Assassin 80

[mhparker]

Hi folks,

I'm surprised that no-one's posted anything on the A80 yet, so here's my WT.

It was a tough puzzle. Rating = 1.5?

Edit: After having now had a chance to take a deeper look, it seems that this puzzle is very similar in difficulty to the A78. So perhaps a 1.25 rating would be more appropriate here?

Assassin 80 Walkthrough

Prelims:

a) 20(3) at R1C5 = {389/479/569/578} (no 1,2)
b) 13(2) at R1C6 = {49/58/67} (no 1..3)
c) 12(4) at R2C7 = {1236/1245}; 1,2 locked for N3
d) 23(3) at R4C8 = {689}, locked for N6
e) 19(3) at R5C1, R6C3, R7C4 and R8C1 = {289/379/469/478/568} (no 1)
f) 10(3) at R6C5 = {127/136/145/235} (no 8,9)
g) 11(4) at R6C8 = {1235}
h) 28(4) at R7C8 = {4789/5689}; 8,9 locked for N9
i) 6(2) at R9C3 = {15/24}

1. Hidden pair (HP) in C7 at R13C7 = {89}, locked for N3
1a. cleanup: R1C6 = {45}

2. Innie/Outie (I/O) diff. C12: R8C3 = R5C2 + 3
2a. -> no 7..9 in R5C2; no 1..3 in R8C3

3. Innie/Outie (I/O) diff. C89: R2C7 = R5C8
3a. -> no 6 in R2C7; no 7 in R5C8

4. 6 in C7 locked in N9 -> not elsewhere in N9
4a. -> 28(4) at R7C8 = {4789} (no 5) (Prelim h), locked for N9

5. Innies N2: R1C46+R3C4 = 9(3) = {135/234}
5a. -> R13C4 = {1..3}
5b. 3 locked in R13C4 for C4 and N2

6. Innies N7: R79C3 = 6(2) = [24/42/51]
6a. cleanup: no 1 in R9C4
6b. Note that due to overlapping 6(2) cages, R7C3 = R9C4

7. 19(3) at R6C3 = {289/469/478/568}
7a. smallest digit (1 of {245}) must go in R7C3
7b. -> R6C34 = {6..9} (no 1..5)

8. Innies N8: R7C6+R9C46 = 10(3) = {127/145/235} (no 6,8,9)

9. 20(3) at R1C5 (Prelim a) = {479/569/578}
9a. -> 20(3) at R1C5 and R1C6 form killer pair (KP) on {45}
9b. -> no 4,5 elsewhere in N2

10. Innie/Outie (I/O) diff. N14: R6C3 = R1C4 + 7
10a. -> R1C4 = {12}, R6C3 = {89}
10b. -> R3C4 = 3 (hidden single, N2 innies (step 5))
10c. split 9(2) at R4C45 = {18/27/45} (no 3,6,9)

11. Innies N14: R126C3 = 19(3) = {289/379/469/478/568} (no 1)

12. 19(3) at R5C1 (Prelim e) and R6C3 form KP on {89}
12a. -> no 8,9 elsewhere in N4
12b. {289} combo for 19(3) at R5C1 blocked by R6C3
12c. -> no 2 in 19(3) at R5C1

13. 16(3) at R3C7 = {169/178/259/268/349/358}
13a. can have only 1 of {89}, which must go in R3C7
13b. -> no 8,9 in R4C6

14. 9 in R4 locked in N6 -> not elsewhere in N6

15. Outies R1234: R5C239 = 12(3)
15a. R5C9 = {68}
15b. -> R5C23 must sum to 4 or 6 = {13/15/24} (no 6,7)

16. Innies N1234: R3C7+R6C3 = 17(2) = {89}
16a. -> no 8,9 in R3C3 (CPE)

17. {89} now unavailable to 17(4) at R3C3
17a. -> 17(4) at R3C3 = {1367/1457/2357/2456}
17b. combining with step 15b, possibilities for R34C3+R5C23 are:
17b1. {67}+{13}: ok
17b2. {47}+{15}: blocked by h6(2) at R79C3 (step 6)
17b3. {2357}: no placement
17b4. {56}+{24}: blocked by h6(2) at R79C3 (step 6)
17c. -> R34C3 = {67}, locked for C3; R5C23 = {13}, locked for R5 and N4
17d. -> R5C9 = 8 (step 15)

18. R4C89 = {69}, locked for R4
18a. -> R34C3 = [67]
18b. cleanup: no 2 in R4C45 (step 10c)

19. 12(3) at R1C3 = {129/138} (no 4,5)
19a. -> R1C4 = 1
19b. cleanup: no 8 in R4C5 (step 10c)

20. Innie N2 (step 5): R1C6 = 5
20a. -> R13C7 = [89]
20b. -> R6C3 = 8 (step 16)
20c. cleanup: split 7(2) at R4C67 = [25]/{34} = {(4/5)..}

21. R4C45 = [81] ({45} blocked by R4C67 (step 20c)
Edit: As Gary points out in his post below, it's actually unnecessary to use R4C67 at this stage, because the 8 of R4 is now locked within the 12(3) cage. Thanks, Gary!

22. 2 in R1 locked in N1 -> not elsewhere in N1
22a. -> split 11(2) at R12C3 = [29] (only possible combo/permutation)

23. Hidden singles (HS) in C3 at R59C3 = [31]
23a. -> R5C2 = 1; R9C4 = 5

24. Outie C123: R6C4 = 6
24a. -> R7C3 = 5
24b. -> R8C3 = 4

25. HS in R5 at R5C1 = 6
25a. -> split 13(2) at R6C12 = {49}, locked for R6 and N4

26. HS in R1 at R1C5 = 9
26a. -> split 11(2) at R2C45 = {47} (last combo), locked for R2 and N2

27. Naked pair (NP) at R3C56 = {28}, locked for R3 and N2
27a. -> R2C6 = 6

28. NP at R4C12 = {25}, locked for R4 and 15(4)
28a. -> split 8(2) at R23C1 = [17] (only possible combo/permutation)

29. HS in C1 at R4C1 = 5
29a. -> R4C2 = 2

30. R23C2 = [85] (hidden singles, R2/C2)

31. Hidden pair (HP) in R1/N3 at R1C89 = {67}
31a. -> R2C9 = 3 (cage sum)

32. HS in C8 at R6C8 = 3
32a. -> R4C67 = [34]

33. R3C89 = [14] (hidden singles, C8/R3)

34. Outie C789: R9C6 = 4
34a. cleanup: no 1,3 in R8C7

35. Innie N8 (step 8): R7C6 = 1
35a. -> split 9(2) at R6C56 = {27} (last combo), locked for R6 and N5

Now all naked singles and 1 cage sum to end.

[gary w]

Ruud's puzzles must be so well crafted...my solution depended critically on exactly the same part of the grid that Mike concentrated on,specifically,



Having established that both r3c7 and r6c3=8/9 then turn attention to the 17(4) cage in N14.Together with the O of r1234 could show that the 17(4) cage must be r34c3={67} and r5c23={13} etc.
I dispensed with step 16 and 20c of Mike's wt ..the latter because there's a hs 8 in r4 at this stage.



My mop-up wasn't as neat as Mike's but it's over now.

Yes,where are all the other guy's wts..be very interesting to see if they use a different approach.

Regards

Gary

[Caida]

Below is my walkthrough for Assassin 80.

I wound up doing a bit trial and error for my step 11 (although I'd rather call it "coloring" of some sort )

I found this one difficult but definitely doable (I managed without peeking at anyone else's walkthrough) -> I haven't really studied the rating system in place yet so hesitate to give it a number.

Cheers,

Caida

I have fixed some errors and added some notes. Thanks Andrew!!


Assassin 80:

Preliminaries:

a. 20(3)n2 = {389/479/569/578} (no 1,2)
b. 12(4)n3 = {1236/1245} (no 7..9) -> 1,2 locked for n3
c. 16(3)n3= {367/457} (no 8,9) -> 7 locked for n3
Note: other combinations {349/358} blocked by 12(4)n3
c1. hidden pair {89} in r13c7 -> no elsewhere in c7
d. 13(2)n23 = [49/58] (no 1..3) -> r1c6 no 6..9
e. 19(3)n4 and n457 and n7 and n8 = {289/379/469/478/568} (no 1)
f. 10(3)n58 = {127/136/145/235} (no 8,9)
g. 23(3)n6 = {689} -> locked for n6
h. 11(4)n69 = {1235} (no 4,6..9) -> I originally tried to take an early elimination of r789c8 no 1,2,3,5 (CPE) here – but it didn’t work – luckily Andrew caught this
i. 6(2)n78 = {15/24} (no 3, 6..9)
j. 28(4)n9 = {4789/5689} (no 1,2,3) -> 8,9 locked for n9 I originally tried to take an early elimination here saying 5689 is blocked by 22(3)n6 – but it didn’t work –Andrew caught this too

adding a couple steps here so I can get to my eliminations:
X1. Innie and Outie c89: r2c7 = r5c8
X2. -> r2c7 no 6
X3. -> r5c8 no 7
X4. -> 7 locked in c7 in n6 -> no 7 elsewhere in c7
X5. -> 7 locked in 28(4)in n9 -> 28(4) = {4789} (no 5,6) -> locked for n9

k&l. steps deleted



1. Innies n2: r14c4+r1c6 = 9(3)
1a. -> min r1c6 = 4 -> max r13c4 = 5 (no 4..9) (can’t have 4 because then r3c4 would be 5)
1b. -> r13c4 = {13/23} -> 3 locked for c4 and n2

2. Innies n7: r79c3 = 6(2) = {24}/[51]
Note that Outie Innie n7: means that r7c3 = r9c4
2a. -> r7c3 no 3,6,7,8,9
2b. -> r9c3 no 5
2c. -> cleanup: 6(2)n78: r9c4 no 1

3. Innie and Outie n14: r6c3 – r1c4 = 7 = [92/81]
3a. -> r6c3 no 2,3,4,5,6,7
3b. -> r1c4 no 3
3c. -> r3c4 = 3 (step 1b)

4. 12(3)n25 = 3{18/27/45} (no 6,9)

5. 16(3)n356 = 9{16/25/34}/8{17/26/35}
5a. -> r4c6 no 8,9

6. 19(3)n457 = [892/874/865/982/964}
6a. -> r6c4 no 2,4,5

7. 19(3)n4: combo {289} blocked by r6c3 -> no 2
7a. -> killer pair {89} in n4 in 19(3)n4 and r6c3 -> no 8,9 elsewhere in n4

8. Innies n8: r7c7+r9c79 = 10(3) = {1[2]7/3[2]5/1[4]5/1[5]4/2[5]3} (no 6,8,9)
9. 20(3)n2 = {479/569/578}
9a. -> killer pair {45} locked in n2 in 20(3)n2 and r1c6 -> no 4,5 elsewhere in n2

10. Outies r1234: r5c239 = 12(3) = {15}[6]/{24}[6]/{13}[8]/{12}[9]
10a. -> r5c23 no 6,7

11. if r6c3 = 8 then r3c7 = 9 and if r6c3 = 9 then r3c7 = 8 here’s how:
11a. r6c3 = 8/9 -> r1c4 = 1/2 (step 3) -> r1c6 = 5/4 (step 1) -> r1c7 = 8/9 (cage sum) -> r3c7 = 9/8 (prelim i);
11b. r3c3 no 8,9 (intersection of both r6c3 and r3c7)
I just looked at Mike's walkthrough and discovered that I could have got this by looking at the Innie and Outie of n1234. I just didn't see that at the time

12. 17(4)n14 = {67}{13}
Note: {74}{15}/{56}{24} blocked by h6(2)n7 (step 2) and {2357} blocked by h12(3)r5 (step 10)
12a. -> {67} locked in r34c3 -> no 6,7 elsewhere in c3
12b. -> {13} locked in r5c23 -> no {13} elsewhere in r5 and n4
12c. -> r5c9 = 8 (step 10)
12d. -> {69} locked in r4c89 -> no 6,9 elsewhere in r4
12e. -> r34c3 = [67]
12f. -> 2 locked in 15(4)n14 in r4 for n4 -> no 2 elsewhere in r4 Andrew pointed out I had missed noting the elimination of 2 from r23c1 at this point

13. 15(4)n14 = {18}{24}/{17}{25}
13a. -> r23c1 no 3,4,5,9
13b. -> 1 locked in r23c1 -> no 1 elsewhere in c1 and n1

14. 16(3)n356 = 9{34}/8{35}
14a. -> r4c67 no 1
14b. -> killer pair {45} locked in r4 in c1267 -> no 4,5 elsewhere in r4

15. pair {18} locked in r4c45 - > no 18 elsewhere in n5

16. 12(3)n12 = 1{29/38}
16b. -> r1c4 = 1
16c. -> r12c3 no 4,5
16d. -> r6c3 = 8 -> r1c6 = 5 -> r1c7 = 8 -> r3c7 = 9 (step 11a)
16e. single: r4c45 = [81]
16f. 12(3)n12 = 1{29} -> {29} locked for c3 and n1

17. h6(2)n7 = [51] (step 2)
17a. -> r9c4 = 5
17b. -> r6c4 = 6
17c. -> r5c23 = [13]
17d. -> r8c3 = 4
17e. -> hidden single: r5c1 = 6
17f. -> r6c12 = {49} locked for r6 and n4
17g. -> hidden single: r4c12 = [52]
17h. -> r23c1 = {17}locked for c1 and n1
17i. -> pair {34} in r1c12 -> locked for r1 and n1
17j. -> pair {58} in r23c2 -> locked for c2
17k. -> hidden single: r12c3 = [29]
17l. -> hidden single: r1c5 = 9

18. 16(3)n3 = {67}[3] 6,7 locked for n3
18a. hidden single: r6c8 = 3
18b. r4c67 = [34]
18c. {125} locked for c9 in r678
18d. -> r3c9 = 4

19. h10(3)n8 (step 8) = [154]
19a. -> r6c56 = {27} -> locked for r6 and n5
19b. singles: r5c456 = [459]; r5c78 = [72]; r7c9 = 2; r2c4 = 7; r23c1 = [17]; r2c78 = [25]; r23c2 = [85]; r2c56 = [46]; r3c8 = 1; r78c4 = [92]

20. 19(3)n8 = [982]
20a. -> all singles and cage sums to finish

[mhparker]

I found this one difficult but definitely doable... I haven't really studied the rating system in place yet so hesitate to give it a number.

Sounds like Caida and Gary (who, unusually, didn't give it a rating himself) are too polite to tell me that I overrated this puzzle...

If so, they may have a point. I took a look back at the A78 (which we rated as 1.25) and that seems to be very much in the same "ball park" as this one. So a 1.25 would probably be nearer the mark here. Would be interesting to have a second opinion on this, though.

BTW: Thanks, Gary, for pointing out that my step 20c was over-complex. I've added a comment to my WT to that effect.

[Afmob]

This was quite a tough assassin. My original walkthrough was quite long since I made lots of little elimination without any real success but this morning I found the key move to crack it.

A80 Walkthrough:

1. N23
a) Innies N3 = 17(2) = {89} locked for C7+N3
b) 16(3) @ N3 = 7{36/45}
c) 13(2) = [49/58]
d) Innies N2 = 9(3) = 3{15/24} because R1C6 = (45) -> R13C4 = 3{1/2} and 3 locked for C4+N2
e) Killer pair (45) locked in 20(3) + R1C6 for N2

2. N124
a) Innies+Outies N14: -7 = R1C4 - R6C3
-> R1C4 = (12) and R6C3 = (89)
b) Hidden Single: R3C4 = 3 @ N2
c) 12(3) @ R3C4 = 3{18/27/45}

3. C789
a) 23(3) = {689} locked for N6
b) Innies+Outies C89: R2C7 = R5C8 = (12345)
c) 6 locked in R789C7 for N9
d) 28(4) = {4789} locked for N9
e) Outies C789 = 12(3): R49C6 <> 8,9 because R1C6 >= 4

4. C123
a) Innies N7 = 6(2) = [24/42/51]
b) 6(2): R9C4 <> 1
c) 19(3) @ R6C3: R6C4 <> 2,4,5 because R7C3 = (245)
d) 19(3) @ R5C1 <> 2 because {289} blocked by R6C3 = (89)
e) Killer pair (89) locked in 19(3) @ R5C1 + R6C3 for N4
f) Innies+Outies C12: 3 = R8C3 - R5C2
-> R8C3 <> 1,2,3 and R5C2 <> 7
g) Outies C123 = 12(3): R6C4 <> 8 because R9C4 <> 1,3
h) 8 locked in R6C123 for N4
i) Innies N14 = 19(3) <> 1

5. R56789
a) Innies N8 = 10(3) <> 6 because R9C4 <> 1,3,6
b) Innies R5 = 12(3): R5C23 <> 6,7 because R5C9 >= 6 (12(3) because of step 5b)

6. N35689
a) 9 locked in R4C89 for N6
b) Outies N5689 = 20(2+1) = {389} -> (89) in R3C7+R6C3 builds pointing pair
-> R3C3 <> 8,9 (20(2+1) because of step 4a)
c) 12(3) @ N9: R9C6 <> 2,3 because R89C7 <> 4,7
d) Innies+Outies N69: R9C6 = R4C7 = (1457)
e) Outies C789 = 12(3): R4C6 <> 4,5 because R9C6 <> 3
f) Outies C789 = 12(3) <> 2 because R19C6 <> 3,6
g) Innies+Outies N5: -8 = R7C6 - (R4C6+R6C4)
-> R4C6 <> 1 because R7C6 <> 2
-> R7C6 <> 3 because there no combo for R4C6+R6C4 = 11(2)
h) Innies N8 = 10(3) = 1{27/45} -> 1 locked for C6+N8
i) Innies N8 = 10(3): R7C6 <> 2 because R9C4 <> 1,7

7. N69
a) 16(3): R4C7 <> 7 because R4C6 <> 1,8
b) Innies+Outies N69: R9C6 = R4C7 = (145)

8. R1234
a) Innies+Outies: -5 = R5C9 - R34C3
-> R34C3 = 11/13(2) = {47/56/67}
b) 17(4) = {1367/1457/2456}

9. C123 !
a) Innies+Outies C12: 3 = R8C3 - R5C2 -> R8C3 <> 9
b) Outies C12 = 20(4) must have {47/56/67} (step 8a)
-> 20(4) = {1478/1568/2567/3467}
c) ! Outies C12 = 20(4) <> {1568} because 8 only possible @ R8C3 and R345C3 @ 17(4) <> 5 (step 9a)
-> Outies = 20(4) = 7{148/256/346} -> 7 locked for C3
d) ! 12(3) <> 8 because it would force Innies N14 to be {289} but that's not
possible since 12(3) can't be {228}
e) 12(3) <> 3 since R1C4 <> 4,5
f) Hidden Single: R5C3 = 3 @ C3
g) 17(4) = {1367} -> {67} locked for C3, R5C2 = 1
h) 12(3) = {129} -> R1C4 = 1, {29} locked for C3+N1
i) 20(4) = {3458} locked for N1 because R3C3 = (67) blocks {3467}
j) 15(4) = {1257} -> {17} locked for N1+C1, {25} locked for R4+N4

10. N47
a) Innies N7 = 6(2) = [51] -> R7C3 = 5, R9C3 = 1
b) 19(3) @ R6C3 = {568} -> R6C3 = 8, R6C4 = 6
c) 20(4) = 46{28/37} -> R8C3 = 4, 6 locked for N7

11. N89
a) Innies N8 = 10(3) = {145} -> R9C4 = 5, R9C6 = 4, R7C6 = 1
b) 12(3) = 4{26/35}, R8C7 <> 3
c) Killer pair (23) locked in 12(3) + R7C9 for N9
d) R7C7 = 6
e) 12(3) = {345} -> R8C7 = 5, R9C7 = 3
f) 11(4) = {1235} -> R7C9 = 2, R8C9 = 1, {35} locked for R6+N6

12. N7
a) Hidden Single: R8C2 = 6
b) 20(4) = {3467} -> R7C2 = 7, R7C1 = 3

13. Rest is singles.

Rating: 1.5, my walkthrough for A78 was longer but I found those moves in A80 were harder to spot and the solving path was quite narrow.

[goooders]

i agree with afmob ,his rating and his reasoning its a 1.5 for me

[Caida]

Sounds like Caida and Gary (who, unusually, didn't give it a rating himself) are too polite to tell me that I overrated this puzzle...

I truely wasn't being polite at all - I am just not familiar with the rating system.

Having looked through the boards to find the definitions of the different ratings I would have to say that A80 is more of a 1.5 rather than a 1.25.

1.25: Harder Assassin. Actually, most recent Assassins seem to have become more difficult than they traditionally used to be. So a rating of "1.25" would be considered the norm now.

1.5: Hard Assassin, having a significantly longer and/or narrower solution path, and/or requiring more advanced techniques.

1.75: Very hard Assassin, but still not hard enough to require a team effort to solve. Does not require any hypotheticals.

bold added by me

I think that if you didn't spot the one key move (which i did unnessecarily through t&e) you weren't able to solve the puzzle - and this makes it a narrow solving path. I think that this was more difficult than "the norm" but it wasn't as crazy difficult as others have been.

Thus my rating of 1.5 for Assassin 80 (which seems to be in agreement with the majority so far).

Cheers,

Caida

[gary w]

Yes,let's go for 1.5.Once you spot where the "weak spot" is it's pretty straightforward but as Caida said it is a narrow solving path..at least I haven't seen any other obvious routes to a solution.

Regards

Gary

[Andrew]

After having now had a chance to take a deeper look, it seems that this puzzle is very similar in difficulty to the A78.

Really?! I found it distinctly harder. Maybe that's because I didn't find the logical breakthrough step but I also needed a hint for A78. Even if I had found it I still think A80 is harder because there is a greater number of difficult steps.

I'm surprised that nobody, apart from Afmob with !s, highlighted what I consider to be the most difficult step. That's Mike's step 17, Caida's step 12 and Afmob's step 9 which did it slightly differently. I thought Afmob's step 9c using the interference between 8 in R8C3 and 5 in R5C2 was really neat!

After coming to a dead halt several days ago I came back to A80 yesterday and looked for things I'd missed before. I found a little more but not enough so I tried a contradiction move which ironically used the feature that I was missing, although I still didn't realise that until I went through the posted walkthroughs and saw what I'd missed.

Congratulations to everyone who managed to solve A80 logically.

For what it's worth, here is my A80 walkthrough.

Prelims

a) R1C67 = {49/58/67}, no 1,2,3
b) R9C34 = {15/24}
c) 20(3) cage in N2 = {389/479/569/578}, no 1,2
d) 19(3) cage in N4 = {289/379/469/478/568}, no 1
e) 19(3) cage at R6C3 = {289/379/469/478/568}, no 1
f) 10(3) cage at R6C5 = {127/136/145/235}, no 8,9
g) 23(3) cage in N6 = {689}, locked for N6
h) 19(3) cage in N7 = {289/379/469/478/568}, no 1
i) 19(3) cage in N8 = {289/379/469/478/568}, no 1
j) 12(4) cage in N3 = {1236/1245}, 1,2 locked for N3
k) 11(4) cage at R6C8 = {1235}
l) 28(4) cage in N9 = {4789/5689}, 8,9 locked for N9

1. 45 rule on N3 2 innies R13C7 = 17 = {89}, locked for N3, clean-up: R1C6 = {45}
[Alternatively Hidden Pair 8,9 in R13C7 for C7.]
1a. Min R3C7 = 8 -> max R4C67 = 8, no 8,9 in R4C6

2. 45 rule on N7 2 innies R79C3 = 6 = {24}/[51], clean-up: no 1 in R9C4

3. 45 rule on N14 1 innie R6C3 = 1 outie R1C4 + 7 -> R6C3 = {89}, R1C4 = {12}

4. 45 rule on N2 3 innies R1C46 + R3C4 = 9 = {135/234} (cannot be {126} because R1C6 only contains 4,5) -> R3C4 = 3, R1C46 = [15/24]
4a. R4C45 = {18/27/45}, no 6,9

5. 20(3) cage in N2 = {479/569/578}
5a. Killer pair 4,5 in 20(3) cage and R1C6, locked for N2
5b. 16(3) cage in N2 = {169/178/268}

6. 19(3) cage in N4 = {379/469/478/568} (cannot be {289} which clashes with R6C3), no 2
6a. Killer pair 8,9 in 19(3) cage and R6C3, locked for N4

7. 45 rule on N69 1 innie R4C7 = 1 outie R9C6, no 6,8,9 in R9C6

8. 19(3) cage at R6C3 = {289/469/478/568}
8a. R7C3 = {245} -> no 2,4,5 in R6C4

9. 45 rule on C12 1 outie R8C3 = 1 innie R5C2 + 3, no 7 in R5C2, no 1,2,3 in R8C3

10. 45 rule on C89 1 outie R2C7 = 1 innie R5C8, no 6 in R2C7, no 7 in R5C8
10a. 7 in N6 locked in R456C7, locked for C7

11. 7 in N9 locked in 28(4) cage = {4789}, 4 locked for N9

12. 45 rule on C123 3 outies R169C4 = 12 = {129/147/156/246}, no 8

13. 45 rule on C789 3 outies R149C6 = 12 = {147/156/246/345} (cannot be {237} because R1C6 only contains 4,5)
13a. 4 of {147/246} must be in R1C6, 3 of {345} must be in R4C6 -> no 4 in R4C6
13b. 5 of {156} must be in R1C6, 3 of {345} must be in R4C6 -> no 5 in R4C6

14. 12(3) cage at R8C7 = {156/237/246/345} (cannot be {147} because 4,7 only in R9C6)
14a. 2 of {237/246} must be in R89C7 -> no 2 in R9C6
14b. 3 of {237/345} must be in R89C7 -> no 3 in R9C6
14c. Clean-up: no 2,3 in R4C7 (step 7)

15. R149C6 (step 13) = {147/156/345} (cannot be {246} because 2,6 only in R4C6), no 2

16. 45 rule on R1234 3 outies R5C239 = 12 = {129/138/156/246} (cannot be {147/237/345} because R5C9 only contains 6,8,9), no 7
16a. 6 of {156/246} must be in R5C9 -> no 6 in R5C23, clean-up: no 9 in R8C3 (step 4)
16b. Max R5C23 (from combinations in step 16) = 6 -> min R34C3 = 11, no 1, no 2 in R3C3

17. 45 rule on R6789 3 outies R5C178 = 15 = {159/249/258/267/348/357/456} (cannot be {168} because 6,8 only in R5C1)
17a. 6,8,9 of {249/348/456} must be in R5C1 -> no 4 in R5C1

18. 45 rule on N8 3 innies R7C6 + R9C46 = 10 = [127/145/154/325/541/721] (cannot be [451] because R9C46 clashes with R9C34), no 2,4,6 in R7C6

19. 45 rule on N5 2 innies R4C6 + R6C4 = 1 remaining outie R7C6 + 8
19a. R6C4 cannot be 7 more than R7C6 -> no 1 in R4C6
19b. R4C6 + R6C4 cannot total 11 -> no 3 in R7C6
19c. R4C6 + R6C4 = [36/39/67/69/76]

20. R7C6 + R9C46 (step 18) = [127/145/154/541/721], 1 locked in R79C6 for C6 and N8

21. 16(3) at R3C7 = {169/178/349/358} (cannot be {367/457} because R3C7 only contains 8,9)
21a. 7 of {178} must be in R4C6 -> no 7 in R4C7, clean-up: no 7 in R9C6 (step 7)

22. 10(3) cage at R6C5 = {127/145/235} (cannot be {136} which clashes with R4C6 + R6C4), no 6

23. 7 in C7 locked in R56C7 for 16(4) cage = {1267/1357/2347}
23a. R5C178 (step 17) = {159/249/267/348/357} (cannot be {258/456} because R5C78 cannot contain {25/45} in 16(4) cage)

24. 9 in R4 locked in R4C89, locked for N6

25. 45 rule on N1234 2 remaining innies R3C7 + R6C3 = 17 = {89}, CPE no 8,9 in R3C3, clean-up: no 2,3 in R4C3 (step 16b)

26. R5C239 (step 16) = {138/156/246}
26a. R5C23 = {13/15/24}
26b. 17(4) at R3C3 = {1367/1457/2456} (cannot be {2357} which clashes with R5C23)
26c. R34C3 = {47/56/67}

27. 45 rule on R1234 4 innies R34C3 + R4C89 = 28 = 4{789}/[56]{89}/[65]{89}/[67]{69}/[74]{89}
27a. -> R34C3 = {47/56}/[67]

28. R1C4 + R6C3 (step 3) = [18/29]
28a. 12(3) cage at R1C3 = {129/147/156/237} (cannot be {138} which clashes with R1C4 + R6C3, cannot be {246} which clashes with R34C3, cannot be {345} because R1C4 only contains 1,2), no 8
28b. 1 of {129} must be in R1C4 (1 cannot be in R12C3 when 2 in R1C4 causes clash between 9 in R12C3 and 9 in R6C3)
28c. -> no 1 in R12C3
[45 rule for N14 3 innies R126C3 = 19, no 1, is a simpler way to do this.]
28d. {156} must be [615] (because [516] clashes with R1C46), no 5 in R1C3, no 6 in R2C3

29. R5C178 = {159/249/267/348/357} (cannot be {258/456} because R5C78 cannot contain {25/45} in 16(4) cage)
29a. 16(4) cage at R5C7 (step 23) = {1267/1357/2347} = {15}[73]/{24}[73]/{34}[72]/{35}[71]/[7216]/[7243]/[73]{15}/[7342]/[75]{13}
29b. -> no 2 in R6C7

There must be a better way to continue from here but I’d struggled to a dead halt.
30. If 12(3) cage at R1C3 = {237} => R12C3 = {37}, R1C4 = 2, R69C4 = [64], R79C3 = [42] => R34C3 = {56}, R5C23 = {24} (steps 26c and 26b) clashes with R79C3 -> no 2 in R1C4
30a. R1C4 = 1, R1C6 = 5 (step 4), R1C7 = 8, R3C7 = 9, clean-up: no 3 in R12C3 (step 28a), no 7,9 in 16(3) cage in N2 (step 5b), no 8 in R4C5 (step 4a), no 7 in R4C6 (step 13), no 5 in R4C7 (step 21), no 4 in R9C4 (step 20), no 2 in R9C3, no 4 in R7C3 (step 2), no 7 in R6C4 (step 12)

31. Naked triple {268} in 16(3) cage in N2, locked for N2

32. R5C3 = 3 (hidden single in C3), R5C29 = [18] (step 26), R34C3 = [67] (step 26b and then step 27a), clean-up: no 4,5 in R12C3 (step 28a), no 2 in R4C45 (step 4a)
32a. Naked pair {29} in R12C3, locked for C3 and N1 -> R6C3 = 8, R7C3 = 5, R6C4 = 6 (step 8), R89C3 = [41], R9C4 = 5, R9C6 = 4, R7C6 = 1 (step 20), R4C6 = 3, R4C7 = 4 (step 21), clean-up: no 5 in R4C5 (step 4a) -> R4C45 = [81]

33. Naked pair {69} in R4C89, locked for R4
33a. Naked pair {25} in R4C12, locked for N4 and 15(4) cage, R23C1 = 8 = {17} (only remaining permutation)
33b. Naked pair {17} in R23C1, locked for C1 and N1
33c. Naked pair {34} in R1C12, locked for R1 and N1
33d. Naked pair {58} in R12C2, locked for C2 -> R4C12 = [52]

34. Naked pair {67} in R1C89, locked for R1 and N3 -> R1C5 = 9, R12C3 = [29], R2C9 = 3 (cage sum), R7C9 = 2
34a. Naked pair {47} in R2C45, locked for R2 -> R23C1 = [17]
34b. Naked pair {25} in R2C78, locked for R2 and N3 -> R23C2 = [85], R2C6 = 6

35. R7C6 = 1 -> R6C56 = 9 = {27} (only remaining combination), locked for R6 and N5, R5C6 = 9, R5C1 = 6, R5C45 = [45], R5C78 = [72], R2C45 = [74], R2C78 = [25]

36. Naked pair {15} in R68C9, locked for C9 and 11(4) cage -> R3C89 = [14], R6C8 = 3

37. R9C6 = 4 -> R89C7 = 8 = [53], R67C7 = [16], R678C9 = [51], R78C4 = [92], R7C5 = 8 (cage sum)

and the rest is naked singles

[mhparker]

After having now had a chance to take a deeper look, it seems that this puzzle is very similar in difficulty to the A78.

Really?! I found it distinctly harder. Maybe that's because I didn't find the logical breakthrough step but I also needed a hint for A78. Even if I had found it I still think A80 is harder because there is a greater number of difficult steps.

There were parallels between the two puzzles, but I am actually expecting the A80 to score higher than the A78. Whether that difference is enough to bring it in line with other 1.5-rated puzzles like the A60 and A78V2 is another matter. It's true that my step 17 (and the equivalent steps in the other WTs) was not so easy to spot for us humans, but stuff like that is a piece of cake for SudokuSolver. Therefore, this is a puzzle for which I'm expecting a disappointingly low SScore. So, whilst I now agree with a rating of 1.5 for this one, I'm only reckoning on an SScore of around 1.3 or so.

Will be interesting to see the next exciting installment of Ed and Richard's SScores to see who's "right". Come on Ed, don't keep us in suspense much longer!

[sudokuEd]

Come on Ed, don't keep us in suspense much longer!

The suspense is driving me crazy too! Thought we had it last week, then someone released A2X-Lite - which came in at 1.87!! Thanks Mike (again ). Been so absorbed the last couple of days that suddenly realized I hadn't logged in here for 23 hours!!

Richard commented to me that it's like squeezing a balloon - you think you have it under control and then it just pops out .

Thanks for your interest.
Cheers
Ed