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 Post subject: Assassin 364
PostPosted: Sat Dec 01, 2018 7:57 am 
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Posts: 1040
Location: Sydney, Australia
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This is the third version of this puzzle. The first two versions were one trick puzzles which just doesn't satisfy anymore. This one makes you work for much longer - and I found it quite difficult to break. Very satisfying to finally get a decent solution! SudokuSolver gives it 1.90 and JSudoku needs to use 6 chains.
code:
3x3::k:5632:5632:3585:3585:3585:5378:4099:4099:4099:7940:5632:5632:3585:3589:5378:4099:5382:5382:7940:1543:1543:4360:3589:5378:2313:2313:5382:7940:0000:0000:4360:4360:2315:2315:0000:5382:7940:7940:5645:5645:5645:5645:5645:0000:0000:4110:7940:2831:2831:3344:3344:3345:3345:0000:4110:1042:1042:5139:2324:3344:3349:3349:0000:4110:4110:6678:5139:2324:5642:3340:3340:0000:6678:6678:6678:5139:5642:5642:5642:3340:3340:
solution:
Code:
+-------+-------+-------+
| 6 7 5 | 3 2 8 | 4 1 9 |
| 3 1 8 | 4 9 6 | 2 5 7 |
| 9 4 2 | 1 5 7 | 6 3 8 |
+-------+-------+-------+
| 2 6 3 | 9 7 4 | 5 8 1 |
| 4 8 7 | 6 1 5 | 3 9 2 |
| 1 5 9 | 2 8 3 | 7 6 4 |
+-------+-------+-------+
| 8 3 1 | 7 6 2 | 9 4 5 |
| 5 2 4 | 8 3 9 | 1 7 6 |
| 7 9 6 | 5 4 1 | 8 2 3 |
+-------+-------+-------+

Cheers
Ed


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 Post subject: Re: Assassin 364
PostPosted: Fri Dec 07, 2018 10:44 pm 
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Here's how I solved it. Steps 8, 19 & 26 are my keys with one of those steps getting no (useful) eliminations!
WT 364:
Prelims courtesy of SudokuSolver
Preliminaries
Cage 4(2) n7 - cells ={13}
Cage 6(2) n1 - cells only uses 1245
Cage 14(2) n2 - cells only uses 5689
Cage 13(2) n9 - cells do not use 123
Cage 13(2) n6 - cells do not use 123
Cage 9(2) n56 - cells do not use 9
Cage 9(2) n8 - cells do not use 9
Cage 9(2) n3 - cells do not use 9
Cage 11(2) n45 - cells do not use 1
Cage 21(3) n2 - cells do not use 123
Cage 20(3) n8 - cells do not use 12
Cage 13(4) n9 - cells do not use 89
Cage 14(4) n12 - cells do not use 9
Cage 26(4) n7 - cells do not use 1

No routine clean-up done unless stated
1. "45" on n7: 1 outie r6c1 = 1
1a. "45" on n3: 1 outie r4c9 = 1

2. 4(2)n7 = {13} only: both locked for n7 and r7

3. "45" on n8: 1 innie r7c6 + 6 = 1 outie r9c7 = [28] only permutation
3a. r7c6 = 2 -> r6c56 = 11 (no 9)
3b. no 7 in 9(2)n8, no 6,8 in r8c5
3c. no 5 in 13(2)n9

4. 21(3)n2: {489} blocked by 14(2)n2 needing an 8 or 9
4a. = {579/678}(no 4)
4b. must have 7: 7 locked for n2 and c6
4c. no 4 in r6c5 (sp11(2))

5. 14(2)r23c5 = {59/68}; 9(2)r78c5 = [81]/[63]/{45}
5a. if 14(2) = {68} -> 9(2) = {45}; or 14(2) = {59}
5b. -> 5 locked in one of those two cages: locked for c5 (combined cages)
5c. no 6 in r6c6 (sp11(2))

6. "45" on n2: 1 innie r3c4 + 4 = 1 outie r1c3
6a. r1c3 = (5678), r3c4 = (1234)
6b. hidden quad 1,2,3,4 in n2 -> r123c4+r1c5 = {1234}
6c. 17(3)r3c4 can only have one of 1,2,3,4 -> no 2,3,4 in r4c45

7. 20(3)r7c4 = {389/479/569/578} = two of 6..9
7a. -> r456c4 must have two of 6..9 for c4
7b. r4c5 = (6789)
7c. sp11(2)r6c56 must have one of 6,7,8
7d. -> killer quad 6,7,8,9 in those four areas: all locked for n5
7e. r4c7 = (456)

Ready for the first advanced step - which totally cracked one of the previous versions of this puzzle
8. 21(3)n2 = {579}/678}
8a.if {579} -> 14(2)n2 = {68} (combined cage)
8b. r9c7 = 8 -> r89c6 + r9c5 = 14 = {149/167/347/356}
8c. if {149} -> 9(2)n8 = [63] only (combined cage)
8d. 9 in c6 in one of 21(3)n2 or r89c6 -> 14(2)n2 = {68} or 9(2)n8 = [63]
8e. must have 6: 6 locked for c5
8f. no 5 in r6c6 (sp11(2))

9. 6 in c5 in 14(2)n2 = {68} or 6 in 9(2)n8 -> no {18} in 9(2)n8 since it would block all 6 in c5 (locking-out cages)
9a. 9(2) = [63]/{45}

10. 8 in n8 only in 20(3) = {389/578}(no 4,6) = 3 or 5
10a. 8 locked for c4

11. 9(2)n8 = [63]/{45} = 3 or 5
11a. killer pair 3,5 with 20(3) (step 10): both locked for n8

12. 3 in c6 only in n5: 3 locked for n5
12a. no 8 in r6c6 (sp11(2))

13. 8 in c6 only in 21(3)n2 = {678} only: 6,8 locked for n2, 6 for c6

14. 14(2)n2 = {59}: both locked for c5
14a. r78c5 = [63]
14b. 13(2)n9 = {49} only: both locked for n9 and r7

15. "45" on n9: 2 remaining innies r78c9 = 11 = [56] only permutation

16. 16(4)r6c1: must have 7 or 8 for r7c1 = [1]{258} only
16a. R7C1 = 8
16b. r8c12 = {25} both locked for r8
16c. r7c4 = 7

17. r8c4 = 8 (hidden single n8)
17a. r9c4 = 5 (cage sum)

18. 17(3)r3c4 must have 6 or 9 for r4c4, and 7 or 8 for r4c5 = {179/368/467}(no 2)
18a. no 6 in r1c3 (IODn2=-4)

Next key step
19. "45" on n2: 3 outies r1c3 + r4c45 = 21 = [5][97]/[7][68]/[8][67]:
19a. note: must have 5 in r1c3 or 6 in r4c4
19b. note2: must have 7 in r1c3 or r4c5
19c. note: no eliminations yet (can take 7 from r4c3 but not important to this optimised solution)

20. 11(2)r6c3: [56] blocked by step 19a
20a. = {29}/[74] = 4 or 9

21. 13(2)n6: {49} blocked by 11(2)n4
21a. = [58]/{67}(no 4,9; no 5 in r6c8)

22. hidden pair 5,6 in r6. 13(2)n6 can only have one of 5,6 -> r6c2 = (56)

23. "45" on r6: 2 remaining innies r6c29 = 9 = [54/63]

24. hidden pair 2,9 in r6 -> 11(2)r6c3 = {29} only

25. "45" on n1: 3 innies r1c3 + r23c1 = 17
25a. but {458} blocked by 6(2)n1 needing 4 or 5
25b. must have 5,7,8 for r1c3 = {278/359/368/467}
25c. note: if has 7 in r1c3 must have {46} in r23c1
25d. note: no eliminations yet

The final crack by removing 6 from r6c2
26. from step 19b. must have 7 in r1c3 or r4c5
26a. if in r1c3 -> r23c1 = {46} (step 25c) -> no 6 in r6c2 (same cage)
26b. if in r4c5 -> 7 in r6 only in 13(2)r6c78 = {67}
26c. -> both places have 6 -> no 6 in r6c2

27. r6c2 = 5
27a. r6c9 = 4 (h9(2)r6c29), r6c56 = [83], r4c5 = 7 -> r34c4 = 10 = [19/46] (no 3)
27b. no 7 in r1c3 (IODn2=-4)
27c. r8c12 = [52]

28. 31(6)r2c1 = {23489}[5]/{23678}[5]
28a. must have 8 -> r5c2 = 8
28b. must have 2 & 3 which are only in c1: locked for c1

29. naked pair {14} in r3c24: both locked for r3

30. naked pair {67} in r6c78: locked for n6
30a. r4c67 = [45], naked pair {19} in r89c6: 1 locked for n8 and c6, r9c5 = 4, r5c6 = 5

31. 22(5)r5c3 must have 1 for n5 = {13459/13567}(no 2)
31a. must have 3: 3 locked for r5
31b. r5c5 = 1

32. naked pair {29} in r5c89: both locked for r5 and n6

Straightforward now.
Cheers
Ed


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 Post subject: Re: Assassin 364
PostPosted: Sat Dec 08, 2018 9:22 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
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Thanks Ed! This was the puzzle that kept on giving. As you wrote - several key steps needed.
As usual I wrote my WT before reading yours. Some similarities and some differences. Where there were similarities we often approached each step from opposite directions.
Assassin 364 WT:
1. Outies n3 -> r4c9 = 1
Outies n7 -> r6c1 = 1
-> 1 in n5/r5 in r5c456

2. 4(3)n7 = {13}
Outies - Innies n8 -> r9c7 = r7c6 + 6
-> [r7c6,r9c7] = [28]

3. Innies n2 -> r1c45 + r23c4 = +10(4) = {1234}
-> 21(3)n2 = {7(68|59)}
Also Since Max r3c4 = 4 -> both r4c45 are Min 5.

4! 1 in c6 only in r589c6
9 in c6 in one of:
a) r123c6 -> 14(2)n2 = {68}
b) r5c6 -> 1 in r89c6
c) r89c6 -> 22(4)n8 = [{149}8]
In none of those cases can 9(2)n8 be {18}
-> 8 in n8 in 20(3)n8
-> 20(3)n8 = {8(57|39)}

5! 6 in c4 only in n5 in r456c4
Also -> 1 in n8 in 22(4)n8
-> 3 in n8 in 20(3) or 9(2)
-> 3 in c6 in n5 in r456c6
-> 13(3) r6c5 from [832] or [742]
If the former -> 14(2)n2 = {59} -> 9(2)n8 = [63] -> 20(3)n8 = {578}
If the latter -> 7 in 20(3)n8 = {578} -> 9(2)n8 = [63] -> 14(2)n2 = {59}
Either way 20(3)n8 = {578}, 9(2)n8 = [63], 14(2)n2 = {59}, 21(3)n2 = {678}, 22(4)n8 = [{149}8]
-> 9 in c4 in n5 in r456c4
Also 5 in c6 in n5 in r45c6
Also -> r46c5 = {78}

(Having read Ed's WT the next step is more complicated than it need be since 13(2)n9 can already only be {49})!

6! Remaining Innies n9 -> r78c9 = +11(2)
3 in r9 only in r9c89
-> 13(4)n9 cannot contain a 6
-> 6 in n9 in 13(2) or H11(2)
-> 6 in r9 in n7 in r9c123
-> 2 in n7 in r8c12
-> 2 in r9 in r9c89
-> 13(4)n9 = [{17}{23}]
-> 13(2)n9 = {49}
-> r78c9 = [56]
-> 16(4)r6c1 can only be [18{25}]
-> 20(3)n8 = [785]
Also 26(4)n7 = {4679} with r8c3 from (49)
Also 22(4)n8 = [{149}8] with r8c6 from (49)

7. (27) already in c6 and 1 already in r4c9 -> 9(2)r4 from {36} or {45}
r4c4 from (69) and r4c5 from (78)
-> 17(3)r3c4 from [197], [368], [467]
The first of these -> r6c56 = [83]
-> In all cases 9(2)r4 can only be {45}

8. 8 in n5 in r46c5 and 3 in n5 in r56c6
-> 11(2)r6 cannot be {38}
Remaining Innies r6 -> r6c29 = +9(2) (No 9)
-> 9 either in 11(2)r6c3 = {29} or 13(2)r6 = {49}
-> 11(2)r6 cannot be {47}
-> 11(2)r6c3 from {29} or [56]

9. 5 in r7c9 prevents H9(2)r6c29 = [45]
-> 4 in n4 in r5c123
-> 4 in n5 only in r46c6
-> 22(4)n8 = [9418]
-> 26(4)n7 = [4{679}]
-> 4 in n4 in r5c12
-> 31(6) = {2349(58|67)}

10. Innies n1 = r1c3 + r23c1 = +17(3)
Innies - Outies n2 -> r1c3 = r3c4 + 4
Since r3c4 from (134) -> r1c3 from (578)
But r1c3 = 7 -> r23c1 = +10(2) for which there is no solution
-> r1c3 from (58)
-> r3c4 from (14)
-> 17(3)r3c4 from [197] or [467]
-> r4c5 = 7
-> 13(3)r6c5 = [832]

(This next step is essentially Ed's Step 19a).

11! 5 in r6 only in r6c23
r1c3 from (58)
If r1c3 = 5 puts 5 in r6c2
If r1c3 = 8 puts 17(3)r3c4 = [467] -> 5 not in r6c3
Either way r6c2 = 5

12. -> r6c9 = 4
-> 9(2)r4 = [45]
Also 13(2)n6 = {67}
-> 11(2)r6 = {29}
Also r8c12 = [52]
Also -> 31(6) = {234589}
-> r5c2 = 8
-> r5c1 = 4
-> r234c1 = {239}
-> Innies n1 can only be r1c3 = 5, r23c1 = {39}
-> r4c1 = 2 and r3c4 = 1
-> r4c4 = 9
-> 11(2)r6 = [92]
-> r5c456 = [615]
Also 6(2)n1 = [42]
etc.


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 Post subject: Re: Assassin 364
PostPosted: Mon Feb 04, 2019 5:59 am 
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Another Assassin which I'd skipped over at the time and had just started when I saw that A369 had been posted, so I continued with it.

As wellbeback said, it kept giving, well at least for a while.

Thanks Ed for pointing out a flawed step and hinting at an alternative way to get the same result.
Here's my walkthrough for Assassin 364:
Prelims

a) R23C5 = {59/68}
b) R3C23 = {15/24}
c) R3C78 = {18/27/36/45}, no 9
d) R4C67 = {18/27/36/45}, no 9
e) R6C34 = {29/38/47/56}, no 1
f) R6C78 = {49/58/67}, no 1,2,3
g) R7C23 = {13}
h) R78C5 = {18/27/36/45}, no 9
i) R7C78 = {49/58/67}, no 1,2,3
j) 21(3) cage at R1C6 = {489/579/678}, no 1,2,3
k) 20(3) cage at R7C4 = {389/479/569/578}, no 1,2
l) 14(4) cage at R1C3 = {1238/1247/1256/1346/2345}, no 9
m) 13(4) cage at R8C7 = {1237/1246/1345}, no 8,9
n) 26(4) cage at R8C3 = {2789/3689/4589/4679/5678}, no 1

Steps Resulting From Prelims and Immediate Placements
1a. Naked pair {13} in R7C23, locked for R7 and N7, clean-up: no 6,8 in R8C5
1b. R3C78 = {18/27/36} (cannot be {45} which clashes with R3C23), no 4,5
1c. 13(4) cage at R8C7 = {1237/1246/1345}, 1 locked for N9
1d. 45 rule on N3 1 outie R4C9 = 1 -> R2C89 + R3C9 = 20 = {389/479/569/578}, no 2, clean-up: no 8 in R4C67
1e. 45 rule on N7 1 outie R6C1 = 1

2. 45 rule on N8 1 outie R9C7 = 1 innie R7C6 + 6 -> R7C6 = 2, R9C7 = 8, clean-up: no 1 in R3C8, no 7 in R4C7, no 5 in R6C8, no 7 in R78C5, no 5 in R7C78
2a. R7C6 = 2 -> R6C56 = 11 = {38/47/56}, no 9
2b. 45 rule on N9 2 remaining innies R78C9 = 11 = [56/65/92] (cannot be {47} which clashes with R7C78), no 4,7, no 3,9 in R8C9
2c. 13(4) cage at R8C7 = {1237/1345} (cannot be {1246} which clashes with R78C9), no 6
2d. 9 in N9 only in R7C789, locked for R7
2e. 45 rule on R6 2 remaining innies R6C29 = 9 = {27/36/45}, no 8,9
2f. 9 in R6 only in R6C34 = {29} or R6C78 = {49} -> R6C34 = {29/38/56} (cannot be {47} locking-out cages), no 4,7
[2 in R6 only in R6C29 and R6C34 would give the same elimination.]

3a. 45 rule on N2 4 innies R1C45 + R23C4 = 10 = {1234}, 4 locked for N2
3b. 45 rule on N2 1 outie R1C3 = 1 innie R3C4 + 4 -> R1C3 = {5678}
3c. 7 in N2 only in 21(3) cage at R1C6, locked for C6, clean-up: no 2 in R4C7, no 4 in R6C5 (step 2a)
3d. 17(3) cage at R3C4 can only contain one of 1,2,3,4 -> no 2,3,4 in R4C45

4. 45 rule on R1 3 innies R1C126 = 2 outies R2C47 + 15
4a. Max R1C126 = 24 -> max R2C47 = 9, no 9 in R2C7

5. 16(4) cage at R1C7 = {1249/1258/1348/1456/2347/2356} (cannot be {1267/1357} which clash with R3C78)
5a. 7 on {2347} must be in R1C789 (R1C789 cannot be {234} which clash with R1C45, ALS block), no 7 in R2C7

6. 26(4) cage at R8C3 = {2789/4589/4679/5678}
6a. 8 of {2789/4589/5678} must be in R8C3 -> no 2,5 in R8C3

7. R23C5 = {59/68}, R78C5 = {45}/[63/81] -> combined cage = {59}[63]/{59}[81]/{68}{45}, 5 locked for C5, clean-up: no 6 in R6C6 (step 2a)
7a. 22(4) cage at R8C6 = {1489/1678/3478/3568}
7b. 7 of {3478} must be in R9C5, 6 of {3568} must be in R9C5 (R89C6 cannot be {56} which clashes with 21(3) cage at R1C6), no 3 in R9C5
7c. 22(4) cage = {1489/1678/3478} (cannot be {3568} because R89C6 = {35} clashes with 21(3) cage at R1C6 + R46C6, killer ALS block), no 5 in R89C6
7d. Consider combinations for R23C5 = {59/68}
R23C5 = {59} => 21(3) cage at R1C6 = {678}, 6,8 locked for C6, R4C6 = {345}, R6C6 = {345} => R89C6 cannot be {34}, ALS block
or R23C5 = {68}, locked for C5 => R78C5 = {45}, 4 locked for N8
-> 22(4) cage = {1489/1678} (cannot be {3478})
, no 3 in R89C6, 1 locked for N8, clean-up: no 8 in R7C5
[My original step 7d was flawed. Thanks Ed for suggesting an alternative way, which I’ve rewritten in my solving style.]
7e. 3 in C6 only in R456C6, locked for N5, clean-up: no 8 in R6C3, no 8 in R6C6 (step 2a)
7f. Killer pair 5,6 in R23C5 and R78C5, locked for C5, clean-up: no 5 in R6C6 (step 2a)
7g. 8 in N8 only in 20(3) cage at R7C4, locked for C4, clean-up: no 3 in R6C3
7h. 20(3) cage = {389/578}, no 4,6
7i. Killer triple 7,8,9 in R23C5, R4C5 and R6C5, locked for C5
7j. 22(4) cage = {1489} (only remaining combination), 4,9 locked for N8, clean-up: no 5 in R78C5
7k. R78C6 = [63], clean-up: no 8 in R23C5, no 7 in R7C78, no 5 in R8C9 (step 2b)
7l. Naked pair {59} in R23C5, locked for C5 and N2
7m. Naked triple {678} in 21(3) cage at R1C6, 6,8 locked for C6, clean-up: no 3 in R4C7
7n. Naked triple {578} in 20(3) cage at R7C4, 5,7 locked for C4, clean-up: no 6 in R6C3
7o. 9 in C4 only in R456C4, locked for N5
7p. Naked pair {49} in R7C78, locked for R7 and N9, clean-up: no 2 in R8C9 (step 2b)
7q. R78C9 = [56], clean-up: no 3,4 in R6C2 (step 2e)
7r. 16(4) cage at R6C1 = {1258} (only possible combination, cannot be {1249} because R7C1 doesn’t contain 4,9) -> R7C1 = 8, R8C12 = {25}, locked for R8 and N7
7s. Naked pair {17} in R8C78, locked for R8 and N9
7t. R2C89 + R3C9 (step 1d) = {389/479/578} (cannot be {569} because 5,6 only in R2C8), no 6
[Note. There is now the Unique Rectangle elimination R6C78 cannot be {49} because R7C78 = {49} but I don’t use that type of step since it doesn’t fully solve a puzzle.]

8. 17(3) cage at R3C4 = [197/368/467] (cannot be {269} because no 7,8 in R4C5, cannot be {278} because no 7,8 in R4C4), no 2 in R3C4, clean-up: no 6 in R1C3 (step 3b)
8a. Consider permutations for 17(3) cage
17(3) cage = [179] => R6C5 = 8, R6C6 = 3 (cage sum) => R4C67 = {45} (only remaining combination)
or 17(3) cage = [368/467] => R4C67 = {45} (only remaining combination)
-> R4C67 = {45}, locked for R4
[This proved to be a key step.]
8b. 4 in N4 only in R5C123, locked for R5
8c. 4 in N5 only in R46C6, locked for C6 -> R89C6 = [91], R9C5 = 4, R8C3 = 4, clean-up: no 2 in R3C2
8d. 4 in N4 only in R5C12, locked for 31(6) cage at R2C1, no 4 in R23C1

9. 45 rule on N1 3 innies R1C3 + R23C1 = 17 = {278/359/368} (cannot be {269} because R1C3 only contains 5,7,8)
9a. 5 of {359} must be in R1C3 -> no 5 in R23C1
9b. 8 of {278} must be in R1C3 -> no 7 in R1C3, clean-up: no 3 in R3C4 (step 3b)
9c. 17(3) cage at R3C4 (step 8) = [197/467] -> R4C5 = 7, R6C5 = 8, R6C6 = 3 (cage sum), R5C6 = 5, R4C67 = [45], clean-up: no 6 in R6C2 (step 2e)
9d. Killer pair 1,4 in R3C23 and R3C4, locked for R3, clean-up: no 8 in R3C8
9e. Killer pair 3,7 in R2C89 + R3C9 (step 7t) and R3C78, locked for N3
9f. 1 in N3 only in 16(4) cage at R1C7 (step 5) = {1249/1456} (cannot be {1258} which clashes with R1C3), no 8, 4 locked for N3
9g. 5 of {1456} must be in R1C8 -> no 6 in R1C8
9h. R2C89 + R3C9 (step 7t) = {389/578}
9i. 5 of {578} must be in R2C8 -> no 7 in R2C8

10. 1,5 in R5 only in 22(5) cage at R5C3 = {12568/13567}, no 9, 6 locked for R5
10a. 8 of {12568} only in R5C3 -> no 2 in R5C3

11. R1C3 + R23C1 (step 9) = {278/359/368}
11a. 4 in N4 only in 31(6) cage at R2C1 = {234589/234679}
11b. {234589} only has one of 2,7, {234679} must have one of 2,7 in R6C2 so cannot have both in R23C1 -> R1C3 + R23C1 = {359/368} (cannot be {278} because cannot have both of 2,7 in R23C1), no 2,7, 3 locked for N1 and 31(6) cage, no 3 in R45C1 + R5C2
[Cracked.]
11c. 31(6) cage = {234589/234679}, 2 locked for N4, clean-up: no 9 in R6C4
11d. R4C4 = 9 (hidden single in N5) -> R3C4 = 1 (cage sum), R1C5 = 2, R1C3 = 5 (step 3b), R3C23 = [42], R6C3 = 9 -> R6C4 = 2, clean-up: no 7 in R3C78, no 4 in R6C78
[Removing any need for the Unique Rectangle.]
11e. Naked pair {67} in R6C78, locked for R6 and N6 -> R16C9 = [94]
11f. Naked pair {36} in R3C78, locked for R3 and N3
11g. Naked pair {78} in R23C9, 8 locked for C9 and N3 -> R2C8 = 5
11h. Naked pair {14} in R1C78, locked for R1 and N3 -> R2C7 = 2
11i. R5C4567 = [6153] -> R5C3 = 7 (cage sum)

and the rest is naked singles.

Rating Comment:
I'll rate my WT for A364 at Easy 1.5 because I used a locking-out cages step and a couple of forcing chains. I wasn't sure how to rate my final breakthrough in step 11b, but not any higher.


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