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Ed
 Post subject: Assassin 441
PostPosted: Fri Sep 01, 2023 8:13 pm 
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Joined: Wed Apr 16, 2008 1:16 am
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Location: Sydney, Australia
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a441x.png
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Note: an X-puzzle so 1-9 cannot repeat on either diagonal.

Assassin 441
I had a hard time with this puzzle. Made logical errors the first few solves even though it got a final solution. Hate that. It resists a long way in so had to go all over the grid. Used one interesting step. It gets 1.85 from SudokuSolver. JSudoku has to grunt a bit which also suggests tough.
triple click code:
3x3:d:k:2816:4609:4609:4609:4354:5379:5379:5380:5380:2816:1797:1797:4354:4354:5379:5380:5380:5382:2816:9735:9735:4354:3592:5379:5379:5382:5382:9735:9735:9735:3592:3592:3337:2826:4107:4107:9735:3084:3084:4621:3854:3337:2826:4107:4107:9735:4621:4621:4621:3854:3337:9743:9743:9743:4368:10001:10001:10001:3854:6930:6930:6930:9743:4368:10001:2579:10001:10001:6930:3092:6930:9743:4368:4368:2579:2325:2325:6930:3092:9743:9743:
solution: 
+-------+-------+-------+
| 2 8 7 | 3 9 4 | 1 6 5 |
| 4 6 1 | 2 5 8 | 7 3 9 |
| 5 3 9 | 1 7 6 | 2 4 8 |
+-------+-------+-------+
| 8 1 4 | 5 2 9 | 6 7 3 |
| 6 9 3 | 7 8 1 | 5 2 4 |
| 7 2 5 | 4 6 3 | 8 9 1 |
+-------+-------+-------+
| 3 5 6 | 9 1 7 | 4 8 2 |
| 9 7 2 | 8 4 5 | 3 1 6 |
| 1 4 8 | 6 3 2 | 9 5 7 |
+-------+-------+-------+
Cheers
Ed
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Andrew
 Post subject: Re: Assassin 441
PostPosted: Sun Sep 10, 2023 9:30 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
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Location: Lethbridge, Alberta, Canada
Thanks Ed for a challenging Assassin! He wrote "It resists a long way in so had to go all over the grid.", so did I. Step 5c was my most interesting step; I used more forcing chains than usual.

Here's how I solved Assassin 441:
Prelims

a) R2C23 = {16/25/34}, no 7,8,9
b) R45C7 = {29/38/47/56}, no 1
c) R5C23 = {39/48/57}, no 1,2,6
d) R89C3 = {19/28/37/46}, no 5
e) R89C7 = {39/48/57}, no 1,2,6
f) R9C45 = {18/27/36/45}, no 9
g) 11(3) cage at R1C1 = {128/137/146/236/245}, no 9
h) 21(3) cage at R2C9 = {489/579/678}, no 1,2,3
i) 39(6) cage at R7C2 = {456789}, no 1,2,3

1a. 45 rule on N3 2 innies R13C7 = 3 = {12}, locked for C7, N3 and 21(5) cage at R1C6, clean-up: no 9 in R45C7
1b. 45 rule in N6789 1 innie R7C5 = 1, R56C5 = 14 = {59/68}, clean-up: no 8 in R9C45
1c. 1 in C6 only in R456C6, locked for N5
1d. 13(3) cage at R4C6 = {139/148/157}, no 2,6
1e. Combined cage R456C6 + R56C5 = {139/68}/{148/59}/{157}{68}, 8 locked for N5
1f. 2 in C6 only in R789C6, locked for N8 and 27(6) cage at R7C6, clean-up: no 7 in R9C45
1g. 45 rule on N69 3 outies R789C6 = 14 contains 2 = {239/248/257}, no 6
1h. 2 in N9 only in R78C9 + R9C89, locked for 38(7) cage at R6C7
1i. 6 in C6 only in R123C6, locked for N2
1j. 45 rule on N23 2 innies R1C4 + R3C5 = 10 = [19]/{28/37}, no 4,5, no 9 in R1C4
1k. 11(3) cage at R1C1 = {128/137/146/245} (cannot be {236} which clashes with R2C23)
1l. 45 rule on N1 4 innies R13C23 = 27 = {3789/4689/5679}, no 1,2
1m. 17(4) cage at R1C5 = {1259/1349/1457/2348/2357} (cannot be {1358} which clashes with R1C4 + R3C5
1n. 14(3) cage at R3C5 = {239/257/347} (cannot be {356} which clashes with R56C5, cannot be {248} = 8{24} which clashes with combined cage R456C6 + R56C5), no 6,8, clean-up: no 2 in R1C4
1o. 4 of {347} must be in R4C4 (cannot be [374/734] which clashes with R1C4 + R3C5 = {37}, CCC), no 4 in R4C5

2a. 45 rule on N69 3 innies R7C78 + R8C8 = 13 = {139/148/157/346}
2b. Hidden killer pair 1,6 in R7C78 + R8C8 and R78C9 + R9C89 for N9, R7C78 + R8C8 contains one of 1,6 -> R78C9 + R9C89 must contain one of 1,6
2c. Since R78C9 + R9C89 contains one of 1,6, 38(7) cage at R6C7 must contain both of them, similarly 2 in R78C9 + R9C89 so 38(7) cage must contain both of 2,5 -> 38(7) cage = {1256789}, no 3,4
2d. 45 rule on N6 3 innies R6C789 = 18 = {189/567}
2e. 2 in N6 only in 16(4) cage at R4C8 = {1249/2347/2356} (cannot be {1258/1267} which clash with R6C789), no 8
2f. Consider combinations for R6C789
R6C789 = {189}, locked for 38(7) cage => R78C9 + R9C89 = {2567}, locked for N9 => R7C78 + R8C8 = {139/148}
or R6C789 = {567}, locked for 38(7) cage => R78C9 + R9C89 = {1289}, locked for N9 => R7C78 + R8C8 = {346}
-> R7C78 + R8C8 = {139/148/346}, no 5,7
2g. 1 of {139/148} must be in R8C8 -> no 8,9 in R8C8
[Eliminations can be made from R6C78, R7C7 and R8C8 using the forcing chain; they don’t seem to help at this stage.]
2h. 39(6) cage at R7C2 = {456789}
2i. Consider combinations for R9C45 = {36/45}
R9C45 = {36}, 6 locked for N8 => 6 in 39(6) cage in R7C23 + R8C2, locked for N7
or R9C45 = {45}, 4 locked for N8 => 4 in 39(6) cage in R7C23 + R8C2, locked for N7
-> R89C3 = {19/28/37}, no 4,6
2j. Hidden killer triple 1,2,3 in 17(4) cage at R7C1 and R89C3 for N7, R89C3 contains one of 1,2,3 -> 17(4) cage must contain two of 1,2,3 = {1259/1268/1349/1367/2348/2357} (cannot be {1457/2456} which only contain one of 1,2,3, cannot be {1358} which clashes with R89C3)

3a. 13(3) cage at R4C6 (step 1d) = {139/148/157}, R56C5 (step 1b) = {59/68}, 14(3) cage at R3C5 (step 1n) = {239/257/347}
3b. Consider combinations for 14(3) cage
14(3) cage = {239}, CPE no 9 in R56C5
or 14(3) cage = {257}, CPE no 5 in R56C5
or 14(3) cage = {347}, 4 locked for N5 => 13(3) cage = {139/157} => R56C5 = {68} (cannot be {59} which clashes with 13(3) cage
-> R56C5 = {68}, locked for C5 and N5, clean-up: no 4 in 13(3) cage, no 3 in R9C4
3c. 4 in N5 only in R456C4, locked for C4, clean-up: no 5 in R9C5
3d. Killer pair 6,8 in R6C5 and R6C789, locked for R6
3e. 6 in N4 only in R4C123 + R5C1, locked for 38(7) cage at R3C2, no 6 in R3C23
3f. 38(7) cage = {1256789/1346789}, 1 locked for N4

4a. 45 rule on R123 3 innies R3C235 = 19 = {289/379/478}, no 5
4b. Hidden killer pair 7,8,9 in R3C235 + R3C89 for R3, R3C235 contain two of 7,8,9 -> R3C89 must contain one of 7,8,9, 21(3) cage at R2C9 contains two of 7,8,9 -> R2C9 = {789}
4c. Killer triple 7,8,9 in R3C235 and R3C89, locked for R3

5a. 38(7) cage at R3C2 (step 3f) = {1256789/1346789}
5b. 45 rule on N1 2 innies R3C23 = 1 outie R1C4 + 9
5c. Consider combinations for R5C23 = {39/48/57}
R5C23 = {39/57} => 8 in N4 only in R4C123 + R5C1, locked for 38(7) cage, no 8 in R3C23 => max R3C23 = 16 => max R1C4 = 7
or R5C23 = {48}, locked for N4 => 8 in 38(7) cage only in R3C23 => caged X-Wing for 8 in R3C23 and R5C23, no other 8 in C23 => 8 in 39(6) cage at R7C2 only in R78C4, locked for C4
-> no 8 in R1C4, clean-up: no 2 in R3C5 (step 1j)
[A long way from solving this Assassin but things get a bit easier after this step.]
5d. 45 rule on N3 3 outies R123C6 = 18 contains 6 for C6 = {468/567} (cannot be {369} which clashes with R1C4 + R3C5), no 3,9

6a. 14(3) cage at R3C5 (step 1n) = {239/257/347} with {347} = [347/743], R9C45 (step 2i) = [54/63]
6b. 45 rule on N7 3 outies R7C4 + R8C45 = 21 = {489/579/678}
6c. Consider combinations for R7C4 + R8C45
R7C4 + R8C45 = {489}, locked for N8 => R9C45 = [63]
or R7C4 + R8C45 = {579}, locked for N8 => R9C45 = [63]
or R7C4 + R8C45 = {678} = {68}7
-> 14(3) cage at R3C5 = {239/257} (cannot be {347}), no 4, 2 locked for R4 and N5
6d. {257} = 7{25}, no 7 in R4C45
6e. R123C6 (step 5d) = {468/567}, 13(3) cage at R4C6 (step 1d) = {139/157}
6f. Consider combinations for 14(3) cage
14(3) cage = {239} = 3{29}/9{23} => 13(3) cage = {157} (cannot be {139} which clashes with R4C45), 5,7 locked for C6
or 14(3) cage = {257} = 7{25}
-> R123C6 = {468}, 4,8 locked for C6 and N2
also 5 in R4C45 or R456C6, locked for N5
6g. 8 in N8 only in R78C4, locked for 39(6) cage at R7C2, no 8 in R7C23 + R8C3
6h. R7C4 + R8C45 = {489/678}, no 5
6i. 4 of {489} must be in R8C5 -> no 9 in R8C5
6j. 7 of {678} must be in R8C5 -> no 7 in R78C4
6k. 39(6) cage = {456789} -> R7C23 + R8C3 = {459/567}, 5 locked for N7
6l. 4 in N5 only in R56C4, locked for 18(4) cage at R5C4
6m. 18(4) cage = {2349/2457}, 2 locked for N4
6n. 38(7) cage at R3C2 (step 3f) = {1346789}, no 5
6o. 45 rule on N56789 3 outies R3C5 + R6C23 = 14 contains 2 in R6C23 = 3{29}/7{25}/9{23}, no 7 in R6C23

7a. 5 in C1 only in 11(3) cage at R1C1 = {245}, locked for N1, 2,4 locked for C1
[Almost there]
7b. R13C23 (step 1l) = {3789}, 3 locked for N1
7c. Naked pair {16} in R2C23, locked for R2
[At long last, I can use a diagonal]
7d. R3C7 = 2 (hidden single on D/) -> R1C7 = 1, clean-up: no 9 in R3C5 (step 1j)
7e. Naked pair {37} in R1C4 + R3C5, locked for N2
7f. R3C235 (step 4a) = {379} (only remaining combination), 7,9 locked for R3, 9 locked for N1 and 38(7) cage at R3C2
7g. Naked triple {378} in 18(3) cage at R1C2, locked for R1
7h. 38(7) cage at R3C2 = {1346789}, 4,8 locked for N4, 4 locked for R4, clean-up: no 7 in R5C7

8a. R13C23 (step 7b) = {3789}
8b. Consider combinations for R5C23 = {39/57}
R5C23 = {39} => caged X-Wing with R13C23, no other 3,9 in C23 => R89C3 = {28}
or R5C23 = {57} => caged X-Wing with R13C23, no other 7 in C23 => R89C3 = {19/28}
-> R89C3 = {19/28}, no 3,7
[Cracked at last. Straightforward from here.]
8c. 3 in N7 only in 17(4) cage at R7C1 = {1349/1367} (cannot be {2348} because 2,4 only in R9C2), 1 locked for N7, clean-up: no 9 in R89C3
8d. Naked pair {28} in R89C3, locked for C3 and N7
8e. R1C2 = 8 (hidden single in R1)
8f. R2C6 = 8 (hidden single in C6)
8g. R6C24 = [24] (hidden pair in R6), 4 placed for D/
8h. R4C3 = 4 (hidden single in C3), clean-up: no 7 in R5C7
8i. R2C3 = 1 (hidden single in C3) -> R2C2 = 6, placed for D\
8j. R7C3 = 6 (hidden single in C3), placed for D/, R5C5 = 8, placed for D\, clean-up: no 3 in R4C7
8k. Naked pair {89} in R78C4, 9 locked for C4, N8 and 39(6) cage at R7C2 -> R8C8 = 4 (step 6b), R9C5 = 3 -> R9C4 = 6, clean-up: no 9 in R8C7, no 8 in R9C7
8l. Naked pair {57} in R78C2, locked for C2, 7 locked for N7, clean-up: no 5,7 in R5C3
8m. Naked pair {39} in R5C23, locked for R5 and N4, clean-up: no 8 in R4C7

9a. R6C5 = 6 -> R6C789 (step 2d) = {189}, locked for N6 and 38(7) cage at R6C7, 1,9 locked for R6)
9b. Naked quad {2567} in R78C9 + R9C89, 5,7 locked for N9
9c. 6 in C7 only in R45C7 = {56}, locked for N6
9d. R2C7 = 7 (hidden single in C7), R2C9 = 9, R3C89 = 12 = {48}, 4 locked for R3 and N3
9e. R1C9 = 5, R2C8 = 3, both placed for D/, R4C89 = [73]

10a. Naked pair {25} in R2C45, locked for R2 and N2 -> R1C5 = 9
10b. Naked pair {25} in R4C45, 5 locked for R4 and N5
10c. R5C4 = 7, R6C6 = 3, placed for D\, R3C5 = 7, R3C3 = 9, placed for D\, R7C7 = 4, placed for D\, R1C1 = 2, placed for D\

and the rest is naked singles, without using the diagonals.
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wellbeback
 Post subject: Re: Assassin 441
PostPosted: Fri Sep 15, 2023 5:27 am 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
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Thanks Ed! Great puzzle. I decided this time to try and find a path with no chains. It took me a loooong time.
Assassin 441 WT:
1. Innies n3 = r13c7 = +3(2) = {12}
-> Outies n3 = r123c6 = +18(3)
-> Remaining Innies c6 = r789c6 = +14(3)
-> Remaining Innie n78 = r7c5 = 1
-> 1 in c6 in r456c6. -> r456c6 = 1 + 12(2) (No (26))
-> 2 in c6 in r789c6. -> r789c6 = 2 + 12(2) (No 6)
-> 6 in c6 in r123c6. -> r123c6 = 6 + 12(2)
-> Remaining Innies n2 = r1c4,r3c5 = +10(2) (Not {46})

2. 38(7)n14 has all (789)
12(2)n4 has one of (789)
-> At least one of (789) in r3c23

3. Values in r3c23 go in c1 in r789c1
Since those 2 values are in a 17(4) -> Those two values are at most +14(2)
-> Remaining Innies n1 = r1c23 are at least +13(2)
-> r1c4 is Max 5
But r1c4 cannot be from (456) (Step 1)
-> r1c4 is Max 3

4! Innies n1 = r13c23 = +27(4)
Since any three of them are at least +18(3) -> no three of them can go in c1 in r789c1
Since the two values from r3c23 are in r789c1 -> the two values in r1c23 must go in c1 in r456c1
-> for the values in r2c23 (= 7(2)), one must be in r456c1 and the other in r789c1
Since 38(7) is missing two values = +7(2) and since we know one of the values in r2c23 is in 38(7) in r456c1 then the other value from r2c23 must also be in the 38(7). Since that value is in r789c1 it can only go in the 38(7) in r4c23.
IOD c1 -> r34c23 = r789c1 + 4
Since all three values in r789c1 are also in r34c23 -> Remaining value in outies c1 (in r34c23) = 4
I.e., 4 in r4c23
(Clarification. r34c23 contains the three values in r789c1 and 4, r34c23 can’t contain 4 twice so no 4 in r789c1 so 4 in c1 only in 11(3)n1. Andrew)

5. -> 3 also in 38(7)
Also 12(2)n4 cannot be {48} -> must be {39} or {57}
-> one of (79) in 38(7) in r3c23
-> 18(3)r1 not [{79}2]
-> Innies n2 = r1c4,r3c5 from [19] or [37]

6. Consider the three values in r789c1. From Steps 3 and 4, two of them are in r3c23 and the other is in r4c23. Call that other value 'X'.
I.e., r4c23 = {4X} and X also in r789c1.
-> None of the values in r789c1 can be 4.
-> 4 in c1 in r123c1.
-> (Since r1c23 is Min +15(2)) 3 in n1 in r3c23
-> Innies n1 = {3789} with 3 in r3c23 and 8 in r1c23
-> 3 in c1 in r789c1

7! Since 1 already in n5 and 4 is in r4c23 and r3c5 from (79) -> 14(3)r3c5 from [7{25}] or [9{23}]
I.e., 2 locked in r4c45
-> 2 in D/ not in n5
Also X (from Step 6) not 2 (I.e., 2 not in r789c1)
Since 39(6)n78 has no (123) -> 2 in D/ not in n7
-> (HS 2 in D/) r3c7 = 2

That basically cracks it!
8. Continuing...
-> r1c7 = 1
-> Innies n2 r1c4,r3c5 = [37]
-> r123c6 = {468}
Also 18(3)r1 = [{78}3]
-> r3c23 = {39}
-> (78) in 38(7) in n4
-> 12(2)n4 = {39}
Also 14(3)r3c5 = [7{25}]
-> 15(3)c6 = [{68}1]
-> 13(3)n5 = <319>
-> r56c4 = {47}
Also -> r789c6 = {257}
-> Innies n89 = r78c4,r8c5 = +21(3) = [{89}4]
-> Innies n7 = {567}
Also (HS 1 in D/) -> r9c1 = 1
-> 17(4)n7 = [{39}14]
-> r4c23 = [14]
-> 7(2)n1 = [61]
-> 11(3)n1 = {245}
-> r456c1 = {678}
-> r6c23 = {25}
Also Innies n6 = r6c789 = +18(3) has a 1 - must be {189}
-> (HS 1 in n9) r8c8 = 1
-> Remaining Innies n9 = r7c78 = +12(2) = {48} (Cannot be {39} since (39) already on D\ in r3c3 and r6c6)
etc.
Now to look at Andrew's!
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Ed
 Post subject: Re: Assassin 441
PostPosted: Sat Sep 16, 2023 3:09 am 
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I've only skimmed the two WTs above but can see my way is very, very different. I put my most interesting step near the beginning even though it doesn't really come in handy until step 22a. Some steps in the middle are pretty cool too I think. Learned a lot from you guys over the years! After Andrew's comment about the diagonals, I think I did forget this was an X-puzzle! It has multiple solutions without the X however, so can't forget them entirely. [edit: thanks to Andrew for going through my WT and making some suggestions]
start to A441:
Preliminaries
Cage 7(2) n1 - cells do not use 789
Cage 12(2) n9 - cells do not use 126
Cage 12(2) n4 - cells do not use 126
Cage 9(2) n8 - cells do not use 9
Cage 10(2) n7 - cells do not use 5
Cage 11(2) n6 - cells do not use 1
Cage 21(3) n3 - cells do not use 123
Cage 11(3) n1 - cells do not use 9
Cage 39(6) n78 - cells ={456789}

1. "45" on n6789: 1 innie r7c5 = 1
1a. -> r56c5 = 14 = {59/68}

2. "45" on n3: 2 innies r13c7 = 3 = {12}: both locked for n3, c7 and 21(5)

3. 1 in c6 only in 13(3)n5: 1 locked for n5
3a. 13(3) = {139/148/157}(no 2,6)

4. "45" on n69: 3 outies r789c6 = 14 and must have 2 for c6
4a. = {239/248/257}(no 6)
4b. 2 locked for n8 and 27(6)

5. 38(7)r6c7 = {1256789/1346789/2345789}
5a. "45" on n6: 4 outies r789c9+r9c8 = 20 and must have 2 for n9
5b. but {2378/2459} both blocked by 12(2)n9 needing one of 3,7,8 or one of 4,5,9
5c. {2369/2468} both blocked by mismatch with 38(7) combos
5d. = {1289/2567}(no 3,4)
5e. -> 38(7) = {1256789} only
5f. -> r6c789 = 18 = {567/189}(no 3,4)

6. r789c6 = 14 (step 4) -> r7c78 + r8c8 = 13 and must have at least one of 3,4 for n9 since the only other place is 12(2) (hidden killer pair)
6a. {139/148/346}(no 5,7)

1st Key step
7. "45" in c7: 3 remaining innies r267c7 = 19
7a. and r6c7 sees all n9 apart from r78c8 so must repeat there
7b. since two cells of the h19(3)c7 and h13(3)n9 now overlap -> the third cells must differ by 6
7c. -> r2c7 - 6 = one of r78c8: = [71]/[9]{3}
7d. ie, r2c7 = (79)
7e. the only other common digits in r6c7 & r78c8 are (689) -> no 5,7 in r6c7, no 4 in r78c8
7f. -> r7c78 + r8c8 (step 6a) = [391/481]/[4]{36}
7g. ie, r7c7 = (34), r8c8 = (136)

8. 21(3)n3: {579} blocked by r2c7
8a. = {489/678}(no 5) = 7 or 9
8b. 8 locked for n3
8c. note: = 6 or 9 but not both

9. deleted

Next key. Nice min/max cascade.
10. 11(3)n1: {236} blocked by 7(2)n1 needing one of those
10a. = {128/137/146/245}
10b. 38(7)r3c2 = {1256789/1346789/2345789}
10c. what is the min. for r4c23?
10d. can't be {12/13} = 3 or 4 since {56789/46789} in r3c23 + r456c1 is blocked by 11(3)
10e. -> min. r4c23 = 5
10f. "45" on c1: 5 outies r34c23 + r9c2 = 21
10g. min. r4c23 + r9c1 = 6 -> max. r3c23 = 15
10h. "45" on r123: 3 innies r3c235 = 19
10i. max. r3c23 = 15 -> min. r3c5 = 4

11. 6 in c6 only in n2: 6 locked for n2

12. "45" on n23: 2 innies r1c4 + r3c5 = 10
12a. = [19/28/37](no 4,5; no 9 in r1c4)

13. 14(3)r3c5
13a. must have 7,8,9 for r3c5 = {239/248/257/347}(no 6) = 4 or 5 or 9
13b. can't have two of 7,8,9, -> no 7,8,9 in r4c45

Another important one
14. 13(3)n5 = {139/148/157}(step 3a)
14a. {148} + {59} in combined cage 13(3)n5 + sp14(2) blocked by 14(3)
14b. -> sp14(2) = {68} only: both locked for n5 and c5
14c. no 2 in r1c4 (h10(2)n2)

15. "45" on n7: 3 outies r78c4 + r8c5 = 21 = {489/579/678}
15a. note: 6 in that cage -> r8c5 = 7
15b. and 6 in r9c4 -> 3 in r9c5
15c. no other place for 6 in n8 -> r89c5 = 7 or 3 (no eliminations yet)

16. 14(3)r3c5 = {239/257/347} (step 13a)
16a. but can't have 4 in r4c4 since r34c5 as [73] blocked by r89c5 (step 15c)

As usual, Andrew saw this next step as CCC
17. "45" on n23: 1 innie r1c4 + 4 = 2 outies r4c45
17a. since 1 innie and 1 outie see each other the 3rd cell must not equal 4 -> no 4 in r4c5 (IOU)
17b. -> 14(3)r3c5 = [9]{23}/[7]{25}
17c. 2 locked for n5 and r4

18. 13(3)n5 = {139/157}(no 4)
18a. -> hidden killer pair 7,9 n5
18b. -> one of 7,9 in r56c4 which must also have 4 for n5
18c. -> r56c4 = {47/49}
18d. 4 locked for c4 and 18(4)

19. 18(4)r5c4 must have {47/49}
19a. but r6c23 can't be {16} since h18(3)n6 needs one of them (step 5f)
19b. -> r6c23 = {23/25}
19c. 2 locked for n4
[edit: from Andrews WT I see I missed the much easier killer pair 6,8 in r6c5789. I may have been able to get rid of a couple of the above steps since the important part was to now be able to get 6 out of r3c23 by doing step 20, which then made step 22 so powerful]

20. 1 and 6 in n4 only in 38(7): locked for that cage

21. "45" on n1: 4 innies r13c23 = 27 = {3789/4689/5679}(no 1,2)

22. "45" r123: 3 innies r3c235 = 19
22a. but {478} blocked by 21(3)n3 can't be [6]{69}
22b. = {379} only: all locked for r3
22c. 3 locked for 38(7)
22d. h27(4)n3 = {3789} only, 3,7,8 locked for n1
22e. -> 8 locked for r1

23. 21(3)n3 = {489/678}
23a. 7,9 only in r2c9 -> r2c9 = (79)
23b. 8 locked for r3

24. naked pair {79} in r2c79: both locked for r2 and n3

25. 17(4)n2 must have 2 for n2 = {1259/2348/2357}
25a. and r9c45 = [63/54]
25b. -> r12c5 as {34} blocked
25c. -> no permutation with 4 possible in r2c5
25d. and [73] blocked from r12c5 which leaves no 6 for n8 (step 15c)
25e. -> no permutation with 3 in r2c5 possible
25f. -> r2c5 = (25)
25g. also 8 blocked from r2c4 since that leaves 3,4 only in r1c5
25h. 17(4) = {1259/2357}(no 4): 5 locked for n2
25i. must have 7,9 -> r1c5 = (79)
25j. and r123c6 = {46}[8]{46} (hidden single 8 n2, hidden pair 4,6): 4 locked for c6

26. 7(2)n1: {25} blocked by r2c5
26a. = {16} only: both locked for n1 and r2

27. naked pair {79} in r13c5: both locked for c5
27a. -> no 6 in r78c4 (step 15c)
27b. -> r9c4 = 6, r9c5 = 3,

28. hsingle 4 in n8 -> r8c5 = 4


Much easier now but ran out of time to get this WT right to the end.
Cheers
Ed
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