.
70. 45 on C9: R9C8 + 9 = R34C9
70a. R9C8 = 3 -->> R34C9 = {39} -->> R23C8 = {68} -->> R56C8 = {27} -->> R56C9 = {15}: No options left for 8(2) in N1
70b. R9C8 = 4 -->> R34C9 = {49} -->> R56C9 = {15} -->> R12C9 = {26} -->> R789C9 = {378}
70c. R9C8 = 5 -->> R34C9 = {59} -->> R23C8 = {48} -->> R56C9 = {24} -->> R789 = {368} -->> R12C9 = {17}
70d. R9C8 = 6 -->> R34C9 = {69} -->> R56C9 = {24}(no {26} in R12C9, only place left for 2) -->> R56C8 = {36} -->> R789C9 = {358/178}
70e.R9C8 = 7 -->> R34C9 = [79] -->> R23C8 = {28/46} -->> R12C9 = {35} -->> R56C9 = {24} -->> R56C8 = {36} -->> R23C8 = {28} -->> R789C9 = {168}
70f. R9C8 = 8 -->> R34C9 = {89} -->> R789C9 = {167/347/356}
71. Conclusions:
71a. R3C9 + R9C8: no 3
71b. 22(4) in R7C9: no {4567}
71c. 8 locked in 22(4) in R7C9 for N9
[Para overlooked that in step 70d R9C8 = 6 is blocked by R56C8 = {36}]RichardAnother move or two.
I'm also assuming 7,9 are gone from r9c3 - no sense ignoring a valid technique that gets rid of a couple of candidates, is there?
[edit - added an extra bit following step 73]71. 45 on c9 - outies r239c8 total 17(3)={278}/{368}/{458} - must use 8 -locked for c8 ({467} blocked by 9(2)n6 no other combos)
72. Can't have an 8 at r8c7 or r9c7 as it eliminates all placements for 8 in c8
72a. r89c7=8 -> r9c8<>8
72b. r89c7=8 -> r456c7<>8 -> r4c9=8 -> r23c8<>8
72c. no places left with 8 in c8
73. 45 on n9 innies total 23
73a. can only have {12569} or {23459} - no 7 so remove 7 from 20(3)n89
73b. {13469} - blocked by 22(4)n9
73c. {14567} - blocked by 22(4)n9
73d. {23567} - blocked by 22(4)n9
73e. {12479} - blocked because r7c78 would need to be {41} -> rest={279} which would require r9c6 being 2 which breaks the 20(4)n89
74. 45 on c89. innies r1478c8=19(4)={1459}/{1369}/{1279} - {1279} requires 7 at r1c8 - so no 2 at r1c8
74a. other combos {15
67}/{
2359}/{
2647}/{
3754} blocked by 9(2)n6
75. adding to step 73: can't have a 3 in r8c8
75a. consider only combo with 3 is {23459} - r7c78={23}/{25}/{34}/{35}
1) 9(2)n6 restricts {23} in r8c78
2) 17(3) from step 71 and 9(2) restrict {24}/{25}/{34}/{35} in r8c78
75b. {23}/{34}/{35} -> no 3 in r8c8
75c. {25} -> {349} in remaining innies. with 2 in r7c8 can only have 9 in r8c8, with 5 in r7c8 can only have 4 or 9 in r8c8
76. {259} now locked in these innies in n9 - not possible in 20(4)RichardJust spotted another couple of steps stemming from the work I just did in n9.
77. 45 on c9 innies total 31.
77a. 6(2)n6 blocks {45679}
77b. 8(2)n3 blocks {35689}
77c. {25789} not possible because there's no 2 left
77c. only combos {16789}/{34789} - no combo with 5, so no 5 in r34c9
78. (cleanup) 45 on n3 - outies= 13 - no 8 in r2c6
79. 45 on n36 - outies r2346c6=18(4)={1269}/{1359}/{1467}/{2349}/{2358}/{2367}/{2457}/{3456}
79a. 45 on c6789 innies r158c6=20(3) - {569} blocked by the 18(4)
79b. only valid combos left in r158c6=20(3) are {389}/{578}/{479} - no 6
Paraquickie
80. 8(2) in R1C9: no [71]
80a. 2 and 5 locked in cages 8(2) + 6(2) in C9.
80b. 6(2) = {15} -->> 8(2) = {26} (needs to use 2)
80c. 6(2) = {24} -->> 8(2) = {35} (needs to use 5)
81. 8 in C89 locked in cages 26(4) R2C8 and 22(4) in R7C9 so both need 8.
81a. 26(4): no {4679}
Ok two more quick simple ones.
82. 24(5) in R1C7 can't have combo of {23/36} because of 8(2) in R1C9.
82a. 24(5): no {13569/23469/23478/23568}
83. 45 on N3: 2 outies = 13; R2C6: no 8.
OK a few more.
84. 26(4) in R2C8: no {3689}: clashes with 8(2) in R1C9.
84a. 26(4): no 3 or 6
84b. Clean up: R9C8: no 6 (step 70), R2C6: no 7
85. 24(5) in R1C7 requires one of 3,6 in N3: no {12489/12579}
85a. (oops, should have seen that before) 1 locked in 24(5) in N3: no {24567}
A few more.
86. 24(5) in R1C7 needs one of {459} in R2C6: no {12678}
86a. 24(5): no 2
87. Combining 45-test N3, step 70 and cage combinations 24(5) in R1C7 and 26(4) in R2C8
87a. R2C6 = 4 --> R4C9 = 9: 8 locked in N1 in 26(4): so 24(5) = {13479} : R3C9: no 7.
87b. R2C6 = 9 -->> 24(5) = {13479}: R3C9: no 7
87c. R2C6 = 5 -->> R4C9 = 8 -->> R3C9 = 9(step 71) : R3C9: no 7.
87d. R3C9: no 7
87e. Clean up: R9C8: no 7(step 70)
One more
88. R4C7: no 1, because of 1's in N3.
88a. R123C7 = 1: R4C7: no 1
88b. R1C8 = 1 -->> R3C6 = 1: R4C7: no 1
RichardWow - Para's been busy!!
Let's notch up the 90 . . .
89. 45 on n6. innies total 30.
89a. can't use any combo with {67} or {26} because of the 9(2)
89b. only combo with a 5 is {15789} - musdt use the 1 at r4c8 - no 5 at r4c8
90. 45 on c8 r1234789c8 total 36 (I know it's a big number but don't panic)!!
90a. can't use 2&3 because of the 9(2)c8
90b. can't use 2&3, 2&5 or 5&6 in r123c8 because of the8(2)n3
90c. only 2 possibilities {1245789} and {1345689}
90d. {1345689} - r23c8 can only be {45} or{58} - so from 90b - r1c8 can't be 6
ParaOk now every cell has at least one digit eliminated.
91. R3C6: no 9
91a. R3C6 = 9 -->> R4C678 = {125} -->> R4C23 = {37} -->> R4C4 = 8 -->> R4C5 = 6: -->> R23C5 = {59}: 2 9's in N2.
Ok some big eliminations and digit number 2.
92. R4C789 = {489/469/468}: can't be {159}
92a. R6C6 = 3 -->> R4C789 = [519] -->> R2C6 = 4 --> R34C6 = [29]: contradiction: 2 9's in R4.
92b. R4C789: no 1,5
92c. 4 locked in R4C789 for R4 and N6
92d. R56C9 = {15} locked for C9
92e. Clean up: R2C4: no 3; R6C4: no 4.
92f. 4 locked in N4 for C3
93. R12C9 = {26} locked for C9 and N3
93a. 22(4) in R7C9 = {3478}: locked for N9
93b. 26(4) in R2C8 = {4589}: 5 locked in R23C8 for C8 and N3
93d. 5 locked in R4 for N5
94. 22(4) in R5C4 = {3469/3478}: 3 (and 4) locked for N5
95. 10(3) in R7C6 needs 3 or 4. Only place is R7C6: R7C6 ={34}
95a. R9C6 = {34} (45 on N9)
95b. {34} locked for N8 and C6 in R79C6
95c. Clean up: R4C9: no 9.
95d. R3C9 = 9 (hidden single in C9)
95e. Naked Pair {34} in R7C69 locked for R7
Ok the puzzle broke open, and ran to the end. So here are the remaining 79 digits.
96. Hidden Pair {12} in R34C6
96a. 17(4) = {1268}: R4C78 = [86]; R4C9 = 4
96b. R2C6 = 9; R7C9 = 3; R7C6 = 4; R9C6 = 3
96c. R9C8 = 4; R8C1 = 4 (both hidden)
96d. R12C1 = [96]; R2C4 = 4; R4C1 = 3(45 on N1)
96e. R12C9 = [62]
97. Naked Pair {27} in R4C23: locked for R4, N4 and 20(4) in R3C4
97a. R4C4 = 5; R4C6 = 1; R3C6 = 2; R6C4 = 2; R4C5 = 9
97b. R56C8 = [27](only possible combination)
97c. R6C6 = 6; R7C8 = 1; R8C8 = 9; R7C7 = 5; R1C8 = 3
97d. R56C2 = [69]; R56C7 = [93]
97e. R7C1 = 2; R7C5 = 6
98. 14(4) in R7C1 = 24{17}: {17} locked in R9C12 for N7 and R9
98a. R7C34 = [97]; R9C4 = 9; R89C9 = [78]
98b. Hidden singles: R8C2 = 3; R3C2 = 4; R1C7 = 4; R2C2 = 5
98c. R3C1 = 1; R9C12 = [71]; R23C7 = [17]; R23C8 = [85]
98d. R3C4 = 6(hidden)
99. 20(3) in R2C5 = [389](only possible combination)
99a. R23C3 = [73]; R1C23 = [28]; R4C23 =[72]
99b. R1C4 = 1; R6C5 = 4; R5C456 = [378]
99c. R56C1 = [58]; R56C3 = [41]; R56C9 = [15]
99d. R1C56 = [57]; R9C5 = 2; R8C456 = [815]; R89C3 = [65]; R89C7 = [26]
and the rest is naked singles, simple elimination and cage sums
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. So, no more V2's for a while - especially with the Cricket World Cup starting. 
. 
I missed that the double isn’t possible because 21(3) cage at R2C1 cannot contain 3. Fortunately this didn’t have much effect; steps 11 to 14 were easy ones which eliminated this double.]