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Assassin 397
http://rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1566
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Author:  Ed [ Wed Jul 01, 2020 6:46 am ]
Post subject:  Assassin 397

Attachment:
a397.JPG
a397.JPG [ 100.4 KiB | Viewed 939 times ]
Note the disjoint cage at r4c9
A killer X so no repeats on the diagonals.

Assassin 397
Couldn't solve this one - but hope it is doable at a score of 1.60 I got a long way in with some interesting interactions, and spent a lot of time on it so can't bear to throw it away. JSudoku has a hard time with 5 complex intersections and even some unusual chains... Look forward to seeing what I couldn't.
this is where I got up to:
Code:
.-------------------------------.-------------------------------.-------------------------------.
| 457       569       59        | 12        268       248       | 17        3458      3458      |
| 34678     3478      478       | 1458      1458      9         | 17        256       26        |
| 1         2         458       | 56        37        37        | 569       4589      4589      |
:-------------------------------+-------------------------------+-------------------------------:
| 489       4789      4789      | 26        26        1         | 35        35        479       |
| 459       34579     26        | 45789     45789     348       | 24689     2679      1         |
| 34579     1         26        | 45789     345789    378       | 24689     2679      247       |
:-------------------------------+-------------------------------+-------------------------------:
| 5689      5689      34578     | 14789     124789    248       | 34569     235679    245678    |
| 24679     4578      34578     | 4789      24789     56        | 234569    1         359       |
| 26        4679      1         | 3         24789     56        | 24569     48        56789     |
'-------------------------------.-------------------------------.-------------------------------'
triple click code:
3x3:d:k:8448:4097:4097:4097:6914:6914:6914:6659:6659:8448:8448:8448:2308:2308:6914:6914:2821:6659:3590:6663:8448:8448:4872:4872:2821:6659:6659:3590:6663:6663:4872:4872:4872:2067:2067:7690:3590:6663:5899:6412:6412:2829:6676:6676:6676:6663:6663:5899:6412:6412:2829:6676:6676:7690:3598:3598:5899:7951:7951:7951:6672:6672:7690:3857:5899:5899:7951:7951:2834:6672:6672:6672:3857:3857:1033:1033:7951:2834:6672:7690:7690:
solution:
Code:
+-------+-------+-------+
| 4 6 9 | 1 8 2 | 7 3 5 |
| 7 3 8 | 5 4 9 | 1 2 6 |
| 1 2 5 | 6 3 7 | 9 4 8 |
+-------+-------+-------+
| 8 9 7 | 2 6 1 | 3 5 4 |
| 5 4 2 | 9 7 3 | 8 6 1 |
| 3 1 6 | 4 5 8 | 2 9 7 |
+-------+-------+-------+
| 9 5 3 | 8 1 4 | 6 7 2 |
| 2 8 4 | 7 9 6 | 5 1 3 |
| 6 7 1 | 3 2 5 | 4 8 9 |
+-------+-------+-------+
Cheers
Ed

Author:  Andrew [ Sun Jul 05, 2020 7:44 am ]
Post subject:  Re: Assassin 397

Thanks Ed for a really challenging Assassin.

Looking at how far you reached, it looks like a key step you missed was my step 8. After that it was slow progress, nibbling away and gradually getting a bit easier. However the final breakthrough took me a while to find; it came to me just after I went to bed so I got up again to add it to my walkthrough and found that the rest was straightforward.

Here is my walkthrough for Assassin 397:
Disjoint 30(5) cage in R4C9 + R67C9 + R9C89

Prelims

a) R2C45 = {18/27/36/45}, no 9
b) 11(2) cage at R2C8 = {29/38/47/56}, no 1
c) R4C78 = {17/26/35}, no 4,8,9
d) R56C6 = {29/38/47/56}, no 1
e) R7C12 = {59/68}
f) R89C6 = {29/38/47/56}, no 1
g) R9C34 = {13}

1a. 45 rule on N8 1 innie R9C4 = 3 -> R9C3 = 1, clean-up: no 6 in R2C5, no 8 in R89C6
1b. 45 rule on N7 2 remaining outies R56C3 = 8 = {26/35}
1c. 45 rule on N47 2 remaining outies R3C12 = 3 = {12}, locked for R3 and N1, clean-up: no 9 in R2C8
1d. 45 rule on N147 2 remaining outies R13C4 = 7 = [16/25]
1e. 45 rule on N3 2 innies R12C7 = 8 = {17/26/35}, no 4,8,9
1f. 45 rule on N3 3 outies R1C56 + R2C6 = 19 = {289/379/469/478/568}, no 1
1g. 45 rule on N5 2 outies R3C56 = 10 = {37/46}
1h. 45 rule on N5 3 innies R4C456 = 9 = {126/135} (cannot be {234} which clashes with R3C56), 1 locked for R4 and N5, clean-up: no 7 in R4C78
1i. R56C6 = {29/38/47} (cannot be {56} which clashes with R4C456), no 5,6
1j. 45 rule on N6 R46C9 = 11 = {29/38/47/56} (cannot be {56} which clashes with R4C78), no 1,5,6
1k. Naked quint {12356} in R4C45678, locked for R4, clean-up: no 8,9 in R6C9
1l. Max R3C1 = 2 -> min R45C1 = 12, no 1,2,3 in R5C1
1m. Naked pair 1,2 in R3C12, CPE no 1,2 in R6C1
1n. R2C45 = {18/27/45} (cannot be [63] which clashes with R3C56), no 3,6
1o. 9 in N2 only in R1C56 + R2C6 = {289/379/469}, no 5
1p. Max R1C4 = 2 -> min R1C23 = 14, no 3,4 in R1C23

2a. R3C1 = 1 (hidden single in C1) -> R45C1 = 13 = {49}/[76/85], no 7,8 in R5C1
2b. R3C2 = 2
2c. 2 in N4 only in R56C3 = {26} (step 1b), locked for C3, 6 locked for N4 and 23(5) cage at R5C3, clean-up: no 7 in R4C1
2d. 2 in N7 only in 15(3) cage at R8C1 = {249/267} (cannot be {258} which clashes with R7C12), no 3,5,8
2e. 45 rule in N7 3 innies R7C3 + R8C23 = 15 must contain 3 for N7 = {348/357}, no 9

3a. R3C4 ‘sees’ all of N1 except for R1C23
3b. R13C4 = [16/25] -> 16(3) cage at R1C2 = [691]/{59}2, 9 locked for R1 and N1
3c. R2C6 = 9 (hidden single in N2), clean-up: no 2 in R56C6, no 2 in R89C6
3d. 45 rule on N3 2 remaining outies R1C56 = 10 = {28/37/46}
3e. 3 in N2 only in R1C56 = {37} or R3C56 (step 1g) = 10 = {37} (locking cages), 7 locked for N2, clean-up: no 2 in R2C45
3f. 2 in N2 only in R1C456, locked for R1, clean-up: no 6 in R2C7 (step 1e)

4a. R12C7 (step 1e) = {17/26/35}, R4C456 (step 1h) = {126/135}
4b. Consider placements for {37} in N2 (step 3e)
R1C56 = {37} => R12C7 = [62] => R4C78 = {35}, locked for R4
or R3C56 = {37}, 3 locked for 19(5) cage at R3C5
-> R4C456 = {126}, 2,6 locked for R4 and N5, 6 locked for 19(5) cage, clean-up: no 4 in R3C56 (step 1g)
4c. Naked pair {37} in R3C56, locked for R3 and N2, clean-up: no 4,8 in R2C8
4d. Naked pair {35} in R4C78, locked for N6, clean-up: no 8 in R4C9 (step 1j)
4e. R12C7 = {17} (cannot be {35} which clashes with R4C7, cannot be [62] which clashes with R1C56), locked for C7 and N3, clean-up: no 4 in R3C7
4f. 8 in R4 only in R4C123, locked for N4
4g. Killer pair 3,7 in R3C6 and R56C6, locked for C6, clean-up: no 4 in R89C6
4h. Naked pair {56} in R89C6, locked for C6, clean-up: no 4 in R1C5 (step 3d)

5a. R4C6 = 1 (hidden single on D/)
5b. R8C8 = 1 (hidden single on D\)
5c. R5C9 = 1 (hidden single in N6)
5d. R6C2 = 1 (hidden single in N4)
5e. Killer pair 2,6 in R13C4 and R4C4, 2 locked for C4
[I ought to have seen step 14a at this point; fortunately the delay wasn’t significant.]

6. 45 rule on N9 3 innies R7C9 + R9C89 = 19 = {289/478/568} (cannot be {379/469} which clash with R46C9), no 3, 8 locked for N9

7a. 2 in R2 only in R2C89, whichever of 5,6 is in R3C4 must be in R1C23 and therefore in R2C89 -> R2C89 = {25/26}, clean-up: no 8 in R3C7
7b. 8 in C7 only in R56C7, locked for N6
7c. 3 in N3 only in R1C89, locked for R1

8a. R46C9 = 11 (step 1j) = {47}/[92]
8b. Consider placement of 9 on D\
9 in R5C5 + R7C7 => no 9 in R3C7 => no 2 in R2C8 => R2C9 = 2 (hidden single in R2)
or R9C9 = 9
-> R46C9 = {47}, locked for C9, N6 and 30(5) disjoint cage at R4C9
8c. 4 in C7 only in R789C7, locked for N9
8d. 9 in R4 only in R4C123, locked for N4, clean-up: no 4 in R4C1 (step 2a)
8e. R7C8 = 7 (hidden single in N9)

9. 2 on D/ only in R2C8 + R9C1, CPE no 2 in R9C8

10. R7C9 + R9C89 (step 6) = 19 = {289/568}
10a. Consider placement for 2 in R2
R2C8 = 2 => R56C8 = {69}, locked for C8 and N6, R56C7 = {28}, locked for C7, R8C1 = 2 (hidden single in C1) => 2 in N9 only in R7C9 + R9C89 = {289} => R9C8 = 8
or R2C9 = 2 => R7C9 + R9C89 = {568} => naked triple {568} in R9C689, 8 locked for N9
-> 8 in R9C89, locked for R9 and N9

11a. 3 in C9 only in R18C9, CPE no 3 in R8C2 using D/
11b. 3 in N7 only in R78C3, locked for C3
11c. Consider placement for 3 in R78C3
R7C3 = 3 => R8C9 = 3 (hidden single in C9)
or R8C3 = 3
->no 3 in R8C7
11d. 3 in R7 only in R7C37, CPE no 3 in R5C5 using both diagonals
11e. 3 in N5 only in R6C5 or in R56C6 = {38} -> no 8 in R6C5 (locking-out cages)

12a. 15(3) cage at R8C1 (step 2d) = {249/267}
12b. Consider permutations for R45C1 = 13 (step 2a) = [85/94]
R45C1 = [85]
or R45C1 = [94] => 15(3) cage = {267}, 6 locked for N7 => R7C12 = [59]
-> 5 in R57C1, locked for C1
Also no 8 in R7C1 -> no 6 in R7C2
12c. 5 in N4 only in R5C12, locked for R5

13. Consider placement for 2 in C4
R1C4 = 2 => R7C6 = 2 (hidden single in C6)
or R4C4 = 2, placed for D\
-> no 2 in R7C7

14a. 6 in C4 only in R34C4, CPE no 6 in R1C1 + R2C2 using D\
14b. Hidden killer pair 2,6 in R4C4 and R7C7 + R9C9 for D\, R4C4 = {26} -> R7C7 + R9C9 must contain one of 2,6
[I hope I’ll be able to use this later.]

15a. 33(6) cage at R1C1 = {345678}, R13C4 (step 1d) = [16/25], 15(3) cage at R8C1 (step 2d) = {249/267}
15b. Consider placement for 6 in C2
R1C2 = 6 => R3C4 = 6
or R9C2 = 6 => R89C1 = {27}, 7 locked for C1 => R1C47 = [17] (hidden pair in R1)
-> R13C4 = [16], R12C7 = [71], clean-up: no 8 in R2C45
[Cracked at last.]
15c. Naked pair {45} in R2C45, locked for R2, 4 locked for N2
15d. R1C1 + R3C3 = [45] (hidden pair in N1), both placed for D\
15e. R5C1 = 5 -> R4C1 = 8 (cage sum)
15f. R789C1 = {269} (hidden triple in C1), 6,9 locked for N7
15g. R3C7 = 9 -> R2C8 = 2, both placed for D/
15h. R2C9 = 6, R3C89 = [48]
15i. R789C1 = [926], R89C6 = [65]
15j. R7C9 + R9C89 (step 10) = [289]
15k. R89C7 = [54], R9C2 = 7, R8C9 = 3, R1C9 = 5, placed for D/
15l. R2C4 = 5 (hidden single in C4)
15m. Naked pair {48} in R8C23, locked for R8 and N7
15n. R19C5 = [82], R5C5 = 7, clean-up: no 4 in R5C6
15o. Naked pair {38} in R56C6, locked for C6 and N5, R6C4 = 4, placed for D/

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

Rating Comment:
I'll rate my walkthrough for A397 at Hard 1.5. Some of my forcing chains were hard to find.

Author:  Ed [ Fri Jul 10, 2020 6:59 am ]
Post subject:  Re: Assassin 397

Turns out this puzzle is not that hard, though did use some chains and one is quite tricky to see (step 22). I missed a hidden single in that diagram above and the rest was very straightforward. Some areas of the puzzle I spent a lot of time on in the diagram above are not even touched in this optimised solution. I haven't finished going through Andrew's WT but can see our ways are quite different for the middle and end. [Thanks to Andrew for some typos and checking my solution]
A397 WT:
Preliminaries courtesy of SudokuSolver

Cage 4(2) n78 - cells ={13}
Cage 14(2) n7 - cells only uses 5689
Cage 8(2) n6 - cells do not use 489
Cage 9(2) n2 - cells do not use 9
Cage 11(2) n5 - cells do not use 1
Cage 11(2) n8 - cells do not use 1
Cage 11(2) n3 - cells do not use 1

This is an optimised solution so not all possible clean-up done. Just what is stated.
1. "45" on n8: 1 innie r9c4 = 3
1a. r9c3 = 1

2. "45" on n47: 2 remaining outies r3c12 = 3 = {12} only
2a. 1 & 2 locked for r3 and n1
2b. r6c1 sees both those -> no 1,2 in r6c1 (Common Peer Elimination CPE)

3. 1 in c1 only in 14(3)r3c1 -> r3c1 = 1
3a. -> r45c1 = 13 = {49/58/67}(no 2,3)
3b. r3c2 = 2

4. 33(6)n1 = {345678}(no 9)
4a. must have 3, locked for n1

5. 9 in n1 only in 16(3) = {169/259}(no 4,7,8)
5a. must have 1,2 -> r1c4 = (12)
5b. 9 locked for r1

6. "45" on n1: 2 outies r13c4 = 7 = [16/25]

7. "45" on r123: 2 remaining innies r3c56 = 10 = {37/46}(no 5,8,9) = 3 or 4, 3 or 6
7a. -> r4c456 = 9 (cage sum)
7b. but {234} blocked by h10(2)
7c. = {126/135}(no 4,7,8,9): note = {26}/{35}
7d. 1 locked for r4 and n5
7e. -> 19(5)r3c5 = {12367/13456}(note: must have 3)
7f. and {36} blocked from 9(2)n2
7g. = {18/27/45}(no 3,6)

8. 8(2)n6 = {26/35}(no 7)
8a. -> with step 7c, 2,3,5,6 locked in r4c45678 for r4

9. r2c6 = 9 (hsingle n2)
9a. no 2 in two 11(2)r58c6

10. "45" on n3: 2 remaining outies r1c56 = 10 = {28/37/46}(no 1,5)
10a. -> r12c7 = 8 (cage sum) = {17/26/35}(no 4,8)
10b. note that {26} in r12c7 must have {37} in r1c56 (no eliminations yet)
10c. and h8(2)r12c7 must have a different combination to 8(2)n6
10d. and since 19(5)r3c5 must have 3
10e. -> {26} blocked from r12c7 since {35} in 8(2)n6 would leave no 3 for the 19(5)
10f. -> r12c7 = {17/35}(no 2,6) = 3 or 7
10g. -> {37} blocked from r1c56
10h. = {28/46}(no 3,7)

11. 3 in n2 only in h10(2)r3c56 = {37}: both locked for r3, 3 for 19(5), 7 for n2
11a. -> r4c456 = {126} only: 2 & 6 locked for r4 and n5
11b. no 2 in 9(2)n2

12. 8(2)n6 = {35} only: both locked for n6

13. h8(2)r12c7: {35} blocked by r4c7
13a. = {17} only: both locked for n3 and c7

14. 2 in n2 only r1: locked for r1

15. r4c6 = 1 (hsingle D/), r8c8 = 1 (hsingle d\)

16. killer pair 2,6 in r134c4: both locked for c4 and 6 for r1c1, r2c2, r3c3 (through d\) (CPE)

17. 2 in n5 in r4c45.
17a. If in r4c5 -> 2 in n8 only in r7c6 -> no 2 in r7c7
17b. or 2 in r4c4 -> no 2 in r7c7 (through d\)
17c. no 2 in r7c7

18. "45" on n7: 2 remaining outies r56c3 = 8 and must have 2 for n4 = {26} only: both locked for 23(5)r5c3: 6 for n4 & c3

19. if 6 in n1 in r2c1 -> 6 on d/ in n3: 6 locked for n3
19a. or 6 in n1 in r1c2 -> 6 in r2 only in r2c89: 6 locked for n3
19b. -> no 6 in r1c8, r3c89

20. 16(3)r1c2 = {169/259} and h10(2)r1c56 = {28/46}
20a. killer single 6: locked for r1

21. 26(5)n3: {24569} blocked since 2 & 6 are only in r2c9
21a. = {23489/34568}
21b. must have 2 or 6 -> r2c9 = (26)

My key step (hopefully clearer now)
22. 6 in n1 in r1c2 or r2c1 -> r1c56 and r2c9 cannot both have 6
22a. -> if r2c9 = 6 -> r3c4 = 6 -> r4c4 = 2
22b. OR r2c9 = 2
22c. -> 2 locked
22d. -> no 2 in r9c9 through d\

[edit: alternatively, and simpler for 22a. if r2c9 = 6 -> 6 in r3 must be in r3c4 (hidden single) -> r4c4 = 2]

This is the step I missed before posting this puzzle. Ooops.
23. r4c4 = 2 (hsingle d\),
23a. r4c5 = 6
23b. r13c4 = [16](h7(2), r12c7 = [71]
23c. r1c23 = 15 = [69]

Much easier now
24. 9(2)n2 = {45} only: both locked for r2, 4 for n2

25. r1c1 + r3c3 = {45} (hpair n1). both locked for d\

26. r45c1 = 13 (cage sum) = {49}/[85](no 7, no 8 in r5c1) = 4 or 5

27. killer pair 4,5 in r145c1: both locked for c1
27a. no 9 in r7c2

28. 15(3)n7 must have 2 for n7 = {249/258/267}
28a. 4 & 5 only in r9c2 -> r9c2 = (457)

29. r56c3 = {26} = 8 -> r7c3 + r8c23 = 15 = {348/357}(no 9)

30. 9 in n7 only in c1: locked for c1
30a. -> h13(2)r45c1 = [85]
30b. r1c1 = 4, r3c3 = 5
30c. no 6 in r2c8

31. r9c1 = 6 (hsingle d/)
31a. -> r8c1 + r9c2 = 9 = [27]
31b. r7c12 = [95]

32. r2c9 = 6 (hsingle n3)
32a. r2c8 = 2 (hsingle n3) -> r3c7 = 9 (placed for d/)

33. naked triple {378} on d\ in cells 2,5,6: all locked for d\, 7 locked for n5
33a. r7c7 = 6, r9c9 = 9

34. r5c5 = 7 (hsingle d/)

35. 11(2)n5 = {38}: both locked for n5 and c6
35a. -> r13789c6 = [27465]

36. naked pair {48} in r39c8: both locked for c8

37. naked pair {35} in r14c8: 3 locked for c8

Lots of naked from here
Cheers
Ed

Author:  wellbeback [ Sat Jul 11, 2020 7:01 pm ]
Post subject:  Re: Assassin 397

Here's how I did it. A number of short chains (as is my wont). Thanks Ed as usual! I found it quite tricky.
Assassin 397 WT:
1. Innies n8 -> 4(2)r9 = [13]
Remaining Outies n47 -> r3c12 = {12}
Whichever of (12) is in r3c2 goes in n4 in r56c3 -> cannot be a 1. I.e., r3c12 = [12]
Remaining outies n7 -> r56c3 = +8(2) = {26}

2. 33(6)r1c1 can only be {345678}
-> 9 in r1c23
Outies n5 = r3c56 = +10(2) = {37} or {46}
-> Remaining Innies n23 = r13c4 = +7(2) from [16] or [25]
Whichever of (56) is in r3c4 goes in r1c23
-> 16(3)r1c2 = [691] or [{59}2]

3. HS 9 in n2 -> r2c6 = 9
-> Remaining outies n3 -> r1c56 = +10(2)

19(5)r3c5 either [{37}{126}] or [(46}{135}]
But the latter case puts r1c56 = {37} (Only place for 3 in n2) and 8(2)n6 = {26}
which leaves no solution for Innies n3 = r12c7 = +8(2)

-> 19(5)r3c5 = [{37}{126}]
-> 8(2)n6 = {35}
-> r1c56 = {28} or {46}
-> r12c7 = {17}

4. HS 1 in D/ -> r4c6 = 1
Innies n9 = +19(3) (No 1)
-> (HS 1 in D\) r8c8 = 1
Innies n6 = +11(2) (No 1)
-> (HS 1 in c9) r5c9 = 1
-> (HS 1 in r6) r6c2 = 1

5. 11(2)n5 from {38} or {47}
-> (379) locked in c6 in r2356c6
-> 11(2)n8 = {56}

6. 2 in r2 only in r2c89
Whichever of (56) is in r3c4 and r1c23 also in r2c89
-> 11(2)n3 from [29] or {56}
-> 3 in n3 in r1c89

7! Innies n6 = r46c9 = +11(2) = {47} or [92]
But the latter case puts 11(2)n3 = [29] which leaves no place for 9 in D\.
-> r46c9 = {47}
-> r56c78 = {2689}
Also -> (HS 7 in n9/c8) r7c8 = 7

8. Innies n9 only from {568} or {289} (I.e., must have an 8)
Whichever of (56) or (29) is in 11(2)n3 must both go in c9 in n9
-> Since at least one must go in innies n9 -> both go in innies n9 in c9
-> r9c8 = 8

9! 2 in n7 only in 15(3)n7 in r89c1
-> 3 in n7 only in Innies n7 = r7c3+r8c23 = +15(3)

3 in n3 in r1c8 or r1c9

Trying 3 in r1c8 puts r8c9 = 3 puts r7c3 = 3
or trying 3 in r1c9 puts r8c3 = 3

-> 3 in n7 in r78c3
Also 3 in D/ only in r1c9 or r7c3
-> 3 in n5 only in r5c6, r6c5, or r6c6

10! Consider the following ...
r134c4 from:
(A) [162] which puts 9(2)n2 = {45} and r1c23 = [69]
(B) [256] which puts 9(2)n2 = {18} and r1c23 = {59} and r2c1 = 6

I.e., 6 in n1 only in r1c2 or r2c1

Also ...
3 in n1 only in r2c12
3 in n4 only in r5c2 or r6c1
3 in n5 only in r5c6, r6c5, or r6c6

So - trying r2c1 = 3 puts r1c2 = 6 -> Option (A)
or trying r2c2 = 3 puts r6c1 = 3 puts 11(2)n5 = [38] puts 8 in n1 in r2c13 -> Option (A)

In either case must be option (A): r134c4 = [162]

11. -> 9(2)n2 = {45}
-> r1c1,r3c3 = {45}
-> (Since r45c1 from {49} or [85]) -> (45) locked in c1 in r145c1
-> 9 in n7 either in 14(2)n7 = [95] or 15(3)n7 = [{29}4]
I.e., 9 in c1 in n7
-> 14(3)c1 = [185]
-> r1c1 = 4 and r3c3 = 5
-> 5 in r1 in r1c89
-> 11(2)n3 = [29]
-> (HS 9 in D\) -> r79c9 = [29]
etc.

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