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 Post subject: Assassin 389
PostPosted: Sun Dec 01, 2019 8:12 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 852
Location: Sydney, Australia
Attachment:
a389.jpg
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Note: a number of cells are not covered by cages. It still has just one solution.

My apologies for posting so quickly after a couple of other, non-Assassin puzzles. I'll cut mine back to one a month so I can try some others'.

Assassin 389
I found this quite hard first time through and it resisted to the very end. Found a bit of a shortcut in an optimised solution but still quite long which has become the norm for a decent, juicy Assassin. Like it! It gets a 1.60. JSudoku has a hard time.
triple click code:
3x3::k:2560:2560:5633:0000:0000:5123:7428:2053:2053:0000:5633:5633:5633:5123:7428:7428:7428:3079:0000:0000:0000:0000:6664:5123:7428:7428:3079:4617:4618:2323:0000:6664:6664:6664:4620:3079:4617:4618:2323:0000:0000:6664:3342:4620:4620:4617:4618:2323:0000:0000:7439:3342:3342:4368:2577:2577:2577:0000:7442:7439:7439:7439:4368:3330:2573:2573:0000:7442:7439:6155:6155:6155:3330:7442:7442:7442:7442:6155:6155:1286:1286:
solution:
Code:
+-------+-------+-------+
| 2 8 4 | 6 5 9 | 3 1 7 |
| 5 6 9 | 3 7 1 | 8 2 4 |
| 1 3 7 | 2 8 4 | 9 6 5 |
+-------+-------+-------+
| 8 5 1 | 9 6 7 | 2 4 3 |
| 7 9 2 | 4 1 3 | 5 8 6 |
| 3 4 6 | 8 2 5 | 1 7 9 |
+-------+-------+-------+
| 6 1 3 | 5 4 2 | 7 9 8 |
| 9 2 8 | 7 3 6 | 4 5 1 |
| 4 7 5 | 1 9 8 | 6 3 2 |
+-------+-------+-------+


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 Post subject: Re: Assassin 389
PostPosted: Tue Dec 10, 2019 6:49 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 852
Location: Sydney, Australia
Here's how I did this one. Many key steps (12, 18, 22, 27). Some are a little complicated but hopefully easy enough to follow. Thanks to Andrew for some corrections.
a389 WT:
Preliminaries
Cage 17(2) n69 - cells ={89}
Cage 5(2) n9 - cells only uses 1234
Cage 8(2) n3 - cells do not use 489
Cage 13(2) n7 - cells do not use 123
Cage 10(2) n1 - cells do not use 5
Cage 10(2) n7 - cells do not use 5
Cage 9(3) n4 - cells do not use 789
Cage 10(3) n7 - cells do not use 89
Cage 20(3) n2 - cells do not use 12

No clean-up done unless stated.
1. 17(2)r6c9 = {89}: both locked for c9

2. 24(5)r8c7 sees all of the 5(2)n9 -> must have exactly one of {14/23} to avoid a clash
2a. = {14568/23568}(no 7,9)

3. 7 and 9 in n9 only in r7: both locked for r7, 7 also locked for 29(5) cage

4. "45" on r89: 1 outie r7c5 + 9 = 2 innies r8c46
4a. since the innie is in the same n8 -> the iod of 9 cannot be in the two innies (IOU) -> no 9 in r8c46

5. "45" on n7: 4 outies r789c5 + r9c4 = 17 and must have 9 for n8 = {1259/1349}(no 6,7,8)
5a. no 9 in r9c23

6. hidden single 7 in n8 -> r8c4 = 7
6a. no 3 in 10(2)n7
6b. no 6 in r9c1

7. "45" on n7: 2 innies r9c23 = 12 = {48/57}(no 1,2,3,6)
7a. -> with step 5, 29(6)r7c5 = {1259-48/1349-57}
7b. must have 4 & 5 -> no 4,5 in r9c6 since it sees all that cage (Common Peer Elimination CPE)

8. 3 in n7 only in r7: locked for r7
8a. -> 10(3) = {136/235}(no 4)

9. killer single 4 in 24(5)r8c7 and 5(2)n9: 4 locked for n9

10. 4 in r7 only in n8: locked for n8

11. "45" on r89: 1 outie r7c5 + 2 = 1 innie r8c6 = [13/46]

Very important step
12. from step 7a, 29(6)r7c5 = {1259-48/1349-57}
12a. {1259-48} must have [12] in r78c5 to avoid {24} in r9 clashing with 5(2)n9
12b. {1349-57} must have 4 in r7c5 since 1 in r7c5 would force 3 into r8c6: ie, two 3 in n8
12c. [4]{139-57} must have 9 in r9c45 to avoid {13} clashing with 5(2)n9
12d. -> r8c5 = (123)
12e. 5 & 9 must be in 29(6): are only in r9: locked for r9

13. 13(2)n7: {58} blocked by r9c23 = 5 or 8
13a. = [94/67]

14. killer pair 4,7 in r9c123: 4 locked for r9 and n7

15. 5(2)n9 = {23}: both locked for r9 and n9

16. 6 in r9 only in r9c67: locked for 24(5) cage

17. "45" on n3: 2 outies r2c6 + r4c9 = 4 = {13/22}

Another key step
18. "45" on n3: 1 outie r2c6 + 8 = 2 innies r23c9
18a. no 8 in r23c9 -> r2c6 <> r3c9
18b. -> r2c6 repeats in n3 only in 8(2) cage
18c. -> from r2c6 + 8 = r23c9 = [1]{36/45}/[2]{37}/[3]{47}
18d. but {36} blocked from r23c9 by 12(3) can't be {363}
18e. {37}[2] in r234c9 blocked by r9c9 = (23)
18f. -> r2c6 + r23c9 = [1]{45}/[3]{47}(no 2 in r2c6)
18g. must have 4: locked for n3 and c9
18h. no 2 in r4c9 (outiesn3=4)
18i. r23c9 = {457}
18j. no 2 in r2c6 -> no 2,6 in 8(2)n3 (from step 18b)

19. 8 & 9 in n3 only in 29(6) = {123689}(no 5,7)

20. deleted

21. hidden single 6 in c9 -> r5c9 = 6
21a.hidden single 2 in c9 -> r9c89 = [32]
21b. r45c8 = 12 = {48/57}(no 1,2,9)
21c. no 5 in r1c9

The VIP step
22. "45" on n6: 3 innies r4c79 + r6c9 = 14 = [419/518/239](r4c7 = (245))
22a. consider sp12(2)r45c8
i. {48} in r45c8 must have 3 in r4c9
ii. or {57} in r45c8 -> r1c89 = [17] -> 12(3)r2c9 = {45}[3]
22b. both ways have 3 in r4c9 -> r4c9 = 3
22c. -> r23c9 = {45}: 5 locked for c9 and n3
22d. and r2c6 = 1 (outiesn3=4)
22e. r8c9 = 1, r1c89 = [17]
22f. and r4c7 + r6c9 = 11 = [29], r7c9 = 8

23. 10(2)n7 = {28} only: both locked for n7 and r8
23a. r8c56 = [36], -> r7c5 = 4 (iodr89=+2)
23b. r9c67 = [86]
23c. r89c1 = [94]
23d. 10(3)n7 = {136}: 1 locked for r7

24. naked pair {25} in r7c46: 5 locked for n8 and r7
24a. -> r7c78 = {79} = 16 -> r67c6 = 7 = {25} only: 5 locked for c6

25. 18(3)r4c1 = {378/567}(no 1,2) = 6 or 8
25a. must have 7: locked for c1 and n4

26. 2 in c1 only in n1: locked for n1
26a. 10(2)n1 = [28/64]

Final cracker
27. 18(3)r4c1 = {378/567} = 6 or 8
27a. -> killer single 8 with 10(2)n1
27b. -> no 8 in r23c1 nor r456c2
27c. -> 8 in c1 only in 18(3)r4c1 = {378} only: 3 locked for n4 and c1

28. 9(3)n4 = {126} only: all locked for n4, 1,6 for c3
28a. r456c2 = {459}: all locked for c2
28b. r1c12 = [28], r9c23 = [75], r8c23 = [28], r7c3 = 3

29. 22(4)r1c3 must have two of {479} for r12c3 and one of (36) for r2c2
29a. = {3469/3478/4567}
29b. but {4567} blocked by r2c1 = (56)
29c. = {3469/3478}(no 2,5)
29d. must have 3: locked for r2
29e. -> r2c24 = {36}/[38]
29f. 4 locked for n1

30. hidden single 2 in n2 -> r3c4 = 2
30a. r7c46 = [52], r6c26 = [45]
30b. r23c8 = [26](both hidden singles c8)

31. hidden single 5 in r1 -> r1c5 = 5
31a. same on 6 -> r1c4 = 6

32. 20(3)n2: {389} blocked by r2c4 = (38)
32a. = {479} only: 7 & 9 locked for n2, 4 for c6
32b. r3c5 = 8

easy now. A couple of cage sums needed
Merry Christmas!
Ed


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 Post subject: Re: Assassin 389
PostPosted: Fri Dec 13, 2019 3:17 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1718
Location: Lethbridge, Alberta, Canada
Thanks Ed for this Assassin. Many of my steps were very similar to Ed's steps, so it would appear that this puzzle has a fairly narrow solving path; that may because there are so many blank cells. Maybe Ed didn't realise that his step 4, my step 6, was also a key step.

Here is my walkthrough for Assassin 389:
Prelims

a) R1C12 = {19/28/37/46}, no 5
b) R1C89 = {17/26/35}, no 4,8,9
c) R67C9 = {89}
d) R89C1 = {49/58/67}, no 1,2,3
e) R8C23 = {19/28/37/46}, no 5
f) R9C89 = {14/23}
g) 20(3) cage at R1C6 = {389/479/569/578}, no 1,2
h) 9(3) cage at R4C3 = {126/135/234}, no 7,8,9
i) 10(3) cage at R7C1 = {127/136/145/235}, no 8,9

1. Naked pair {89} in R67C9, locked for C9

2. 45 rule on N7 2 innies R9C23 = 12 = {39/48/57}, no 1,2,6

3. 45 rule on N6 3 innies R4C79 + R6C9 = 14
3a. Min R6C9 = 8 -> max R4C79 = 6, no 6,7,8,9 in R4C79
3b. 18(3) cage at R4C8 = {279/369/378/459/468/567} (cannot be {189} which clashes with R6C9), no 1

4. 45 rule on N3 2(1+1) outies R2C6 + R4C9 = 4 = [13/22/31]
4a. R4C79 + R6C9 = 14 (step 3)
4b. Max R46C9 = 12 -> min R4C7 = 2
4c. 12(3) cage at R2C9 = {147/156/237/246/345}
4d. 1 of {147/156} must be in R4C9 -> no 1 in R23C9

5. 45 rule on C6789 3 outies R234C5 = 21 = {489/579/678}, no 1,2,3

6. 45 rule on R89 2 innies R8C46 = 1 outie R7C5 + 9 -> no 9 in R8C46 (IOU)
6a. Max R8C46 = 15 -> max R7C5 = 6
6b. Min R8C46 = 10, no 9 in R8C46 -> no 1 in R8C46

7. 24(5) cage at R8C7 = {14568/23568} (cannot be {12489/12579/12678/13479/13569/13578/23469/23478/24567} which clash with R9C89), no 7,9
[Alternatively, and simpler, 24(5) cage and R9C89 ‘see’ each other so form combined 29(7) cage = {1234568}, no 7,9.]
7a. 7,9 in N9 only in R7C789, locked for R7, 7 locked for 29(5) cage at R6C6
7b. Hidden killer pair 1,2 in 10(3) cage at R7C1 and R8C23, 10(3) cage contains one of 1,2 -> R8C23 must contain one of 1,2 = {19/28}, no 3,4,6,7
7c. 7 in N7 only in R89C1 = {67} or R9C23 = {57} -> R89C1 = {49/67} (cannot be {58}, locking-out pairs)
7d. 18(3) cage at R4C1 = {189/378/459/567} (cannot be {279/369/468} which clash with R89C1), no 2
7e. 9(3) cage at R4C3 = {126/234} (cannot be {135} which clashes with 18(3) cage at R4C1), no 5, 2 locked for C3 and N4, clean-up: no 8 in R8C2
7f. 18(3) cage at R4C2 = {189/378/459/567} (cannot be {369/468} which clash with 9(3) cage)

8. R1C89 = {17/26/35}, 12(3) cage at R2C9 (step 4c) = {147/156/237/246/345}
8a. Consider permutations for R2C6 + R4C9 (step 4) = [13/22/31]
R2C6 + R4C9 = [13] => 1 in N3 only in R1C89 = {17} => 12(3) cage = {45}3
or R2C6 + R4C9 = [22] => 2 in N3 only in R1C89 = [26] => 12(3) cage = {37}2
or R2C6 + R4C9 = [31] => 3 in N3 only in R1C89 = {35} => 12(3) cage = {47}1
-> R1C89 = {17/35}/[26], no 6 in R1C8, no 2 in R1C9
and R23C9 = {37/45/47}, no 2,6
8b. 12(3) cage = {147/237/345}
8c. 8,9 in N3 only in 29(6) cage at R1C7 = {123689/124589}, no 7

[I ought to have spotted this immediately after step 7. Step 6 is clearly a key step.]
9. 9 in N8 only in R8C5 + R9C56, locked for 29(6) cage at R7C5, clean-up: no 3 in R9C23 (step 2)
9a. Killer pair 4,7 in R89C1 and R9C23, 4 locked for N7
9b. 3 in N7 only in 10(3) cage at R7C1, locked for R7
9c. 3 in N9 only in R89C789, CPE no 3 in R9C6
9d. 29(6) cage at R7C5 contains 9 = {124589/134579} (cannot be {123689/124679/234569} because R9C23 only contains {48/57}), 1 locked for N8
9e. 29(6) cage = {124589/134579}, CPE no 4,5 in R9C6
9f. 29(6) cage = {124589/134579} -> R78C5 + R9C45 = {1259/1349}, no 6,7,8
9g. R8C4 = 7 (hidden single in N8), clean-up: no 6 in R9C1
9h. Combined 29(7) cage R8C789 + R9C6789 = {1234568}, 1,4,5 locked for N9
9i. 4 in R7 only in R7C456, locked for N8
9j. 6 in R9 only in R9C67, locked for 24(5) cage at R8C7

10. 8 in N7 only in R8C3 + R9C23 -> killer X-Wing for 8 in R8C3 + R9C23 and 24(5) cage at R8C7, no other 8 in R89
[Then I found that step wasn’t necessary because of …]
10a. R8C46 = R7C5 + 9 (step 6), R8C4 = 7 -> R8C6 = R7C5 + 2 -> R7C5 = {14}, R8C6 = {36}
10b. 29(5) cage at R6C6 contains 7 for R7C78 = {23789/25679/34679/35678} (cannot be {14789} because R8C6 only contains 3,6), no 1

11. Consider combinations for 29(6) cage at R7C5 (step 9f) = {124589/134579}
29(6) cage = {124589} => R9C23 = {48} => R8C23 = {19}, 9 locked for R8
or 29(6) cage = {134579} => R9C23 = {57}, R7C5 = 4, R8C5 + R9C45 = 1{39}/3{19} (cannot be 9{13} which clashes with R9C89)
-> no 9 in R8C5
11a. 9 in R8 only in R8C123, locked for N7, clean-up: no 4 in R8C1
11b. 4 in N7 only in R9C123, locked for R9, clean-up: no 1 in R9C89
11c. Naked pair {23} in R9C89, locked for R9 and N9
11d. 29(6) cage = {124589/134579}
11e. 2 of {124589} must be in R8C5, 5,7 of {134579} must be in R9C23 -> no 5 in R8C5
11f. 29(6) cage = {124589/134579}, 5 locked for R9

12. 12(3) cage at R2C9 (step 8b) = {147/345} (cannot be {237} which clashes with R9C9), no 2, 4 locked for C9 and N3
12a. 3 of {345} must be in R4C9 -> no 3 in R23C9
12b. Killer pair 1,5 in 12(3) cage at R8C9, locked for C9, clean-up: no 3,7 in R1C8
12c. R2C6 + R4C9 (step 4) = {13}, no 2
12d. Naked pair {13} in R2C6 + R4C9, CPE no 1,3 in R4C6
12e. 29(6) cage at R1C7 (step 8c) = {123689/124589}, 2 locked for N3, clean-up: no 6 in R1C9

[Reworked from here. First time through I carelessly overlooked {57} for R45C8.]
13a. R5C9 = 6 (hidden single in C9) -> R45C8 = 12 = {39/48/57}
13b. R9C9 = 2 (hidden single in C9) -> R9C8 = 3, clean-up: no 9 in R45C8
13c. Consider combinations for R45C8 = {48/57}
R45C8 = {48}, locked for C8 => R8C89 = {15}
or R45C8 = {57}, 5 locked for C8 => R1C89 = [17], R23C9 = {45}, 5 locked for C9 => R8C9 = 1
-> 1 in R8C89, 1 locked for R8 and N9, clean-up: no 9 in R8C23
13d. R8C23 = [28], R8C56 = [36], R8C1 = 9 -> R9C1 = 4, R9C67 = [86], R7C5 = 4 (step 10a), clean-up: no 8 in R1C1, no 1,6 in R1C2
13e. Naked pair {25} in R7C46, 5 locked for R7 and N8
13f. Killer triple 1,4,5 in R1C8, R45C8 and R8C8, locked for C8

14a. 18(3) cage at R4C1 (step 7d) = {378/567}, no 1, 7 locked for C1 and N4, clean-up: no 3 in R1C2
14b. Killer pair 3,6 in 18(3) cage at R4C1 and 9(3) cage at R4C3 (step 7e), locked for N4
14c. 18(3) cage at R4C2 (step 7f) = {189/459}, 9 locked for C2, clean-up: no 1 in R1C1
14d. R1C12 = [28/64] (cannot be [37] which clashes with R1C9)
14e. Killer pair 4,8 in R1C2 and 18(3) cage at R4C2, locked for C2

15a. R234C5 (step 5) = {579/678}, 7 locked for C5
15b. 7 in N25 only in 20(3) cage at R1C6 and 26(5) cage at R3C5 -> both must contain 7
15c. 20(3) cage at R1C6 = {479/578}, no 3,6, 7 locked for N2
15d. 8 of {578} must be in R2C5 -> no 5 in R2C5

16a. 18(3) cage at R4C1 (step 14a) = {378/567}
16b. Consider permutations for R1C12 = [28/64]
R1C12 = [28] => 8 in N4 only in 18(3) cage at R4C1 = {378}
or R1C12 = [64] => 18(3) cage at R4C1 = {378}
-> 18(3) cage at R4C1 = {378}, 3,8 locked for C1 and N4
16c. 18(3) cage at R4C2 (step 14c) = {459} (only remaining combination), 4,5 locked for C2, 4 locked for N4, R1C2 = 8 -> R1C1 = 2, R9C23 = [75]
16d. Naked triple {126} in 9(3) cage at R4C3, 1,6 locked for C3 -> R7C3 = 3
16e. 6 in R1 only in R1C45, locked for N2

17. 29(5) cage at R6C6 must contain 6,7 = {25679/35678} (cannot be {34679} because R7C6 only contains 2,5), no 4, 5 locked for C6
17a. 9 of {25679} must be in R7C78 -> no 9 in R6C6
17b. 20(3) cage at R1C6 = {479} (only remaining combination, cannot be {578} because 5,8 only in R2C5), 4,9 locked for N2, 4 locked for C6

18a. R4C79 + R6C9 = 14 (step 3)
18b. R4C79 + R6C9 = [239/419] (cannot be [518] which clashes with R45C8) -> R4C7 = {24}, R6C9 = 9 -> R7C9 = 8
18c. R7C78 = {79} = 16, R8C6 = 6 -> R67C6 = 7 = {25}, 2 locked for C6
18d. R25C6 = {13} (hidden pair in C6)
18e. 2 in N2 only in R23C4, locked for C4 -> R7C4 = 5, R67C6 = [52], R6C2 = 4

19a. 7 in N5 only in R4C56, locked for R4, clean-up: no 5 in R5C8 (step 13c)
19b. 26(5) cage at R3C5 must contain 7 = {14579/14678/23579/23678} (cannot be {13679} because 1,3 only in R5C6, cannot be {24578} because 2,4 only in R4C7)
19c. R4C7 = {24}, R5C6 = {13} -> R3C5 + R4C56 = {579/678}
19d. R3C5 = {58} -> no 8 in R4C5

20. R12C3 = {479} -> 22(4) cage at R1C3 R12C3 + R2C24 = {47}[38]/{49}[18]/{49}[63] (R12C3 cannot be {79} because R2C24 cannot total 6), 4 locked for N1, R2C4 = {38}
20a. Killer pair 1,3 in R2C24 and R2C6, locked for R2
20b. R3C4 = 2 (hidden single in N2)
20c. 1 in N2 only in R1C45 + R2C6, CPE no 1 in R1C7

[Repeat the forcing chain in step 13 now that I’ve reduced R4C79 + R6C9.]
21. R4C79 + R6C9 (step 18b) = [239/419]
21a. Consider combinations for R45C8 (step 13c) = {48}/[57]
R45C8 = {48}, locked for N4 => R4C79 = [23], R1C9 = 7 => R1C8 = 1
or R45C8 = [57] => R1C8 = 1
-> R1C89 = [17]
21b. Naked pair {45} in R23C9, 5 locked for C9 and N3 -> R48C9 = [31], R2C6 = 1 (step 4 or hidden single in N2), R5C6 = 3, R4C7 = 2 (step 18b)
21c. R1C457 = [653] (hidden triple in R1), R3C5 = 8 -> R4C56 = [67] (step 19c)
21d. R237C7 = [897] -> R6C7 = 1, R6C45 = [82], R6C8 = 7 -> R5C7 = 5 (cage sum)

and the rest is naked singles.

Rating Comment:
I'll rate my WT for A389 at Easy 1.5. I used some short forcing chains and the same IOU that Ed used.


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 Post subject: Re: Assassin 389
PostPosted: Fri Dec 20, 2019 4:41 am 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 208
Location: California, out of London
Excellent puzzle. Thanks Ed :applause:
We all started off in very similar ways - but after that although we had equivalent steps we did them in very different orders.
For example - the resolution of Innies n6 which in Ed's WT was his VIP Step 22 (very nice!) was my penultimate step 17.
Merry Christmas all!
Corrections & clarifications thanks to Ed and Andrew.

Assassin 389 WT:
1. 17(2)r6c9 = {89}
All cells in 24(5)n9 and 5(2)n9 see each other.
-> Between them they are the values (1234568)
-> (79) in n9 in r7c789

2. Innies n7 = r9c23 = +12(2)
-> Outies n7 = that part of 29(6) in n8 = +17(4)
IOD r89 -> r8c4 + r8c6 = r7c5 + 9
-> 9 not in r8c46
-> 9 in n8 in r8c5 or r9c45
Since Outies n7 = +17(4) -> HS 7 in n8 -> r8c4 = 7

3. 29(6)n78 = +12(2) in n7 and +17(4) in n8
(A) In n7 = {48}, n8 = {1259}
(B) In n7 = {57}, n8 = {1349}

If (A) this puts 5(2)n9 = {23}

If (B) this puts 10(3)n7 = {136} which puts Outies n7 = [4{139}]
Also puts 13(2)n7 = {49}
-> At least one of (14) in r9 in r9c145
-> (again) 5(2)n9 = {23}

Either way 5(2)n9 = {23}
-> 24(5)n9 = {14568}

4. 3 in n7 only in 10(3)
-> 3 in n8 only in r8c56

5. Outies n3 = r2c6 + r4c9 = +4(2) = [13], [22], or [31]
Since 8 already in c9 -> 12(3)c9 cannot have two of its values sum to 4.
-> Whichever of (123) is in r2c6 cannot go in 12(3)c9 -> must go in 8(2)n3
-> Outies n3 cannot be [22] since that leaves no place for 2 in c7
-> Outies n3 = {13}
-> 8(2)n3 = {17} or {35}

6! Innies n6 = r4c79,r6c9 = +14(3) = [518], [419], or [239]
-> The only way r2c6 can be 1 is if 17(2)c9 = [98] which puts r9c6 = 8
-> Either r2c6 = 3 or r9c6 = 8 or both.

7. Consider Step 3 Option (A): 29(6) in n7 = {48} and in n8 = {1259}
This puts r8c6 = 3 and r9c6 as not-8 which contradicts Step 6.
-> r9c23 = {57}

8. -> 10(3)n7 = {136}
-> Outies n7 = [43{19}]
-> 13(2)n7 = [94]
-> 10(2)n7 = {28}
Also 6 in n9 in 24(5) -> (HS 6 in n8) r8c6 = 6
-> r9c7 = 6
Also HP r7c46 = {25}
-> r9c6 = 8
-> r8c789 = {145}
-> r7c789 = {789}

9. Trying r7c78 = {78} puts innies n6 = [518] and r2c6 = 3 which leaves no solution for 29(5)r6c6.
-> r7c78 = {79}, 17(2)c9 = [98], r67c6 = {25}
Also r4c79 = [23] or [41]

10. 9 in n4 only in 18(3)r4c2
-> 9 in n1 in r123c3
-> 10(2)n1 not {19}

4 already in c1 -> 10(2)n1 not [46]
Either 4 in 18(3)r4c2 -> 10(2)n1 not [64]
or 4 in 9(3)n4 (= {234}) -> 18(3)r4c1 = {567} -> 10(2)n1 not [64]

(Step 5) 8(2)n3 either {17} or {35} -> 10(2)n1 not {37}

-> 10(2)n1 = {28}

11. -> r18c2 = {28}
-> 8 in n4 only in c1
-> 10(2)n1 = [28]
-> 10(2)n8 = [28]
-> (12) in n4 in c3
-> 9(3)n4 = {126}
-> 18(3)r4c2 = {459}
-> 18(3)r4c1 = {378}
-> 10(3)n7 = [{16}3]
Also r9c23 = [75]
-> r123c3 = {479}
Also 5 in n1 in r23c1
Also 3 in n1 in r23c2

12. Outies c6789 = r234c5 = +21(3)
-> 2 in c5 only in r56c5
-> r67c6 = [52]
-> r7c4 = 5
-> 5 in n2 in r123c5
Also 2 in n2 in r23c4
Also 18(3)r4c2 = [{59}4]

13. Possibilities for n3
(A) Outies n3 = [13], 8(2)n3 = {17}, 12(3)c9 = [{45}3]
(B) Outies n3 = [31], 8(2)n3 = {35}, 12(3)c9 = [{47}1]
-> 4 in n3 in r23c9
-> 29(6)n3 = {123689}
-> 6 in n3 in r23c8
-> HS 6 in c9 -> r5c9 = 6
-> HS 2 in c9 -> 5(2)n9 = [32]

14. Outies c6789 = r234c5 = +21(3) = {678} or {579}
7 cannot go in r3c5 since that leaves no place for 7 in c6
6 in r1 only in r1c45
-> 8 not in r2c5 since r234c5 cannot be [876] or [867]

15. -> 20(3)n2 = {479} with 4 in r13c6. (Only remaining possibility).
-> Whichever of (13) is in r4c9 goes in c6 in r5c6
-> Whichever of (79) is in r2c5 goes in c6 in r4c6
-> 7 in n5 in r4c56

16. r123c3 = {479} -> r2c4 cannot be 2
-> (HS 2 in n2) r3c4 = 2

17! Trying r4c79 = [41] puts r5c6 = 1 puts 9(3)n4 = [621]
This leaves no place for 2 in r4
-> r4c79 = [23]

18. -> r25c6 = [13]
Also r23c9 = {45} and 8(2)n3 = {17}
-> (HS 5 in r1) r1c5 = 5
-> r1c4 = 6 and r234c5 = [786] and r4c6 = 7
-> r13c6 = {49} and r2c4 = 3
Also r3c3 = 7
-> r12c3 = {49} and r2c2 = 6
Also 18(3)r4c1 = [873] and 9(3)n4 = [126]
etc.


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 Post subject: Re: Assassin 389
PostPosted: Tue Dec 31, 2019 12:45 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 852
Location: Sydney, Australia
A389 felt like quite a milestone puzzle since it resisted for so long. The fact that all three WTs use all the same but many areas of the puzzle in the solutions, does indeed suggest a very narrow solving path. However, it doesn't make for a longest average WT, A382 gets that honour.

Decided to try and compare A389 to the recent Assassins so made up this table. I found a different solver order in Jsudoku which means it uses chains far less (only A383 in this list below) so not a very discriminating criteria. Used 'complex intersections' instead which includes the sort of sneaky innie/outie difference blocks we humans like to use. I usually look at the 'complex interesection' part of the summary log when I'm trying to decide if JSudoku is suggesting a possible interesting Assassin.

Anyway, make of this table what you will. Nothing leaps out but found it interesting anyway. Note: the JS column is v1.5b2, using a non-default solver order, and the number refers to the number of 'complex intersections' routine

+-----------------------------------+----------+------------+-------+----------+---------------+-------+--------------+
| Puzzle WT sub-steps #| Andrew | wellbeback | Ed | Average | Andrew rating | Score | JSudoku |
+-----------------------------------+----------+------------+-------+----------+---------------+-------+--------------+
| Assassin 380 | 104 | 85 (!!)| 76 | 88 | H1.50 | 1.90 | 3 |
| Assassin 381 | 76 | 63 (!!)| 89 | 76 | 1.50 | 1.60 | 0 |
| Assassin 382 | 106 | 105 (!!!!)| 101 | 104 | 1.50 | 1.75 | 3 |
| Assassin 383 | 97 | 86 (!!!!)| 107 | 97 | H1.50 | 2.00 | 4 + 1 chain|
| Assassin 384 | 88 | 46 (!!!)| 79 | 71 | H1.50 | 1.95 | 4 |
| Assassin 385 | 91 | 78 (!)| 58 | 76 | 1.50 | 2.10 | 4 |
| Assassin 386 | 105 | 67 (!!!)| 72 | 81 | 1.50 | 1.45 | 6 |
| Assassin 387 | 77 | 75 ()| 81 | 78 | H1.25 | 1.55 | 5 |
| Assassin 388 | 99 | no WT | 64 | 82 | 1.50 | 1.55 | 5 |
| Assassin 389 | 110 | 104 (!!)| 89 | 101 | E1.50 | 1.60 | 2 |
+-----------------------------------+----------+------------+-------+----------+---------------+-------+--------------+

Cheers
Ed


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 Post subject: Re: Assassin 389
PostPosted: Tue Dec 31, 2019 6:29 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1718
Location: Lethbridge, Alberta, Canada
An interesting post by Ed. I thought that he'd stopped looking at my ratings! ;)

Any puzzle which I rate H1.25 or higher is definitely an Assassin! These days my ratings are mostly in the 1.5 range.

I'd better make a New Year's Resolution to update my tables in the Archive post. However full archives won't be extended beyond their current state, since all the posts are on this forum.

Note that one reason why my WTs often have more steps is that I don't try to optimise them although I do re-work for errors and occasionally when I spot something which I really ought to have seen earlier.

wellbeback's solving approach tends to give shorter WTs!

Happy New Year to all!

Andrew


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