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 Post subject: Assassin 398
PostPosted: Sat Aug 01, 2020 9:16 am 
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Posts: 1040
Location: Sydney, Australia
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Assassin 398

I found a fairly quick way to crack this but couldn't find a way to also make a harder end. It's very interesting though which is the main thing. SudokuSolver gives it 1.70 JSudoku has no trouble and only uses 1 "complex intersection" which is why I first tried it.

triple click code:
3x3::k:6144:6144:9217:7426:7426:7426:7426:7426:0000:6144:9217:9217:1027:7940:7940:0000:7940:7940:6144:6144:9217:1027:8965:8965:7940:8965:8965:8198:9217:9217:8198:775:8965:8965:8965:6152:8198:8198:8198:8198:775:4873:1802:1803:6152:4876:6413:3342:3342:4111:4873:1802:1803:6152:4876:6413:3342:4111:4111:4873:4873:3856:6152:4876:6413:6413:4625:4625:3858:3858:3856:1811:4876:1044:1044:4625:3349:3349:3858:3858:1811:
solution:
Code:
+-------+-------+-------+
| 5 6 2 | 8 4 7 | 9 1 3 |
| 1 8 4 | 3 5 9 | 6 2 7 |
| 9 3 7 | 1 6 2 | 8 5 4 |
+-------+-------+-------+
| 4 9 6 | 5 2 3 | 7 8 1 |
| 7 2 8 | 6 1 4 | 5 3 9 |
| 3 5 1 | 7 9 8 | 2 4 6 |
+-------+-------+-------+
| 2 7 5 | 4 3 6 | 1 9 8 |
| 8 4 9 | 2 7 1 | 3 6 5 |
| 6 1 3 | 9 8 5 | 4 7 2 |
+-------+-------+-------+

Cheers
Ed


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 Post subject: Re: Assassin 398
PostPosted: Thu Aug 06, 2020 4:47 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks Ed for this latest Assassin.

I didn't find a quick way to crack it so wonder what I missed. For me it was fairly slow going, feeling harder than appears from the steps in my walkthrough.

I was slightly surprised that SudokuSolver gave the same score when I added a disjointed cage for the two blank cells, so they weren't the reason for its score.

Here is my walkthrough for Assassin 398:
Prelims

a) R23C4 = {13}
b) R45C5 = {12}
c) R56C7 = {16/25/34}, no 7,8,9
d) R56C8 = {16/25/34}, no 7,8,9
e) R78C8 = {69/78}
f) R89C9 = {16/25/34}, no 7,8,9
g) R9C23 = {13}
h) R9C56 = {49/58/67}, no 1,2,3

1a. Naked pair {13} in R12C4, locked for C4 and N2
1b. Naked pair {12} in R45C5, locked for C5 and N5
1c. Naked pair {13} in R9C23, locked for R9 and N7, clean-up: no 4,6 in R8C9

2a. 45 rule on whole grid 2 innies R1C9 + R2C7 = 9 = {18/27/36/45}, no 9
2b. 45 rule on C123 3 outies R456C4 = 18 = {459/468/567}
2c. 45 rule on R56789 4 outies R4C1459 = 12 = {1236/1245}, no 7,8,9, 1,2 locked for R4
2d. 6 of {1236} must be in R4C4 -> no 6 in R4C19
2e. 45 rule on N1 2 outies R4C23 = 15 = {69/78}
2f. 45 rule on N1456789 3 innies R4C678 = 18 = {369/378/459} (cannot be {468/567} which clash with R4C23)

3a. 1 in N8 only in R78C6
3b. 45 rule on N8 4 innies R7C456 + R8C6 = 14 = {1238/1247/1256/1346}, no 9
3c. 45 rule on N8 1 outie R6C5 = 2 innies R78C6 + 2
3d. Min R78C6 = 3 -> min R6C5 = 5
3e. Max R6C5 = 9 -> max R78C6 = 7, no 7,8 in R78C6
3f. 3 in N5 only in R456C6, locked for C6
3g. R78C6 cannot total 4 -> no 6 in R6C5
3h. 18(3) cage at R8C4 = {279/378/459/567} (cannot be {369} which clashes with R456C4, cannot be {468} which clashes with R9C56)
3i. R7C456 + R8C6 = {1238/1256/1346} (cannot be {1247} which clashes with 18(3) cage), no 7
3j. R7C456 + R8C6 = {1256/1346} (cannot be {1238} = [83]{12} because 16(3) cage at R6C5 = [583] clashes with R456C4), no 8, 6 locked for N8, clean-up: no 7 in R9C56
3k. Hidden killer pair 2,3 in R7C456 + R8C6 and 18(3) cage for N8, R7C456 contains one of 2,3 -> 18(3) cage must contain one of 2,3 = {279/378}, no 4,5
3l. 3 of {378} must be in R8C5 -> no 8 in R8C5
3m. 16(3) cage at R6C5 = {259/268/349/367} (cannot be {358} = [853] which clashes with R7C456 + R8C6, cannot be {457} = 7{45} which clashes with R7C456 + R8C6)
3n. 7,8,9 only in R6C5 -> R6C5 = {789}
3o. 2 of {259} must be in R7C4 -> no 5 in R7C4
3p. 3 of {349} must be in R7C5 -> no 4 in R7C5
3q. R6C5 = {789} -> R78C6 = 5,6,7 = {14/15/16}, no 2 in R78C6
3r. 2 in N8 only in R789C4, locked for C4

4a. 45 rule on C1234 2 innies R17C4 = 1 outie R8C5 + 5
4b. R8C5 = {379} -> R17C4 = 8,12,14 = [62/84/86] -> R1C4 = {68}
4c. R456C4 (step 2b) = {459/567} (cannot be {468} which clashes with R1C4), no 8, 5 locked for N5

5a. 15(4) cage at R8C6 = {1248/1257/1347/1356/2346} (cannot be {1239} = [13]{29} which clashes with R89C9), no 9
5b. 15(4) cage = {1257/1347/1356/2346} (cannot be {1248} because 1{248} clashes with R89C9 and [41]{28} clashes with R9C56), no 8
5c. 15(4) cage = {1257/1347/2346} (cannot be {1356} = [13]{56} which clashes with R89C9)
5d. 45 rule on N9 2 innies R7C79 = 1 outie R8C6 + 8
5e. Hidden killer pair 8,9 in R7C79 and R78C8 for N9, R78C8 contains one of 8,9 -> R7C79 must contains one of 8,9
5f. R8C6 = {1456} -> R7C79 = 9,12,13,14 = {18/39/48/49/58/59} (cannot be {68} which clashes with R89C8), no 2,6,7
5g. Consider combinations for 15(4) cage
15(4) cage = {1257} => R8C9 = 3 (CPE)
or 15(4) cage = {1347/2346} => R8C7 = 3
-> 3 in R8C79, locked for R8 and N9
5h. 18(3) cage at R8C4 = {279} (only remaining combination), 2,9 locked for N8, clean-up: no 4 in R9C56
5i. Naked pair {58} in R9C56, locked for R9, 5 locked for N8, clean-up: no 2 in R8C9
5j. R7C5 = 3 (hidden single in N8) -> R6C5 + R7C4 = 13 = [76/94], no 8 in R6C5
5k. R1C4 = 8 (hidden single in C4), clean-up: no 1 in R2C7 (step 2a)
5l. Killer pair 7,9 in R456C4 and R6C5, locked for N5
5m. 15(4) cage = {1257/1347/2346} = [15]{27}/[13]{47}/[63]{24} (cannot be [43]{26} which clashes with R89C9) -> R8C6 = {16}, R8C7 = {35}
5n. 4 in N8 only in R7C46, locked for R7
5o. 4 in N9 only in R9C789, locked for R9
5p. R9C5 = 8 (hidden single in C5) -> R9C6 = 5
5q. Naked pair {79} in R68C5, locked for C5
5r. Naked triple {456} in R123C5, 4,6 locked for N2
5s. R4C678 (step 2f) = {369/378/459}
5t. 4 of {459} must be in R4C6 -> no 4 in R4C78
5u. 2 in N9 only in R9C789, locked for R9
5v. R8C4 = 2 (hidden single in N8)

6a. R456C4 (step 4c) = {459/567}
6b. Consider combinations for 15(4) cage at R8C6 (step 5m) = [15]{27}/[13]{47}/[63]{24}
15(4) cage = [15]{27}/[13]{47}, 7 locked for R9 => R9C4 = 9
or 15(4) cage = [63]{24} => R7C4 = 4
-> R456C4 = {567}, 6,7 locked for C4 and N5, R68C5 = [97], R79C4 = [49], clean-up: no 8 in R7C8
6c. 19(4) cage at R5C6 = {1468} (only possible combination, cannot be {1369/1459} because 1,5,6,9 only in R7C67) -> R56C6 = {48}, R7C67 = [61], R4C6 = 3 (hidden single in N5), R8C6 = 1, clean-up: no 6 in R56C7, no 6 in R9C9
6d. 15(4) cage = [15]{27}/[13]{47}, 7 locked for R9 and N9 -> R7C8 = 9, R8C8 = 6, R7C9 = 8 (hidden single in N9), R9C1 = 6, clean-up: no 1 in R56C8
6e. Naked pair {35} in R8C79, 5 locked for R8
6f. Naked quad {2345} in R56C78, locked for N6 -> R4C9 = 1, R45C5 = [21], clean-up: no 8 in R2C7 (step 2a)
6g. R4C678 (step 2f) = 18, R4C6 = 3 -> R4C78 = 15 = {78}, locked for R4 and 35(7) cage at R3C5, 7 locked for N6 -> R56C9 = [96] , clean-up: no 3 in R2C7 (step 2a)
6h. Naked pair {69} in R4C23, locked for N4 and 36(6) cage at R1C3, 6 locked for R4 -> R4C14 = [45]
6i. 36(6) cage at R1C3 = {156789/246789}, no 3, 7,8 locked for N1

7a. 7 in C9 only in R12C9, locked for N3, clean-up: no 2 in R1C9 (step 2a)
7b. R3C3 = 7 (hidden single in R3C3)
7c. 8 in N1 only in R2C23, locked for R2, clean-up: no 1 in R1C9 (step 2a)
7d. R3C7 = 8 (hidden single in R3) -> R4C78 = [78]
7e. R9C8 = 7 (hidden single in R9)
7f. R1C7 = 9 (hidden single in C7)
7g. R2C7 = 6 (hidden single in C7) -> R1C9 = 3 (step 2a), R8C9 = 5 -> R9C9 = 2, R23C9 = [74]
7h. R1C6 = 7 (hidden single in N2)
7i. R1C467 = [879] = 24 -> R1C58 = 5 = [41], R2C568 = [592], R3C568 = [625]

8a. R1C2 = 6 (hidden single in N1), R4C23 = [96], R3C1 = 9 (hidden single in N1)
8b. R89C1 = [86] = 14 -> R67C1 = 5 = [32], R7C3 = 5
8c. R7C2 + R8C23 = [749] = 20 -> R6C2 = 5

and the rest is naked singles.

Rating Comment:
I'll rate my WT for A398 at Easy 1.5. I used a fair amount of combination analysis, also a couple of short forcing chains.


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 Post subject: Re: Assassin 398
PostPosted: Mon Aug 10, 2020 7:25 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
And thanks as always for the WT Andrew. You made a lot more out of your key areas than I could, even though I looked hard there, hence, had to find a different way (steps 9 & 10, & 15). Glad you mentioned about the blanks cells. I decided to leave them in as a distraction even though they weren't much use (though I managed to get one nice step with them (22b)). Thanks to Andrew for a couple of typos.

A398 WT:
Preliminaries from SudokuSolver
Cage 3(2) n5 - cells ={12}
Cage 4(2) n2 - cells ={13}
Cage 4(2) n7 - cells ={13}
Cage 15(2) n9 - cells only uses 6789
Cage 7(2) n9 - cells do not use 789
Cage 7(2) n6 - cells do not use 789
Cage 7(2) n6 - cells do not use 789
Cage 13(2) n8 - cells do not use 123
Note: no clean-up done unless stated.
1. 4(2) cages in both n2 and n7 = {13}: both locked for their column, row and nonet

2. 3(2)n5 = {12}: both locked for c5 and n5

3. "45" on c123: 3 outies r456c4 = 18 = {459/468/567} = 6 or 9, 5 or 8 (no eliminations yet)

4. "45" on n8: 1 outie r6c5 - 2 = 2 innies r78c6
4a. -> min. r6c5 = 5
4b. -> max. r78c6 = 7 (no 7,8,9)

5. 3 in c5 only in r78c5 in n8: locked for n8
5a. {69}[3] blocked from 18(3)n8 by h18(3) (step 3)
5b. -> if 3 in r8c5 in 18(3) = {78}[3] only
5c. and if 3 in r7c5 in 16(3): {58}[3] blocked by h18(3) (step 3)
5d. -> with 3, 16(3) = {49/67}[3]
5e. -> 3 in r78c5 must also have 4 or 6 or 8 in one of r6c5 + r789c4 (no eliminations yet)

6. h18(3)r456c4: {468} blocked since it would leave no 3 in n8
6a. = {459/567}(no 8)
6b. 5 locked for c4 and n5
[note: steps 3-5 could have been excluded and this step 6 would still work but with more combinations. However, I only like to work with 4 combos, and its a bit simpler to break a big step into bits.]

7. "45" on n1: 2 outies r4c23 = 15 = {69/78}(no 1...5)

8. "45" on r5..9: 4 outies r4c1459 = 12 = {1236/1245}(no 7,8,9)
8a. 1 and 2 locked for r4
8b. 6 in {1236} must be in r4c4 -> no 6 in r4c19

The killer steps
9. "45" on n7: 4 outies r6c1234 = 16 = h16(4)
9a. r4c4 sees all n4 apart from r6c123 so must repeat there
9b. -> r46c4 are part of the h16(4) and also h18(3)r456c4. Since an 18(3) can't fit into a 16(4)
9c. -> r5c4 cannot be in the h16(4) -> must repeat in n4 in r4c23
9d. -> r5c4 from (679)

10. h18(3)r456c4 = {45}[9]{45}/{567}(no 9 in r6c4)
10a. when {45} in r46c4 -> h16(4) must have {45} (step 9a) -> the other two cells = 7 = {16} only
10b. also with {45}[9]{45} in r456c4, 9 also in r4c23 in the h15(2) = {69} (step 9c), but 6 is also in r6c123! -> {45}[9]{45} blocked
10c. -> r456c4 = {567} only: 6 & 7 both locked for c4 and n5

11. 18(3)n8 cannot have 3 (step 5b)
11a. -> r7c5 = 3 (hsingle)
11b. -> r6c5 + r7c4 = 13 = [94]

12. 18(3)n8 ={29}[7]: only valid combination, 2 & 9 locked for c4 and n8
12a. r1c4 = 8

13. naked triple {456} in r123c5: all locked for n2, 5 & 6 for c5
13a. r9c56 = [85]

14. "45" on r4...9, including h15(2)r4c23 -> 3 innies r4c678 = 18
14a. -> "45" on n6: 2 outies r4c6 + r7c9 = 11 = [38/47]

15. r7c9 = (78) and sees all 7 & 8 in n6 apart from r4c78 so must repeat there
15a. -> {78} blocked from h15(2)r4c23
15b. -> r4c23 = {69}: both locked for r4, n4 and 36(6)
15c. -> r5c4 = 6 (step 9c)
15d. r46c4 = [57]

16. h12(4)r4c1459 = {1245} only: 4 locked for r4
16a. r4c6 = 3
16b. -> r7c9 = 8 (outies n6 = 11)

17. r56c6 = {48} = 12 -> r7c67 = 7 = {16} only: 6 locked for r7
17a. r78c8 = [96] only permutation
17b. r7c67 = [61], r8c6 = 1

18. 32(6)r4c1 = {2478}[56]
18a. 2,4,8 all locked for n4, 8 also for r5
18b. r67c3 = 6 (cage sum) = [15]
18c. r9c23 = [13]

19. r9c1 = 6 (hsingle n7)
19a. -> r678c1 = 13 = [328]
19a. -> r67c2 = [57], r45c1 = [47]

20. r9c789 = {247}: 2 & 4 locked for n9, 2 for r9
20a. r89c4 = [29]

21. two 7(2) cages n6 = [34/52], all locked for n6
21a. r456c9 = [196], r56c6 = [48]

22. "45" on whole grid -> 2 zero cells in n3 = 9 (no 8,9)
22a. "45" on c9 including the h9(2) -> 1 outie r2c7 + 5 = 2 innies r23c9
22b. -> no 5 in r23c9 since the difference can't be 0 (Innie/Outie difference unequal, IOU)

23. r4c78 = {78}: both locked for 35(7) cage

24. 31(5)r2c5 must have 9 which is only in r2c6 + r3c7
24a. r3c6 sees both those -> no 9 in r3c6
24b. r3c6 = 2, r3c9 = 4
24c. r3c58 = 11 (cage sum) = [65]

25. r56c56 = [5324]

26. "45" on c9: 2 remaining innies r12c9 = 10 = {37}: both locked for n3
26a. r2c7 = 6 -> r1c9 = 3 (h9(2))

27. 31(5)r2c5 = {25789}(no 1,4)
27a. -> r2c59 = [57]

easy now
Cheers
Ed


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 Post subject: Re: Assassin 398
PostPosted: Wed Aug 26, 2020 4:20 am 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Thanks as always Ed. here's how I did it. It is not optimized - I found afterwards that if I had done one of my later steps much earlier it would have simplified things. (See comment in WT). Haven't looked through the other WTs yet. Will try and do that soon.
Corrections & clarifications thanks to Andrew.
Assassin 398 WT:
1. 4(2)n7 = {13}
3(2)n5 = {12}
4(2)n2 = {13}
-> 1 in n8 in r78c6

2. IOD n8 -> r6c5 = r78c6 + 2
-> r6c5 is Min 5 and three more than the other value in r78c6 (along with the 1 already there)
-> 3 in n5 in c6
-> 3 in n8 in r78c5

3. Outies n1 = r4c23 = +15(2) = {69} or {78}
32(6)n45 cannot be missing two values that sum to 15.
-> One of the values in r4c23 must also be in r5c4. I.e., r5c4 from (6789)

4. Outies n23 = r4c678 = +18(3)
-> Remaining Innies r4 = r4c1459 = +12(4) = {12(36|45)}
-> r4c4 from (456)

5. Outies c123 = r456c4 = +18(3)
This can only be one of:
(A) [468]
(B) [567]
(C) <576> (<XYZ> means [XYZ] or its reverse [ZYX])
(D) <486>
(E) <495>

6. Outies n7 = r6c1234 = +16(4)
Whatever is in r4c4 must go in n4 in r6c123 (I.e., be one of the +16(4))
-> Outies c123 cannot be <495> since that puts r4c23 = {69} and r6c1234 = {1456} with 6 in r6c123
-> 5(E) is eliminated

7! Consider case where Outies n1 r4c23 = {78}
This puts r56c9 = {78}
and puts r47c9 = {36} or {45}
and puts the 7(2)s in n6 = {16} and {25}
and puts one of 5(C) or 5(D)

But 5(C) is eliminated since r6c1234 would have to be {56(14|23)} which both contradict the 7s in n6
and 5(D) is eliminated since r6c1234 would have to be {4615} which contradicts the {16} in n6

-> Only 5(A) and 5(B) left
-> r5c4 = 6
-> r4c23 = {69}

8. -> r4c678 = {378} and r4c1459 = {1245}
Also 9 in n6 in r56c9
Also 6 in n6 in r6
Also 32(6)n45 can only be {245678} with r4c4 from (45) and r5c4 = 6 and (78) in r5
-> (13) in n4 in r6
Also 24(5)n1 = {369(15|24)} and 36(6)n14 in n1 = {78(15|24)}

9. Outies n236 = r4c6 + r7c9 = +11(2)
These cannot be [74] since that puts r4c78 = {38} and no place for 4 in n6
-> r4c6,r7c9 = {38} and 7 in r4c78

10. Trying r4c6 = 8 puts r456c4 = [567] puts r56c6 = [34] and r6c5 = 9
But this puts r78c6 = {16} which leaves no solution for 19(4)r5c6
-> r4c6,r7c9 = [38]
-> r4c78 = {78}
-> 15(2)n9 = {69}

11! Remaining IOD n9 -> r7c7 = r8c6
-> r7c7 cannot be 3
-> 3 in n9 in r8c79
-> (HS 3 in n8) r7c5 = 3
-> r456c4 cannot be [468] since that leaves no solution for 16(3)r6c5
-> r456c4 = [567]

(I noticed later that we can get the result of Step 11 (r7c5 = 3 and r456c4 = {567}) immediately after step 6!
I.e., r456c4 = {567} or {468} -> No solution for 18(3)n8 with 3 in r8c5 -> r7c5 = 3, etc.
That would make subsequent steps to it much different and much easier!


12. Continuing...
-> r6c123 = {135}
Also (Since r6c5 is now min 8) -> 19(3)r6c5 = [943]
-> r78c6 = {16}
-> 19(4)r5c6 = [4861]
-> r8c6 = 1
-> 15(2)n9 = [96]
-> r9c1 = 6

13. 7 in c9 only in r12c9
-> (HS 7 in r3) r3c3 = 7
-> (HS 8 in r3) r3c7 = 8
-> r4c78 = [78]
-> r9c8 = 7

14. Innies whole puzzle = r1c9 + r2c7 = +9(2)
-> (HS 9 in n3) r1c7 = 9
Since 31(5) must contain a 9 -> 9 in n2 in r2c6
-> 9 in n1 in r3c12
Also 9 in n8 in r89c4
-> 13(2)n8 = [85]
-> r8c79 = {35}
-> r9c79 = {42}
-> 18(3)n8 = [279]
-> r8c123 = {489}
-> r7c123 = {257}

15. Also HS 7 in c6 -> r1c6 = 7
Also NS 8 in c4 -> r1c4 = 8
-> 29(5)r1 must be [84791]
-> HS 7 in c9 -> r2c9 = 7
Also HS 6 in c7 -> r2c7 = 6
-> r1c9 = 3
-> Remaining Innie c9 -> r3c9 = 4
Also 1 in c9 only in r45c9
-> 24(4)c9 = [1968]
-> 7(2)n9 = [52] and r89c7 = [34]
etc.


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