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Assassin 395
http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1556
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Author:  Ed [ Fri May 01, 2020 6:18 am ]
Post subject:  Assassin 395

Attachment:
a395.JPG
a395.JPG [ 99.98 KiB | Viewed 1268 times ]
Note: it has a disjoint 30(5) at r2c5689 + r3c9

Assassin 395

Really happy with my solution for this puzzle. Many enjoyable steps but of course, some are not easy to find. Quite a short walk-through with my first key step available very early. SudokuSolver gives it 1.75 and JSudoku has a very hard time.
triple click code:
3x3::k:5376:5376:3585:3585:3585:3585:5378:5378:5378:5376:2307:1540:1540:7685:7685:5378:7685:7685:3334:2307:7943:7943:5128:5128:5128:5128:7685:3334:3334:7943:7943:3081:3081:4618:4618:4618:2571:2571:7943:6924:6924:3081:4618:4109:4109:3086:3086:7943:6924:3087:3087:3600:4109:2833:3086:3090:3090:6924:4883:3087:3600:3600:2833:1812:9493:9493:9493:4883:4883:2582:2582:2071:1812:9493:9493:9493:9493:3864:3864:3864:2071:
solution:
Code:
+-------+-------+-------+
| 9 4 2 | 1 3 8 | 7 5 6 |
| 8 6 1 | 5 4 7 | 3 2 9 |
| 5 3 7 | 2 9 6 | 1 4 8 |
+-------+-------+-------+
| 7 1 9 | 6 5 4 | 8 3 2 |
| 2 8 4 | 9 7 3 | 5 6 1 |
| 6 5 3 | 8 2 1 | 4 9 7 |
+-------+-------+-------+
| 1 7 5 | 3 6 9 | 2 8 4 |
| 4 2 6 | 7 8 5 | 9 1 3 |
| 3 9 8 | 4 1 2 | 6 7 5 |
+-------+-------+-------+
Cheers
Ed

Author:  wellbeback [ Sun May 03, 2020 6:50 pm ]
Post subject:  Re: Assassin 395

Thanks Ed. I enjoyed this puzzle a lot. I should think my early key step is the same as yours which gave it a good start, but, as you say, there was still some work to do after that which involved further interesting steps. Hope you're keeping safe & well!
Assassin 395 WT:
1. Outies r89 = r7c5 = 6
Outies n69 = r9c7 = 2
Remaining innies n578 = r4c4 + r7c1 = +7(2)
-> r7c1 is Max 5

2! 37(7)n78 = (489) + two of the pairs (17), (26), (35)
-> (26) are both in 37(7) or both not in
If (26) not in 37(7) this puts 7(2)n7 = {16} and r7c1 = 2
But this puts r4c4 = 5 which leaves no place for 5 in n4
-> (26) both in 37(7) in r89c23

3. -> 7(2)n7 = {34}
-> 12(2)n7 = {57}
-> r7c1 = 1 and r4c4 = 6
-> r89c23 = {2689}
Also 1 in n8 in 37(7)
-> 27(7) = [(2689}{147}]
-> 19(3)n8 = [6{58}]
-> r7c46 = {39}
-> r7c789 = {248}
-> 15(3)r9 = [2{67}]
-> 10(2)n9 = {19}
-> 8(2)n9 = [35]
-> 7(2)n7 = [43]
-> r8c23 = {26} and r9c23 = {89}
Also r8c4 = 7 and r9c45 = {14}

4. Innies c12 = r789c2 = +18(3)
This can only be r789c2 = [729]
-> r789c3 = [568]

5. r7c46 = {39}
Either:
a) r7c46 = [39] -> r6c56 = [21], 12(3)r4c5 = {345}, 27(4) = [<879>}3]
b) r7c46 = [93] -> 12(3)r4c5 = [2{19}], 12(3)r6c5 = [{45}3], 27(4) = [<873>9]

Either way -> r5c5 = 7 and r567c4 = {389}

6. 11(2)r6c9 either [74] or [92]
-> Innies n6 = r6c79 = +11(2) either [47] or [29]
But the latter case puts r6c56 = {45} and r6c12 = [83] which leaves no value for r6c4
-> 11(2)r6c9 = [74]
-> 14(3)r6c7 = [4{28}]
-> 12(3)r6c5 = [219]
-> 12(3)r4c5 = {345}
-> r567c4 = [{89}3]

7. Outies n3 = r23c56 = +26(4)
Since (25) in n2 in r123c4 -> Outies n3 = {3689} or {4679}

8! Outies r12 = r3c29 = +11(2)
Given previous placements this can only be [38] or [56]
But since r3c9 'sees' all outies n3 -> r3c9 cannot be 6
-> r3c29 = [38] and outies n3 = {4679}
-> r2c2 = 6
-> r23c6 = [76] and r23c5 = {49}
-> 14(4)r1 = [{12}{38}]
-> r123c4 = {125}
-> r9c45 = [41]

9. HS 8 in r2 -> 21(3)n1 can only be [948]
-> 21(4)n3 = [{57}63]
Also HS 5 in n1 -> 13(3)r3c1 = [571]
Also 4 in n4 in r45c4
-> HS 6 in n4 -> r6c12 = [65]
-> 10(2)n4 = [28]
-> r456c3 = {349}
-> r3c34 = [72]
-> r1c34 = [21]
-> r2c34 = [15]

10. Also r56c4 = [98]
-> 8 in n6 in r4c78
-> HS 9 in n6 -> r6c8 = 9
-> r456c3 = [943]
etc.

Author:  Andrew [ Wed May 06, 2020 10:46 pm ]
Post subject:  Re: Assassin 395

Well done wellbeback for spotting that early key step! I didn't so I found it a much harder puzzle.

Here is my walkthrough for Assassin 395:
Disjoint 30(5) cage R2C5689 + R3C9

Prelims

a) R23C2 = {18/27/36/45}, no 9
b) R2C34 = {15/24}
c) R5C12 = {19/28/37/46}, no 5
d) R67C9 = {29/38/47/56}, no 1
e) R7C23 = {39/48/57}, no 1,2,6
f) R89C1 = {16/25/34}, no 7,8,9
g) R8C78 = {19/28/37/46}, no 5
h) R89C9 = {17/26/35}, no 4,8,9
j) 21(3) cage at R1C1 = {489/579/678}, no 1,2,3
j) 19(3) cage at R7C5 = {289/379/469/478/568}, no 1
k) 14(4) cage at R1C3 = {1238/1247/1256/1346/2345}, no 9
l) 27(4) cage at R5C4 = {3789/4689/5679}, no 1,2

1a. 45 rule on R89 1 outie R7C5 = 6, clean-up: no 5 in R6C9
1b. R7C5 = 6 -> R8C56 = 13 = {49/58}
1c. 27(4) cage at R5C4 = {3789/4689/5679}, CPE no 9 in R4C4

2a. 45 rule on R1 2 outies R2C17 = 11 = {47/56}/[83/92], no 1,8,9 in R2C7
2b. 45 rule on R12 2 outies R3C29 = 11 = [29]/{38/47/56}, no 1 in R3C2, no 1,2 in R3C9, clean-up: no 8 in R2C2
2c. 45 rule on N6 2 innies R6C79 = 11 = {29/38/47}/[56], no 1,6 in R6C7
2d. 45 rule on N69 1 outie R9C6 =2, clean-up: no 5 in R8C1, no 6 in R8C9
2e. R9C6 = 2 -> R9C78 = 13 = {49/58/67}, no 1,3

3a. 45 rule on N3 4 outies R23C56 = 26 = {2789/3689/4589/4679/5678}, no 1
3b. 45 rule on N23 2 outies R12C3 = 1 innie R3C4 + 1
3c. Min R12C3 = 3 -> min R3C4 = 2

4. 45 rule on N1234 2(1+1) outies R4C4 + R7C1 = 7 = {25/34}/[61], no 7,8,9, no 1 in R4C4
[I missed R4C4 + R7C1 cannot be [52] when R4C4 ‘sees’ all remaining 5s in N4. This forces 37(7) cage at R8C2 to contain both of 2,6 in N7 and leads to a much simpler solution.]

5a. 37(7) cage at R8C2 = {1246789/1345789/2345689} must contain 4,8,9
5b. R8C4 + R9C45 cannot contain both of 4,8 (which clashes with R8C56) -> R89C23 must contain at least one of 4,8 -> R7C23 = {39/57} (cannot be {48} which clashes with R89C23), no 4,8
5c. 8 in N7 only in R89C23, locked for 37(7) cage
5d. 45 rule on N9 3 remaining innies R7C789 = 14 = {149/158/239/248} (cannot be {257} which clashes with R89C9), cannot be {347} which clashes with R7C23), no 7, clean-up: no 4 in R6C9, no 7 in R6C7 (step 2c)
[First time through I then made a careless elimination which wasn’t valid, so have re-worked with a heavier step to reach the same result.]
5e. Consider combinations for R89C9 = {17/35}/[26]
R89C9 = {17/35} => R7C789 = 14 = {149/239/248} (cannot be {158} which clashes with R89C9)
or R89C9 = [26] => 37(7) cage at R8C2 = {1345789} (only remaining combination), R78C1 = [26] (hidden pair in N7) => R9C1 = 1, R8C4 = 1 (for the 37(7) cage) then
either R7C23 = {39}, locked for N7 => 9 in 37(7) cage in R9C45, 9 locked for R9 => R9C78 = {58}, locked for N9 => R7C789 = {149}
or R7C23 = {57}, locked for R7 => R7C789 = {149}
-> R7C789 = {149/239/248}, no 5, clean-up: no 6 in R6C9, no 5 in R6C7 (step 2c)
5f. 45 rule on N78 3 remaining innies R7C146 = 13 = {139/148/157/238} (cannot be {247} which clashes with R7C789)
5g. R9C78 (step 2e) = {58/67} (cannot be {49} which clashes with R7C89), no 4,9
5h. 9 in R9 only in R9C2345, locked for 37(7) cage at R8C2

6a. R7C789 (step 5e) = {149/239/248}
6b. Consider combinations for R8C56 = {49/58}
R8C56 = {49} => 4 in N9 only in R7C789 = {149/248}
or R8C56 = {58} => 8 in R7 only in R7C789 = {248}
-> R7C789 = {149/248}, no 3, 4 locked for R7 and N9, clean-up: no 3 in R4C4 (step 4), no 8 in R6C9, no 3 in R6C7 (step 2c), no 6 in R8C78
6c. 6 in N9 only in R9C789, locked for R9, clean-up: no 1 in R8C1
6d. Hidden killer pair 5,6 in R89C9 and R9C78 for N9, R9C78 contains one of 5,6 -> R89C9 must contain one of 5,6 = [26]/{35}, no 1,7

7a. R7C146 (step 5f) = {139/157/238}
7b. Consider combinations for R7C23 = {39/57}
R7C23 = {39}, 9 locked for N7 => 9 in 37(7) cage at R8C2 in R9C45, locked for N8 => R8C56 = {58}, 5 locked for N8
or R7C23 = {57}, 5 locked for R7
-> no 5 in R7C46
7c. R7C146 = {157} must be [571], no 7 in R7C6
7d. 5 in R7 only in R7C123, locked for N9, clean-up: no 2 in R8C1
7e. 37(7) cage at R8C2 (step 5a) must contain both or neither of 2,6
7f. Consider placement of 2,6 in N7
R789C1 = [261] => R7C46 = {38}, 3 locked for R7 => R7C23 = {57}
or R8C23 = {26}, 6 locked for N7 => R89C1 = {34}, 3 locked for N7 => R7C23 = {57}, 5 locked for N7 => R7C1 = 1
-> R7C1 = {12}, R7C23 = {57}, 7 locked for R7 and N7, 1 in R79C1, locked for C1 and N7, clean-up: no 2,4 in R4C4 (step 4), no 9 in R5C2
7g. R7C146 = {139/238} -> R7C46 = {38/49}, 3 locked for N8
7h. 9 in N7 only in R9C23, locked for R9
7i. Consider combinations for R8C56 = {49/58}
R8C56 = {49} => R7C46 = {38} => R9C45 = {15/17} (cannot be {57} which clashes with R9C78), 1 locked for R9 => R89C1 = {34}
or R8C56 = {58}, R7C46 = {39} => R7C1 = 1, R89C1 {34}
-> R7C1 = 1, R89C1 = {34}, locked for C1 and N7 clean-up: no 7 in R2C7 (step 2a)
7j. R7C1 = 1 -> R7C46 = {39}, 9 locked for R7 and N8, R4C4 = 6 (step 4), clean-up: no 6,7 in R5C2, no 2 in R6C9, no 9 in R6C7 (step 2c), no 4 in R8C56
7k. Naked pair {58} in R8C56, locked for R8, 5 locked for N8
7l. Naked pair {26} in R8C23, 2 locked for R8, R8C9 = 3 -> R9C9 = 5, R89C1 = [43], clean-up: no 6,8 in R3C2 (step 2b), no 1,3 in R2C2, no 8 in R6C9 (step 2c), no 8 in R7C9, no 7 in R8C78, no 8 in R9C78
7m. R8C4 = 7 (hidden single in R8)
7n. 27(4) cage at R5C4 = {3789} (only remaining combination) -> R5C5 = 7, R567C4 = {389}, locked for C4, 8 locked for N5, clean-up: no 3 in R5C2
7o. 1 in N1 only in R123C3, locked for C3
7p. 2,5 in C4 only in R123C4, locked for N2
7q. 12(3) cage at R6C5 = {129/345}
7r. R7C6 = {39} -> R6C56 = [21]/{45}
7s. Killer pair 2,4 in R6C56 and R6C7, locked for R6
7t. R7C1 = 1 -> R6C12 = 11 = {56}/[83]
7u. 12(3) cage at R4C5 = 2{19}/{345}, no 1,9 in R4C5
7v. 2 in N5 only in R46C5 -> 12(3) cage at R4C5 = 2{19} or 12(3) cage at R6C5 = [219] -> 1,9 in R4567C6, locked for C6 (locking-cages)

8a. R23C56 (step 3a) = {3689/4679}
8b. 6,7 of {4679} must be in R23C6 -> no 4 in R23C6
8c. Hidden killer pair 7,8 in R1C56 and R23C56 for N2, R23C56 contains one of 7,8 -> R1C56 must contain one of 7,8
8d. 14(4) cage at R1C3 = {1238/1247}, no 5,6, 1,2 locked for R1
8e. One of 7,8 in R1C56 -> no 7,8 in R1C3
8f. 7 of {1247} must be in R1C6 -> no 4 in R1C6
8g. 4 in C6 only in R456C6, locked for N5

9a. 45 rule on C12 3 innies R789C2 = 18 = [729] (only possible permutation) -> R789C3 = [568], clean-up: no 1 in R2C4, no 4,9 in R3C9 (step 2b), no 8 in R5C1
9b. 1 in N2 only in R1C45, locked for R1
9c. 31(6) cage at R3C3 contains 6 and 7,9 for C3 = {135679/234679}, 3 locked for C3
9d. 45 rule on N1 4 innies R12C3 + R3C13 = 15 = {1239/1248/1257} (cannot be {1347/1356/2346} which clash with R23C2), no 6
9e. 3 of {1239} must be in R3C3 -> no 9 in R3C3
9f. 5 of {1257} must be in R3C1 -> no 7 in R3C1

10a. 13(3) cage at R3C1 = {157/238/247/256} (cannot be {139/148/346} because 1,3,4 only in R4C2), no 9
10b. 3 of {238} must be in R4C2 -> no 8 in R4C2
10c. 7 of {157} must be in R4C1, 5 of {256} must be in R4C2 -> no 5 in R4C1
10d. 9 in N1 only in R12C1, locked for C1, clean-up: no 1 in R5C2
10e. R4C2 = 1 (hidden single in N4) -> R34C1 = [57], clean-up: no 4 in R23C2
10f. R23C2 = [63] -> R3C9 = 8 (step 2b)
10g. R2C1 = 8 (hidden single in R2) -> R2C7 = 3 (step 2a), R1C123 = [942], R2C3 = 1 -> R2C4 = 5, R6C12 = [65]
10h. R6C5 = 2 -> R67C6 = 10 = [19], R6C7 = 4 -> R6C9 = 7 (step 2c), R7C9 = 4
10i. R5C9 = 1 (hidden single in C9) -> R56C8 = 15 = [69]

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A395 at Hard 1.5 for the two-stage forcing chain in step 5e.

Author:  Ed [ Sun May 10, 2020 8:08 am ]
Post subject:  Re: Assassin 395

3 quite different solutions! Even though wellbeback and me started in a similiar way, we saw things quite differently after that (my step 8,12,13). Wish I'd found his step 8 line 3.
I enjoyed Andrew's WT (though skipped most of 5e). Love the way he keeps finding neat steps in the same areas...So, a very interesting puzzle in the end! (Thanks Andrew and wellbeback for some typos).

I've just finished a major project so have some time again to get Assassins ready. Will plan for June 1 so if someone else wants to do a mid-month one, feel free!
WT:
Preliminaries courtesy of SudokuSolver
Cage 6(2) n12 - cells only uses 1245
Cage 7(2) n7 - cells do not use 789
Cage 8(2) n9 - cells do not use 489
Cage 12(2) n7 - cells do not use 126
Cage 9(2) n1 - cells do not use 9
Cage 10(2) n9 - cells do not use 5
Cage 10(2) n4 - cells do not use 5
Cage 11(2) n69 - cells do not use 1
Cage 21(3) n1 - cells do not use 123
Cage 19(3) n8 - cells do not use 1
Cage 27(4) n58 - cells do not use 12
Cage 14(4) n12 - cells do not use 9

1. "45" on n69: 1 outie r9c6 = 2
1a. -> r9c78 = 13 = {49/58/67}(no 1,3)

2. "45" on r89: 1 outie r7c5 = 6
2a. -> r8c56 = 13 = {49/58}(no 3,7)

3. "45" on n1234: 2 outies r4c4 + r7c1 = 7 (no 7,8,9, no 1 in r4c4)

Key step, wellbeback's step 3 came at this the opposite way. Andrew's note at step 4 gives yet another way to see this.
4. 5 in r4c4 must repeat in n4 only in r6c12, but from step 3, 2 is in r7c1, but this makes 12(3) = {255} -> blocked
4a. -> no 5 in r4c4, no 2 in r7c1

5. 2 in n7 only in r8: locked for r8
5a. 8(2)n9 = {17/35}(no 6) = 3 or 7

6. 10(2)n9: {37} blocked by 8(2)n9 (step 5a)
6a. = {19/46}(no 2,3,7,8) = 4 or 9

7. h13(2)r9c78: {49} blocked by 10(2)n9
7a. = {58/67}(no 4,9)

8. 37(7)r8c2 must have 8 -> r8c56 + r9c78 cannot both be {58}
8a. -> {46} blocked from 10(2)n9
8b. = {19} only: both locked for n9 and r8
8c. 8(2)n9 = {35}: both locked for c9 and n9
8d. r9c78 = {67} both locked for r9, 7 for n9
8e. 11(2)r6c9 = [92/74]
8f. naked triple {248} in r7c789: 4,8 locked for r7
8g. 14(3)r6c7 must have 8 for n9 in r7c78 = {248} only -> r6c7 = (24)
8h. r8c56 = 13 = {58}: both locked for r8 and n8
8i. r89c9 = [35]

9. 4 in n8 only in 37(7) cage: locked for that cage
9a. r89c1 = [43] only place for 4 in n7
9b. r8c4 = 7

10. 12(2)n7 = {57}: 5 locked for n7
10a. r7c1 = 1 -> r4c4 = 6 (h7(2))

11. "45" on c12: 3 innies r789c2 = 18 = [729] only permutation
11a. r789c3 = [568]

12. 27(4)r5c4 = {3789} only -> r5c5 = 7
12a. r567c4 = {389}: all locked for c4, 8 locked for n5

13. "45" on r12: 1 innie r2c2 + 2 = 1 outie r3c9 = [46/57/68], r2c2 = (456), r3c9 = (678), r3c2 = (345)
13a. -> 30(5) disjoint, r2c5: {34689/45678} blocked by iodr12=-2
13b. {25689} blocked by 6(2)r2c3 = 2 or 5
13b. = {15789/24789/35679} = 4 or 5
13c. must have 9: locked for r2

14. killer pair 4,5 in 30(5)r2c5 + 6(2)r2c3: both locked for r2
14a. r23c2 = [63], r3c9 = 8 (iodr12=-2)

15. r2c1 = 8 (hsingle r2)
15a. -> r1c12 = 13 = [94] only
15b. 1 outie r1 -> r2c7 = 3

16. 10(2)n4 = [28], r6c12 = [65] (cage sum), 13(3)r3c1 = [571]

17. r456c3 = {349} = 16 + 6 in r4c4 = 22 -> r3c34 = 9 = [72]

18. 14(4)r1c3 = {1238/1256}(no 7)
18a. must have 2 -> r1c3 = 2, r2c34 = [15], r1c4 = 1

19. r23c5 = {49}(naked pair): both locked for c5 and n2
19a. r23c6 = [76]

20. naked triple {249} in r247c9: 4 & 9 locked for c9
20a. r67c9 = [74], r15c9 = [61]
20b. -> r56c8 = 15 = [69]

easy now. singles and cage sums
Cheers
Ed

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