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 Post subject: Assassin 401
PostPosted: Sun Nov 01, 2020 5:17 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
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Found a "zero" version of this with much higher SSscore and JS stats for which my solution works. However, in the interests of a wide range of WTs have decided to stick with the vanilla. Will be interested to know what I missed. I found this puzzle difficult. SudokuSolver gives it 1.65 and JSudoku uses 3 complex intersections.
triple click code:
3x3::k:3072:3072:3072:8961:5378:5378:3587:3587:3587:3076:3076:8961:8961:6405:5378:5378:3846:3846:4359:2312:2312:8961:6405:6405:6405:3846:6921:4359:4359:2312:8961:6405:6405:6921:6921:6921:3338:3338:3338:8961:9739:9739:3340:6921:6925:5390:5390:9739:9739:9739:9739:3340:2575:6925:5390:2320:4369:4369:9739:6418:787:2575:6925:5390:2320:4369:3860:3860:6418:787:4373:6925:2582:2582:4369:3860:6418:6418:4373:4373:6925:
solution:
Code:
+-------+-------+-------+
| 6 2 4 | 9 3 7 | 5 8 1 |
| 9 3 7 | 8 1 5 | 6 2 4 |
| 8 1 5 | 4 6 2 | 3 9 7 |
+-------+-------+-------+
| 2 7 3 | 5 9 4 | 8 1 6 |
| 4 8 1 | 2 7 6 | 9 5 3 |
| 5 6 9 | 3 8 1 | 4 7 2 |
+-------+-------+-------+
| 7 5 6 | 1 4 8 | 2 3 9 |
| 3 4 8 | 7 2 9 | 1 6 5 |
| 1 9 2 | 6 5 3 | 7 4 8 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 401
PostPosted: Wed Nov 04, 2020 10:53 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks Ed for your latest Assassin. No need to post the harder "zero" version; this was challenging enough after an early breakthrough, which I found using a wellbeback type step.

Thanks Ed for pointing out that my step 1h didn't make sense. Looks like I did a copy and paste from an earlier step and forgot to finish making the necessary changes.
Here is my walkthrough for Assassin 401:
Prelims

a) R2C12 = {39/48/57}, no 1,2,6
b) R56C7 = {49/58/67}, no 1,2,3
c) R67C8 = {19/28/37/46}, no 5
d) R78C2 = {18/27/36/45}, no 9
e) R78C7 = {12}
f) R9C12 = {19/28/37/46}, no 5
g) 9(3) cage at R3C2 = {126/135/234}, no 7,8,9

1a. 45 rule on N69 1 outie R3C9 = 7
1b. Naked pair {12} in R78C7, locked for C7 and N9, clean-up: no 8,9 in R6C8
1c. 45 rule on N3 2 remaining innies R23C7 = 9 = {36/45}
1d. 14(3) cage at R1C7 = {149/158/239/248} (cannot be {356} which clash with R23C7), no 6
1e. 45 rule on N8 2 remaining innies R7C45 = 5 = {14/23}
1f. Killer pair 1,2 in R7C45 and R7C7, locked for R7, clean-up: no 7,8 in R8C2
1g. 45 rule on R1 3 innies R1C456 = 19 = {379/469/478/568} (cannot be {289} which clashes with 14(3) cage, no 1,2
1h. Hidden killer pair 1,2 in 12(3) cage at R1C1 and 14(3) cage for R1, 14(3) cage contains one of 1,2 -> 12(3) cage must contain one of 1,2 -> no 9 in 12(3) cage
1i. 45 rule on C9 3 remaining innies R124C9 = 11 = {128/146/236/245}, no 9

2a. 45 rule on C123 2 innies R26C3 = 1 outie R7C4 + 15 -> R26C3 = 16,17 = {79/89}, 9 locked for C3
2b. R26C3 = 16,17 -> R7C4 = {12} -> R7C5 = {34} (step 1e)

3. 45 rule on R1234 2 outies R5C48 = 7 = {16/25/34}, no 7,8,9

4a. Consider placements for R7C4 = {12}
R7C4 = 1 => R78C7 = [21] => 1 in N7 only in R9C12 = {19}
or R7C4 = 2 => R78C7 = [12] => 2 in N7 only in R9C12 = {28}
-> R9C12 = {19/28}
[Or, as wellbeback would put it, whichever of 1,2 is in R7C4 must be in R8C7 and also in R9C12.]
4b. R78C2 = {36/45}/[72] (cannot be [81] which clashes with R9C12), no 8 in R7C2, no 1 in R8C2
4c. 45 rule on C12 4 outies R1345C3 = 13 = {1237/1246/1345}, no 8, 1 locked for C3
4d. 45 rule on N78 3 (2+1) innies R78C1 + R7C5 = 14
4e. Consider permutations for R7C45 = [14/23] (step 1e)
R7C45 = [14] => R9C12 = {19} (step 4a)
or R7C45 = [23] => R78C1 = 11, no 1 in R8C1 => 1 in N7 only in R9C12 = {19}
-> R9C12 = {19}, locked for R9, 9 locked for N7
4f. Step 4a must also work in the reverse direction, otherwise there would be no place for 2 in N7 -> R7C4 = 1, R7C5 = 4, R78C7 = [21], clean-up: no 6 in R5C8 (step 3), no 6 in R6C8, no 5 in R8C2
[Alternatively a neat way to get this result is
Consider placement for 2 in N7
2 in R78C12 => grouped X-Wing with R78C7 = {12}, no other 2 in R78
or 2 in R789C3
-> no 2 in R7C4, …]
4g. R7C4 = 1 -> R26C3 (step 2a) = 16 = {79}, 7 locked for C3
4h. 8 in C3 only in R789C3, locked for N7
4i. 17(4) cage at R7C3 contains 1,8 = {268}1/{358}1, no 4
4j. R78C2 = [54/72] (cannot be {36} which clashes with 17(4) cage), no 3,6
4k. R7C5 = 4 -> R78C1 = 10 = {37}/[64], no 2,5, no 6 in R8C1
4l. R78C1 = 10 -> R6C12 = 11 = {29/38/56} (cannot be {47} which clashes with R78C1), no 1,4,7
4m. 4 in N7 only in R8C12, locked for R8
4n. 15(3) cage at R8C4 = {258/267/357}, no 9
4o. 9 in N8 only in R78C6, locked for C6
4p. 38(7) cage at R5C5 contains 4 so must contain 3,8, both locked for N5, clean-up: no 4 in R5C8 (step 3)

5a. 4 in C4 only in 35(6) cage at R1C4 = {245789/345689}, 5,8 locked for C4, 8 locked for N2
5b. 15(3) cage at R8C4 (step 4n) = {267/357} (cannot be {258} because 5,8 only in R8C5), no 8, 7 locked for N8
5c. 5 of {357} must be in R8C5 -> no 3 in R8C5
5d. 45 rule on C1234 2 innies R6C34 = 1 outie R8C5 + 10
5e. R6C34 cannot total 17 -> no 7 in R8C5
5f. Min R8C5 = 2 -> min R6C34 = 12, no 2 in R6C4
5g. 7 in N8 only in R89C4, locked for C4
5h. R8C5 = {256} -> R6C34 = 12,15,16 = [93/96/79], 9 locked for R6 and 38(7) cage at R5C5, clean-up: no 4 in R5C7, no 2 in R6C12 (step 4l)
5i. 21(4) cage at R6C1 (steps 4k and 4l) = {38}[64]/{56}{37}, CPE no 3,6 in R45C1

[At this stage I originally found combined cage R2356C7, with simplification of 17(3) cage at R8C8, but this proved to be unnecessary after my next step, so I’ve omitted it.]

6a. R1C456 (step 1g) = {379/469/478/568}, R23C7 (step 1c) = {36/45}
6b. R1C456 = {379/469/568} (cannot be {478} = [874] because no combination for 21(4) cage at R1C5)
6c. Consider placement of 6 in N1
6 in 12(3) cage at R1C1 => R1C456 = {379}
or 6 in R3C123 => no 3 in R2C7 => R1C456 = {379/469} (cannot be {568} = 8{56} because no combination for 21(4) cage at R1C5)
-> R1C456 = {379/469}, no 5,8, 9 locked for R1 and N2
6d. 9 in N3 only in 15(3) cage at R2C8 = {159/249}, no 3,6,8, 9 locked for C8, clean-up: no 1 in R6C8
6e. 9 in C7 only in R45C7, locked for N6
6f. 8 in N3 only in 14(3) cage at R1C7 = {158/248}, no 3,6, 8 locked for R1
6g. R23C7 = {36} (hidden pair in N3), locked for C7, clean-up: no 7 in R56C7
6h. R9C7 = 7 (hidden single in C7) -> R89C8 = 10 = [64], clean-up: no 3 in R6C8
6i. R8C4 = 7 (hidden single in N8) -> R8C5 + R9C4 = 8 = [26/53], no 2 in R9C4, clean-up: no 3 in R7C1 (step 4k)
6j. R8C5 = {25} -> R6C34 (step 5h) = 12,15 = [93/96] -> R6C3 = 9, R2C3 = 7, clean-up: no 5 in R2C12
6k. Naked pair {36} in R69C4, locked for C4

7a. Hidden killer pair 8,9 in R2C12 and R3C1, R2C12 contain one of 8,9 -> R3C1 = {89}
7b. 45 rule on N1 1 remaining innie R3C1 = 1 outie R4C3 + 5, R3C1 = {89} -> R4C3 = {34}
7c. 9(3) cage at R3C2 = {135/234} (cannot be {126} which doesn’t contain one of 3,4), no 6
7d. 6 in N1 only in 12(3) cage at R1C1 = {156/246}, no 3
7e. R1C56 = {37} (hidden pair in R1) -> R1C4 = 9 (step 6c)
7f. R1C56 = {37} = 10 -> R2C67 = 11 = [56], R3C7 = 3
7g. 9(3) cage = {135/234} -> R4C3 = 3 -> R3C1 = 8
7h. R8C1 = 3 (hidden single in N7) -> R7C1 (step 4k) = 7, R7C2 = 5 -> R8C2 = 4
7i. R78C1 = [73] = 10 -> R6C12 = 11 = [56] -> R69C4 = [36], R8C5 = 2 (cage sum)
7j. R3C1 = 8 -> R4C12 = 9 = [27]
7k. R2C12 = [93] (hidden pair in N1) -> R9C12 = [19], R5C123 = [481]
7l. R23C5 = [16] -> R2C8 = 2
7m. R6C8 = 7 -> R7C8 = 3, R5C8 = 5, R5C7 = 9 -> R6C7 = 4

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A401 at 1.5. That's mainly for step 6c.


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 Post subject: Re: Assassin 401
PostPosted: Tue Nov 10, 2020 4:17 am 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Andrew knows me so well ... :)

Thanks for another fine puzzle Ed! Here's how I did it.
Missing steps & clarifications thanks to Ed
and corrections thanks to Andrew


Assassin 401 WT:
1. IOD c123 -> r26c3 = r7c4 + 15
-> r26c3 = {79} and r7c4 = 1 or
r26c3 = {89} and r7c4 = 2
-> 9 locked in c3 in r26c3

2. (As Andrew forecasted)
3(2)n9 = {12}
-> Whichever of (12) is in r7c4 goes in r8c7 and in r9c12

3. IOD n7 -> r78c1 = r7c4 + 9
-> r8c1 cannot be 9
Also r7c1 cannot be 9 since that would put r7c4 = r8c1 (contradicting step 2)
-> 9 in n7 only in r9c12
-> 10(2)n7 = {19}
-> r7c4 = 1
-> r26c3 = {79}
Also 3(2)n9 = [21]

4. Remaining Innie n8 = r7c5 = 4
-> 15(3)n8 cannot contain a 9
-> 9 in n8 in r78c6

5. Outies c12 = r1345c3 = +13(4) (No 8)
-> 8 in c3 in r789c3
-> Remaining Innies n7 = r78c1 = +10(2) = {37} or [64]
-> Remaining Outies n7 = r6c12 = +11(2) cannot be {47}

6. 9 in c3 in r2c3 or r6c3.
If the former -> 9 in c4 in r6c4
-> 9 in r6c34
-> r6c12 cannot be {29}
-> r6c12 = {56} or {38}

7. Since r7c5 = 4 -> 38(7)n5 = {34789(16|25)}
-> (38) in 38(7) in n5

8! Outies n69 -> r3c9 = 7
Outies r6789 = r5c5679 = +25(4)
Trying r6c12 = {38} puts r5c56 = {38} puts 27(5)r3c9 = [7{3458}]
But this leaves no solution for Outies r6789 = r5c5679 = +25(4)
-> r6c12 = {56}
-> r78c1 = {37}
-> 9(2)n7 = [54]
-> r789c3 = {268}
Also -> r6c12 = [56]

9! Given:
Outies c12 = r1345c3 = {1345} with 5 in r13c3
6 in n1 in r13c1
7 in n1 in r2c3 or r1c2
12(2)n1 from [93] or [48]
r2c3 from (79)

Either r1c3 = 5 -> 7 in n1 can only be in r2c3
Or r3c3 = 5 -> 9(3)r3c2 = {135} -> r1c3 = 4
-> r1c3 from (45)
But in that latter case 12(2)n1 = [93] which again puts r2c3 = 7

Either way r2c3 = 7
-> r6c3 = 9

Alternate Step 9.
I discovered a much simpler way to get to the end conclusion of Step 9 above.
6 in r6c2 -> 7 not in r5c7
-> 7 in n6 only in r6c78
-> r26c3 = [79]


10. 9 in n1 only in r23c1
-> 10(2)n7 = [19]
-> No solution for 17(3)r3c1 with r3c1 = 9
-> 12(2)n1 = [93]
Also since r1c3 from (45) -> (HS 8 in n1) r3c1 = 8
-> 17(3)r3c1 = [827]

-> (HS 6 in c1) r1c1 = 6
Also -> r1c23 = [24] or [15], and r3c23 = [15] or [24]
-> r4c3 = 3
-> 13(3)n4 = [481]


11. Outies r1234 = r5c48 = +7(2) = {25}
-> 5 not in 38(7)n5
-> 38(7)n5 = {1346789}
-> (HS 2 in n5) r5c4 = 2
-> r5c8 = 5

12. Also (18) in n5 in r6, and in n6 in r4
Also 2 in n6 in r6
-> 27(5)r3c9 = [7{168}5]
-> 13(2)n6 = [94]
-> (HS 7 in n6) 10(2)r6c8 = [73]
-> r56c9 = [32]
-> r789c9 = {589}

13. Also remaining Innies n3 = r23c7 = +9(2) = [63]
Also 7 in n2 in r1c56
-> 21(4)n2 = [{37}56]
-> Remaining Innie r1 -> r1c4 = 9
Also (HS 5 in c7) r1c7 = 5
-> 14(3)n3 = [581]
-> 12(3)r1c1 = [624]
etc.


Last edited by wellbeback on Thu Dec 03, 2020 8:52 pm, edited 1 time in total.

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 Post subject: Re: Assassin 401
PostPosted: Sat Nov 14, 2020 10:08 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
My WT is quite different to the other two though we all started in very similar fashion. Really nice chains from Andrew (step 6c) and wellbeback (step 8) to give easier solutions than mine. [Big thanks to Andrew for checking my solution including some typos.]

WT:
Preliminaries courtesy of SudokuSolver
Cage 3(2) n9 - cells ={12}
Cage 12(2) n1 - cells do not use 126
Cage 13(2) n6 - cells do not use 123
Cage 9(2) n7 - cells do not use 9
Cage 10(2) n7 - cells do not use 5
Cage 10(2) n69 - cells do not use 5
Cage 9(3) n14 - cells do not use 789

1. "45" on c123: 2 innies r26c3 - 15 = 1 outie r7c4
1a. -> r26c3 = 16/17 = {79/89}
1b. 9 locked for c3
1c. r7c4 = (12)

2. 3(2)n9 = {12}: both locked for c7 and n9

These next couple of steps made this an Assassin for me. Didn't enjoy the later part.
3. r7c4 sees all 1 or 2 in r9 apart from the 10(2)n7 so must repeat there
3a. 10(2) = {19/28}(no 3,4,6,7)

4. "45" on n7: 1 outie r7c4 + 9 = 2 innies r78c1
4a. but [91][1] and [92][2] blocked since r7c4 repeats in 10(2)n7
4b. and no 9 in r8c1 since the innie/outie difference can't be 0 (IOU)
4c. -> no 9 in r78c1

5. 9 in n7 only in 10(2) = {19}: both locked for r9, 1 for n7
5a. -> r7c4 = 1 (step 3)
5b. r78c6 = [21]
5c. and r36c3 = 16 (iodc123=+15) = {79} only: 7 locked for c3

6. "45" on n8: 1 remaining innie r7c5 = 4

7. 15(4)n8 = {258/267/357}(no 9)

8. 9 in n8 only in c6: locked for c6

9. "45" on n69: 1 outie r3c9 = 7

10. "45" on n3: 2 remaining innies r23c7 = 9 = {36/45}(no 8,9)

11. 38(7)r6c3 = {136789/235789}[4]
11a. must have 3 & 8: both locked for n5

12. 35(6)r2c3 must have 8: locked for n2, c4

13. 14(3)n3: {356} blocked by r23c7 needing 5 or 6
13a. = {149/158/239/248}(no 6)
13b. = 4 or 5 or 9

Another key step
14. 21(4)r1c5 = {1479/1569/2379/2469/3459/3567}
14a. ie, has 9 in r1c5, or is {3567}
14b. -> if h9(2)r23c7 = {45} -> 14(3)n7 = {239} -> 21(4) = {67}[35]
14c. but "45" on r1: 3 innies r1c456 = 19 (no 1), so can't have both {67} in r1c56
14d. -> {45} blocked from r23c7
14e. -> r23c7 = {36}: both locked for n3, c7 and neither are in r2c5 which sees both those cells (Common Peer Elimination, CPE)

15. 25(6)r2c5 = {123469/124567}
15a. must have 4: locked for c6
15b. must have 9 or have both 1 & 5

16. 21(4)r1c5 = {1569/2379/3567}
16a. but {1569} as [9516] only, clashes with 25(6)
16b. = {2379/3567}(no 1)

17. 1 in n2 only in 25(6)r2c5 -> no 1 in r4c56

18. 1 in n5 only in 38(6)r6c3 = {136789}[4}(no 2,5)
18a. must have 6 in n5: locked for n5

19. 25(6)r2c5 = {123469/124567}
19a. must have 6: locked for r3

20. 6 in n1 only in 12(3) = {156/246}(no 3,7,8,9)

21. 3 & 7 in r1 only in h19(3)r1c456 = {379} only, 9 locked for r1, all locked for n2

This step was probably available at step 16 but couldn't see it....
22. 25(6)r2c5 = {123469/124567}
22a. 21(4)r1c5 = {2379/3567}
22b. but {2379} as [9723] only, blocked by 25(6)r2c5 needing 9 or both 2,7
22c. = {3567} only (no 9)
22d. -> r1c4 = 9 (hsingle n2)
22e. r1c56 = {37} -> r2c56 = [56], r3c7 = 3, r26c3 = [79]

23. "45" on n1: 1 remaining innie r3c1 - 5 = 1 outie r4c3 = [83/94]

24. 9(3)r3c2 = {135/234}
24a. must have 3 -> r4c3 = 3 -> r3c1 = 8
24b. -> r3c23 = 6; but {24} blocked by r3c4 = (24)
24d. -> r3c23 = {15} only: both locked for r3 and n1

25. naked triple {246} in r3c456: 2 & 4 locked for n2 and r3
25a. r2c45 = [81]
25b. r3c8 = 9
25c. r2c89 = {24}: both locked for n3, 4 for r2

26. r4c5 = 9 (hsingle n5)
26a. 25(6)r2c5 = {123469}(no 7)

27. r4c12 = 9 (cage sum)
27a. but {45} blocked by r4c46 = {245}
27b. = {27} only: both locked for r4, n4
27c. r4c46 = [54], r35c4 = [42], r4c7 = 8

28. "45" on n7: 2 remaining outies r6c12 = 11 = {56} only: both locked for r6, n4 and 21(4) cage
28a. -> r78c1 = 10 (cage sum) = {37} only: both locked for c1, n7

29. 9(2)n7 = [54] only permutation

30. r56c7 = [94] only permutation
30a. r19c7 = [57]
30b. -> r89c8 = 10 (cage sum) = [64] only permutation

31. r9c4 = 6 (hsingle c4)
31a. -> r8c45 = 9 = [72] only permutation

easy now
Cheers
Ed


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