SudoCue Users

This one started easily but then kept me thinking. Nice to see that innie/outie differences were back again after Assassin 29 which I was surprised to find could be solved without using them.

Here is my walkthrough, as always effectively in the order that I solved the puzzle.

1. R1C34 = {79}, no other 7,9 in R1

2. R1C67 = {58}, no other 5,8 in R1

3. R9C34 = {12}, no other 1,2 in R9

4. R9C67 = {34}, no other 3,4 in R9

5. 23(3) cage in N7 = {689}, no other 6,8,9 in N7

6. 11(3) cage in N1, no 9

7. R34C1 = {19/28/37/46}, no 5

8. R67C1 = {19/28/37/46}, no 5, no 1,2,4 in R6C1

9. R34C9 = {49/58/67}, no 1,2,3

10. R67C9 = {18/27/36/45}, no 9

11. 16(5) cage in N214 = {12346}

12. 17(5) cage in N698 = 123{47/56}, no 8,9

13. 31(5) cage in N478 must contain 9 in R6C2 or R8C4

14. 22(3) cage in N698 = 9{58/67}

15. 20(3) cage in N2, no 1,2

16. 20(3) cage in N9, no 1,2

17. 45 rule on R6789 3 innies R6C456 = 7 = {124}, no other 1,2,4 in R6 or N5, no 5,7,8 in R7C9 (step 10)

18. 45 rule on R1234 3 innies R4C456 = 22 = 9{58/67}, 9 locked for R4 and N5, no 1 in R3C1 (step 7), no 4 in R3C9 (step 9)

19. R5C456 = 3{58/67}, 3 locked for R5 and N5

20. 45 rule on C5 3 innies R456C5 = 12, min. R56C5 = 4, max. R4C5 = 8

21. 45 rule on C6789 3 innies R456C6 = 20, no 1,2 in R6C6 -> [974] -> R6C45 = {12}, R5C45 = {36}, no other 6 in R5, R4C45 = {58}, no other 5,8 in R4, no 2 in R3C1 (step 7), no 5,8 in R3C9 (step 9)

22. R9C6 = 3, R9C7 = 4, no 5 in R6C9 (step 10)

23. R456C5 = 12 (step 20), only valid combinations = [561]/[831] -> R6C5 = 1, R6C4 = 2 -> R9C34 = [21], no 8 in R6C1 (step 8)

24. 2 in C5 locked in N8, no other 2 in N8, 13(3) cage in N8 = 2{47/56}, no 8,9

25. 9 in C5 locked in N2, no other 9 in N2 -> R1C34 = [97], no 1 in R4C1 (step 7)

26. 7 in C5 locked in N8, 13(3) cage = {247} -> R9C5 = 7, R78C5 = {24}, no other 4 in C5 or N8

27. 5 in R9 locked in R9C89, no other 5 in N9

28. 2 in 16(5) cage in N214 locked in R234C2, no other 2 in C2

29. 9 in 22(3) cage in N698 locked in R67C7, no other 9 in C7

30. 45 rule on R9 2 remaining outies R8C19 = 15 = {69}/[87], no 3,8 in R8C9

31. 45 rule on C1 2 outies R19C2 - 9 = 1 innie R5C1, max. R19C2 = 15 -> max. R5C1 = 6 = {1245}

32. 45 rule on C9 2 outies R19C8 - 9 = 1 innie R5C9, max. R19C8 = 15 -> max. R5C9 = 6 = {1245}

33. 12(3) cage in N3, max. R1C89 = 10 -> no 1 in R2C9, also there is no valid combination with 3 in R2C9

34. 17(5) cage in N698 (step 12) = {12356}, no 7, 1,2 in 17(5) cage locked in N9, no other 1,2 in N9 -> R67C9 = {36} (step 10), no other 3,6 in C9, no 7 in R34C9 (step 9) -> R34C9 = [94], no 6 in R3C1 (step 7)

35. R2C5 = 9 (hidden single in N2), R13C5 = [38]/[56], R1C6 and R3C5 both {58}, no other 5,8 in N2

36. Naked triple {568} in R178C6, no other 6 in C6

37. Naked triple {346} in R235C4, no other 6 in C4

38. R8C9 = 7 (naked single) -> R9C89 = {58}, no other 8 in R9 or N9, R9C12 = {69}, R8C1 = 8, no 2 in R4C1 (step 7)

39. R7C7 = 9 (hidden single in N9), no 7 in R7C6 -> no 6 in R6C7

40. R8C4 = 9 (hidden single in N8), no 9 in R6C2

41. R5C8 = 9 (hidden single in N6), R5C79 = {12}, no other 1,2 in R5 or N6

42. R6C7 = 8 (hidden single in N6), R7C6 = 5, R1C67 = [85]

43. R8C6 = 6 (hidden single in N8), no other 6 in 17(5) cage, R7C8 + R8C78 = {123}, no other 3 in N9, R6C8 = 5, R9C89 = [85], R7C9 = 6, R6C9 = 3, no 7 in R7C1 (step 8)

44. Killer pair 1,2 in R15C9 -> R2C9 = 8, R1C89 = [31], R5C9 = 2, R5C7 = 1, R78C8 = {12}, no other 1,2 in C8, R8C7 = 3

45. R1C5 = 6 (naked single), R3C5 = 5, R4C45 = [58], R5C45 = [63]

46. R1C2 = 4 (naked single), R1C1 = 2, R2C1 = 5

47. R5C1 = 4 (naked single), no 6 in R4C1 (step 7), no 6 in R6C1 (step 8), R5C23 = {58}

48. R34C1 = {37}, no other 3,7 in C1 -> R67C1 = [91], R9C12 = [69]

49. Naked triple {467} in R234C8, no other 6,7 in 20(5) cage -> R2C67 = [12]

50. Naked triple {346} in R2C234, no other 3,4,6 in R2 or in 16(5) cage -> R34C2 = [12]

and the rest is naked and hidden singles, simple elimination and cage totals for 18(3) and 31(5) cages in N478

Happy Christmas to all Assassin solvers and especially to Ruud for his wonderful efforts in composing these puzzles.

Assassin 30 came out a little too easy for my liking. But have made a very very difficult V2 for it. However, this V2 does not have the same final solution as the original - just a similiar cage layout. My thanks to JC over at DJape for teaching how to make this type of V2.

This V2puzzle requires many contradiction moves between nonets. Please post if you'd like some help with the next contradiction move.

This is Bullseye 3 (aka Assassin30V2.1)
3x3::k:4352:4352:4098:4098:512417971543:4352:5642:5642:5642:5124:6670:6670:66702578:56421556:5124:5399:5399:66702578:56421078210782:5399:6670285228527694650:4650:465079821078210782:4147:59402605:79823119:4922:4147:4147:59404415:7982:7982:7982:4922:5940:5940:59404415:44152378:492218693655:

Just did this killer for the first time.

That center nonet is interesting. Putting together all hidden cages gives you a puzzle usually called microsums. Rules for that puzzle are as follows: Place 1 to 9 in the grid exactly once. The numbers outside the grid indicate the sum of the digits in that row and column.

The puzzle then looks like this. And gives you 5 digits and 2 naked pairs. Very nice way to start of a killer puzzle.
Hope this is clear and you find it interesting.
Using this in the puzzle, it solves pretty easily.

greetings

Para

p.s.

you could also use this for Ed's V2. This then give you these options.

Hello,

Could I interest anyone in a tag solution for Assasin 30v2-1?? I haven't seen a walkthrough for it.

I'll start

I've got 11 steps done (and a really onerous step 12)


Any further steps (or suggestions/corrections to these ones) would be wonderful!!!

Caida


Assassin 30V2.1

Preliminaries
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: no 4 (combo {479} blocked by r1c4)
2a. -> 5 locked in 20(3)n2 for c5
2b. -> 4 locked in 19(3)n8 for c5
2c. killer pair {79} locked for n2 in 20(3) and r1c4
2d. cleanup: 7(2)n89: r9c7 no 3
2e. cleanup: 9(2)n78: r9c3 no 5

3. 17(3)n1
3a. -> min r1c12 = 9 -> max r2c1 = 8 (no 9)
3b. -> max r1c12 = 14 -> min r2c1 = 3 (no 1,2)
3c. -> r2c1 no 6 (no way to make 11(2) in r1c12 without 6)

4. Innies r1234: r4c456 = 16(3) = {169/178/259/268}
4a. -> r4c46 no 4

5. Innies r6789: r6c456 = 13(3) = {148/157/247/256}
5a. -> r6c46 no 9

6. Innies c1234: r456c4 = 19(3) = {469/478/568}
6a. Innies c6789: r456c6 = 20(3) = {479/569/578}
6b. from steps 4, 5, 6 and 6a. the only possible combinations for n5 are: [529 637 814/925 736 418]
6c. -> second possibility is blocked by r1c4 (actually it is blocked because of step 6 and step 6a)
So: n5 = [529 637 814] (working from r4c4 going left to right then down)
6d. -> 2 locked in 6(3)n124 in r3
6e. cleanup: 9(3)n78: = {27}/[63/81] -> r9c3 no 1,3,4
6f. cleanup: 10(2)n12: no 8; r3c1 no 1
6g. cleanup: 13(2)n36: r3c9 no 4,8
6h. cleanup: 10(2)n47: r7c1 no 2,6,9
6i. cleanup: 9(2)n69: r7c9 no 1,5,8

more explanation for why other alternatives for n5 won't work:

If n5 = [925 637 418]
-> 16(2)n12 = [97]
-> 20(3)n2 = {569}
-> r2c4 = 8
-> r3c6 = 4
-> r34c7 = [98]
-> 18(3)n6 = {459}
-> 11(3)n4 = {128}
-> r4c3 = 3
-> r4c8 = 1
-> 9(2)n89 = [63]
-> 13(2)n46 = {67}
9(2)n69 no longer possible

If n5 = [826 439 715]
-> 16(2)n12 = [79]
-> 20(3)n2 = {578)
-> r2c4 = 6
-> 18(3)n6 = {567}
-> 11(3)n4 = {128}
-> r4c123 = [753]
-> r3c1 = 3
-> 17(3)n1 = {458}
-> r3c2 = 6
this places 2 6s in 21(5)n124


7. 21(3)n236 = [894]/ [6]{78}/[8]{67}
7a. -> r3c7 = {6..9} (no 4,5)

8. Outties and Innie c9: r5c9 less r19c8 = 3
8a. -> min r19c8 = 3 -> min r5c9 = 6 (no 1,2,4,5)
8b. -> max r5c9 = 9; -> max r19c8 = 6 (no 6..9)
8c. -> r5c9 = 8/9; r19c8 = either 5(2) or 6(2) = {23}/[14/15/24]
8d. -> r8c9 (correction should be r9c8) no 1

9. 18(3)n6 = {189/459} no 2
9a. -> 2 locked for r5 in 11(3)n4
9b. -> 9 locked in 18(3)n6 for r5
9c. cleanup: 10(2)n47: r7c1 no 8

10. 12(3)n478: min r7c3 = 3
10a. -> r8c34 no 9

11. killer pair {68} locked in n2 for 20(3) and r3c6
11a. ->r2c6 no 6,8 (I could have also seen the quadruplet {1234} locked for n2)
11b. hidden single: r2c4 = 4 (should have seen this much earlier)

12. 8 locked in n4 in either 22(5) or 11(3);
12a. combinations for 22(5) = 4-{1368}

Other combinations (4-{1269/1278/1359/1368/2358/2367})blocked by following:

12b. if 22(5) = 4{1269} -> 2 locked for n1 in 22(5) -> r3c4 = 2 -> r34c3 = {13} -> both r34c3 and 22(5) need a 1 -> 1 locked in n4 in r4c23 -> 11(3) = {245} -> 8 eliminated from n4;

12c. if 22(5) = 4(1278} -> 2 locked for n1 in 22(5) -> r3c4 = 2 -> r34c3 = {13} -> both r34c3 and 22(5) need a 1 -> 1 locked in n4 in r4c23 -> 11(3) = {245} -> 8 locked in 22(5) in r4c2 -> {127} locked in 22(5) in n1 -> r34c3 = [31] -> r1c34 = [97] -> r34c1 = [46] -> remaining numbers in n1 don’t work for 17(3)

12d. 22(5) = 4{1359} -> r4c23 = {13} -> 11(3)n4 = 128 -> blocked by r4c23

12e. 22(5) = 4{2358} -> 2 locked for n1 in 22(3) -> r3c4 = 2 -> r34c3 = [13] -> r4c2 = 8 -> 11(3)n4 = {245} -> 3 and 5 locked for n1 in 22(3) -> 17(3)n1 = {467} -> r34c1 = [91] -> blocked by r1c3

12f. 22(5) = 4{2367} -> 11(3)n4 = {128} -> 2 locked for n1 in 22(3) -> r34c3 = [13] -> r1c34 = [97] -> 3 locked for n1 in 22(3) -> 18(3)n6 = {459} -> 4 locked in n4 in r4c1 -> r34c1 = [64] -> r4c2 = 6 -> 20(3)n2 = {569} -> r3c6 = 8 -> r34c7 = {67} -> blocked by r3c1 and r4c3



I think there should be some easy next steps (unless there is a better step 12) making all eliminations surrounding 22(5)n124 = 4{1368}

Here's a few more steps:

Also - I've made some typo corrections to my steps above.



13. Single: r3c3 = 2
13a. r1c2 = no 6,8 (CPE with 22(5)n124
13b. {13} locked in n1 in 22(5) -> r4c2 no 1

14. 5 locked in n1 in 17(3)
14a. 17(3)n1 = {458} (no 3,6,7) -> {458} locked for n1
14b. -> r4c2 = 8
14c. -> r3c8 = 4 (hidden single)
14d. {136} locked in n1 in 22(5)
14e. cleanup: 10(2)n14: r4c1 no 4,6,7
14f. -> 6 and 7 locked in n4 in r6c123 -> no 6,7 elsewhere in r6
14g. -> triple {235} locked in r6 for n6 -> no 2,3,5 elsewhere in n6 or r6

15. 11(3)n4 = {245}
15a. 18(3)n6 = {189}
15b. cleanup: 10(2)n47: r7c1 no 7
15c. 13(2)n36: r3c9 no 5
15d. 9(2)n69: r7c9 no 2,3
15e. 7(2)n78: r9c4 no 7

16. Hidden single: r3c5 = 5

Caida wrote:Could I interest anyone in a tag solution for Assasin 30v2-1??

I hope I can join in on Sunday (after I finish Para's toroidal JK)....have a vague idea how I originally 'solved' Bullseye3, but would never do that sort of move now. Have really lost my nerve with hypotheticals. So, would love to find a satisfying solution.

Caida wrote:6b. from steps 4, 5, 6 and 6a. the only possible combinations for n5 are: [529637814/925736418]

Can you take another look at step 6c and give some more explanation? From looking at Para's tiny puzzle for that n5, there seems to be some other options for n5, [925 637 418] or [826 439 715], maybe more.

But even getting to Para's candidates for n5 is quite difficult from memory. Hmm.

Cheers
Ed

sudokuEd wrote:I hope I can join in on Sunday

Yay!!!

sudokuEd wrote:Can you take another look at step 6c and give some more explanation? From looking at Para's tiny puzzle for that n5, there seems to be some other options for n5, [925 637 418] or [826 439 715], maybe more.

Oh dear. I know that I eliminated them - but can't for the life of me figure out how.

Below my step 6 above I have added why the two combinations ([925 637 418] or [826 439 715]) won't work. I will try and figure out how I eliminated them before. I don't remember doing such a long complicated path.

I've been going in circles on this particular puzzle for ages and ages now and would love to have a solution!

Cheers,

Caida

Hi all

Here's my walk-through for Bullseye 3. Had to resort one trick and a "Richard-style 45-tests" but happy with the general result. Those big 45-tests are pretty cool to do, but a monster to analyze, so you have to carefully set your restrictions to get the proper results.

[edit] As Afmob wondered about the rating and Mike mentioned something about Ed doing a rating bit for everything under 50 i'll add my rating advice. I feel it is 1.75 somewhere in the vicinity of A60RP-lite. Some tricky moves and careful analysis did the trick just like A60RP-Lite.

Walk-through Bullseye 3

1. R5C5 = 3

2. 20(3) at R1C5 = {479/569/578} = {8|9..}: no 1,2

3. R1C67 and R9C67 = {16/25/34}: no 7,8,9

4. R34C1 and R67C1 = {19/28/37/46}: no 5

5. 6(3) at R3C3 = {123}

6. 21(3) at R3C6 = {489/579/678}: no 1,2,3

7. R34C9 = {49/58/67}: no 1,2,3

8. 11(3) at R5C1 = {128/146/245}: no 7

9. R67C9 and R9C34 = {18/27/36/45}: no 9

10. 19(3) at R7C5 = {469/478/568}: {289} blocked by 20(3) at R1C5: no 1,2

11. R1C34 = {79} -> locked for R1

12. 6(3) at R1C8 = {123} -> locked for N3
12a. Clean up: R1C6: no 4,5,6
12b. Naked Triple {123} in R1C689 -> locked for R1

13. 17(3) at R1C1 = {46}[7]/{68}[3]/{458}: R2C1: no 1,2,6,9

14. Hidden Pair {12} in R46C5 -> R46C5 = {12} -> locked for N5

15. 45 on R5: 2 innies: R5C46 = 13 = {49/58/67}
15a. 18(3) at R5C7 = {189/279/459/567}: {468} blocked by 11(3) at R5C1

16. 45 on R1234: 3 innies: R4C456 = 16 = {6[1]9/7[1]8/5[2]9/6[2]8}: no 4

17. 45 on R6789: 3 innies: R6C456 = 13 = {4[1]8/5[1]7/4[2]7/5[2]6}: no 9

18. 45 on C1234: 3 innies: R456C4 = 19 = {469/478/568} = [964]/{7[4]8}/{5[6]8}: {78}[4] blocked by R6C456(needs 7 or 8 when R6C4 = 4); others blocked by step 15: R46C4: no 6; R5C4 = {46}
18a. 11(3) at R5C1 = {128/245}: {146} blocked by R5C4: no 6; 2 locked for R5 and N4
18b. 2 in 6(3) at R3C3 locked for R3
18c. Clean up: R5C6 = {79}; R34C1: no 8; R7C1: no 8

19. 45 on C6789: 3 innies: R456C6 = 20 = [974]/[695]/{5[7]8}: [794] blocked by R5C46; [596] blocked by R6C456(needs 5 when R6C6 = 6): R6C6: no 6,7; R4C6: no 7

20. R4C456 = {5[2]9}/[826]: {6[1]9} blocked by R5C46; [718] blocked by R456C6(when R4C6 = 8, they also need 7): R4C5 = 2; R4C4 = {589}; R4C6 = {569}
20a. R6C5 = 1
20b. R456C4 = [964/568/847]: R6C4 = {478}
20c. R456C6 = [974/695/578]
20d. clean up: R7C1: no 9; R7C9: no 8

21. 20(3) at R1C5 = {569/578} = {7|9..},{6|7..}: {479} blocked by R1C4: no 4; 5 locked for C5 and N2
21a. 4 in C5 locked for N8
21b. Killer Pair {79} in R1C4 + 20(3) at R1C5 -> locked for N2
21c. Clean up: R9C3: no 5; R9C7: no 3

22. 45 on R1: 2 outies and 1 innie: R1C5 + 1 = R2C19: R2C19 = 6/7/9 = [42/51/43/52/81/72]: R2C1: no 3
22a. 17(3) at R1C1 = {458}/{46}[7]: 4 locked for N1
22b. Clean up: R4C1: no 6

23. 45 on C9: 2 outies and 1 innie: R5C9 = R19C8 + 3: Min R19C8 = 3 -> Min R5C9 = 6: no 1,4,5; Max R5C9 = 9 -> Max R19C8 = 6: R9C8: no 6,7,8,9

24. 12(3) at R6C3 = [9]{12}/{138/147/156/237/246/345}: R7C34: no 9

25. 21(3) at R3C6 = [4]{89}/[6]{78}/[8]{49}/[8]{67}: R34C7: no 5

26. 20(3) at R1C5 = [5]{69/78}/[6]{59}/[8]{57}

27. 45 on R1: R1C5 + 1 = R2C19 = [5]-[51]/[6]-[43/52]/[8]-[72/81]: [5]-[42] blocked by 17(3) at R1C1
27a. 45 on R1: 4 outies: R2C19 + R23C5 = 21 = [51]-{69/78}/[43]-{59}/[52]-[95]/[81]-{57}: [72]-{57} blocked by R1C34(leaves no 7 in R1): R2C1: no 7
27b. 17(3) at R1C1 = {458} -> locked for N1

28. 19(3) at R7C5 = {469/478} = {8|9..}; R456C4 = [964/568/847] = {8|9..}
28a. Killer y-wing on 9 in N2: R789C5(89) and R456C4(89) see all 9's in N2, so can't both use 9, one of them needs an 8 -> eliminate 8 from all peers of R456C4 and R789C5: R789C4: no 8
28b. R9C3: no 1

29. 45 on N236: 6 innies: R123C4 + R6C789 = 24: Min R6C789 = 9 -> Max R123C4 = 15
29a. R12C4 = [98/96/78] blocked by step 29; R12C4 = [76] blocked by 20(3) at R1C5: R2C4: no 6,8
29b. 8 in C4 locked in R46C4 for N5
29c. R456C4 = [568/847]: R4C4: no 9; R6C4: no 4
29d. 9 in N5 locked within R45C6 for C6
29e. R456C6 = [974/695]: R4C6: no 5

30. 45 on N2: 6 innies: R123C4 + R123C6 = 25 = {123469/123478}
30a. When {123478}: R1C4 = 7: No 4 in C6 as this is blocked by R6C46 = [75/84]: R2C4 = 4
30b. When {123469}: Can't have both {46} in C6 as this is blocked by R456C6 = [695/974]: R2C4 = 4
30c. Conclusion: R2C4 = 4
30d. R5C46 = [67]; R4C46 = [59]; R5C46 = [84]
30e. Clean up: R3C1: no 1; R3C9: no 4,8; R7C1: no 2,6; R7C9: no 1,5; R9C3: no 3,4

31. Hidden Triple {123} in N2: R12C6 + R3C4 = {123}

32. 4 in N1 locked for R1
32a. Clean up: R1C6: no 3
32b. 3 in R1 lockd for N3

33. 21(3) at R3C6 = [6]{78}/[894]/[8]{67}: R3C7: no 4
33a. R3C8 = 4(hidden)

34. 26(5) at R2C6 = 4-[1]{59}[7]/[1]{678}/[2]{89}[3]/[2]{59}[6]/[2]{578}/[3]{568}: R4C8: no 1
34a. 1 in R4 locked for N4

35. 11(3) at R5C1 = {245} -> locked for R5 and N4
35a. R4C2 = 8(hidden)
35b. 6 and 9 in N4 locked for R6
35c. Clean up: R3C1: no 6; R3C9: no 5; R7C9: no 3

36. R3C5 = 5(hidden)
36a. Naked Pair {68} in R1C5 + R3C6 -> locked for N2

37. 6 in N1 locked within 22(5) at R2C2: 22(5) = 48{136}: R2C23 + R3C2 = {136} -> locked for N1
37a. R3C3 = 2
37b. 2 in C4 locked for N8
37c. Clean up: R4C1: no 7; R9C4: no 7; R9C7: no 5

38. Naked Pair {13} in R4C13 -> locked for R4 and N4
38a. Naked Triple {467} in R4C789 -> locked for N6
38b. Clean up: R7C1: no 7; R7C9: no 2

39. Hidden Pair {13} in R3C24 -> R3C2 = {13}
39a. 6 in N1 locked for R2

40. 45 on R12: 2 outies: R3C2 + R4C8 = 7 = [16]
40a. R3C4 = 3; R4C13 = [31]; R3C1 = 7; R1C34 = [97]
40b. R12C5 = [69]; R1C67 = [25]; R23C6 = [18]; R2C9 = 2
40c. R1C12 = [84]; R2C1 = 5; R8C4 = 9(hidden); R6C1 = 9(hidden)
40d. R7C1 = 1; R79C4 = [21]; R9C3 = 8; R8C2 = 9{hidden)
40e. Clean up: R7C9: no 7; R9C67: no 6

41. 17(3) at R8C1 = 9{26} -> R89C1 = {26} -> locked for C1 and N7
41a. R5C123 = [425]

42. 31(5) = 9{4567} -> R6C2 = 6; R8C3 = 4; R78C2 = {57} -> locked for N7
42a. R2C23 = [36]; R67C3 = [73]

43. Killer Pair {46} in R34C9 + R7C9 -> locked for C9

44. 14(3) at R8C9 = {257}: {128} blocked as R8C9 only cell with {18}: R9C8 = 2; R89C9 = {57} -> locked for C9 and N9

And the rest is all naked singles.

greetings

Para

Hi Caida. If Para can do it, so can we . Gee it's good to be playing tag again .

Caida wrote: Bullseye 3 (aka A30V2.1)

Preliminaries
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: no 4 (combo {479} blocked by r1c4)
2a. -> 5 locked in 20(3)n2 for c5
2b. -> 4 locked in 19(3)n8 for c5
2c. killer pair {79} locked for n2 in 20(3) and r1c4
2d. cleanup: 7(2)n89: r9c7 no 3
2e. cleanup: 9(2)n78: r9c3 no 5

3. 17(3)n1
3a. -> min r1c12 = 9 -> max r2c1 = 8 (no 9)
3b. -> max r1c12 = 14 -> min r2c1 = 3 (no 1,2)
3c. -> r2c1 no 6 (no way to make 11(2) in r1c12 without 6)

4. Innies r1234: r4c456 = 16(3) = {169/178/259/268}
4a. -> r4c46 no 4

5. Innies r6789: r6c456 = 13(3) = {148/157/247/256}
5a. -> r6c46 no 9

6. Innies c1234: r456c4 = 19(3) = {469/478/568}
6a. Innies c6789: r456c6 = 20(3) = {479/569/578}

An alternative way in n5 from here.
13. "45" r5: r5c46 = h13(2) = {49/58/67} = [4/6/8..]
13a. {468} blocked from 18(3)n6

14. from steps 1 & 6: In n5 a Very hidden cage Vh26(4)r46c46 = {4589/4679/5678}
14a. 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]
14b. 6 in Vh26(4) in {4679/5678} must be in c6 (see iv & v) -> no 6 in r46c4
14c. 7 in Vh26(4) in {4679/5678} must be in c4 (see iv & v) -> no 7 in r46c6
14d. r5c4 = {4678}(no 59)(step 14a.i..v)
14e. r5c6 = {5679}(no 48)(step 14a.i..v)

15. Same thing with r46 in n5. Very hidden Vh26(4)r46c46 = {4589/4679/5678}
15a. 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]

16. In summary, h16(3)r4 & h13(3)r6 = {259-148/268-157} = {5[2]9-4[1]8}/[826-715] = 8{..}
16a. r46c5 = [21]
16b. 8 locked for n5 -> h13(2)r5c46 = {49/67}(no 5)
16c. h16(3)r4 = [2]{59}/[826](no 7)(no 8 in r4c6)
16d. h13(3)r6 = 1{48/57}(no 6) = {4[1]8}/[715](no 5 in r6c4)

Using Caida's steps where possible
7. 21(3)n236 = {489/678}(no 5) ({579} blocked by no digits in r3c6)

8. Outties and Innie c9: r5c9 less r19c8 = 3
8a. -> min r8c19 = 3 -> min r5c9 = 6 (no 1,2,4,5)
8b. -> max r5c9 = 9; -> max r19c8 = 6 (no 6..9)

9. 18(3)n6 = {189/459/567}(no 2)({279} blocked by h13(2)r5c46)
9a. -> 2 locked for r5 in 11(3)n4 = 2{18/45}(no 6)
9b. 2 locked for n4
9c. -> no 8 in r37c1

10. 2 must be in 6(3)n1 and is only in r3: 2 locked for r3
10a. -> no 8 in r4c1

So should be here.
I see that Para's 'microsum' marks has a few less in it. Must be something I missed. Please make the next step no 17.

Para wrote: Ed doing a rating bit for everything under 50 i'll add my rating advice

Yes I am planning to, so thanks for this. If anyone does one of these early puzzles, let us know what you think of it.

Cheers
Ed

Hi Ed,

I loved the way you got in to n5 - made so much more sense!!

Continuation of Tag for 30v2-1

First off: missing a couple eliminations from your picture

1 in r6 locked in c5 -> no 1 elsewhere in r6
-> r7c1 no 9
-> r7c9 no 8


reiterating step 16 here so I can "see" it: based on steps above n5 = [925 637 418 / 529 637 814 / 826 439 715]
Note: [529 637 418 / 529 736 418 / 529 736 814 / 925 637 814] -> conflict with step 6 and 6a

17. r5c4 no 7
17a r5c6 no 6
This gets us to Para’s ‘microsum’ marks above

A couple of very minor steps - nothing groundbreaking that I can see just yet.

18. 12(3)n478: min r7c3 = 3
18a. -> r7c34 no 9
19. Outies and Innie r1: r1c19 – r1c5 = 1
19a. -> min r1c5 = 5 -> min r2c19 = 6 -> r2c1 no 3

20. 17(3)n1 = {458/467} -> 4 locked in n1, not elsewhere in n1
20a. r4c1 no 6

21. Not sure how to write this but here goes:
21a. 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
21b. -> r789c4 no 8 (either 8 is in r456c4 or it is in r789c5)
21c. -> r9c3 no 1

I don't know how to do a picture of remaining candidates - I do my puzzles in excel - I think if I just copy and paste I'll get a big mess.

Caida


Testing putting in candidates:
It is really ugly - I'll check if Andrew has a better way!

Caida wrote: I don't know how to do a picture of remaining candidates - I do my puzzles in excel - I think if I just copy and paste I'll get a big mess.

I'd ask Andrew, he's also an excel solver.

Para

Trying again.

I knew that r2c1 could not be 7 - but my logic was not fully formed before.

Continuation of Tag for Assassin 30v2-1

22. r2c1 <>7 here’s why:
22a. -> if r2c1 = 7 then r1c5 = 8 (step 19) and r1c45 = [97] (only option available -> this puts both 8 and 9 in 20(3) -> not possible as 20(3)n2 = {578/569}
22b. 17(3)n1 = {458} -> locked for n1

Still looking for something substantial.

Another contradiction move

Continuation of Tag for Assassin 30v2-1

23. r4c4 <> 8 here’s why
23a. -> if r4c4 = 8 then n5 = [826 439 715] and 11(3)n4 = {128} and r4c2 = 5 and r3c4 = 9 now only place for 9 in n1 is in 22(5)n124 -> This means that 22(5) = 9 + 5 (from r6c4) + 8(3). This 8(3) can’t be 125 (already used the 5) and it can’t be 134 (no 4 available) -> no possible combination for 22(5)n124.
23b. -> r4c46 = {59} -> locked for n5 and r4
23c. -> r5c46 = [67]
23d. -> r6c46 = {48} -> locked for r6
23e. cleanup: r3c1 no 1
23f. r9c3 no 3
23g. r3c9 no 4,8
23h. r7c1 no 2,6
23i. r3c7 no 4 (needs both 8 and 9 in r3c6 and r4c7)
23j. r7c9 no 1,5
23k. 9 in n4 locked in r6 -> no 9 elsewhere in r6

24. 4 locked in r3 in r3c68
24a. -> r2c6 no 4 (CPE)


Candidate list (sorry can't figure out how to format better):

Hi Caida,
hope you don't mind me being a party pooper and leave aside step 23. Too much like t&e for me - can't 'see' it in my head. Found something else to make a bit of headway.

Continuing from step 22
25. innies c1: r12589c1 = h25(5) and must have 5 and must have {45/48/58} for r12c1
i. = {12589}
ii. = {14569}
iii. = {14578}
iv. = {23578} Blocked: Like this. = {58}[2]{37}. But {37} is not possible in 17(3)n7
v. = {24568}
vi. = {34567} Blocked: like this. = {45} but no 3,6,7 in r5c1
25a. in summary h25(5)c1 = {12589/14569/14578/24568}(no 3)

26. 3 in c1 must be in 1 of 10(2) cages in c1 -> 7 must be in 1 of 10(2)s for c1. 7 locked for c1.

27. h25(5)c1 = {12589/14569/24568} & r12c1 = {45/48/58}
27a. 5 in {12589} must be in r12c1 = {58} only
27b. 5 in {14569} must be in r12c1 = {45} only
27c. 5 in {24568} must be in r12c1 = {45/58} OR it is in r5c1 for {48}{256}. It can't be in r89c1 as this would have r89c1 = {56} which is not possible in a 17(3)cage
27d. -> no 5 in r89c1

28. 17(3)n7 = {179/269/368/467}(no 5) ({278} is blocked by {28} in r89c1 which leaves nothing for r5c1 in innies c1 step 27c;{458} must have {48} in c1 which clashes with r12c1)
28a. 7 in {179} must be in r9c2 -> no 1 r9c2
28b. 7 in {467} must be in r9c2 -> no 4 r9c2

Ed wrote:hope you don't mind me being a party pooper and leave aside step 23. Too much like t&e for me - can't 'see' it in my head.

No worries. I can understand why you’d want to avoid it. It is very much like t&e – I was getting desperate.

How about a move looking at c9?

29. Consider options for innies of c9: r1258c9 = 23(5)
29a. Some prework before getting started – important to remember the following items:
29a1. r12c9 must contain {12/13/23}
29a2. r1c8 contains the other set of the 6(3): = {3/2/1}
29a3. if there is a 1 in c8 of n3 or n9 then there is a 1 in 18(3)n6 (r5c7 = 1)
29a4. if r5c7 contains a 1 then it must also have either an 8 or 9 in r5c9 (18(3)n6 = [1]{89})
29a5. 13(2)n36 = {49/58/67}
29a6. 9(2)n69 = {18/27/36/45}
29b. h23(5)c9 = {12389/12578/13568}
here are explanations why other options don’t work:
29b1. h23(5)c9 <> {12569/13469/13478/14567/23468} all block all options for 13(2)n36 (step 29a5)
29b2. h23(5)c9 <> {12479) -> r12c9 = {12} (step 29a1) -> r1c8 = 3 (step 29a2) -> r89c9 = {49}(only option for 14(3)n9) -> r5c9 = 7(only one left in h23(5)) -> r9c8 = 1(completion of 14(3)n9) -> r5c7 = 1 (step 29a3) -> all combinations for 18(3)n6 blocked
29b3. h23(5)c9 <> {23567} -> r12c9 = {23} (step 29a1) -> r1c8 = 1 (step 29a2) -> r5c7 = 1 (step 29a3) -> no {89} possible for 18(3)n6 -> all combinations for 18(3)n6 blocked
29b4. h23(5)c9 <> {23459} -> -> r12c9 = {23} (step 29a1) -> r1c8 = 1 (step 29a2) -> r5c7 = 1 (step 29a3) -> 9(2)n69 = [81] (step 29a6) -> no {89} possible for 18(3)n6 -> all combinations for 18(3)n6 blocked
29c. 1, 8 locked for c9 in r12589c9 -> no 1,8 elsewhere in c9
29d. -> r34c9 no 5
29e. -> r12589c9 no 4


Step 23+24 are still valid if ever you want to use them