SudoCue Users

I didn't think that I was the first to post a walkthrough for this assassin since it's quite late .
This killer was strange since it took me a long time to find the key moves although they were quite easy to see (step 5a, 8a).

Note that step 5a could be applied right after step 1a though in a different form and probably more easier way:

(Triple click to see)

1b. Consider placement of Outies C89 = 3(2) = {12} -> R12C9+R23C8 <> 1,2:
- i) R1C7 = 1 -> R4C7 @ 26(5) = 2 -> 26(5) has no 1,2 @ N3
- ii) R1C7 = 2 -> R4C7 @ 26(5) = 1 -> 26(5) has no 1,2 @ N3

A86 Walkthrough:

1. C6789
a) Outies C89 = 3(2) = {12} locked for C7
b) 9(2) = [18/27]
c) Innies = 5(2) = {14/23}
d) 8(3) = 1{25/34} -> 1 locked for N9

2. R1+C56
a) Killer pair (12) locked in 6(2) + 9(2) for R1
b) 17(4) must have 1 or 2 and it's only possible @ R2C5 -> R2C5 = (12)
c) Innies+Outies R1: -13 = R2C5 - R1C19 -> R1C19 <> 3,4
d) 3 locked in R1C456 @ 17(4) for N2
e) 5(2): R4C5 <> 2
f) Killer pair (12) locked in R2C5 + 5(2) for C5
g) 2 locked in R23C5 for N2
h) Innies C6789 = 5(2): R9C6 = (12)
i) Killer pair (12) locked in 17(4) @ N8 + R9C6 for C6+N8
j) 2 locked in 16(4) @ C4 -> R5C3 <> 2

3. C1234
a) Outies C12 = 9(2) -> R4C3 = (4578)
b) Innies C1234 = 12(2) <> 6
c) Innies N7 = 14(4) <> 9
d) 11(2): R6C3 <> 2

4. R789
a) Outies R89 = 17(4) must have 1 or 2 and it's only possible @ R7C6 -> R7C6 = (12)
b) Naked pair (12) locked in R79C6 for C6
c) 17(4) <> 12{59/68} since (12) only possible @ R7C6
d) Innies N9 = 25(4) <> 2 because R89C7 @ 17(4) can't have two of (689)
e) 2 locked in 8(3) -> 8(3) = {125} locked for N9
f) 12(2) <> 7
g) 13(2): R6C7 <> 8
h) Naked triple (125) locked in R9C689 for R9
i) 5 locked in R9C89 for N9

5. C89 !
a) ! R2C9+R23C8 @ 26(6) <> 1,2 because together with 8(3) = {125} 26(6) would build
a generalized X-Wing for C89 which would leave no 1,2 @ N6
b) 1 locked in 26(6) for R4+N6
c) Hidden pair (12) locked in R1C7+R3C9 for N3 -> R3C9 = (12)
d) 17(3) = {179/269/278} -> R45C9 <> 2,3,4,5
e) Naked pair (12) locked in R38C9 for C9
f) R9C9 = 5

6. C45
a) 1 locked in R56C4 @ 16(4) for C4; R5C3 <> 1
b) 16(4) = 12{49/58/67} <> 3
c) 5(2): R3C5 <> 4
d) Innies C1234 = 12(2): R1C4 <> 7

7. R123
a) Naked pair (12) locked in R3C59 for R3
b) Innies+Outies R1: -13 = R2C5 - R1C19; R2C5 = (12)
-> R1C19 = 14/15(2) = {59/68/69} <> 7 because {78} blocked R1C8 = (78)
c) 17(4) <> 6 because {1367} impossible since (67) only possible @ R1C5
d) 6 locked in R1C19 -> R1C1 <> 5 (step 7b)

8. C123 !
a) ! R7C1 <> 1 because it sees all 1 of N4
b) 1 locked in R8C13 for R8
c) R8C9 = 2, R9C8 = 1, R9C6 = 2, R7C6 = 1, R3C9 = 1, R1C7 = 2 -> R1C8 = 7

9. R123
a) R3C5 = 2 -> R4C5 = 3
b) 6(2) = {15} locked for R1+N1
c) 17(4) = {1349} -> R2C5 = 1, {49} locked for R1+N2
d) 26(6) = 1256{39/48} -> R4C7 = 1, R4C8 = 2; 5 locked for C8+N3; 6 locked for N3
e) 17(3) = {179} -> 7,9 locked for C9+N6

10. C456
a) Innies C6789 = 5(2) = {23} -> R1C6 = 3
b) Innies C1234 = 12(2) = [48/93]
c) 20(3) <> 4 because {479} blocked by R1C5 = (49)
d) 20(3) = 5{69/78} -> 5 locked for C5

11. C456
a) Outies N3 = 14(2) = {68} locked for C6+N2
b) 26(4) @ N2 = {3689} -> {39} locked for C7+N3
b) 23(4) = 57{29/38} -> 5,7 locked for C4; R23C3 = [29/38/83]
c) 26(4) @ N5 must have 6 xor 8 and it's only possible @ R5C7 -> R5C7 = (68)
d) 17(4) @ N8 = {1457} -> R8C6 = 5, {47} locked for C7+N9
e) 26(4) @ N5 = {4679} -> R5C7 = 6, {479} locked for N5

12. R456
a) 21(4) = {3468} -> R7C9 = 6
b) R6C7 = 5
c) 20(3) = 5{69/78} -> R5C5 = 5
d) 16(4) = {1267} because R456C4 <> 4,9 -> R5C3 = 7
e) Hidden pair (12) locked in R56C4 for N5 -> R56C4 = {12}
f) R4C4 = 6, R6C5 = 8 -> R7C5 = 7

13. N78
a) Outies N8 = 7(2) = [16/34/43]
b) 18(3) <> 4,6 because 6{39/48} blocked by Killer pairs (36,46) of Outies N8
c) 18(3) = 8{19/37} -> 8 locked for N7
d) 13(2) = {49} locked for C2+N7
e) 18(3) = {378} -> 3 locked for N7
f) 15(4) must have 2 xor 5 and it's only possible @ R7C1 -> R56C2+R6C1 <> 2

14. Rest is singles.

Rating: 1.25. This assassin could be solved quite quickly if you find the right moves.

Thanks for getting the thread going, Afmob!

This was a difficult puzzle, because there are 2 critical moves (Afmob's steps 5a and 8a = my steps 1a and 8a) that are difficult to spot. If you don't get both of them, you'll have an exceedingly hard time solving this Assassin!

As if to prove me wrong, Gary has just done it using only the first of these two "critical" moves! He used two advanced moves instead. Well done, Gary!

In that sense, it was similar to my Maverick 2, which also had an extremely narrow initial solving path, except the M2 remained difficult, whereas this puzzle is relatively straightforward (once those 2 critical moves have been found...).

I'm glad Afmob rated this puzzle as 1.25, because that's the rating I was going to give it. However, because of the narrow solving path, some people may find it subjectively harder than that, or have difficulty completing the puzzle at all.


Assassin 86 Walkthrough

Prelims

a) 6(2) at R1C2 = {15/24}
b) 9(2) at R1C7 = {18/27/36/45} (no 9)
c) 26(6) at R1C9 = {1(23479/23569/23578/24568/34567)}
d) 26(4) at R2C6 and R4C6 = {2789/3689/4589/4679/5678} (no 1)
e) 5(2) at R3C5 = {14/23}
f) 20(3) at R5C5 = {389/479/569/578} (no 1,2)
g) 11(2) at R6C3 = {29/38/47/56} (no 1)
h) 13(2) at R6C7 and R7C2 = {49/58/67} (no 1..3)
i) 12(2) at R7C8 = {39/48/57} (no 1,2,6)
j) 8(3) at R8C9 = {125/134}; 1 locked for N9

1. Outies C89: R14C7 = 3(2) = {12}, locked for C7
1a! -> no 1,2 in R23C8+R12C9 (CPE)
1b. cleanup: no 1..6 in R1C8

2. Hidden pair (HP) on {12} at R1C7+R3C9
2a. -> R3C9 = {12}
2b. 17(3) at R3C9 = {179/269/278}
2c. -> R45C9 = {6789}

3. R3C9 and R4C7 must contain the same digit, because they both see R1C7
3a. -> this digit (1 or 2) must be constrained to C8 within N9
3c. -> R9C8 = {12}

4. 6(2) at R1C2 and R1C7 form killer pair (KP) on {12}
4a. -> no 1,2 elsewhere in R1

5. 17(4) at R1C4 = {1349/1358/1367/1457/2348/2357/2456} = {(1/2)..}
(Note: {1259/1268} unplaceable)
5a. {12} only available in R2C5
5b. -> R2C5 = {12}

6. 1 in 26(6) at R1C9 (Prelim c) locked in R4C78 for R4 and N6
6a. cleanup: no 4 in R3C5

7. 1 in N5 locked in C4 -> not elsewhere in C4
7a. no 1 in R5C3

8. 1 in N4 locked in R56C12
8a. -> no 1 in R7C1 (CPE)

9. Hidden single (HS) in R7 at R7C6 = 1

10. Innies C6789: R19C6 = 5(2) = [32] (last combo/permutation)

11. Naked single (NS) at R9C8 = 1
11a. -> R3C9 = 1, R4C7 = 1 (both step 3)
11b. -> R1C78 = [27], R34C5 = [23]
11c. -> R2C5 = 1
11c. cleanup: no 4 in R1C23; no 5 in R78C8

12. split 7(2) at R89C9 = [25] (last combo/permutation)
(Note: {34} blocked by 12(2) at R7C8)
12a. cleanup: no 8 in R6C7

13. 7 unavailable for 26(6) at R1C9 (Prelim c) = {1256(39/48)}
13a. -> killer single (KS) at R4C8 = 2
13b. 5 locked in R23C8 for C8 and N3
13c. 6 locked for N3

14. split 16(2) at R45C9 = {79}, locked for C9 and N6
14a. cleanup: no 4,6 in R7C7

15. Naked pair (NP) at R1C23 = {15}, locked for R1 and N1

16. split 13(2) at R1C45 = {49} (last combo), locked for R1 and N2

17. Outies N3: R23C6 = 14(2) = {68} (last combo), locked for C6, N2 and 26(4)
17a. -> R23C7 = 12(2) = {39} (last combo), locked for C7 and N3
17b. cleanup: no 4 in R6C7

18. 9 in C8/N9 locked in 12(2) at R7C8 = {39} (no 4,8), locked for C8 and N9

19. HS in C9/N6 at R6C9 = 3
19a. cleanup: no 8 in R7C3

20. NP at R23C4 = {57}, locked for C4 and 23(4)
20a. -> R23C3 = 11(2) = [29]/{38}
20b. -> R2C3 = {238}, R3C3 = {389}

21. Split 16(3) at R8C6+R89C7 = {457} (no 6,8,9) (last combo)
21a. -> R8C6 = 5; R89C7 = {47}, locked for C7 and N9

22. R67C7 = [58]
22a. -> R5C7 = 6, R7C9 = 6
22b. cleanup: no 7 in R8C2

23. R12C9 = [84]
23a. -> R1C1 = 6

24. HS in R4 at R4C4 = 6

25. HS in C5/N5 at R5C5 = 5
25a. -> split 15(2) at R67C5 = [87] (last combo/permutation)
25b. cleanup: no 4 in R6C3; no 3 in R7C3; no 6 in R8C2

26. R56C8 = [84]

27. HP in N5 at R56C4 = {12} (no 4,9)
27a. -> R5C3 = 7 (cage sum)
27b. cleanup: no 4 in R7C3

28. R45C9 = [79]

29. R456C6 = [947]

30. 19(4) at R7C4 = {1(369/468)} (all other combos unplaceable)
30a. -> KS at R8C3 = 1

31. R1C23 = [15]
31a. cleanup: no 6 in R6C3

32. HS in C3 at R9C3 = 6 (alternatively, outie N8)

33. Innies N7 = 7(2) = [52] (last combo/permutation)
33a. -> R6C3 = 9
33b. cleanup: no 8 in R8C2

Now it's all just singles and cage sums.

This was a nice puzzle based on a number of killer 1/2 pairs,viz;



1.r7c14=1/2 and with 6(2) cage r1c23 1/2 fixed at r1c237 in r1
2. Thus r3c9=1/2.26(6) must contain a 1 and this must be at r4c7/8.3.In r1 3 must be in N2..cannot go elsewhere.R4C5<>2
4.So r4c5=3/4 and r3c5=1/2.
5.Combo analysis on 17(4) cage N2 -> r2c5 also=1/2.
6.r19c6=5 and r1c6 therefore=3/4 so r9c6=1/2
7.r789c6+r7c9=25 and contains a 6.So cannot also contain a 5.If 5 in the 12(2) cage N9 -> r1c7=1->r3c9=2->r9c8=2..contradiction.Thus the 8(3) cage N9={125}
8.Now if r9c6=1 ->r8c9=1->r3c9=2 ->r1c7=1 then cannot place the 4 (see 6.) at r1c6.So r9c6=2 r8c9=2 and now many placements are possible.
9.r1c19={68}
10.r4c3={48} and so r3c1={48}
11.The 33(6) cage N14 doesn't (after placements) contain a 1.So other two missing digits total 11.Combo analysis shows that this 11 pair must go in r23c3.Thus r23c4=12,r23c6=14 (={68} and it's pretty much all over now.



I'ld rate it about 1.25 again really based on time to complete..not very scientific I'll admit.!!Took me about 1.5 hours.

Regards

Gary

Mike wrote:This was a difficult puzzle, because there are 2 critical moves (Afmob's steps 5a and 8a = my steps 1a and 8a) that are difficult to spot. If you don't get both of them, you'll have an exceedingly hard time solving this Assassin!

I'll agree with most of that. It took me a very long time to spot my step 2a, which is clearly the key move for this puzzle. Congratulations to Ruud for creating a puzzle that required this key move! I didn't spot the other move that Mike identified. It clearly speeds up the solution, particularly when used with his step 3, but doesn't seem to be critical. In Afmob's walkthrough steps 4a and 4b speeded up the solution.

In his introduction to A73V1.5, Mike wrote:The A86 is not very suitable for the creation of variants, so let's wind the clock back a little...

Having spent a lot of time before I found step 2a, I know that it would have been very difficult to solve A86 without that step. Therefore any variant would almost certainly need to provide a similar step or it would risk becoming an Unsolvable.

Here is my walkthrough for A86. Not one of my best. I only modified it as required after finding step 2a. As well as taking a long time to find that step, I'd also taken time to find step 19 and a modified version of step 22, both before I found step 2a.

Thanks Mike for the comments, which I've added below, and for corrections to steps 9 and 22; I've also corrected a typo in step 32a.

Mike's comments about some of my steps, in the later message where he analyses Gary's steps, are appreciated!

Prelims

a) R1C23 = {15/24}
b) R1C78 = {18/27/36/45}, no 9
c) R34C5 = {14/23}
d) R67C3 = {29/38/47/56}, no 1
e) R67C7 = {49/58/67}, no 1,2,3
f) R78C2 = {49/58/67}, no 1,2,3
g) R78C8 = {39/48/57}, no 1,2,6
h) R567C5 = {389/479/569/578}, no 1,2
i) 8(3) cage in N9 = 1{25/34}, 1 locked for N9
j) 26(4) cage at R2C6 = {2789/3689/4589/4679/5678}, no 1
k) 26(4) cage at R4C6 = {2789/3689/4589/4679/5678}, no 1
l) 26(6) cage at R1C9 = {123479/123569/123578/124568/134567}, must contain 1

1. 45 rule on C12 2 outies R14C3 = 9 = [18/27]/{45}, no 1,2,3,6,9 in R4C3

2. 45 rule on C89 2 outies R14C7 = 3 = {12}, locked for C7, clean-up: R1C8 = {78}
2a. R1C9 + R2C89 + R3C8 can 'see' R14C7 -> CPE no 1,2 in R1C9 + R2C89 + R3C8
2b. 1 in 26(6) cage locked in R4C78, locked for R4 and N6, clean-up: no 4 in R3C5
2c. R3C9 = {12} (hidden pair R1C7 + R3C9 for N9)
2d. R345C9 = {179/269/278}, no 3,4,5
2e. 2 of {269/278} must be in R345C9 -> no 2 in R45C9

3. Killer pair 1,2 in R1C23 and R1C7, locked for R1
3a. 1 in R789C6 locked in C6, locked for N8

4. 45 rule on C1234 2 innies R19C4 = 12 = {39/48/57}, no 2,6

5. 45 rule on C6789 2 innies R19C6 = 5 = [32/41]

6. 45 rule on R1 2 innies R1C19 = 1 outie R2C5 + 13
6a. Min R1C19 = 14, no 3,4
6b. Max R1C19 = 17 -> max R2C5 = 4

7. 3 in R1 locked in R1C456, locked for N2, clean-up: no 2 in R4C5
7a. 17(4) cage at R1C4 = {1349/1358/1367/2348/2357}
7b. 1,2 only in R2C5 -> R2C5 = {12}
7c. Naked pair {12} in R23C5, locked for C5 and N2
7d. Naked pair {12} in R3C59, locked for R3

8. Hidden killer triple 7,8,9 in R1C19, R1C45 and R1C8 for R1 -> R1C19 must contain one of 7,8,9
8a. R2C6 = {12} -> R1C19 = 14,15 (step 6) = {59/68/69} (cannot be {78} because of step 8), no 7
[Simpler was "cannot be {78} which clashes with R1C8".]

9. 45 rule on R89 2 innies R8C28 = 2 outies R7C46 + 8, max R8C28 = 17 -> max R7C46 = 9, no 9, no 8 in R7C6

10. 6 in N9 locked in R7C79 + R89C7
10a. 45 rule on N9 4 innies R7C79 + R89C7 = 25 = {2689/3679/4678}, no 5, clean-up: no 8 in R6C7

11. 45 rule on N7 4 innies R7C13 + R89C3 = 14 = {1238/1247/1256/1346/2345}, no 9, clean-up: no 2 in R6C3
11a. 18(3) cage in N7 = {189/279/369/378/459/567} (cannot be {468} which clashes with R7C13 + R89C3)

12. 45 rule on N89 5 innies R7C4579 + R8C4 = 33, max R7C4579 = 30 -> min R8C4 = 3

13. 45 rule on N9 2 innies R7C79 = 2 outies R78C6 + 8
13a. Min R78C6 = 5 (cannot be {12} which clashes with R9C6, cannot be {13} which clashes with R19C6) -> min R7C79 = 13, no 2,3 in R7C9

14. 2 in N9 locked in 8(3) cage = {125}, locked for N9, clean-up: no 7 in R78C8

15. Naked triple {125} in R9C689, locked for R9, clean-up: no 7 in R1C4 (step 4)
15a. 18(3) cage in N7 (step 11a) = {189/279/369/378/459/567}
15b. 5 of {459} must be in R8C1 -> no 4 in R8C1

16. Naked triple {125} in R389C9, locked for C9
[Mike: At this point, 5 of both R9 and C9 is locked in 8(3)N9 -> R9C9 = 5.
Another example of my "killer brain" being blind, as well as the very long time I took before I saw step 2a.]

17. 17(4) cage at R1C4 (step 7a) = {1349/1358/2348/2357} (cannot be {1367} because 6,7 only in R1C5), no 6
17a. 6 in R1 locked in R1C19 = {68/69} (step 8a), no 5
[Alternatively Killer pair 4,5 in R1C23 and R1C456, locked for R1]
17b. 17(4) cage at R1C4 = {1349/2348/2357} (cannot be {1358} which clashes with R1C19 which must be {68} when R2C5 = 1)
17c. 7 of {2357} must be in R1C5 -> no 5 in R1C5

18. 17(4) cage at R7C6 = {1349/1358/1367/1457/2348/2357/2456} (cannot be {1259/1268} which clash with R9C6)
18a. Cannot be {1358} because {13} in R78C6 clashes with R19C6 and {38} in R89C7 clashes with R7C79 + R89C7
18b. -> 17(4) cage at R7C6 = {1349/1367/1457/2348/2357/2456}
18c. 8,9 of {1349/2348} must be in R89C7 because R89C7 cannot be {34} (step 10a) -> no 8,9 in R8C6

19. Killer pair 1,2 in R78C6 and R9C6, locked for C6 and N8, clean-up: no 7 in R7C6 (step 9)

20. R7C79 + R89C7 (step 10a) = {3679/4678}
20a. R789C7 cannot be {367} which clashes with R6C7 -> no 9 in R7C9, clean-up: no 4 in R7C7 (step 13a), no 9 in R6C7
20b. R789C7 cannot be {467} which clashes with R6C7 -> no 8 in R7C9
20c. No 6 in R7C6, CPE R7C6 'sees' all cells of R7C79 + R89C7
[That has been there since step 13a, when it would have eliminated 6,7, but I only spotted it here.]

21. 19(4) cage at R7C4 = {1369/1378/1459/1468/1567/2359/2368/2458/2467/3457} (cannot be {1279} because 1,2 only in R8C3)
21a. Only combination without 1,2 is {3457} -> no 6,8 in R8C3

22. 1,2 in N5 locked in R123C4, locked for 16(4) cage -> no 1,2 in R5C3
22a. 16(4) cage at R4C4 = {1249/1258/1267}, no 3

23. R7C79 + R89C7 (step 20) = {3679/4678}
23a. 17(4) cage at R7C6 (step 18b) = {1349/1367/1457/2348/2357/2456}
23b. R89C7 cannot be {67} because R78C6 = {13} clashes with R19C6
23c. -> R89C7 must contain 3 or 4 -> no 4 in R7C9

24. Naked quad {6789} in R1457C9, locked for C9

25. 7 in C9 locked in R457C9, CPE no 7 in R56C8

26. Killer pair 6,7 in R45C9 and R7C9, locked for C9
26a. R1C1 = 6 (hidden single in R1)
[Mike: You could have dispensed with step 25 and had KP on {67} eliminating 6,7 from R56C8, too.
Wow! A CPE killer pair. I’ve never thought of that and don’t remember it ever appearing on the forum.]

27. 26(6) cage at R1C9 = {123479/123569/123578/124568} (cannot be {134567} because R1C9 only contains 8,9)
27a. 1,2 only in R4C78 -> R4C78 = {12}, locked for R4 and N6
27b. R1C9 = {89} -> no 8,9 in R23C8
27c. 7 in C8 locked in R123C8, locked for N3

28. 16(4) cage at R4C4 (step 22a) = {1249/1258/1267}
28a. 1,2 only in R56C4 -> R56C4 = {12}

29. 21(4) cage at R5C8 = {3468/3567} (cannot be {3459} because R7C9 only contains 6,7), no 9

30. 9 in C8 locked in R78C8 -> R78C8 = {39}, locked for C8 and N9, clean-up: no 4 in R6C7

31. 21(4) cage at R5C8 = {3468/3567} -> R6C9 = 3, R2C9 = 4, clean-up: no 8 in R7C3

32. 26(6) cage at R1C9 (step 27) = {124568} (only remaining combination) -> R1C9 = 8, R1C78 = [27], R3C9 = 1, R4C78 = [12], R34C5 = [23], R2C5 = 1, clean-up: no 4 in R1C23, no 6 in R45C9 (step 2d)
32a. Naked pair {56} in R23C8, locked for C8 and N3 -> R9C8 = 1, R9C6 = 2, R89C9 = [25]
32b. Naked pair {79} in R45C9, locked for C9 and N6 -> R7C9 = 6, clean-up: no 5 in R6C3, no 7 in R8C2
32c. Naked pair {48} in R56C8, locked for N6

33. 1,4 locked in 17(4) cage at R7C6 (step 18) = {1457} (only remaining combination), no 3,6,8
33a. Naked pair {47} in R89C7, locked for C7 and 17(4) cage -> R7C7 = 8, R6C7 = 5, R5C7 = 6, clean-up: no 7 in R4C4 (step 22a), no 5 in R8C2
33c. Naked pair {15} in R78C6, locked for C6 and N8

34. R5C7 = 6 -> R456C6 = 20 = {479} (only remaining combination), locked for C6 and N5 -> R1C6 = 3, clean-up: no 4,9 in R5C3 (step 22a)

35. Naked pair {15} in R1C23, locked for R1 and N1
35a. Naked pair {49} in R1C45, locked for N2
35b. Naked pair {68} in R23C6, locked for N2
35c. Naked pair {57} in R23C4, locked for C4 and 23(4) cage, clean-up: no 8 in R5C3 (step 22a)

36. R5C5 = 5 (hidden single in C5), R5C3 = 7, R4C4 = 6 (step 22a), R45C9 = [79], R5C6 = 4, R46C6 = [97], R56C8 = [84], R6C5 = 8, R7C5 = 7 (cage sum), clean-up: no 3,4 in R7C3, no 6 in R8C2

37. 6 in C5 locked in R89C5, 20(4) cage in N8 = {2369/2468}
37a. Only other 8,9 in R8C4 -> R8C4 = {89}

38. R23C4 = {57} = 12 -> R23C3 = 11 = [29]/{38}, no 4, no 9 in R2C3

39. 19(4) cage at R7C4 (step 21) = {1369/1468} (cannot be {1459} because 1,5 only in R8C3) -> R8C3 = 1
39a. R7C4 = {34} -> no 3,4 in R9C3

40. R78C6 = [15], R1C23 = [15], R7C3 = 2, R6C3 = 9

41. Naked pair {38} in R23C3, locked for C3 and N1 -> R4C3 = 4, R9C3 = 6, R4C2 = 5 (cage sum), R4C1 = 8, clean-up: no 8 in R8C2

42. R4C1 = 8 -> R35C1 = 6 = [42], R56C4 = [12], R5C2 = 3, R6C12 = [16], R7C1 = 5

43. Naked pair {49} in R78C2, locked for C2 and N7 -> R23C2 = [27], R2C1 = 9, R9C2 = 8, R23C7 = [39], R23C3 = [83], R23C6 = [68], R23C4 = [75], R23C8 = [56]

44. R8C4 = 8 (hidden single in C4), R7C4 = 4 (step 39)

and the rest is naked singles

Hi folks,

So, another hectic forum week draws to a close . Time to discuss a few moves made in some of the walkthroughs.

Andrew wrote:It took me a very long time to spot my step 2a, which is clearly the key move for this puzzle.

Yes, we all required this move. It seems to be pretty critical. To prove that, I removed the 26(6)n36 cage, and let SudokuSolver and JSudoku try to solve this modified puzzle. This way, they could readily determine that R14C7 = {12}, but could no longer perform the CPE move eliminating the {12} from R12C9+R23C8. The result was that neither program could now complete the puzzle individually, although manually synchronizing the candidates in JSudoku with those of SudokuSolver then allowed JSudoku (using several chains) to successfully solve the grid. However, although I haven't yet analyzed the steps in detail, it looked like pretty tough going...

Incidentally, I found out that the two chains Gary used to break this puzzle were (in the re-wored form presented below), both capable of being expressed as AICs. To illustrate this, however, I would like to apply them to Andrew's WT, starting from his step 10a, due to ease of following the lead-up steps.

Grid state after Andrew's step 10a:
From this position, Gary's first chain can be applied:

(5)r8c9+r9c89=(5,7)r78c8-(7=8)r1c8,(1)r1c7-(1=2)r3c9-(2)r789c9=(2)r9c8

[Note: Alternative explanation for those unfamiliar with Eureka notation:
(a) R8C9+R9C89 <> {5..} => R78C8 = {5..} (strong link, N9)
(b) -> R78C8 = {7..} (neutral link, combinations 12(2))
(c) -> R1C8 <> 7 (weak link, C8)
(d) -> R1C8 = 8 (strong link, bivalue cell R1C8)
(e) -> R1C7 = 1 (neutral link, combinations 9(2))
(f) -> R3C9 <> 1 (weak link, N3)
(g) -> R3C9 = 2 (strong link, bivalue cell R3C9)
(h) -> R789C9 <> 2 (weak link, C9)
(i) -> R9C8 = 2 (strong link, N9)]

=> if 8(3)n9 does not contain a 5, it must contain a 2 (in r9c8)
-> {134} combo for 8(3)n9 blocked
-> 8(3)n9 = {125}, locked for N9
-> cleanup: no 7 in R78C8

Note: Andrew locked the 2 in 8(3)n9 without using chains by considering the overlapping cages 17(4) at R78C6+R89C7 and the 25(4) N9 innies at R789C7+R7C9. R78C6 must sum to 5 or higher, because {12/13} are both blocked by R19C6. Hence R7C79 must sum to at least 13, removing both of {23} from R7C9 and thus locking the 2 of N9 into the 8(3) cage = {125}. Afmob used a similar approach, but had already reduced R7C6 to {12}. Therefore, Gary's first AIC did not really provide a significant benefit compared to the classical Killer approach used by Andrew and Afmob. I'm basically just including it out of interest, and to show how AICs are actually not that uncommon on this forum, even though they are often not recognized and labelled as such.

New grid position:
From this position, Gary's second chain can be applied:

(2=1)r9c6,(4)r1c6-(4=1)r1c23-(1=2)r1c7-(2=1)r3c9-(1)r89c9=(1)r9c8-(1=2)r9c6

[Note: Alternative explanation for those unfamiliar with Eureka notation:
(a) R9C6 <> 2 => R9C6 = 1 (strong link, bivalue cell R9C6)
(b) -> R1C6 = 4 (neutral link, combinations innies C6789)
(c) -> R1C23 <> 4 (weak link, R1)
(d) -> R1C23 = 1 (strong link, combinations 6(2))
(e) -> R1C7 <> 1 (weak link, R1)
(f) -> R1C7 = 2 (strong link, bivalue cell R1C7)
(g) -> R3C9 <> 2 (weak link, N3)
(h) -> R3C9 = 1 (strong link, bivalue cell R3C9)
(i) -> R89C9 <> 1 (weak link, C9)
(j) -> R9C8 = 1 (strong link, 8(3))
(k) -> R9C6 <> 1 (weak link, R9)
(l) -> R9C6 = 2 (strong link, bivalue cell R9C6)]

=> if R9C6 does not contain a 2, it must contain a 2 (contradiction)
-> R9C6 = 2

Note: This second AIC of Gary's provided a useful (although more complex) alternative to the CPE move for N4 eliminating the 1 from R7C1, which Afmob used. Andrew avoided both by considering the overlapping cages 13(2) at R67C7 and the 25(4) N9 innies at R789C7+R7C9. The N9 innies must contain both of {67}, and they cannot both be within R89C7 (which must sum to at most 12). Furthermore, they cannot be split 1:1 between R7C7 and R89C7 because R89C9 would then clash with R6C7. Therefore, 1 of {67} must be in R7C9. This was IMO a critical move in Andrew's WT.

That's all for today, folks. Roll on A87!

Thanks for the analysis Mike..it really is fascinating and much appreciated.It's nice to see procedures that have been rather "stumbled" through being clarified and given a logical underpinning.
I agree with you too that placing <>1 at r1c7 and so at r7c6 was easier and neater than the route I took.Well done to those who saw it.

Best regards

Gary