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HATMAN
 Post subject: HS X 20
PostPosted: Thu Dec 18, 2014 1:54 pm 
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Human Solvable X 20



It is exactly six months since my last human solvable (and I've probably attempted over 10).

I sorted this one out over the last couple of days. I'm posting it now for Christmas as I travel to Kenya at the weekend and am not sure of another chance to post it.

Both solvers give up completely. Apart from the HS bits I found it relatively easy.

Merry Christmas all.





Image

JS Code:
3x3:d:k:7707:21:28:4880:4880:4880:29:30:5401:31:7707:3339:3339:5140:3338:3338:5401:32:33:3084:7707:5140:5140:5140:5401:3337:34:5137:3084:7707:7707:5140:2586:5401:3337:6158:5137:3858:3858:3352:3352:3352:2586:2586:6158:5137:3334:3858:3858:4631:4886:4886:3335:6158:35:3334:5139:4631:4631:4886:36:3335:6158:37:5139:3333:3333:4886:3336:3336:38:39:5139:40:41:6927:6927:6927:6927:42:43:


Solution:

934865127
256714983
187239465
843651279
519427638
762398541
475182396
398546712
621973854


thanks for correction Andrew
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Afmob
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PostPosted: Fri Dec 19, 2014 9:42 pm 
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Thanks for this Killer, HATMAN! A typical one-trick pony which is either unsolvable or easy as pie.

HS 20X:
1. N1379+D/
a) ! Innies of whole grid = 46(4+4+4+5) which implies
- 1) R1C23+R23C1 = 10/11(4) = 123{4/5} -> 1,2,3 locked for N1
- 2) R1C78+R23C9 = 10/11(4) = 123{4/5} -> 1,2,3 locked for N3
- 3) R78C1+R9C23 = 10/11(4) = 123{4/5} -> 1,2,3 locked for N7
- 4) R7C7+R89C89 = 15/16(5) = 1234{5/6} -> 1,2,3,4 locked for N9
b) Hidden triple (123) locked in R4C6+R5C5+R6C4 @ D/ for N5; R4C6+R5C5+R6C4 = (123)
c) Using R4C6+R5C5+R6C4 = (123): Outie D/ = R4C7 = 2
d) 10(3) = 1{36/45} -> CPE: R5C5+R4C9 <> 1
e) 13(3) = {247} since R5C46 <> 2,3 and 6{25/34} blocked by Killer pairs (46,56) of 10(3) -> R5C5 = 2; 4,7 locked for R5+N5
f) Hidden Single: R6C3 = 2 @ N4
g) 15(4) = {1239} -> 9 locked for R5+N4; CPE: R6C1 <> 3

2. N4+D/
a) 20(3) @ N4 = {578} locked for C1+N4
b) 20(3) @ D/ = 9{47/56} since R9C1 = (469) -> 9 locked for N7+D/
c) 13(2) = [67] since R6C2 = (46) -> R6C2 = 6, R7C2 = 7
d) 20(3) @ D/ = {569} locked for N7+D/
e) 21(4) = {2478} -> 4,7,8 locked for N3

3. N1379
a) Innies of whole grid = 46(4+4+4+5) with R1C78+R23C9 = {1235} = 11(4) locked for N3 which implies
- 1) R1C23+R23C1 = 10(4) = {1234} locked for N1
- 2) R78C1+R9C23 = 10(4) = {1234} locked for N7
- 3) R7C7+R89C89 = 15(5) = {12345} locked for N9
b) R8C3 = 8 -> R8C4 = 5, R8C2 = 9, R9C1 = 6, R1C1 = 9
c) 12(2) = [84] -> R3C2 = 8, R4C2 = 4
d) R2C2 = 5
e) 13(2) @ N1 = {67} locked for R2

4. Rest is singles without considering diagonals.

Rating:
1.0 - 1.25?. I used very large Innies.


Last edited by Afmob on Tue Dec 23, 2014 7:59 am, edited 2 times in total.

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HATMAN
 Post subject: Re: HS X 20
PostPosted: Sat Dec 20, 2014 6:03 pm 
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HS X 20 Hard

Afmob in response to your challenge here is a harder version.

Image

JS code:
3x3:d:k:7707:10:12:8200:8200:8200:13:16:5401:17:7707:6411:6411:5140:8200:8200:5401:18:26:6411:7707:5140:5140:5140:5401:3337:28:8984:6411:7707:7707:5140:5910:5401:3337:6158:8984:8984:8984:5910:5910:5910:5910:5910:6158:8984:3334:8984:8984:4631:8213:8213:3335:6158:29:3334:5139:4631:4631:8213:30:3335:6158:31:5139:3333:3333:8213:8213:8213:32:33:5139:34:35:6927:6927:6927:6927:36:37:
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Afmob
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PostPosted: Sat Dec 20, 2014 11:44 pm 
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First, I didn't challenge you. Second and more, importantly, here are the codes which even work for SumoCue:
HS 20X:
3x3:d:k:7680:1:2:4867:4867:4867:6:7:5384:9:7680:3339:3339:5133:3342:3342:5384:17:18:3091:7680:5133:5133:5133:5384:3353:26:5147:3091:7680:7680:5133:2592:5384:3353:6179:5147:3877:3877:3367:3367:3367:2592:2592:6179:5147:3374:3877:3877:4657:4914:4914:3380:6179:54:3374:5176:4657:4657:4914:60:3380:6179:63:5176:3393:3393:4914:3396:3396:70:71:5176:73:74:6987:6987:6987:6987:79:80:

HS 20X Hard:
3x3:d:k:7680:1:2:8195:8195:8195:6:7:5384:9:7680:6411:6411:5133:8195:8195:5384:17:18:6411:7680:5133:5133:5133:5384:3353:26:8987:6411:7680:7680:5133:5920:5384:3353:6179:8987:8987:8987:5920:5920:5920:5920:5920:6179:8987:3374:8987:8987:4657:8242:8242:3380:6179:54:3374:5176:4657:4657:8242:60:3380:6179:63:5176:3393:3393:8242:8242:8242:70:71:5176:73:74:6987:6987:6987:6987:79:80:
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Andrew
 Post subject: Re: HS X 20
PostPosted: Sun Dec 21, 2014 5:16 am 
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Thanks HATMAN for a fun puzzle. :D

As Afmob said, it's a one-trick pony and once the HS bits have been found the rest is straightforward.

Here is my walkthrough for Human Solvable X 20:
Prelims

a) R2C34 = {49/58/67}, no 1,2,3
b) R2C67 = {49/58/67}, no 1,2,3
c) R34C2 = {39/48/57}, no 1,2,6
d) R34C8 = {49/58/67}, no 1,2,3
e) R67C2 = {49/58/67}, no 1,2,3
f) R67C8 = {49/58/67}, no 1,2,3
g) R8C34 = {49/58/67}, no 1,2,3
h) R8C67 = {49/58/67}, no 1,2,3
i) 19(3) cage at R1C4 = {289/379/469/478/568}, no 1
j) 20(3) cage at R4C1 = {389/479/569/578}, no 1,2
k) 10(3) cage at R4C6 = {127/136/145/235}, no 8,9
l) 20(3) cage at R7C3 = {389/479/569/578}, no 1,2
m) 27(4) cage at R9C4 = {3789/4689/5679}, no 1,2

1. 10(3) cage at R4C6 and 13(3) cage at R5C4 form combined 23(6) cage because these cages “see” each other = {123458/123467}, no 9
1a. 10(3) cage = {127/136/145/235} -> 13(3) cage = {148/238/247/346}, no 5

2. 15(4) cage at R5C2 and 13(3) cage at R5C4 don’t quite form a combined cage because R6C3 doesn’t “see” the 13(3) cage
2a. R5C23 + R6C4 cannot be {123/124} which clash with 13(3) cage -> no 8,9 in R6C3
[Can probably make similar eliminations from R6C4 and possibly more from R6C3 because combined 23(6) cage, step 1, must have at least three of 1,2,3,4 in R5.]

3. 27(4) cage at R9C4 = {3789/4689/5679}, 9 locked for R9

4. 45 rule on N2 2 innies R2C46 = 1 outie R4C5 + 6, min R2C46 = 9 -> min R4C5 = 3

5. 45 rule on D/ 3 innies R4C6 + R5C5 + R6C4 = 1 outie R4C7 + 4
5a. Min R4C6 + R5C5 + R6C4 = 6 -> min R4C7 = 2
[Also IOU, R5C5 + R6C4 cannot total 4 so R5C5 + R6C4 cannot be {13}. At this stage I can’t see how to use this.]

[A typical HATMAN feature is the ‘ring’ of 13(2) cages and a 12(2) cage in R2, R8, C2 and C8 but this doesn’t seem to be any help at this stage.

After a few conventional steps it was time to look for Human Solvable step(s). I looked at 45 rule on C456 but 6(4+2) outies = 2 innies R4C46 + 33 but this didn’t seem to lead to anything. Then I found …]
6. 45 rule for N24568 16(5+5+2+4) outies in N1379 = 114
6a. Max for 5 cells = 35, max for 4 cells = 30 and max for 2 cells = 17, but only if not limited by other cages
6b. Min R7C2 + R8C3 = 14 (from steps 6 and 6a), no 4 in R7C2 and R8C3, clean-up: no 9 in R6C2 + R8C4
6c. Max R7C2 + R8C3 = 15 (cannot be {79/89} which clash with 20(3) cage at R7C3)
[Alternatively max 5 cells in N7 = 35 -> max R7C2 + R8C3 = 15.]
6d. R7C2 + R8C3 + 20(3) cage = 34,35 = {46789/56789}, no 3, 6,7,8,9 locked for N7
6e. R7C2 + R8C3 = 14,15 -> 14(5+5+4) outies in N139 = 99,100 -> min 4 outies = 29, min 5 outies = 34
6f. 5 outies in N1 R1C1 + R23C23 = 34,35 = {46789/56789}, no 1,2,3, 6,7,8,9 locked for N1, clean-up: no 9 in R4C2
6g. 5 outies in N3 R1C9 + R23C78 = 34,35 = {46789/56789}, no 1,2,3, 6,7,8,9 locked for N3
6h. 4 outies in N9 R7C89 + R89C7 = 29,30 = {5789/6789}, no 1,2,3,4, 7,8,9 locked for N9, clean-up: no 9 in R6C8 + R8C6

7. R4C6 + R5C5 + R6C4 = {123} (hidden triple on D/), locked for N5
7a. R4C6 + R5C5 + R6C4 = {123} = 6 -> R4C7 = 2 (step 5)

8. 10(3) cage at R4C6 = {136/145}, no 7
8a. 10(3) cage = {136/145} -> 13(3) cage at R5C3 = {238/247} using step 1a but cannot be {238} because 2,3 only in R5C2 -> 13(3) cage = {247} -> R5C5 = 2, placed for D\, R5C46 = {47}, locked for R5 and N5
8b. 10(3) cage = {136}, 6 locked for R5 and N6, clean-up: no 7 in R37C8
8c. 10(3) cage = {136}, CPE no 1,3 in R4C9

9. R6C3 = 2 (hidden single in R6) -> 15(4) cage at R5C2 = {1239} (only remaining combination), 9 locked for N4 and R5
9a. R5C19 = {58} (hidden pair in R5)
9b. 20(3) cage at R4C1 = {578} (only remaining combination), locked for C1 and N4, clean-up: R3C2 = {89}, R7C2 = {79}

10. 20(3) cage at R7C3 = {569} (only remaining combination, cannot be {479} which clashes with R7C2) -> R9C1 = 6, placed for D/, R7C3 + R8C2 = {59}, locked for N7 and D/, R7C2 = 7 -> R6C2 = 6, R8C3 = 8 -> R8C4 = 5, R8C2 = 9, R7C3 = 5, R3C2 = 8 -> R4C2 = 4, R2C2 = 5, placed for D\
10a. R23C3 = {67} (hidden pair in N1), R1C1 = 9 (hidden single in N1), placed for D\
10b. R2C3 = {67} -> R2C34 = {67}, locked for R2
10c. R6C6 = 8, R4C4 = 6, placed for D\, R3C3 = 7, R4C3 = 3 (cage sum), R2C34 = [67]
10d. R1C9 + R2C8 + R3C7 = [784], R2C67 = [49], clean-up: no 5 in R6C8
10e. R3C8 = 6 (hidden single in N3) -> R4C8 = 7
10f. R5C46 = [47] -> R8C67 =[67]
[HATMAN’s ‘ring’ has now proved to be very useful.]
[Afmob pointed out that it’s naked singles at this stage. Checking for them manually, it wasn’t obvious to me until after step 11 that I’d reached naked singles. Afmob checks for singles with software, probably copying and pasting from his software worksheet.]

11. R2C46 = R4C5 + 6 (step 4), R2C46 = [74] = 11 -> R4C5 = 5, R6C5 = 9, R67C8 = [49]
11a. 20(3) cage at R4C1 = [857]
11b. Naked pair {13} in R7C7 + R8C8, locked for N9 -> R9C9 = 4, R8C9 = 2, R9C8 = 5
11c. R45C9 = [98], R7C9 = 6 -> R6C9 = 1 (cage sum)

and the rest is naked singles, without using the diagonals.

All I'll say about rating is that it's hard to know what rating to give the HS bits. I'll agree with Afmob's rating.
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HATMAN
 Post subject: Re: HS X 20
PostPosted: Sun Dec 21, 2014 5:42 am 
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Afmob

I meant that in jest. But it was a one and a half trick pony which is needed for the harder version

On Sumocue I go through computers pretty quickly and have not had the code for about two computers.
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Afmob
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PostPosted: Sun Dec 21, 2014 12:07 pm 
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Looking at SudokuSolver it seems that my hardest step was a bit OTT but it was still fun the use. Otherwise the clean-up is a little bit longer than in my first walkthrough.
HS 20X Hard:
1. N1379+D/
a) ! Innies of whole grid = 46(4+4+4+5) which implies
- 1) R1C23+R23C1 = 10/11(4) = 123{4/5} -> 1,2,3 locked for N1
- 2) R1C78+R23C9 = 10/11(4) = 123{4/5} -> 1,2,3 locked for N3
- 3) R78C1+R9C23 = 10/11(4) = 123{4/5} -> 1,2,3 locked for N7
- 4) R7C7+R89C89 = 15/16(5) = 1234{5/6} -> 1,2,3,4 locked for N9
b) Hidden triple (123) locked in R4C6+R5C5+R6C4 @ D/ for N5; R4C6+R5C5+R6C4 = (123)
c) Using R4C6+R5C5+R6C4 = (123): Outie D/ = R4C7 = 2
d) 23(6) = 1234{58/67} -> R5C5 = 2; 4 locked for R5; CPE: R4C9 <> 1,3
e) ! X-Wing (2) locked in R19C28 for R19
f) Hidden Single: R6C3 = 2 @ C3
g) 35(7) = 258{1379/1469/3467} -> 5,8 locked for N4

2. N13479+D\ !
a) ! Innies+Outies N4: -10 = R6C4 - (R4C23+R6C2): R6C4 = (13) -> R4C23+R6C2 = 11/13(3) = 46{1/3} since {13}[7/9] blocked by R4C6 = (13) -> 4,6 locked for N4
b) 13(2) @ N4 = [49/67]
c) 20(3) = 5{69/78} since {479} blocked by R7C2 = (79) -> 5 locked for N7+D/
d) 21(4) = 24{69/78} -> 4 locked for N3
e) Innies of whole grid = 46(4+4+4+5) with R1C78+R23C9 = {1235} = 11(4) locked for N3 which implies
- 1) R1C23+R23C1 = 10(4) = {1234} locked for N1
- 2) R78C1+R9C23 = 10(4) = {1234} locked for N7
- 3) R7C7+R89C89 = 15(5) = {12345} locked for N9
f) Naked quad (1234) locked in R2378C1 for C1
g) Killer pair (79) locked in 13(2) @ N4 + 20(3) for N7
h) 13(2) @ N8 = [67/85]
i) 1,3 locked in R7C7+R8C8+R9C9 @ D\ for N9
j) ! Consider placement of 1 in N3 -> R9C9 <> 1
- i) 1 locked in R23C9 @ N3 for C9
- ii) 1 locked in R1C78 @ N1 for R1 -> 1 locked in R23C1 @ N1 for C1 -> 1 locked in R9C23 @ N7 for R9

3. N13679
a) 1 locked in R9C23 @ R9 for N7
b) 1 locked in R23C1 @ C1 for N1
c) 1 locked in R1C78 @ R1 for N3
d) 1 locked in 24(4) @ C1 = {1689} for N6
e) Hidden Single: R1C9 = 7 @ C9, R7C2 = 7 @ N7 -> R6C2 = 6
f) 20(3) = {569} locked for D/+N7
g) 23(6) = 1234{58/67} -> R4C6 = 1; 3 locked for R5+N6
h) R8C3 = 8 -> R8C4 = 5, R8C2 = 9
i) 21(4) = {2478} -> 8 locked for N3

4. N246
a) Naked pair (34) locked in R4C23 for R4
b) 13(2) @ N3 = [67] since R3C8 = (69) -> R3C8 = 6, R4C8 = 7
c) R2C7 = 9
d) 25(4) = {4678} -> R4C2 = 4, R3C2 = 8; 6,7 locked for R2
e) 30(5) = {35679} -> R4C3 = 3, R3C3 = 7, R2C2 = 5; 6,9 locked for D\
f) 32(5) = {45689}

5. Rest is singles without considering diagonals.

Rating:
Easy 1.5. I used a small forcing chain (X-Cycle?).


Last edited by Afmob on Tue Dec 23, 2014 8:01 am, edited 1 time in total.

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Andrew
 Post subject: Re: HS X 20
PostPosted: Mon Dec 22, 2014 5:41 am 
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Thanks HATMAN for the Hard version. My first impression was that combining five pairs of cages would make it a lot harder, but it didn't prove to be that much harder (no, that's not a challenge! anyway there's another of your recent puzzles which I haven't yet worked out how to get started), although it was somewhat longer to finish.

Here is my walkthrough for Human Solvable 20 X Hard:
Thanks Afmob for pointing out that my step 11b didn't work until one combination had been eliminated. I've inserted an extra step to do that. I've also made various detail changes to later steps, including removing some unnecessary singles.

Prelims

a) R34C8 = {49/58/67}, no 1,2,3
b) R67C2 = {49/58/67}, no 1,2,3
c) R67C8 = {49/58/67}, no 1,2,3
d) R8C34 = {49/58/67}, no 1,2,3
e) 20(3) cage at R7C3 = {389/479/569/578}, no 1,2
f) 27(4) cage at R9C4 = {3789/4689/5679}, no 1,2
g) 32(5) cage at R1C4 = {26789/35789/45689}, no 1
h) 23(6) cage at R4C6 = {123458/123467}, no 9

Steps resulting from Prelims
1a. 27(4) cage at R9C4 = {3789/4689/5679}, 9 locked for R9
1b. 32(5) cage at R1C4 = {26789/35789/45689}, CPE no 8,9 in R2C45

[I’ll stick with the same Human Solvable steps that I used for the original puzzle, even though Afmob’s 17(4+4+4+5) innies for the whole grid is slightly simpler.]

2. 45 rule for N24568 16(5+5+2+4) outies in N1379 = 114
2a. Max for 5 cells = 35, max for 4 cells = 30 and max for 2 cells = 17, but only if not limited by other cages
2b. Min R7C2 + R8C3 = 14 (from steps 2 and 2a), no 4 in R7C2 and R8C3, clean-up: no 9 in R6C2 + R8C4
2c. Max R7C2 + R8C3 = 15 (cannot be {79/89} which clash with 20(3) cage at R7C3)
[Alternatively max 5 cells in N7 = 35 -> max R7C2 + R8C3 = 15.]
2d. R7C2 + R8C3 + 20(3) cage = 34,35 = {46789/56789}, no 3, 6,7,8,9 locked for N7
2e. R7C2 + R8C3 = 14,15 -> 14(5+5+4) outies in N139 = 99,100 -> min 4 outies = 29, min 5 outies = 34
2f. 5 outies in N1 R1C1 + R23C23 = 34,35 = {46789/56789}, no 1,2,3, 6,7,8,9 locked for N1
2g. 5 outies in N3 R1C9 + R23C78 = 34,35 = {46789/56789}, no 1,2,3, 6,7,8,9 locked for N3
2h. 4 outies in N9 R7C89 + R89C7 = 29,30 = {5789/6789}, no 1,2,3,4, 7,8,9 locked for N9, clean-up: no 9 in R6C8

3. R4C6 + R5C5 + R6C4 = {123} (hidden triple on D/), locked for N5
3a. 45 rule on D/ 3 innies R4C6 + R5C5 + R6C4 = 1 outie R4C7 + 4, R4C6 + R5C5 + R6C4 = {123} = 6 -> R4C7 = 2

4. 23(6) cage at R4C6 = {123458/123467} -> R5C5 = 2, placed for D\
4a. 23(6) cage = {123458/123467}, 4 locked for R5
4b. 23(6) cage = {123458/123467}, CPE no 1,3 in R4C9
4c. 1,3 on D\ only in R7C7 + R8C8 + R9C9, locked for N9

5. 2 in R6 only in 35(7) cage at R4C1 = {1235789/1245689/2345678}, 5,8 locked for N4, clean-up: no 5,8 in R7C2

6. R7C2 + R8C3 = 14,15 (steps 2b and 2c) = [68/69/78/96] (cannot be [95] which clashes with 20(3) cage at R7C3), no 5,7 in R8C3, clean-up: no 6,8 in R8C4

7. Hidden killer pair 4,5 in 21(4) cage at R1C9 and 20(3) cage at R7C3 for D/, 20(3) cage contains one of 4,5 -> 21(4) cage must contain one of 4,5
7a. R1C9 + R23C78 contains one of 4,5 (step 2g) which must be in 21(4) cage -> no 4,5 in R2C7 + R3C8, clean-up: no 8,9 in R4C8

8. R6C4 “sees” all of N4 except for R4C23, R6C4 = {13} -> one of R4C23 must contain the same value (cannot be in both, which would clash with R4C6)
8a. Killer pair 1,3 in R4C23 and R4C6, locked for R4
8b. Hidden killer pair 1,3 in R4C23 and 35(7) cage at R4C1 for N4, R4C23 contains one of 1,3 -> 35(7) cage must contain one of 1,3 in N4
8c. 35(7) cage contains one of 1,3 in N4 and R6C4 = {13} -> 35(7) cage (step 5) = {1235789} (only remaining combination, other combinations only contain one of 1,3), no 4,6, 7,9 locked for N4, clean-up: no 6 in R7C2
8d. R7C2 + R8C3 = (step 6) = [78/96], clean-up: no 4 in R8C4
[It’s technically simpler just to use hidden killer pairs, rather than the “clone” which I initially saw for the first part of step 8.]

9. 20(3) cage at R7C3 = {569/578} (cannot be {479} which clashes with R7C2), no 4, 5 locked for N7 and D/
9a. 4 on D/ only in 21(4) cage at R1C9, locked for N3

10. N24568 16(5+5+2+4) outies in N1379 = 114 (step 2)
10a. 5 outies in N3 = {46789} = 29, 2 outies in N7 = [78/96] (step 8d) = 15 -> 5 outies in N1 = 35 and 4 outies in N9 = 30
10b. 5 outies in N1 = 35 = {56789}, no 4, 5 locked for N1
10c. 4 outies in N9 = 30 = {6789}, no 5, 6 locked for N9, clean-up: no 8 in R6C8
10d. Naked quad {1234} in R2378C1, locked for C1
10e. R6C3 = 2 (hidden single in R6)
10f. Killer pair 1,3 in 35(7) cage at R4C1 and 23(6) cage at R4C6, locked for R5

11. 30(5) cage at R1C1 must contain three of 6,7,8,9
11a. Hidden killer quad 6,7,8,9 in 30(5) cage and R6C6 for D\, 30(5) cage can only contain three of 6,7,8,9 which must be on D\ -> R6C6 = {6789}, no 6 in R4C3
[Afmob pointed out that step 11b wasn’t yet valid, because 30(5) cage could still be {34689}, so I’ve inserted an extra step to eliminate that combination.]
11aa. Hidden killer pair 8,9 in 30(5) cage at R1C1 and 25(4) cage at R2C3 for N1, 25(4) cage must contain at least one of 8,9 which must be in N1 -> 30(5) cage cannot contain more than one of 8,9 in N1 -> 30(5) cage = {15789/35679/45678} (cannot be {34689} = {689}[34] which has both of 8,9 in N1), 5,7 locked for D/
11b. 30(5) cage contains one of 1,3,4, R4C3 = {134} -> no 4 in R4C4
11c. 6 in N4 only in R46C2, locked for C2
11d. Naked triple {134} in R149C3, locked for C3
11e. Naked quad {5789} in R456C1 + R5C3, locked for N4
11f. Naked triple {134} in R7C7 + R8C8 + R9C9, locked for N9
11g. Deleted, unnecessary after inserting step 11aa.

12. 6 in C1 only in R19C1, CPE no 6 in R1C9 using D/
12a. 6 in C9 only in 24(4) cage at R4C9 = {1689/3678/4569}
12b. Hidden killer triple 7,8,9 in R1C9 and 24(4) cage for C9, 24(4) cage must contain two of 7,8,9 -> R1C9 = {789}, 24(4) cage = {1689/3678} (cannot be {4569} which only contains one of 7,8,9), no 4,5, 8 locked for C9
12c. 24(4) cage = {1689/3678} -> R6C9 = {13}
12d. Naked pair {13} in R6C49, locked for R6

13. R9C9 = 4 (hidden single in C9)
13a. 1 in R9 only in R9C23, locked for N7
13b. 1 in C1 only in R23C1, locked for N1
13c. 1 in R1 only in R1C89, locked for N3

14. R6C9 = 1 (hidden single in C9) -> 24(4) cage at R4C9 (step 12c) = {1689}, locked for C9
14a. R6C4 = 3, R4C6 = 1, R5C2 = 1, R9C3 = 1 (hidden single in R9)

15. R1C9 = 7, placed for D/, clean-up: no 6 in R4C8
15a. 21(4) cage at R1C9 contains 2,4,7 = {2478} -> R2C8 + R3C7 = {48}, locked for N3 and D/
15b. Naked triple {569} in 20(3) cage at R7C3, locked for N7 -> R7C2 = 7, R6C2 = 6, R8C3 = 8, R8C4 = 5, R8C2 = 9, R8C9 = 2, R9C8 = 5, R9C1 = 6, R7C3 = 5, clean-up: no 8 in R7C8

16. Naked quad {3789} in 27(4) cage at R9C4, 3 locked for R9 and N8 -> R9C2 = 2
16a. Naked pair {34} in R1C23, locked for R1 and N1
16b. R1C78 = [12] (hidden pair in N3)

17. 6 in R1 only in R1C456, locked for N2 and 32(5) cage at R1C4 -> R2C7 = 9
17a. Naked triple {568} in R1C456, locked for R1 and N2, R2C6 = 4 (cage sum), R2C8 = 8, R2C2 = 5, R3C8 = 6 -> R4C8 = 7, R1C1 = 9, placed for D\, R3C3 = 7, R6C6 = 8, R4C4 = 6, R4C3 = 3 (cage sum)

18. R2C3 = 6, R34C2 = [84] -> R2C4 = 7 (cage sum)

and the rest is naked singles, without using the diagonals.

Rating Comment:
I'll rate my walkthrough for HS 20 X Hard at Hard 1.25. I used a hidden killer quad. I didn't take account of the "clone", step 8, in my rating since that result could be obtained using hidden killer pairs.
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