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 Post subject: Assassin 390
PostPosted: Wed Jan 01, 2020 7:58 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 852
Location: Sydney, Australia
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1-9 cannot repeat on the diagonals. Note: r9c1 belongs to the 26(6)r7c2

Great puzzle to bring in the new year! SSscore = 1.65

triple click code:
3x3:d:k:4608:4608:4609:4609:6402:6402:9475:4356:4356:2565:4608:4609:4358:6402:6402:9475:3079:4356:2565:5128:4609:4358:4358:6402:9475:3079:3079:5128:5128:5128:2825:5642:2571:9475:9475:9475:3852:5389:5128:5642:2825:2571:9475:8206:1039:3852:5389:5389:5642:5642:8206:8206:8206:1039:3852:6672:6672:6672:6672:8206:8206:1809:1809:3858:3858:6672:7699:7699:7699:6932:6932:6932:6672:3858:7699:7699:7699:2837:2837:6932:6932:
solution:
Code:
+-------+-------+-------+
| 1 9 5 | 3 7 4 | 6 2 8 |
| 3 8 6 | 2 1 5 | 9 4 7 |
| 7 2 4 | 9 6 8 | 1 3 5 |
+-------+-------+-------+
| 5 1 3 | 6 4 9 | 7 8 2 |
| 2 7 9 | 8 5 1 | 4 6 3 |
| 4 6 8 | 7 3 2 | 5 9 1 |
+-------+-------+-------+
| 9 5 2 | 4 8 7 | 3 1 6 |
| 8 3 1 | 5 9 6 | 2 7 4 |
| 6 4 7 | 1 2 3 | 8 5 9 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 390
PostPosted: Mon Jan 06, 2020 9:11 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1718
Location: Lethbridge, Alberta, Canada
Thanks Ed. An excellent Assassin for the new year! Quite a lot of early progress but then it resists right to the end!

Here is my walkthrough for Assassin 390:
Prelims

a) R23C1 = {19/28/37/46}, no 5
b) 11(2) cage at R4C4 = {29/38/47/56}, no 1
c) R45C6 = {19/28/37/46}, no 5
d) R56C9 = {13}
e) R7C89 = {16/25/34}, no 7,8,9
f) R9C67 = {29/38/47/56}, no 1
g) 21(3) cage at R5C2 = {489/579/678}, no 1,2,3

1a. Naked pair {13} in R56C9, locked for C9 and N6, clean-up: no 4,6 in R7C8
1b. 45 rule on N36 3 innies R5C8 + R6C78 = 20 = {479/569/578}, no 2
1c. 2 in N6 only in R4C789 + R5C7, locked for 37(7) cage at R1C7
1d. 45 rule on N5 1 innie R6C6 = 2, placed for D\, clean-up: no 9 in 11(2) cage at R4C4, no 8 in R45C6, no 9 in R9C7
1e. 45 rule on N2 1 innie R1C4 = 3, clean-up: no 8 in R5C5
1f. 45 rule on N1 1 remaining innie R3C2 = 2, clean-up: no 8 in R23C1
1g. 45 rule on N4 1 remaining outie R7C1 = 9, R5C1 = 2 (hidden single in N4) -> R6C1 = 4 (cage sum), clean-up: no 1,6 in R23C1
1h. Naked pair {37} in R23C1, locked for C1 and N1
1i. 21(3) cage at R5C2 = {579/678}, 7 locked for N4

2a. 45 rule on R89 2 innies R8C3 + R9C1 = 7 = [16/25/61]
2b. 45 rule on R789 2 remaining innies R7C67 = 10 = {37/46}
2c. R7C89 = [16/25/52] (cannot be [34] which clashes with R7C67)
2d. 45 rule on N9 2 innies R79C7 = 11 = [38/47/65/74]
2e. 45 rule on N78 2 remaining innies R79C6 = 10 = {37/46}
2e. R5C8 + R6C78 (step 1b) = {569/578} (cannot be {479} which clashes with R7C67), no 4, 5 locked for N6
2f. 4 in N6 only in R4C789 + R5C7, locked for 37(7) cage at R1C7
2g. 45 rule on N3 3 innies R123C7 = 16 = {169/178/358} (cannot be {367} which clashes with R79C7)
2h. 12(3) cage at R2C8 = {129/147/237/246/345} (cannot be {138/156} which clash with R123C7), no 8

3. Hidden killer pair 1,3 in R6C45 and R6C9 for R6, R6C9 = {13} -> R6C45 must contain one of 1,3
3a. 22(4) cage at R4C5 = {1489/1579/3478/3568} (cannot be {1678} which clashes with R45C6, cannot be {3469} which clashes with 11(2) cage at R4C4, cannot be {4567} which doesn’t contain 1 or 3)
3b. 1,3 in R6C45 -> no 1,3 in R4C5 + R5C4
3c. 3 of {3568} must be in R6C5 -> no 6 in R6C5
[I can see interactions between 11(2) cage at R4C4, R45C6 when it’s {37/46}, R79C6 and R79C7 but there don’t seem to be any eliminations from them yet.]

[I’ve been a bit slow to use 37(7) cage at R1C7.]
4a. 2,4 in N4 only in 37(7) cage at R1C7 = {1246789/2345689}, CPE no 6,8,9 in R6C7
4b. Combined half cage 37(7) cage + R6C7 = {1246789}5 (cannot be {2345689}7 which clashes with R79C7 because 3,5,7 all in C7) -> R6C7 = 5, 37(7) cage = {1246789}, no 3, 1 locked for C7 and N3, clean-up: no 6 in R7C7 (step 2d), no 4 in R7C6 (step 2b), no 6 in R9C6
4c. 3 in N3 only in 12(3) cage at R2C8 (step 2h) = {345} (only remaining combination, cannot be {237} = [237] which clashes with R3C1), 3 locked for C8, 4,5 locked for N3
4d. R123C7 (step 2g) = {169} (only remaining combination, cannot be {178} which clashes with R79C7), 6,9 locked for C7, N3 and 37(7) cage at R1C7
4e. R56C8 = {69} (hidden pair in N6), locked for C8, 6 locked for 32(6) cage at R5C8, clean-up: no 4 in R7C7 (step 2b), no 4 in R9C6 (step 2e), no 7 in R9C7
4f. Naked pair {37} in R7C67, locked for R7
4g. Naked pair {37} in R79C6, locked for C6 and N8

5. Hidden killer pair 3,7 in 15(3) cage at R8C1 and R9C3 for N7, R9C3 can only contain one of 3,7 -> 15(3) cage must contain at least one of 3,7 = {348/357} (cannot be {168/456} which don’t contain 3 or 7), no 1,6, 3 locked for C2 and N7
5a. 15(3) cage = 5{37}/8{34}, no 5,8 in R89C2
5b. 18(4) cage at R1C1 = {189/468} (cannot be {459} = 5{49} which clashes with 15(3) cage), no 5, 8 locked for N1
5c. 5 in N1 only in R123C3, locked for C3
5d. R4C3 = 3 (hidden single in R4)
5e. 8 in R3 only in R3C456, locked for N2

6. Consider combinations for 21(3) cage at R5C2 = {579/678}
21(3) cage = {579} = 5{79}, 9 locked for R6 => R6C8 = 6
or 21(3) cage = {678}, locked for N4 => 6 in R4 only in R4C456
-> no 6 in R6C4

7. 2 on D/ only in R1C9 + R7C3, CPE no 2 in R7C9, clean-up: no 5 in R7C8

8. Consider placement for 3 in C7
R7C7 = 3 => R9C7 = 8 (step 2d) => 8 in N6 only in R4C89
or R8C7 = 3, R5C5 = 3 (hidden single on D\) => R4C4 = 8
-> 8 in R4C489, locked for R4, also no 8 in R8C7
8a. 20(5) cage at R3C2 = {12359/12368}
8b. 8 of {12368} must be in R5C3 -> no 6 in R5C3

9. R1C4 = 3 -> R123C3 = 15 and contains 5 for N1 = {159/456}
9a. 20(5) cage at R3C2 (step 8a) = {12359/12368} -> R4C12 + R5C3 = {159}/{16}8
9b. Consider placements for 9 in N1
9 in R12C2, locked for C2 => R4C12 = {15/16}, R5C3 = {89}
or 9 in R123C3 = {159}, locked for C3 => R5C3 = 8
-> R5C3 = {89}, R4C12 = {15/16}, 1 locked for R4, clean-up: no 9 in R5C6
9c. 9 in R4 only in R4C56, locked for N5
[At this stage I originally analysed 21(3) cage at R5C2 but it wasn’t very helpful then, so I’ve omitted it

10. 22(4) cage at R4C5 (step 3a) = {1579/3478/3568} (cannot be {1489} which clashes with R45C6)
10a. Consider placement of 7 in C3
R6C3 = 7 => 22(4) cage = {3478/3568} (cannot be {1579} because 5,7,9 only in R4C5 + R5C4)
or R9C3 = 7, R79C6 = [73], R7C7 = 3 (step 2d), placed for D\ => R6C5 = 3 (hidden single in N5)
-> 22(4) cage = {3478/3568}, R6C5 = 3, R56C9 = [31], R5C6 = 1 (hidden single in N5) -> R4C6 = 9, placed for D/, clean-up: no 8 in R4C4
10b. R7C7 = 3 (hidden single on D\), R67C6 = [73] -> R9C7 = 8
10c. R8C2 = 3 (hidden single in N7), placed for D/ -> R3C8 = 3 (hidden single in N3), R23C1 = [37]
10d. 8 in N5 only in R56C4, locked for C4
10e. 8 on D\ only in R1C1 + R2C2, locked for N1
10f. 7 in N7 only in R9C23, locked for R9

11. Consider combinations for R8C3 + R9C1 (step 2a) = [16/25/61]
R8C3 + R9C1 = {16}, locked for N7
or R8C3 + R9C1 = [25], R8C1 = 8 => R9C2 = 4 (cage sum), R9C3 = 7 (hidden single in N7)
-> R9C3 = {247}
11a. 1,6 in N7 only in R7C12 + R8C3 + R9C1, locked for 26(6) cage at R7C2

12. 17(3) cage at R2C4 = {179/269} (cannot be {278} because 2,7 only in R2C4, cannot be {458/467} which clash with R123C6), no 4,5,8, 9 locked for N2
12a. 2,7 only in R2C4 -> R2C4 = {27}
12b. Naked triple {169} in R3C457, locked for R3
12c. R3C6 = 8 (hidden single in R3)
12d. Killer pair 4,5 in R3C3 and 11(2) cage at R4C4, locked for D\

13. 18(3) cage at R1C1 = {189/468}, 26(6) cage at R7C2 = {124568}, R8C3 + R9C1 (step 2a) = [16/25/61]
13a. Consider permutations for R7C89 = [16/25]
R7C89 = [16] => 1 on D\ only in R1C1 + R2C2 => 18(3) cage = {189}
or R7C89 = [25] => 2 in 26(6) cage only in R8C3 => R9C1 = 5, R8C12 = [83], R9C2 = 4 (cage sum) => 18(3) cage = {189}
-> 18(3) cage = {189}, locked for N1, 9 locked for C2
13b. Naked triple {456} in R123C3, 4,6 locked for C3
13c. 1 in C3 only in R78C3, locked for N7

14. Consider placements for R8C1 = {58}
R8C1 = 5 => R9C1 = 6 => R9C9 = 9, placed for D\ => R1C1 + R2C2 = {18}
or R8C1 = 8 => R1C1 = 1, R12C2 = [98]
-> R1C1 + R2C2 = {18}, 1 locked for N1 and D\, R1C2 = 9, R8C8 = 7, placed for D\, clean-up: no 4 in 11(2) cage at R4C4
[Cracked at last; the rest is straightforward.]
14a. Naked pair {56} in 11(2) cage at R4C4, locked for N5 and D\ -> R9C9 = 9, R3C3 = 4, R3C9 = 5, R2C8 = 4, R7C9 = 6 -> R7C8 = 1
14b. R8C3 = 1, R9C1 = 6 (hidden pair in N7), 6 placed for D/, R123C7 = [691], R12C3 = [56], R12C6 = [45], 11(2) cage at R4C4 = [65], R3C45 = [96], R2C4 = 2 (cage sum)
14c. Naked pair {45} in R78C4, locked for C4, 4 locked for N8, R9C345 = [712], R7C5 = 8, R8C56 = [96] -> R8C4 = 5 (cage sum), R7C3 = 2, placed for D/
14d. R18C1 = [18], R1C5 = 7, R1C9 = 8, placed for D/

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

Rating Comment:
I'll rate my walkthrough for A390 at 1.5. I used quite a lot of short forcing chains.


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 Post subject: Re: Assassin 390
PostPosted: Sat Jan 11, 2020 1:25 am 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 208
Location: California, out of London
Thanks Ed! Many moves that were quite straightforward in principle were well hidden. Excellent!
(Corrections thanks to Andrew & Ed)
Assassin 390 WT:
1. Innies n5 -> r6c6 = 2
Innies n2 -> r1c4 = 3
Innies n12 -> r3c2 = 2
Outies n124 -> r7c1 = 9
-> 15(3)r5c1 = [249]
-> 21(3)n4 = {7(68|59)}
-> 20(5)r3c2 = [2{13(59|68)}]
Also 10(2)n1 = {37}

2. Remaining Outies n36 = r7c67 = +10(2) = {37} or {46}
Innies n78 = r79c6 = +10(2)
-> r9c6 = r7c7 (from 3467)

3. 4(2)n6 = {13}
Innies n36 = r5c8 + r6c78 = +20(3)
This cannot be {389}
Also since r7c67 = {37} or {46} -> Innies n36 cannot be {479}
-> Innies n36 = {569} or {578}

4. Trying r7c67 = {46} ...
... puts Innies n36 = {578} and outies n3 = {2469}, and
... puts Innies n3 = r123c7 = {178} or {358}, and
... puts innies n9 = r79c7 = +11(2) = [65] or [47]
-> (578) all locked in c7 in r12379 which leaves no value for r6c7
-> r7c67 = {37}

5. -> Innies n36 = {569}
-> Outies n3 = {2478}
-> Innies n3 = r123c7 = {169}
-> 3 in n3 in r23c8
-> 8 in n3 in 17(3)
Also r6c7 = 5 and r56c8 = {69}

6. 7(2)n9 = [16] or {25}
Remaining Innies r7 = r7c2345 = +19(4) = {48(25|16))
Outies r7 = r8c3 + r9c1 = +7(2) = {16} or [25]
The values in those two outies also go in 7(2)n9

7. Innies n78 = r79c6 = +10(2) = {37}
Whichever of (37) is in r7c7 goes in n8 in r9c6 -> goes in n7 in r8c2
I.e., r8c2 from (37)

8! 3 in D\ only in r5c5 or r7c7
Trying r7c67 = [37] puts 7 in r8c2 and 11(2)n5 = [83]
This leaves no place for 8 in c1
-> r7c67 = [73]
-> 11(2)r9c6 = [38]
-> r8c2 = 3
Also 8 in n6 in r4c89
Also 3 in n3 in r3c8
-> 10(2)r2c1 = [37]

9. 11(2)n5 from {47} or {56}
-> 10(2)n5 = {19}
-> (HS 1 in r6) 4(2)n6 = [31]
-> (HS 3 in n4) r4c3 = 3
-> (HS 3 in n5) r6c5 = 3
-> 8 in n5 in r56c4

10! All the values from 17(3)n2 must go in c6 in r456789c6
-> Only possible solutions for 17(3)n2 are [2{69}] and [7{19}]
-> (169) locked in r3 in r3c457 with 9 in r3c45
-> 9 in n3 in r12c7
-> (HS 9 in D/) 10(2)n5 = [91]
-> 1 in n4 in r4c12
-> 20(5)r3c2 = [2{15}39] or [2{16}38]

Also at least one of (27) in r1 in r1c89
-> 17(3)n3 = {278} and 12(3)n3 = <435>

11. 15(3)n7 from [834] or [537]
If the former -> 4 in n1 in r123c3 -> r123c3 = {456} and 18(3)n1 = {189}
If the latter -> r8c3,r9c1 = {16} -> 7(2)n9 = {16} -> 1 in D\ in 18(3)n1
-> in both cases r123c3 = {456} and 18(3)n1 = {189}
-> (HS 8 in r3) r3c6 = 8

12! Whichever of (45) is in r3c3 goes in n3 in r2c8 -> nowhere else in the diagonals
r8c3,r9c1 = [25] or [16]
-> Whichever of (56) is in r9c1 goes in n9 in r7c89 -> only place for it in D\ is in r4c4
-> 11(2)n5 = {56}

13. -> r3c3 = 4 and 12(3)n3 = [435]
Also r5c5,r9c1 = {56} -> r3c7 = 1
-> 17(3)n2 = [2{69}]

14. 7 in n7 only in r9c23
1 in c3 only in r89c3
But 1 in r9c3 puts 15(3)n7 = [537] which leaves no solution for r8c3,r9c1 = +7(2)
-> HS r8c3 = 1
-> r9c1 = 6
-> 7(2)n9 = [16]
Also 11(3)n5 = [65]
-> 20(5)r3c2 = [2{15}39]
-> r56c8 = [69]
-> 21(3)n4 = <768>
Also r3c45 = [96]

15. Whichever of (78) is in r6c3 can only go in D/ in r1c9
-> (HS 2 in D/) r7c3 = 2
-> r1c8 = 2
-> r12c9 = {78}
-> r4c8 = 8 and 7 in r45c7
-> (HS 7 in D\) r8c8 = 7
-> NS r9c8 = 5

16. HP r1c1,r2c2 = {18}
-> r1c2 = 9
-> (NS on D\) r9c9 = 9
-> r8c79 = {24}
-> (HS 2 in c5) r9c5 = 2
-> (HS 9 in c5) r8c5 = 9
-> (HS 8 in c5) r7c5 = 8
-> 15(3)n7 = [834]
etc.


Last edited by wellbeback on Sun Feb 09, 2020 8:12 pm, edited 1 time in total.

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 Post subject: Re: Assassin 390
PostPosted: Tue Jan 21, 2020 7:10 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 852
Location: Sydney, Australia
Thanks for the 'excellent' ratings from both Andrew and wellbeback. Their summaries exactly say what I felt about this puzzle, "Quite a lot of early progress but then it resists right to the end!", and "Many moves that were quite straightforward in principle were well hidden." Really enjoyed both the WTs. I also found some cool steps. Really enjoyed steps 6, 14, 18, 23, 25 & 33!

WT for a390:
Preliminaries
Cage 4(2) n6 - cells ={13}
Cage 7(2) n9 - cells do not use 789
Cage 10(2) n5 - cells do not use 5
Cage 10(2) n1 - cells do not use 5
Cage 11(2) n89 - cells do not use 1
Cage 11(2) n5 - cells do not use 1
Cage 21(3) n4 - cells do not use 123

1. "45" on n2: 1 innie r1c4 = 3
1a. -> 1 innie n1, r3c2 = 2
1b. 1 outie n4, r7c1 = 9
1c. -> r56c1 = 6 and must have 2 for n4 = {24} only: both locked for c1 and 4 for n4
1d. -> 10(2)n1 = {37} only: both locked for c1 and n1

2. "45" on n36789: 1 outie r6c6 = 2, placed for d\
2a. r56c1 = [24]
2b. no 8 in 10(2)n5, no 9 in 11(2)n5

3. 4(2)n6 = {13}: both locked for c9 and n6

4. "45" on r789: 2 innies r7c67 = 10 = {37/46}(no 1,5,8) = 4 or 7
4a. -> r5c8 + r6c78 = 20 (cage sum) but {479} blocked by r6c67
4b. = {569/578}(no 4)
4c. must have 5: locked for n6

5. "45" on c789: 2 remaining outies r79c6 = 10 = {37/46}(no 5,8,9)
5a. r9c7 = (4578)

6. "45" on r89: 2 innies r8c3 + r9c1 = 7 = {16}/[25](no 3,4,7,8; no 5 in r8c3)
6a. h7(2)r8c3r9c1 sees all r7 apart from r7c6789.
6b. Can't both repeat in r7c67 since its a h10(2) -> must both repeat in r7c89 since its also a 7(2)
6c. -> 7(2)n9 = [16]/{25}(no 3,4; no 6 in r7c8)
6d. (alternatively, h10(2)r7c67 must have one of 3,4 for r7 -> {34} blocked from 7(2)n9
6e. but note: 6b turns out to be really useful later - see step 18)

7. "45" on n9: 2 innies r79c7 = 11
7a. but [65] blocked by 7(2)n9 needs one of them (step 6.)
7b. = [38]/{47}(no 5,6)
7c. when 3 in r7c7 -> 7 in r7c6 (h10(2))
7d. -> r79c7 = {47} or r7c67 = [73] -> 7 locked: no 7 in r6c7
7e. no 4 in r7c6, no 6 in r9c6

8. 2 & 4 in n6 only in 37(7): locked for that cage
8a. 37(7) = 24689{17/35}
8b. -> no 6,8,9 in r6c7 (Common Peer Elimination (CPE))
8c. r6c7 = 5
8e. -> 37(7) = {1246789}(no 3)

9. "45" on n3: 3 innies r123c7 = 16
9a. but {178} blocked by r79c7 (step 7b)
9b. = {169} only, locked for c7, n3 and 37(7) cage

10. 6 in n6 only in r56c8 in split15(2) = {69} only: both locked for c8, 6 for 32(6) cage
10a. -> r7c67 = {37}: both locked for r7
10b. no 7 in r9c7 (h11(2)r79c7)
10c. no 4 in r9c6

11. naked pair {37} in r79c6: both locked for c6 and n8

12. deleted

13. 3 in n3 only in r23c8 in 12(3): 3 locked for c8
13a. 12(3) = {237/345}(no 8)

14. 3 in r8 in r8c27 and 3 on d\ in r5c5 or r7c7 -> one of r5c5 or r8c2 must be 3: locked for d/ (turbot fish)
14a. hsingle 3 in n3 -> r3c8 = 3
14b. r23c1 = [37]
14c. 12(3)n3 = [3]{45} only: 4 and 5 locked for n3

15. r123c6 must have at least one of 5 or 8 for c6 since the only other spot is r8c6 (hidden killer pair)
15a. -> 25(5) = {12589/14569/14578/24568}
15b. must have 5: locked for n2

16. 17(3)n2: {278} blocked by 2 & 7 only in r2c4
16a. = {179/269/467}(no 8)
16b. 2 & 7 only in r2c4 -> r2c4 = (27)

17. 2 on d/ only in r1c9 and r7c3; r7c9 sees both of those -> no 2 in r7c9 (CPE)
17a. no 5 in r7c8

18. r9c1 and r7c9 cannot both be 5 since they see all 5 in n3
18a. -> r8c3 + r9c1 (h7(2)) cannot be [25] since it forces 5 in r7 into r7c9
18b. must be {16}: both locked for n7 and 26(6) cage
18c. 7(2)n9 must be [16](hidden single r7)

19. r123c3 = 15: but {168} blocked by r8c3 = (16)
19a. = {159/456}(no 8)
19b. must have 5: locked for n1 and c3

20. killer pair 1,6 in r123c3 and r8c3: both locked for c3

21. 1 in n4 only in r4: locked for r4
21a. no 9 in r5c6

22. 15(3)n7 = {348/357}
22a. -> no 5 or 8 in r89c2
22b. 3 locked for c2 and n7
(can also reduce r8c2 t0 (37) since r7c7 = r9c6; but don't think it helps much so not included. Surely wellbeback will use it though!! edit: he did!!)

23. 17(3)n2 = {179/269/467} = 4 or 9 in r3
23a. consider 5 on d/
23b. If 5 in r2c8 -> 4 in r3c9 -> 17(3)n2 has 9 in r3 -> no 9 in r3c7
23c. or 5 on d/ is in r5c5 -> 6 in r4c4 -> 10(2)r4c6 = [91] -> no 9 in r3c7
23d. -> no 9 in r3c7
23e. killer pair 1,6 in r3c457: both locked for r3

24. r6c45 must have 1 or 3 in r6 since the only other place is r6c9
24a. -> 22(4) must have 1 or 3: can't have both since 1+3+8+9=21
24b. -> no 1 or 3 elsewhere in 22(4)
24c. -> r4c3 = 3 (hsingle r4)

I've changed how I express this next step using less explanation. Hope the reasoning is still clear.
25. if 1 in c2 in r12c2 -> 18(3) = [8]{19}
25a. or 1 in c2 in r4c2
25b. -> 9 blocked from r4c2 (locking-out cages)

26. 9 on d/ only in n5: locked for n5
26a. -> r4c6 = 9 (hsingle r4), r5c6 = 1
26b. r56c9 = [31]
26c. r8c2 = 3 (hsingle d/), r6c5 = 3 (hsingle n5), r7c7 = 3 (hsingle d\), r79c6 = [73], r9c7 = 8
26d. r3c6 = 8 (hsingle r3)

27. 5 on d/ only in r2c8 + r5c5, r8c8 sees both those -> no 5 in r8c8 (Common Peer Elimination CPE)

28. 11(2)n5: {47} blocked by r8c8 = (47)
28a. = {56} only: both locked for n5 and d\

29. hsingle 5 in r3 -> r3c9 = 5, r2c8 = 4 (placed for d/, r8c8 = 7

30. hidden pair 1 & 8 on d\ in r1c1 + r2c2 = {18} in 18(3) = {189} -> r1c2 = 9, 1 locked for n1
30a. r3c3 = 4 (placed for d\), r9c9 = 9

31. r12c3 = {56}: 6 locked for c3
31a. r8c3 = 1, r9c1 = 6 (placed for d/)
31b. r5c5 = 5, r4c4 = 6

32. 6 in n4 only in 21(3) = {678}, 7 & 8 locked for n4
32a. r5c3 = 9, r56c8 = [69]

A nice elimination to finish it off
33. r6c3 + r6c4 = {78} -> no 8 in r7c3 (CPE), both also locked for r6
33a. r7c3 = 2, placed for d/

easy now.
Cheers
Ed


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