Here is a simplified version of the Bullseye 3 tag solution. It uses 1 hypothetical so should be rated a 2.0. However, Para's solution is much better (just leave out step 27 since it's not needed) because he found a "45" that we missed. Great work Para
. He uses a 'trick' chain move to unlock it - so I would give a 2.0 rating for Para's WT as well.
Great to have this one solved properly.
Bullseye 3 simplified tag WT
Preliminaries
0. r5c5 = 3
a) 6(3)n124 and n3 = {123} (no 4..9) -> {123} locked for n3
b) 16(2)n12 = {79} (no 1..6,8) -> {79} locked for r1
c) 10(2)n14 and n47 = {19/28/37/46} (no 5)
d) 20(3)n2 = {479/569/578} (no 1,2)
e) 7(2)n23 = [16/25/34] (no 7,8) -> r1c6 no 4,5,6; -> {123} locked for r1 in c689
e1) 7(2)n89 = {16/25/34} (no 7..9)
f) 21(3)n236 = {489/579/678} (no 1..3)
g) 13(2) n36 = {49/58/67} (no 1..3)
h) 11(3)n4 = {128/146/245} (no 7,9)
i) 9(2)n46 and n78 = {18/27/36/45} (no 9)
j) 19(3)n8 = {289/469/478/568} (no 1)
1. Innies c5: r46c5 = 3(2) = {12} (no 4..9)
1a. -> {12} locked for n5 and c5
2. 20(3)n2 = {569/578}(no 4) (combo {479} blocked by r1c4)
2a. -> 5 locked in 20(3)n2 for c5 & n2
2b. -> 4 locked in 19(3)n8 for c5 & n8
2c. cleanup: 7(2)n89: r9c7 no 3
2d. cleanup: 9(2)n78: r9c3 no 5
3. killer pair {79} locked for n2 in 20(3) and r1c4
4. 17(3)n1
4a. min r1c12 = 9 -> max r2c1 = 8 (no 9)
4b. max r1c12 = 14 -> min r2c1 = 3 (no 1,2)
4c. -> r2c1 no 6 (no valid permutations)
5. Innies r1234: r4c456 = h16(3) & must have 1/2 = {169/178/259/268}
5a. -> r4c46 no 4
6. Innies r6789: r6c456 = h13(3) = {148/157/247/256}
6a. -> r6c46 no 9
7. Innies c1234: r456c4 = h19(3) = {469/478/568}
7a. Innies c6789: r456c6 = h20(3) = {479/569/578}
8. "45" r5: r5c46 = h13(2) = {49/58/67} = [4/6/8..]
8a. {468} blocked from 18(3)n6
9. from steps 1 & 8: In n5 a Very hidden cage Vh26(4)r46c46 = {4589/4679/5678}
9a. Each combo must have exactly 2 numbers overlap with the 2 hidden cages in c4 & 6
i. {4589} -> h19(3)c4 & h20(3)c6 = {568-479} -> the 2 left-over candidates must be in r5c46 = 13 = [67]
ii. ........-> h19(3)c4 & h20(3)c6 = {469-578} -> the 2 left-over candidates must be in r5c46 = 13 = [67]
iii.........-> h19(3)c4 & h20(3)c6 = {478-569} -> the 2 left-over candidates must be in r5c46 = 13 = [76]
iv. {4679} -> h19(3)c4 & h20(3)c6 = {478-569} -> the 2 left-over candidates must be in r5c46 = 13 = [85]
v. {5678} -> h19(3)c4 & h20(3)c6 = {478-569} -> the 2 left-over candidates must be in r5c46 = 13 = [49]
10. 6 in Vh26(4) in {4679/5678} must be in c6 (see 9iv & v) -> no 6 in r46c4
10a. 7 in Vh26(4) in {4679/5678} must be in c4 (see 9iv & v) -> no 7 in r46c6
10b. r5c4 = {4678}(no 59)(step 9a.i..v)
10c. r5c6 = {5679}(no 48)(h13(2)r5c46)
11. Same thing with r46 in n5. Very hidden Vh26(4)r46c46 = {4589/4679/5678}
11a. Each combo must have exactly 2 numbers overlap with the 2 hidden cages in r4 & r6
i. {4589} -> h16(3)r4 & h13(3)r6 = {259-148} = {5[2]9-4[1]8}
ii. {4679} Blocked: Like this: h16(3)r4 & h13(3)r6 = {169-247} = [916-724] but r46c4 = [79] clashes with r1c4
iii. {5678} -> h16(3)r4 & h13(3)r6 = {178-256} = [718-526] BUT BLOCKED: h19(3)c4 cannot have [75]
iv.....................................= {268-157} = [826-715]
12. In summary, h16(3)r4 & h13(3)r6 = {259-148/268-157} = 8{..}
12a. r46c5 = [21]
12b. 8 locked for n5 -> h13(2)r5c46 = {49/67}(no 5)
12c. h16(3)r4 = [2]{59}/[826](no 7)(no 8 in r4c6)
12d. h13(3)r6 = 1{48/57}(no 6) = {4[1]8}/[715](no 5 in r6c4)
12e. clean-ups:no 8 in r3c1
12f. no 9 in r7c1
12g. no 8 in r7c9
12h. 2 in 6(3)n1 must be in r3: 2 locked for r3
12i. no 8 in r4c1
13. 21(3)n236 = {489/678}(no 5) ({579} blocked by no digits in r3c6)
14. Outties and Innie c9: r5c9 less r19c8 = 3
14a. min r19c8 = 3 -> min r5c9 = 6 (no 1,2,4,5)
14b. max r5c9 = 9 -> max r19c8 = 6 (no 6..9)
Now the trick that Caida spotted. Maybe this is one for
Sudoku SolverV3
15. Outies c9: r159c8 + r5c7 = 15 AND empty rectange/generalized X-Wing on 1 for c89 between 6(3)n3 & r9c8 IF r19c8 = 1 -> min r19c8 = 5. Like this.
15a. If r19c8 = [12/13] = 3/4 -> 1 in c9 must be in n9 -> 1 in n6 must be in r5c7 which is in outies of c9. But since outies of c9 = 15 and 3/4 + 1 = 4/5 ->
r5c8 would have to be 10/11! therefore r19c8 must be min of 5 = [14]/{23}
15b. If r19c8 = [21/31] = 3/4 -> generalized X-Wing on 1 with 6(3)n3 -> 1 in n6 must be in r5c7. But since outies c9 = 15, this would force r5c8 = 10/11.
Therefore, r19c8 cannot be [21/31]
15b. -> min r19c8 = 5 -> max r5c78 = 10 (outies c9)
15c. -> min r5c9 = 8 (cage sum) (no 6,7)
15d. min r19c8 = 5 -> min r9c8 = 2
16. 18(3)n6 must have 8/9 = {189/459}(no 2,6,7) ({279} blocked by h13(2)r5c46)
16a. = 9{18/45}: 9 locked for r5 & n6
16b. no 4 in r5c4 (h13(2)r5c46) or r3c9
16c. 9 in {489} in 21(3)n2 must be in r3c7 -> no 4 in r3c7
17. NP {67} in r5c46: both locked for r5 & n5
18. h16(3)r4c456 = [2]{59}: all locked for r4 & n5
18a. no 1 r3c1
18b. no 4 or 8 in r3c9
19. NP {48} in r6c46: both locked for r6
19a. no 2 or 6 in r7c1
19b. no 1 or 5 in r7c9
20. 2 locked for r5 in 11(3)n4 = 2{18/45}
20a. 2 locked for n4
20b. no 8 in r7c1
20c. min r6c3 = 3 -> max r7c34 = 9 (no 9)
21. h19(3)r456c4 = {469/568}
21a. r5c4 = 6
21b. r5c6 = 7
21c. no 3 in r9c3
22. deleted
23. Outies and Innie r1: r2c19 – r1c5 = 1
23a. min r1c5 = 5 -> min r2c19 = 6 -> r2c1 no 3
24. 17(3)n1 = {458/467} -> 4 locked in n1, not elsewhere in n1
24a. r4c1 no 6
Para calls this next move a killer Y-wing in his WT
25. r456c4 must contain either 8 or 9 – if they contain a 9 then r123c5 must contain 9 and r789c5 contains an 8; otherwise r456c4 contains an 8
25b. -> r789c4 no 8 (either 8 is in r456c4 or it is in r789c5)
25c. -> r9c3 no 1
(note: Para's "45" on n236 eliminates 8 from r2c4 which would have now left hidden single 8 in r6c4 for us)
Now a very productive hypothetical - which means we should rate this puzzle as a 2.0
26. "45" i/o c9: r19c8 = 5/6 = {23}/[14/15/24] -> r1289 = 14/15 (cage sums = 20) AND must have 1 for c9
26a. r19c8 = [23] -> r1289c9 = 15: r12c9 = {13} = 4 -> r89c9 = 11 = {29/56} ({47} blocked by [1/3/4/7] needed by 9(2)c9)
26b. r19c8 = [32] -> r1289c9 = 15: r12c9 = {12} = 3 -> r89c9 = 12 = {39/57} ({48} blocked by [4/8] needed by 9(2) & 13(2)c9)
26c. r19c8 = [14] -> r1289c9 = 15: r12c9 = {23} = 5 -> r89c9 = 10 = {19} ({46} blocked - r89 must have 1 for c9)
26d. r19c8 = [15] -> r1289c9 = 14: r12c9 = {23} = 5 -> r89c9 = 9 = {18} ({45} blocked - 1 must be in r89 for c9)
26e. r19c8 = [24] -> r1289c9 = 14: r12c9 = {13} = 4 -> r89c9 = 10 = {28} ({46} blocked by 4 in r9c8)
27. In summary, r1289c9 = 1{356/239/238/257}(no 4)
27a. 14(3)n9 = {149/158/239/248/257/356} ({347} blocked by no {37} possible in r89c9 steps 26a..e.)
28. {459} is blocked from 18(3)n6 by 4s in c9. Like this.
28a. 4 in r7c9 -> 5 in r6c9 -> {459} blocked from 18(3)
28b. 4 in r4c9 -> {459} blocked from 18(3)
28c. 18(3)n6 = {189}: all locked for r5 & n6
28d. no 5 or 8 in r3c9
28e. 8 must be in 21(3)n2: only in r3: 8 locked for r3
29. 11(3)n4 = {245}: all locked for n4
29a. no 6 in r3c1
29b. no 8 in r7c1
30. r4c2 = 8 (hsingle n4)
31. r6c4 = 8 (hsingle c4)
32. 8 in n1 only in 17(3) = {458}: all locked for n1
32a. 8 locked for c1
32b. 17(3)n7: no 8 = {179/269/359/467}
33. r2c4 = 4 (hsingle c4)
33b. 4 in n1 only in r1: 4 locked for r1
33c. no 3 r1c6
33d. 3 in r1 only in c89: no 3 in r2c9
34. r4c4 = 5 (cage sum h19(3)c4)
34a. no 4 in r9c3
35. r46c6 = [94]
36. 22(5)n1 must have 4 & 8 (& no 5) = 48{127/136}(no 9) = 1{..} = [2/3..]
36a. 1 required in 22(5)n1 only in n1: 1 locked for n1
37. Killer pair 2/3 between 22(5)n1 & r3c3: no 3 in r3c1
37a. no 7 in r4c1
38. NP {13} in r4c13: both locked for r4 and n4
38a. no 7 in r7c1
39. NTriple {467} in r4c789: all locked for n6
39a. no 2,3 in r7c9
40. r3c8 = 4 (hsingle r3)
41. NP {79} in n1 in r3c1 & r1c3: 7 locked for n1
42. 22(5)n1 = 48{136}(no 2): 3 locked for n1
43. r3c3 = 2
43a. no 7 in r9c4
44. 2 in n2 only in c6: locked for c6
44a. no 5 in r9c7
45. hidden pair {13} in r3c24:no 6 in r3c2
45a. 6 in n1 only in r2: 6 locked for r2
46. hidden triple {123} in n2: no 8 in r2c6
47. r3c5 = 5 (hsingle r3)
48. NP {68} in n2: no 8 in r2c5
49. "45" c1: 4 outies r159c2 & r5c3 = 20
49. max r15c2 + r5c3 = [545] = 14 -> min r9c2 = 6
50. 17(3)n7 must have 1 of 6/7/9 for c1 and 1 of 6/7/9 in r9c2 -> must have 2 of {679}
50a. = {179/269/467}(no 3,5)
51. 3 in c1 in 1 of 10(2) cages -> 7 locked for c1
No way to go - time for some creativity from Peter
52. 26(5)n236 must contain a 4 and only 1 of {123}
52a. 26(5)n236 = {14678/24569/24578}(no 3)
i. {14579} blocked: r2c6 = 1 -> r1c67 = [25] -> can’t place 5 in 26(5)
ii. {34568} blocked: r2c6 = 3 -> r3c4 = 1 -> r1c67 = [25] -> can’t place 5 in 26(5)
52b. r2c6 no 3
52c. single: r3c4 = 3: no 6 in r9c3
52d. r4c3 = 1
52e. r4c1 = 3: no 7 in r6c1
52f. r3c1 = 7: no 3 in r7c1, no 6 in r4c9
52g. r1c34 = [97]
52h. r12c5 = [69](cage sum)
52i. r3c6 = 8
52j. r1c7 = 5
52k. r1c6 = 2
52l. r1c2 = 4
52m. r1c1 = 8
52n. r2c1 = 5
52o. r2c6 = 1: no 6 in r9c7
52p. r2c9 = 2: no 7 in r7c9
52q. r3c2 = 1
53. Naked pair {78} locked for 26(5)n236 in r2c78
54. r3c8 = 6
55. r8c4 = 9 (hsingle c4)
56. r6c1 = 9 (hsingle n4)
57. r7c1 = 1
57a. r7c4 = 2
57b. r9c34 = [81]:no 6 in r9c6
58. r67c3 = [73/64] = [3/6..]
58a. r7c3 = {34}
59. Killer pair {36} in r267c3: both locked for c3
60. 31(5)n478: = [9]{4567}(no 2,3) (only permutation without 8)
61. r7c3 = 3 (hsingle n7), r6c3 = 7(cage sum), r6c2 = 6, r2c23 = [36], NP {57} at r78c2: both locked for c2 & n7
r59c2 = [29], r5c13 = [45], r8c3 = 4
62. 16(3)n689 = [259/268/358/367]
r7c7 = {789}
63. 14(3)n9 = {257/356}(no 1,8)
63a. = 5{..}: 5 locked for n9
64. r1c9 = 1 (hsingle c9)
Rest is singles. Whew