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 Post subject: Assassin 348
PostPosted: Sat Mar 31, 2018 11:32 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 749
Location: Sydney, Australia
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a348.JPG
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This is an X killer. 1-9 cannot repeat on either diagonal.

Assassin 348
Happy Easter! Have very fond memories from Easter 2006, see here

I used a couple of interesting moves in A348: with one I'd call advanced. Perhaps that step has a more conventional way but I couldn't see it. JSudoku came up with this pattern but SudokuSolver has a very hard time with it giving it a 1.90. Normally, I'd ditch it straight away, but when I checked "Deduce all moves", JS had no trouble with it! Glad I kept it and tried it. Really like it.

code: paste into solver:
3x3:d:k:8192:8192:8192:8192:2817:2817:1538:1538:1538:5123:8192:3076:8192:2565:2822:2822:3335:3592:5123:5123:3076:3076:2565:5385:5385:3335:3592:4874:5123:3076:5643:7436:7436:5385:5389:5389:4874:5643:5643:5643:7436:4110:4110:4110:5389:4874:4874:4111:7436:7436:4110:4624:4113:5389:3858:3347:4111:4111:1556:4624:4624:4113:4113:3858:3347:2837:2837:1556:8470:4624:8470:4113:2583:2583:2583:2584:2584:8470:8470:8470:8470:
solution:
Code:
+-------+-------+-------+
| 4 7 8 | 9 6 5 | 3 1 2 |
| 5 3 6 | 1 2 4 | 7 9 8 |
| 2 9 1 | 3 8 7 | 5 4 6 |
+-------+-------+-------+
| 7 4 2 | 5 3 6 | 9 8 1 |
| 8 6 9 | 2 7 1 | 4 3 5 |
| 3 1 5 | 4 9 8 | 6 2 7 |
+-------+-------+-------+
| 6 5 3 | 8 1 9 | 2 7 4 |
| 9 8 4 | 7 5 2 | 1 6 3 |
| 1 2 7 | 6 4 3 | 8 5 9 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 348
PostPosted: Wed Apr 04, 2018 5:22 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1587
Location: Lethbridge, Alberta, Canada
Thanks Ed for your latest Assassin. I'm glad that you decided to keep it. :D At first it seemed hard but then I found a couple of key steps, after which the rest was easy.

Here is my walkthrough for Assassin 348:
Prelims

a) R1C56 = {29/38/47/56}, no 1
b) R23C5 = {19/28/37/46}, no 5
c) R2C67 = {29/38/47/56}, no 1
d) R23C8 = {49/58/67}, no 1,2,3
e) R23C9 = {59/68}
f) R78C1 = {69/78}
g) R78C2 = {49/58/67}, no 1,2,3
h) R78C5 = {15/24}
i) R8C34 = {29/38/47/56}, no 1
j) R9C45 = {19/28/37/46}, no 5
k) R1C789 = {123}
l) 21(3) cage at R3C6 = {489/579/678}, no 1,2,3
m) R9C123 = {127/136/145/235}, no 8,9
n) 12(4) cage at R2C3 = {1236/1245}, no 7,8,9

Steps resulting from Prelims
1a. Naked triple {123} in R1C789, locked for R1 and N3, clean-up: no 8,9 in R1C56 + R2C6
1b. R23C8 = {49/67} (cannot be {58} which clashes with R23C9), no 5,8
1c. Killer pair 6,9 in R23C8 and R23C9, locked for N3, clean-up: no 2,5 in R2C6
1d. R78C2 = {49/58} (cannot be {67} which clashes with R78C1), no 6,7
1e. Killer pair 8,9 in R78C1 and R78C1, locked for N7, clean-up: no 2,3 in R8C4
1f. R23C5 = {19/28/37} (cannot be {46} which clashes with R1C56, no 4,6

2a. 45 rule on R1 2 outies R2C24 = 4 = {13}, locked for R2, clean-up: no 8 in R2C7, no 7,9 in R3C5
2b. 45 rule on N3 2 innies R23C7 = 12 = [48/57/75], no 4 in R3C7
2c. R23C7 = 12 -> R34C7 cannot total 12 (CCC) -> no 9 in R3C6
2d. Hidden killer triple 1,2,3 in R2C4, R23C5 and R3C4 for N2, R2C4 = {13}, R23C5 must contain one of 1,2,3 -> R3C4 = {123}
2e. 45 rule on N7 2 innies R78C3 = 7 = [16]/{25/34}, no 6,7 in R7C3, no 7 in R8C3, clean-up: no 4 in R8C4
2f. Max R7C3 = 5 -> min R6C3 + R7C4 = 11, no 1 in R6C3 + R7C4
2g. R78C3 = 7 -> R67C3 cannot total 7 (CCC) -> no 9 in R7C4
2h. 45 rule on R9 2 outies R8C68 = 8 = {17/26/35}, no 4,8,9

3a. 45 rule on C1234 2 innies R69C4 = 10 = {19/28/37/46}, no 5
3b. 45 rule on C6789 2 innies R14C6 = 11 = {47/56}
3c. Combined half-cage R14C6 + R2C6 = {47}6/{56}4/{56}7, 6 locked for C6, clean-up: no 2 in R8C8 (step 2h)

4. R1C56 = {47/56}, R2C67 = {47}/[65] -> combined cage R1C56 + R2C67 = {47}[65]/{56}{47}, CPE no 7 in R2C5, clean-up: no 3 in R3C5
4a. 3 in N2 only in R23C4, locked for C4, clean-up: no 7 in R69C4 (step 3a), no 3,7 in R9C5
4b. Killer pair 1,2 in R23C5 and R78C5, locked for C5, clean-up: no 8,9 in R9C4, no 1,2 in R6C4 (step 3a)
4c. 3 in C5 only in R456C5, locked for N5

[Step 5 probably isn’t needed, but I saw it before I saw step 6.]
5. 45 rule on C1234 1 outie R9C5 = 1 innie R6C4, 45 rule on C6789 1 outie R1C5 = 1 innie R4C6 -> 29(5) cage at R4C5 must contain the same numbers as R14569C5
5a. R14569C5 = 29(5) = {34679/35678} (cannot be {34589} which clashes with R78C5)

6. R23C5 = {28}[91], R78C5 = {15/24} -> combined cage R2378C5 = {28}{15}/[91]{24} -> R9C45 = {46} (cannot be [19/28] which clash with combined cage), locked for R9 and N8, clean-up: no 8,9 in R6C4 (step 3b), no 2 in R78C5, no 5 in R8C3, no 2 in R7C3 (step 2e)
[Cracked. The rest is straightforward.]
6a. Naked pair {15} in R78C5, locked for C5 and N8, clean-up: no 6 in R1C6, no 9 in R2C5, no 5 in R4C6 (step 3b), no 6 in R8C3, no 1 in R7C3 (step 2e), no 3,7 in R8C8 (step 2h)
6b. Naked pair {28} in R23C5, locked for C5 and N2
6c. Naked pair {13} in R23C4, locked for C4
6d. Naked pair {46} in R69C4, locked for C4
6e. R1C4 = 9 (hidden single in N2), clean-up: no 2 in R8C3
6f. 8 in R1 only in R1C123, locked for N1

7. 1 in N7 only in 10(3) cage at R9C1 = {127}, locked for R9 and N7, clean-up: no 8 in R78C1
7a. Naked pair {69} in R78C1, locked for C1 and N7, clean-up: no 4 in R78C2
7b. Naked pair {58} in R78C2, locked for C2 and N7
7c. Naked pair {34} in R78C3, locked for C3
7d. 12(4) cage at R2C3 = {1236} (only remaining combination) -> R3C4 = 3, R2C24 = [31], 3 placed for D\
7e. Naked triple {126} in R234C3, locked for C3 -> R9C3 = 7
7f. 9 in C3 only in R56C3, locked for N4
7g. R3C2 = 9 (hidden single in N1), clean-up: no 4 in R2C8, no 5 in R2C9
7h. R3C2 = 9 -> R23C1 + R4C2 = 11 = {146/245}, no 7, CPE no 4 in R456C1
7i. 6 of {146} must be in R4C2 -> no 1 in R4C2
7j. 7 in N1 only in R1C12, locked for R1, clean-up: no 4 in R1C56, no 4,7 in R4C6 (step 3b)
7k. R1C56 = [65], R1C3 = 8, R9C45 = [64], R4C6 = 6, R6C4 = 4, both placed for D/ -> R7C3 = 3, placed for D/, R8C3 = 4 -> R8C4 = 7
7l. Naked quad {3589} in R9C6789, locked for 33(6) cage at R8C6 -> R8C6 = 2 -> R8C8 = 6 (step 2h), placed for D\, R78C1 = [69], R7C4 = 8, R7C3 = 3 -> R6C3 = 5 (cage sum), R7C2 = 5, R8C2 = 8, placed for D/, R79C6 = [93]
7m. R45C4 = {25}, R5C3 = 9 -> R5C2 = 6 (cage sum)
7n. R5C5 = 7, placed for both diagonals, R1C1 = 4, placed for D\, R1C2 = 7
7o. R2C8 = 9 -> R3C8 = 4
7p. R3C67 = [75] -> R4C7 = 9 (cage sum), R2C67 = [47], R9C789 = [859]
7q. R7C5 = 1 -> R7C7 = 2, placed for D\
7r. R6C7 = 6 (hidden single in C7), R7C67 = [92] -> R8C7 = 1 (cage sum), R1C7 = 3
7s. R56C6 = {18}, R5C7 = 4 -> R5C8 = 3 (cage sum)
7t. R7C89 = [74], R8C9 = 3 -> R6C8 = 2 (cage sum)
7u. R1C8 = 1 -> R1C9 = 2, placed for D/

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

Rating Comment:
I'll rate my walkthrough for A348 at Hard 1.25 for steps 4 and 6; maybe it should just be 1.25.


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 Post subject: Re: Assassin 348
PostPosted: Wed Apr 11, 2018 6:41 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 749
Location: Sydney, Australia
Wonderful solution Andrew! Though did skip step 5 since I couldn't understand it. Not required anyway. I can't see a way to block up Andrew's path without killing my own. Would like to have made him work a bit.

A348 WT:
Had to work harder but found one interesting step (11).
Preliminaries courtesy of SudokuSolver
Cage 6(2) n8 - cells only uses 1245
Cage 14(2) n3 - cells only uses 5689
Cage 15(2) n7 - cells only uses 6789
Cage 13(2) n3 - cells do not use 123
Cage 13(2) n7 - cells do not use 123
Cage 10(2) n2 - cells do not use 5
Cage 10(2) n8 - cells do not use 5
Cage 11(2) n2 - cells do not use 1
Cage 11(2) n23 - cells do not use 1
Cage 11(2) n78 - cells do not use 1
Cage 6(3) n3 - cells ={123}
Cage 21(3) n236 - cells do not use 123
Cage 10(3) n7 - cells do not use 89
Cage 12(4) n124 - cells do not use 789

1. "45" on n7: 2 innies r78c3 = 7 = [16]{25/34}(no 7,8,9; no 6 in r7c3)=[1/4/5..]

2. 10(3)r9c1: {145} blocked by r78c3
2a. = {127/136/235}(no 4) = [3/7..]

3. 10(2)r9c4: {37} blocked by 10(3)n7 (step 2a)
3a. = {19/28/46}(no 3,7)

4. 6(3)n3 = {123}: all locked for r1 and n3
4a. no 8,9 in 11(2)r1c5

5. "45" on r1: 2 outies r2c24 = 4 = {13} only: both locked for r2 and neither are in r4c4 since it sees both outies through D\
5a. no 8 in r2c67

6. "45" on n3: 2 innies r23c7 = 12 = [48]{57}(no 6,9; no 4 in r3c7)
6a. r2c6 = (4,6,7)

7. "45" on c6789: 2 innies r14c6 = 11 = {47/56}(no 1,2,3,8,9)

8. "45" on n3: 2 outies r3c6 + r4c7 - 9 = 1 innie r2c7
8a. since one innie sees one outie the remaining outie can't equal the IOD of 9 -> no 9 in r3c6 (IOU)

9. 11(2)r1c5 = {47/56} and r2c6 = (467) -> 6 locked for n2 in one of those 3 cells
9a. no 4 in 10(2)n2

10. 21(3)r3c6 = {489/579/678}: must have 6,9 which are only in r4c7 -> r4c7 = (69)

The Almost locked set of {45678} = 30 in r1234c6 is what got my attention now. This is a combined outies + hidden cage step
11. 3 outies n3 = 20 + h11(2)r14c6 (step 7) combine to sum to 31: ie, r1234c6 + r4c7 = 31
11a. = {4567}+[9]/{4678}+[6]
11b. must have 4,6,7: all locked for c6
(another way to see this: "45" on n69: 5 outies r56789c6 - 14 = 1 innie r4c7. Outies must have {1239} = 15 for c6 -> one remaining outie + 1 = r4c7 -> can only have 5,8 in one remaining outie-> no 4,6,7)

12. 7 in n8 only in r78c4: locked for c4

13. hidden killer triple 1,2,3 in n2: 10(2) has one of 1,2,3 -> r3c4 = (123)

14. "45" on n23: 1 outie r4c7 + 4 = 3 innies r123c4
14a. -> r123c4 = 10/13 = [532/832/913/931](no 4)
14b. must have 3: locked for n2 and c4
14c. no 7 in 10(2)n2

15. killer pair 1,2 in 10(2)n2 and 6(2)n8: both locked for c5
15a. no 8,9 in r9c4

16. 1 in c5 only in 10(2)n2 = [91] or in 6(2)n8 -> [19] blocked from 10(2)n8
16a. 10(2)n8 = [28]{46}(no 1,9) = [2/4..]

17. 6(2)n8: {24} blocked by 10(2)n8
17a. r78c5 = {15} only: both locked for c5 and n8

18. 10(2)n2 = {28} only: both locked for n2: 8 for c5

19. r123c4 = [9]{13} only (step 14a): 1 locked for c4
19a. r4c7 = 9 (step 14a)

20. 10(2)n8 = {46} only: both locked for n8 and r9

21. 11(2)r8c3 = [38/47]
21a. -> h7(2)r78c3 = {34} only: both locked for n7 and c3

22. 10(3)n7 = {127} only: all locked for r9 and 7 for n7
22a. 15(2)n7 = {69} only: both locked for c1 and n7
22b. 13(2)n7 = {58}: both locked for c2

23. 16(3)r6c3 must have 3,4 (but can't have both) for r7c7 = {358/367/457}(no 1,2,9)
23a. must have 5,6 which are only in r6c3 -> r6c3 = (56)

24. naked pair {78} in r78c4: 8 locked for c4 and n8

25. 5 & 8 in r9 only in n9: locked for n9

26. "45" on r9: 2 outies r8c68 = 8 = [26] only permutation: 6 locked for D\

27. r789c6 = {239} = 14 -> "45" on n78: 1 remaining outie r6c3 +9 = r789c6
27a. r6c3 = 5

28. naked triple {126} in r234c3: all locked for c3 and no 1 in r3c4

29. "45" on r123: 2 remaining outies r4c23 = 6 = [42] only permutation

30. r789c6 = 14 -> from "45" on n9: 2 remaining outies r6c78 = 8 = {17}[62](no 3,4,8)

on from there. Remember the Diagonals!
Cheers
Ed


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 Post subject: Re: Assassin 348
PostPosted: Mon Apr 30, 2018 12:42 am 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 158
Location: California, out of London
Thanks Ed. I found this much easier than the previous puzzle. Nowhere near a 1.90 for sure! Apologies for the delay in posting. I actually wrote most of it a couple of days after you posted, but then got distracted with work and forgot to finish it. Keep up the good work!

Assassin 348 WT:
1. 6(3)r1 = {123}
Outies r1 = +4(2) -> r2c24 = {13}
Innies n3 -> r23c7 = +12(2) = {57} or [48]. (Cannot be [84] since 11(2)r2c6 cannot be {38}).
-> r34c7 cannot be +12(2)
-> r3c6 cannot be 9.

2. In n7 one of (123) in r78c3 and the other two in r9c123.
-> 10(3)n7 from {127}, {136}, or {235}
-> 10(2)n8 cannot be {37}

3. 6(2)n8 from {15} or {24}
If the former -> 10(2)n2 cannot be {19}
If the latter -> 10(2)n8 = {19} -> 10(2)n2 cannot be {19}
-> HS 9 in n2 -> r1c4 = 9
-> 1 in n2 in r23c4

4. 11(2)r1c5 from {47} or {56}
-> 10(2)n2 cannot be {46}
Innies n3 -> r23c7 = +12(2) = {57} or [48].
-> 11(2)r2c6 from [65], [47], [74]
In no case can 10(2)n2 be {37}
-> 10(2)n2 = {28}
-> 6(2)n8 = {15}

5. Also 3 in n2 only in r23c4
-> r23c4 = {13}
-> Remaining innies n2 -> r23c6 = +11(2)
-> Remaining outies n3 -> r4c7 = 9

6. Innies c1234 = r69c4 = +10(2) -> r6c4 = r9c5
r69c4 cannot be {37} or {19} and 10(2)r9c4 cannot be {28}
-> r69c4 = {46} and 10(2)r9c4 = {46}

7. Innies n7 -> r78c3 = +7(2)
This cannot be {16) since neither 1 or 6 can go in r8c3.
-> 1 in n7 in r9c123
-> 10(3)n7 = {127}
-> 15(2)n7 = {69}
-> 13(2)n7 = {58}
-> r78c3 = {34}

8. Since 12(4)r2c3 must contain one of (34)
-> r3c4 = 3
-> r2c24 = [31]
Also r234c3 = {126}
Remaining outies n1 = r4c23 = +6(2)
-> 6 in r23c3
-> HS 6 in r1 -> 11(2)r1c5 = [65]
-> r1c123 = {478}
-> 5 in n1 in 20(4)
-> r4c23 = [42]
-> r23c3 = [61]
-> 20(4)n1 = [{25}94]
-> 10(3)n7 = [127]
-> r1c123 = [478]

9. Also 10(2)n8 = [64]
-> r6c4 = 4
Also remaining Innie c6789 -> r4c6 = 6
->r456c5 = {379}

10. HP r56c3 = {59}
-> Remaining Innie n4 -> r5c2 = 6
-> 19(4)n4 = [{378}1]
Trying r5c3 = 5 gives no solution for 22(4)r4c4
-> r56c3 = [95]
-> r45c4 = [52]
-> r56c6 = [18]
-> r5c78 = {34}
-> Remaining Innies n6 -> r6c78 = {26}
Also HS 5 in r5 -> r5c9 = 5
-> 21(4)n6 = [{18}57]
-> 19(4)n4 = [7831]
-> r456c5 = [379]

Also 14(2)n3 = [86]
-> 13(2)n3 = [94]
-> 11(2)r2c6 = [47]
-> r3c67 - [75]
Also r4c89 = [81]
Also r5c78 = [43]
Also HS 5 in c8 -> r9c8 = 5

11. 4 in c9 in r78c9
-> No solution for 16(4)r6c8 if r6c8 = 6
-> r6c78 = [62]
-> 6 not in 16(4)r6c8
-> HS 6 in c8 r8c8 = 6
-> Remaining outie r9 -> r8c6 = 2
-> HS 2 in r7 -> r7c7 = 2
-> 18(4)r6c7 = [6921]
-> r9c69 = [39]
-> 16(4)r6c8 = [27{34}]
-> r7c34 = [38]
-> r8c34 = [47]
-> r78c9 = [43]
-> 13(2)n7 = [58]
-> 6(2)n8 = [15]
Finally 6(3)r1 = [312]


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