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 Post subject: Assassin 394
PostPosted: Wed Apr 01, 2020 8:17 am 
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Posts: 1040
Location: Sydney, Australia
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X-puzzle so 1-9 cannot repeat on the diagonals

Assassin 394
I found this quite hard and had to use a couple of chains and some big combo work. However, it has some interesting interactions that kept me entertained! It gets 1.65. JSudoku uses 3 "complex intersections".

triple click code:
3x3:d:k:4864:4864:4864:5889:5889:5889:1026:4867:4867:7940:2053:2053:2053:6918:5889:1026:6407:4867:7940:7940:7940:6918:6918:6407:6407:6407:5640:7940:7940:5385:8458:6918:6407:6407:5640:5640:3083:5385:8458:8458:8458:2316:2316:4365:5640:3083:5385:5385:8458:8458:8206:4365:4365:5640:3083:5385:8206:8206:8206:8206:4879:4879:4879:4880:4880:8206:2321:2321:3090:4879:3091:3091:4880:4628:4628:4628:4628:3090:3605:3605:3605:
solution:
Code:
+-------+-------+-------+
| 3 9 7 | 2 8 4 | 1 6 5 |
| 4 2 5 | 1 6 9 | 3 7 8 |
| 6 8 1 | 7 5 3 | 2 4 9 |
+-------+-------+-------+
| 7 5 3 | 6 9 1 | 8 2 4 |
| 9 1 8 | 5 4 2 | 7 3 6 |
| 2 4 6 | 3 7 8 | 9 5 1 |
+-------+-------+-------+
| 1 7 9 | 4 3 6 | 5 8 2 |
| 5 6 2 | 8 1 7 | 4 9 3 |
| 8 3 4 | 9 2 5 | 6 1 7 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 394
PostPosted: Sat Apr 04, 2020 10:21 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks Ed for another interesting Assassin; yes it was entertaining at times when I found some of the interactions. It was the sort of puzzle where, at times I moved away to other areas, my steps 3 and 6, but then came back to where I'd been working before and made further progress.

Here is my walkthrough for Assassin 394:
Prelims

a) R12C7 = {13}
b) R5C67 = {18/27/36/45}, no 9
c) R8C45 = {18/27/36/45}, no 9
d) R89C6 = {39/48/57}, no 1,2,6
e) R8C89 = {39/48/57}, no 1,2,6
f) 19(3) cage at R1C1 = {289/379/469/478/568}, no 1
g) 19(3) cage at R1C8 = {289/379/469/478/568}, no 1
h) 8(3) cage at R2C2 = {125/134}
i) 19(3) cage at R8C1 = {289/379/469/478/568}, no 1
j) 27(4) cage at R2C5 = {3789/4689/5679}, no 1,2

1a. 8(3) cage at R2C2 = {125} (cannot be {134} which clashes with R2C7), locked for R2 -> R12C7 = [13]
1b. 45 rule on R1 2 remaining outies R2C69 = 17 = {89}, locked for R2
1c. 45 rule on C789 (or N36) 3 outies R345C6 = 6 = {123}, locked for C6, R5C7 = {678}, clean-up: no 9 in R89C6
1d. R8C45 = {18/27/36} (cannot be {45} which clashes with R89C6), no 4,5
1e. 45 rule on R9 2 innies R9C16 = 13 -> R9C1 = {5689}
1f. 45 rule on R89 2 innies R8C37 = 6 = [15/24/42]
1g. 45 rule on N78 1 outie R6C6 = 2 innies R7C12, no 9 in R7C12
1h. 27(4) cage at R2C5 = {3789/4689/5679}, CPE no 9 in R1C5
1i. 3,4 of {3789/4689} must be in R2C5 + R3C56 (R3C56 cannot be {89} which clashes with R2C6), no 3,4 in R4C5

2a. 45 rule on R9 3 outies R8C126 = 18 = {279/468/567} (cannot be {369} because no 3,6,9 in R8C6, cannot be {378} which clashes with R8C89, cannot be {459} which clashes with R8C37), no 3
2b. 7 of {279} must be in R8C6, 7 of {567} must be in R8C6 (R8C12 cannot contain both of 6,7) -> no 7 in R8C12
2c. {279} must be {29}7 with R9C1 = 8 (cage total), 6 of {468/567} must be in R8C12 -> no 6 in R9C1, clean-up: no 7 in R9C6 (step 1e), no 5 in R8C6
2d. 9 in R8 only in R8C126 = {279} = {29}7
or in R8C89 = {39}
-> no 7 in R8C89 (locking-out cages), clean-up: no 5 in R8C89
2e. 7 in R8 only in R8C456, locked for N8
2f. 45 rule on N78 7 innies R7C123456 + R8C3 = 32 form 32(7) cage because they ‘see’ each other = {1234589/1234679/1235678}, 3 locked for R7
2g. R7C123456 + R8C3 = {1234589/1234679} (cannot be {1235678} because 4,9 in R7C789 clashes with R8C89), 9 locked for R7 and 32(6) cage
2h. Max R6C6 = 8 -> max R7C12 (step 1g) = 8, no 8 in R7C12
2i. R7C789 must contain 5,8 or 6,7 and one of 1,2,4
2j. 14(3) cage at R9C7 = {158/167/239} (cannot be {149/347} which clash with R8C89, cannot be {257/356} which clash with R7C789, cannot be {248} which clashes with R9C16), no 4

3a. 45 rule on N6 2 innies R45C7 = 1 outie R3C9 + 6
3b. R45C7 cannot be [26/28] which eliminate 2 from C6 and cannot be [27] because no 3 in R3C9 -> no 2 in R4C7
3c. Min R45C7 = 10 -> min R3C9 = 4

4a. Consider combinations for R8C37 (step 1f) [15/24/42]
R8C37 = [15] => 1 in R7 must be in R7C89 => 14(3) cage at R9C7 (step 2j) = {239}, locked for N9 => R8C89 = {48}
or R8C37 = {24}
-> 4 in R8C3789, locked for R8
4b. R8C126 (step 2a) = {279/567} -> R8C6 = 7, R9C6 = 5, R9C1 = 8 (step 1e), placed for D/, clean-up: no 2 in R8C45

5a. R6C6 = R7C12 (step 1g)
5b. R7C123456 + R8C3 (step 2g) contains 4
5c. Either R7C12 contains 4 => R6C6 greater than 4
or R6C6 ‘sees’ 4 in R7C3456+ R8C3
-> no 4 in R6C6
5d. R6C6 = {68} -> R7C12 = 6,8 = {15/24/17/35} (cannot be {26} which clashes with R8C12), no 6
5e. Whichever of 6,8 is in R6C6 must also be in R7C89, R7C789 cannot contain both of 6,8 (which clashes with R7C3456, there’s no 6,8 in R7C12 + R8C3) -> 32(6) cage at R6C6 must contain both of 6,8 = {135689/234689} (cannot be {245678} because R7C3456 must contain 9 for R7), no 7, 3 locked for R7
5f. R7C789 cannot contain both of 6,8 -> no 6,8 in R7C7

6a. 45 rule on N2356 2 innies R2C4 + R6C6 = 1 outie R5C3 + 1
6b. Min R2C4 + R6C6 = 7 -> min R5C3 = 6
6c. Max R2C4 + R6C6 = 10 -> no 5 in R2C4
6d. 5 in R2 only in R2C23, locked for N1

7a. R8C37 (step 1f) = [15/24/42]
7b. R7C12 (step 5d) = {15/17} (cannot be {24} because 2 in R7C12 + 5 in R8C35 = [15] clash with R8C12), no 2,4, 1 locked for R7 and N7, clean-up: no 5 in R8C7
[And the rest is fairly straightforward.]
7c. Naked pair {24} in R8C37, locked for R8, clean-up: no 8 in R8C89
7d. Naked pair {39} in R8C89, locked for R8 and N9, clean-up: no 6 in R8C45
7e. Naked pair {56} in R8C12, locked for N7
7f. Naked pair {17} in R7C12, 7 locked for R7 and N7
7g. R7C12 = {17} = 8 -> R6C6 (step 1g) = 8, placed for D\, R2C69 = [98]
7h. R7C8 = 8 (hidden single in N9) -> R7C79 + R8C7 = 11 = {245}, no 6, 2 locked for N9
7i. Naked triple {167} in 14(3) cage at R9C7, 1,6 locked for R9
7j. 1 on D/ only in R4C6 + R5C5 + R6C4, locked for N5, clean-up: no 8 in R5C7
7k. Naked pair {67} in R59C7, locked for C7
7l. R4C7 = 8 (hidden single in N6)
7m. R2C4 + R6C6 = R5C3 + 1 (step 6a), R6C6 = 8 -> R2C4 + R5C3 = [18/29]
7n. R2C6 = 9 -> R1C456 = 14 = {248/347/356} (cannot be {257} because R1C6 only contains 4,6)
7o. R1C6 = {46} -> no 4,6 in R1C45
7p. Killer triple 1,2,3 in R1C456, R2C4 and R3C6, locked for N2
7q. 27(4) cage at R2C5 = {5679} (cannot be {4689} which clashes with R1C6) -> R3C5 = 9, R2C5 + R3C45 = {567}, locked for N2, 5 locked for R3
7r. R1C6 = 4 -> R1C45 = {28}, locked for R1, 2 locked for N2
7s. R2C4 = 1 -> R5C3 = 8 (step 7m), R8C45 = [81], R1C45 = [28], R3C6 = 3, R5C6 = 2 -> R5C7 = 7
7t. R2C9 = 8 -> R1C89 = 11 = {56}, 6 locked for R1 and N3
7u. Naked triple {379} in 19(3) cage at R1C1, 7,9 locked for N1
7v. R3C6 + R4C67 = [318] = 12 -> R2C8 + R3C78 = 13 = {247}, 4,7 locked for N3, 2 locked for R3, 7 locked for C8
7w. R6C7 = 9 (hidden single in C7) -> R56C8 = 8 = [35/53/62]
7x. Killer pair 5,6 in R1C8 and R56C8, locked for C8
7y. Naked pair {56} in R1C9 + R8C2, locked for D/

8a. R3C7 + R8C3 = [29] (hidden pair on D/)
8b. 3 on D/ only in R5C5 + R6C4, locked for N5
8c. R8C7 = 4, R7C7 = 5, placed for D\
8d. R3C2 = 8 (hidden single in N1), R2C1 + R3C13 = {146} = 11 -> R4C12 = 12 = {57}, locked for R4 and N4
8e. R1C3 = 7 (hidden single in C3)
8f. R38C9 = [93], R8C8 = 9, placed for D\, R1C1 = 3, placed for D\, R5C5 = 4, placed for both diagonals, R4C4 = 6, placed for D\, R2C8 = 7, R3C138 = [614], R8C1 = 5 -> R8C2 = 6, placed for D/

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

Rating Comment:
I'll rate my WT for A394 at 1.5.


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 Post subject: Re: Assassin 394
PostPosted: Fri Apr 10, 2020 9:09 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Really good WT by Andrew. Love 2f! 5c is very good. That elimination was really important for my solution too. 5e has really nice logic to get down the combo list. Wish I'd done that. Great vision with 7b. [Thanks to Andrew for some corrections and suggestions.]

A394 WT:
Preliminaries courtesy of SudokuSolver
Cage 4(2) n3 - cells ={13}
Cage 12(2) n9 - cells do not use 126
Cage 12(2) n8 - cells do not use 126
Cage 9(2) n8 - cells do not use 9
Cage 9(2) n56 - cells do not use 9
Cage 8(3) n12 - cells do not use 6789
Cage 19(3) n1 - cells do not use 1
Cage 19(3) n3 - cells do not use 1
Cage 19(3) n7 - cells do not use 1
Cage 27(4) n25 - cells do not use 12

1. "45" on r1: 1 innie r1c7 + 16 = 2 outies r2c69
1a. -> r1c7 = 1, r2c7 = 3, r2c69 = 17 = {89}, both locked for r2

2. 8(3)r2c2 = {125}: all locked for r2

3. "45" on n36: 3 outies r345c6 = 6 = {123} only, all locked for c6
3a. 25(6)r2c8 must have 1 which is only in r34c6 -> 1 locked for c6
3b. r5c7 = (67)
3c. no 9 in 12(2)n8

4. "45" on n2, 2 innies r2c4 + r3c6 + 5 = 1 outie r4c5
4a. -> max. 2 innies = 4 (no 5)
4b. -> min. r4c5 = 8
4c. 5 must be in 8(3)r2c2, only in n1, locked for n1

5. 4 in n2 in 27(4)r2c5 = {4689} = {46} in n2
5a. or 4 in n2 in 23(4) -> 6 in 23(4) must also have 4 (Locking-out cages)
5b. -> 23(4) = {2489/2579/3479/3578/4568}
5c. but {3479} blocked by 3/4/7 needed by 27(4)r2c5 in n2
5d. 5b. -> 23(4) = {2489/2579/3578/4568}
5e. note: if it has 9 it must also have 2

Now for some fun stuff
6. "45" on r89: 2 innies r8c37 = 6 = [15]/{24}
6a. r8c3 sees all r7 except r7c789 so must repeat there
6b. -> 19(4)r7c7 must have {15/24}
6c. = {1459/1567/2458/2467}(no 3)
6d. but {1459} blocked by 12(2)n9 which needs one of 4,5,9
6e. = {1567/2458/2467}(no 9)

7. "45" on r789: 1 outie r6c6 = 2 innies r7c12
7a. -> no 9 in r7c12
7b. -> 9 in r7 only in 32(6)r6c6 -> no 9 in r6c6
7c. -> max. r7c12 = 8 (no 8)

This feels like hypotheticals to me. Usually only like to work with 4 combos.
8. 32(6)r6c6 = {125789/134789/135689/145679/234689/235679}
8a. must have 1 or 3
8b. -> r7c12 cannot be {13} -> from step 7, no 4 in r6c6

9. from step 5d. 23(4)n2 = {2489/2579/3578/4568}
9a. but [68] in r12c6 -> 12(2)n8 = {57}, but this is blocked by r6c6 = (5678)
9b. -> no 6 in r1c6 (since only combo with 6 is {4568}) [Andrew suggested that another way to see this is from '23(4)n2 cannot contain 6 and 9'. Take your pick.]

10. 6 in c6 only in r67c6 in 32(6) cage, locked for that cage
10a. = {135689/145679/234689/235679}
10b. and no 6 in r7c7 since it sees both those cells

11. 33(6)r4c4 must have 4 for n5
11a. and can't have both 2 & 3 because r5c6 = (23)
11b. = {145689/245679/345678}
11c. note: must have 4,5,6

Loved finding this step. Andrew finds these all the time
12. from step 7. r6c6 = 2 innies r7c12
12a. -> the value in r6c6 cannot repeat in r7c12 -> can only be in r7 in r7c89
12b. 6 in c6 only in r67c6
12c. if in r6c6 -> from step 11c -> r5c3 = 6 -> r5c7 = 7; also 6 must be in r7c89 (step 12a.) -> r7c89 + r78c7 ={67}/{24} only
12d. or 6 in c6 in r7c6 -> 19(4)n9 = {2458} only
12e. -> 19(4)n9 = {2458/2467}(no 1)
12f. must have 2 & 4: both locked for n9
12g. must have 5 or 7 -> {57} blocked from 12(2)n9
12h. -> 12(2)n9 = {39} only: both locked for n9 and r8

13. 1 in n9 only in r9 in 14(3) = {158/167}
13a. 1 locked for r9

14. 1 in n8 only in r78c56: r8c3 sees all those -> no 1 in r8c3 (Common Peer Elimination, CPE)

15. h6(2)r8c37 = {24} only: both locked for r8

16. 9(2)n8 = {18} only: both locked for n8, 8 for r8

17. 12(2)n8 = {57} only: both locked for n8 and c6

18. 19(3)n7 must have two of 5,6,7 for r8c12 = {568} only
18a. r9c1 = 8, placed for d/
18b. 5 & 6 locked for n7, 5 for r8
18c. r89c6 = [75]

19. naked triple {167} in r9c789: 6 & 7 locked for r9 and n9
19a. -> no 6 in r6c6 (step 12a)
19b. -> r6c6 = 8: placed for d\
19c. -> r7c12 = 8 (step 12) = {17} only: both locked for n7

20. r4c5 = 9
20a. r12c6 = [49], r7c6 = 6, r2c9 = 8, r7c8 = 8 (hsingle n9), r4c7 = 8 (hsingle n6)

21. "45" on n2: 2 innies r2c4 + r3c6 = 4 = [13]
21a. r5c67 = [27], r4c6 = 1, r9c7 = 6
21b. r8c45 = [81]

22. "45" on n6: 1 outie r3c9 = 9
22a. r8c89 = [93]: 9 placed for d\

23. "45" on n5: 1 outie r5c3 = 8

24. r1c89 = 11 = {56} only: both locked for r1, n3

25. naked pair {56} in r1c9 + r8c2: both locked for d/

26. r7c3 = 9 (hsingle d/), r3c7 = 2 (hsingle d/)

27. "45" on n1: 2 remaining outies r4c12 = 12 = {57} only: both locked for r4, n4, and 7 for 31(6)r2c1

28. r1c45 = 10 (cage sum) = [28] only

29. r6c1 = 2 (hsingle c1), r5c1 = 9 (hsingle c1) -> r7c1 = 1 (cage sum)

easy now
Cheers
Ed


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 Post subject: Re: Assassin 394
PostPosted: Thu Apr 16, 2020 8:59 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Here's how I did it. The important step in all our WTs (I think) is to reduce the possibilities for innies r89 - and we all did that in different ways. A such, this puzzle is a good example illustrating our different strengths.
Assassin 394 WT:
1. 8(3)r2 contains a 1
-> 4(2)n3 = [13]
-> 8(3)r2 = {125}
Also -> Remaining Outies r1 -> r2c69 = +17(2) = {89}

2. IOD n2 -> Either:
(a) [r2c4,r3c6,r4c5] = [{12}8], or
(b) [r2c4,r3c6,r4c5] = [139]
-> 5 in r2c23

3. Outies n36 = r345c6 = +6(3) = {123}
-> 12(2)n8 = {57} or {48}

4. Innies r89 = r8c37 = +6(2) = [15], [24], or [42]
Whatever goes in r8c3 goes in r7 in n9 in r7c789
-> The remaining two values in r7c789 = +13(2)
That H13(2) cannot be {49} since 19(4) = {1459} leaves no solution for 12(2)n9
-> Either (58) or (67) in r7c789
-> 12(2)n9 cannot be {57}

5. 1 in r8 only in r8c345.
Either:
(a) r8c37 = {24} puts 9(2)n8 = {18} puts 12(2)n8 = {57}
(b) r8c37 = [15] puts 19(4)n9 = [{167}5] puts (48) in n9 either both in r8 or both in r9 which puts 12(2)n8 = {57}
Either way -> 12(2)n8 = {57}

!6. Trying r8c37 = [15] puts 19(4)n9 = [{167}5] puts 32(6)r6c6 = [65{389}1]
But that leaves no solution for 9(2)n8
-> r8c37 = {24}

7. -> 9(2)n8 = {18} (HS 1 in r8)
-> 12(2)n9 = {39}
-> 19(3)n7 can only be [{56}8]
-> 12(2)n8 = [75]
-> 19(4)n9 = {2458} with (58) in r7
-> 14(3)n9 = {167}

8. Only possible solution for 32(6) = [8{23469}] with 2 or 4 in r8c3
-> r7c12 = {17}
Also -> r2c69 = [98]
-> r2c4,r3c6 = [13] and r4c5 = 9
Also (Since 25(6) must contain a 1) -> r45c6 = [12]

9. Remaining Outie n5 -> r5c3 = 8
-> Remaining outies n1 -> r4c12 = {57}
etc.


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