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 Post subject: Assassin 352
PostPosted: Thu Jun 14, 2018 7:12 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 793
Location: Sydney, Australia
Attachment:
a352.jpg
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This is an X puzzle so 1-9 cannot repeat on either diagonal.

NOTE: a disjoint 8(3) cage at r4c1+r67c2

Assassin 352
Nice easy start and then lots of areas to work in. I found it a good challenge but all normal Assassin stuff. Just well hidden from me. Won't even give the SSscore for this one. JSudoku has no problem.

code:
3x3:d:k:5632:5632:5632:1537:2306:2306:7939:7939:5380:5637:5637:5632:1537:3078:3078:7939:2823:5380:5637:5637:6152:6152:7939:7939:7939:2823:5380:2072:3081:6152:7178:7178:7178:7178:5380:5380:4107:3081:6152:7178:4876:5645:5645:5645:5390:4107:2072:1296:1296:4876:3857:5645:5390:5390:4107:2072:3346:4876:4876:3857:5645:4371:5390:3860:3860:3346:2325:6166:3857:4371:4371:3351:3860:2575:2575:2325:6166:6166:6166:4371:3351:
solution:
Code:
+-------+-------+-------+
| 8 6 3 | 5 7 2 | 4 9 1 |
| 7 9 5 | 1 4 8 | 2 6 3 |
| 4 2 1 | 9 3 6 | 7 5 8 |
+-------+-------+-------+
| 3 5 8 | 6 9 4 | 1 2 7 |
| 9 7 6 | 8 2 1 | 3 4 5 |
| 1 4 2 | 3 5 7 | 9 8 6 |
+-------+-------+-------+
| 6 1 9 | 4 8 3 | 5 7 2 |
| 2 8 4 | 7 1 5 | 6 3 9 |
| 5 3 7 | 2 6 9 | 8 1 4 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 352
PostPosted: Sat Jun 23, 2018 8:55 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 793
Location: Sydney, Australia
Guessing there won't be any other WTs for this puzzle until after England win the World Cup.

A352 WT
23 steps:
Preliminaries courtesy of SudokuSolver
Cage 5(2) n45 - cells only uses 1234
Cage 6(2) n2 - cells only uses 1245
Cage 12(2) n4 - cells do not use 126
Cage 12(2) n2 - cells do not use 126
Cage 13(2) n7 - cells do not use 123
Cage 13(2) n9 - cells do not use 123
Cage 9(2) n8 - cells do not use 9
Cage 9(2) n2 - cells do not use 9
Cage 10(2) n7 - cells do not use 5
Cage 11(2) n3 - cells do not use 1
Cage 8(3) n47 - cells do not use 6789

No routine clean-up done unless stated.
1. "45" on n1: 1 innie r3c3 = 1; Placed for D\
1a. no 4 in r6c4

2. "45" on c123: 2 outies r36c4 = 12 = [93] only permutation
2a. 3 placed for D/
2b. no 8 in r3c8
2c. r6c3 = 2
2d. r3c34 = 10 -> r45c3 = 14 = {68} only: both locked for n4 and c3
2e. no 2 in r2c8
2f. no 4 in 12(2)n4

3. 13(2)n7 = {49} only: both locked for n7 and c3
3a. 10(2)n7 = {37} only: both locked for r9 and n7

4. "45" on n4: 1 remaining outie r7c1 + 1 = 2 remaining innies r4c1 + r6c2
4a. since 1 innie sees one outie they cannot be equal -> no 1 in r6c2 (IOU)

5. 8(3)r4c1 = {125/134}: can't have both 4,5 -> no 4,5 in r3c1 nor r7c2
5a. must have 1 -> no 1 in r789c1

6. "45" on n7: 2 innies r7c12 = 7 = [61/52]

7. 16(3)r5c1 must have 5,6 for r7c1 = {169/367/457}
7a. but {37}[6] blocked by 12(2)n4 = 3 or 7
7b. = {169/457}(no 3)
7b. can't have both 5,6 -> no 5 in r56c1

8. 5 in n4 only in c2: locked for c2
8a. 5 in c3 only in 22(4)n1: 5 locked for n1
8b. must have one of 3,7 for r12c3
8c. = {2578/3568/4567}(no 9)
8d. can't have both 3,7 -> no 3,7 in r1c12

9. 9 in n1 only in r2: locked for r2
9a. no 2 in r3c8

10. "45" on n2: 2 remaining innies r3c56 = 9
10a. but {45} blocked by 6(2)n2 = 4 or 5
10b. = {27/36}(no 4,5,8)

11. 31(7)r1c7 must have {27/36} for r3c56 = {124789/134689/135679/234679/23568} = 3 or 8 (I usually like to get combos down to 4 only before I look at them so this is a big combo step for me)
11a. -> [83] blocked from 11(2)n3 since it sees all 8 and 3 in 31(7)
11b. 11(2) = {47/56}(no 3,8) = 4 or 6

12. "45" on c9: 2 outies r46c8 = 10
12a. but {46} blocked by 11(2)n3
12b. = {19}[28/37](r4c8 = (1239), r6c8 = (1789)

13. "45" on n23: 2 remaining outies r4c89 = 9 = [18/27/36](no 9, r4c9 = (678)
13a. no 1 in r6c8 (h10(2)r46c8)

14. 4 in r4 only in 28(5)r4c4 -> no 4 in r5c4

15. 9(2)n8: {45} blocked by 6(2)n2
15a. = {18}/[72](r8c4 = (178), r9c4 = (128)

16. killer pair 1,2 in r1289c4: locked for c4

17. "45" on r1234: 1 innie r4c1 + 18 = 3 outies r5c234
17a. = [1]+[5]{68}/[3]+[7]{68}
17b. -> 12(2)n4 = {57} only: both locked for n4 and c2
17c. r4c1 = 3 (hsingle n4)
17d. -> r5c2 = [7] (step 17a.), r4c2 = 5
17e. naked pair {68} in r5c34: both locked for r5

18. r6c2 = 4, r7c1 = 1 (cage sum), r7c1 = 6 (h7(2)r7c12)
18a. naked pair {19} in r56c1: 9 locked for c1
18b. r2c2 = 9 (hsingle n1): placed for D\

19. r9c23 = [37], r12c3 = 8 -> r1c12 = 14 = [86] only permutation, 8 placed for D\, r3c2 = 2, r8c2 = 8; place for D/

20. h9(2)r3c56 = {36} only: both locked for n2 and r3 and no 3,6 elsewhere in 31(7)r1c7

21. r3c9 = 8 (hsingle r3), h9(2)r4c89 = [27] only permutation, r6c8 = 8 (h10(2)r46c8)

22. r12c9 = 4 (cage sum) = [13] only permutation; 1 placed for D/

23. r7c9 = 2 (hsingle in c9) -> r56c9 = 11 (cage sum) = [56] only permutation

remember the diagonals and all very easy now.
Cheers
Ed


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 Post subject: Re: Assassin 352
PostPosted: Tue Jun 26, 2018 8:40 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 179
Location: California, out of London
Thanks Ed! Here is my WT. Combo work is not my forté and my initial solution path used quite a bit of it, so I spent some time looking for an alternative path but without success.

Comparing our WTs showed a lot of similarities you'll see.

Keep 'em coming!

Assassin 352 WT:
1. Innies n1 -> r3c3 = 1
-> Min r6c3 = 2
-> Max r6c4 = 3.
Outies c123 -> r36c4 = +12(2) can only be [93]
-> r6c3 = 2

2. Also r45c3 = {68}
-> 13(2)n7 = {49}
-> 10(2)n7 = {37}

3. Innies n7 = r7c12 = +7(2) (from [61] or {25})
-> r4c1 + r7c2 cannot be +7(2)
-> r6c2 cannot be 1
-> Disjoint cage 8(3)r4c1 from [152] or [341]

4. Innies c3 = r129c3 = [{35}7] or [{57}3]
If the former -> 22(4)n1 = [{68}{35}] -> HS 1 in n2 -> 6(2)n2 = {15}
If the latter -> 12(2)n2 = {48} -> 6(2)n2 = {15}
Either way 6(2)n2 = {15} and 12(2)n2 = {48}
Remaining innies n2 = r3c56 = +9(2)
-> 9(2)n2 and H9(2)r3c56 are from {27} and {36}

5. Also 9(2)n8 = [72]
-> r457c4 = {468}

6. 22(4)n1 from [{68}{35}], [{46}{57}], or [{28}{57}]
-> 9 in n1 in r2c12
-> 9 in n3 in r1c789

7. r1c56 and r3c56 = {2367}
-> None of (2367) in r1c78
8 already in r2 -> r3c8 cannot be 3.
Also 3 already in D/ -> 3 in n3 only in r2c79 or r3c9

8. Remaining innies n3 -> r123c9 = +12(3)
Trying 9 in r1c9 puts r123c9 = [912]
Puts r3c56 = {36}
Leaves no place for 3 in n3
-> 9 in r1c78

9. Trying either 3 or 6 in 31(6) in n3 puts r3c56 = {27} and the other cells in 31(6) in n3 as {3469}
But this leaves no solution for 11(2)n3
-> 3 in n3 in r23c9
-> 6 in n3 in 11(2)
Since 5 already in r12 in n12
-> 11(2)n3 = [65]

10. Remaining outies n3 -> r4c89 = +9(2)
-> Max r4c8 = 8
Outies c9 -> r46c8 = +10(2) (no 5)
-> Min r6c8 = 2
Since (23) already in r6 -> Min r6c8 = 4
Since 6 already in c8 -> r46c8 cannot be {46} -> Min r6c8 = 7
Since 3 already in r23c9 -> r4c8 cannot be 3 -> r6c8 cannot be 7
-> r6c8 from (89)
-> r4c89 from [27] or [18]
Whichever of (12) is in r4c8 must go in c9 in r7c9
-> r56c9 = +11(2)

11. 7 in n9 only in r7
-> 13(2)n9 from {49} or {58}
-> 6 in c9 only in r56c9
-> r56c9 = {56}
-> 13(2)n9 = {49}

12! r4c8 and r7c9 are both 1 or both 2
But in disjoint cage 8(3)r4c1 - 1 must be in r4c1 or r7c2
-> r4c8 = r7c9 = 2
-> r6c8 = 8 and r4c9 = 7
-> 31(6) in n3 must be {2749}
-> r3c56 = {36}
-> 9(2)n2 = {27}
-> r23c7 = {27}
-> r1c78 = {49}
-> r123c9 = [138]

Easy from here...
E.g.,
13. Disjoint 8(3) can only be [341]
-> Innies n7 -> r7c1 = 6
Also 12(2)n4 = {57}
-> 10(2)n7 = [37]
-> r12c3 = [35]
-> r1c12 = [86]
Also -> r56c1 = {19}
-> HS 9 in n1 -> r2c2 = 9
-> (D\) -> 13(2)n9 = [94]
-> 13(2)n7 = [94]
-> Remaining innies r89 -> r8c6 = 5
-> 15(3)n7 = [285]
-> Remaining innies n9 = r79c7 = [58]
-> HS r8c7 = 6
-> 24(4)n8 = [1{69}8]
-> r89c8 = [31]
etc.


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 Post subject: Re: Assassin 352
PostPosted: Fri Jul 06, 2018 2:07 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1626
Location: Lethbridge, Alberta, Canada
As Ed is already aware I posted a walkthrough with a faulty step last month, deleting it the following day when I realised my error. I've now reworked my later steps to make the elimination I'd originally thought that I'd made.

Here is my reworked walkthrough for Assassin 352:
This is a Killer-X. There is a disjoint 8(3) cage at R4C1.

Prelims

a) R12C4 = {15/24}
b) R1C56 = {18/27/36/45}, no 9
c) R2C56 = {39/48/57}, no 1,2,6
d) R23C8 = {29/38/47/56}, no 1
e) R45C2 = {39/48/57}, no 1,2,6
f) R6C34 = {14/23}
g) R78C3 = {49/58/67}, no 1,2,3
h) R89C4 = {18/27/36/45}, no 9
i) R89C9 = {49/58/67}, no 1,2,3
j) R9C23 = {19/28/37/46}, no 5
k) 8(3) disjoint cage at R4C1 = {125/134}

Steps resulting from Prelims
1a. R1C56 = {18/27/36} (cannot be {45} which clashes with R12C4), no 4,5
1b. R89C4 = {18/27/36} (cannot be {45} which clashes with R12C4), no 4,5

2. 45 rule on N1 1 innie R3C3 = 1, placed for D\, clean-up: no 4 in R6C4, no 9 in R9C2
2a. 45 rule on C3 2 outies R36C4 = 12 = [93] -> R6C3 = 2 (cage sum), 3 placed for D/, clean-up: no 3 in R2C56, no 2 in R2C8, no 8 in R3C8, no 6 in R89C4, no 8 in R9C2
2b. R3C34 = [19] = 10 -> R45C3 = 14 = {68}, locked for C3 and N4, clean-up: no 4 in R45C2, no 5,7 in R78C3, no 2,4 in R9C2
2c. Naked pair {49} in R78C3, locked for C3 and N7, clean-up: no 1,6 in R9C2
2d. Naked pair {37} in R9C23, locked for R9 and N7, clean-up: no 2 in R8C4 no 6 in R8C9
2e. Killer pair 1,2 in R12C4 and R89C4, locked for C4
2f. Killer pair 4,5 in R12C4 and R2C56, locked for N2
2g. Killer pair 3,7 in R45C2 and R9C2, locked for C2
2h. 5 in C3 only in R12C3, locked for N1
2i. 5 in C3 only in R12C3 = {35/57} = 8,12 -> R1C12 = 10,14 = {28/46/68}, no 3,7,9
2j. 5 in R3 only in R3C789, locked for N3, clean-up: no 6 in R3C8
2k. 9 in N1 only in R2C12, locked for R2, clean-up: no 2 in R3C8
2l. 3,6 in N2 only in R13C56, CPE no 3,6 in R1C78
2m. 6 in C4 only in R457C4, CPE no 6 in R56C5
2n. 3 on D\ only in R7C7 + R8C8, locked for N9
2o. 3 in R7C7 + R8C8, CPE no 3 in R5C8
2p. 45 rule on N2 2 remaining innies R3C56 = 9 = {27/36}, no 8
2q. 45 rule on N4 4(2+2) outies R3C56 + R4C89 = 18, R3C56 = 9 -> R4C89 = 9 = {18/27/36/45}, no 9 in R4C89
2r. 45 rule on C9 2 outies R46C8 = 10 = [19/28/37]/{46}, no 5,7,8 in R4C8, no 1,5 in R6C8, clean-up: no 1,2,4 in R4C9
2s. 3 in R1 only in 22(4) cage at R1C1 = [8635] or in R1C56 = {36}, 6 locked for R1 (locking cages)

3. 8(3) disjoint cage at R4C1 = {125/134} = {15}2/[34]1, no 4 in R4C1, no 5 in R7C2
3a. Killer pair 3,5 in 8(3) disjoint cage and R45C2, locked for N4
3b. 45 rule on N7 2 innies R7C12 = 7 = [52/61]
3c. Combined half cage R4C1 + R6C2 + R7C12 = [1552/3461] (cannot be [5152]), no 5 in R4C1, no 1 in R6C2
3d. 1 in N4 only in R456C1, locked for C1
3e. 5 in N4 only in R456C2, locked for C2
3f. 15(3) cage at R8C1 = {168/258}
3g. 1 of {168} must be in R8C2 -> no 6 in R8C2
3h. 6 in N7 only in R789C1, locked for C1, clean-up: no 4 in R1C2 (step 2i)

4. 45 rule on R89 2 outies R7C38 = 1 innie R8C6 + 11
4a. Min R7C38 = 12 -> no 1,2 in R7C8
4b. Max R7C38 = 17 -> max R8C6 = 6

5. 45 rule on R1234 2 outies R5C34 = 2 innies R4C12 + 6, R4C12 are both odd -> R5C3 is even -> R5C4 must be even = {468}
5a. Max R5C34 = 14 -> max R4C12 = 8, no 9 in R4C2, clean-up: no 3 in R5C2
5b. 3 in N4 only in R4C12, locked for R4, clean-up: no 6 in R4C89 (step 2q), no 4,7 in R6C8 (step 2r)

[I made a careless omission in my original step 5 so this is my new key step. Probably more complicated than the way that Ed solved this Assassin (this comment was made before I went through his WT); this step was triggered by thinking of a way to eliminate my omission, I actually realised this step while watching a DVD.]
6. Consider placement for 4 in N2
4 in R12C4, locked for C4
or in R2C56 = {48}, locked for R2 => R23C8 = [65/74] => R46C8 (step 2r) = [19/28] (cannot be [46] which clashes with R23C8) => 4 in R4 in R4C4567, locked for 28(5) cage at R4C4 => no 4 in R5C4
-> no 4 in R5C4
[Now things are fairly straightforward]
6a. Naked pair {68} in R5C34, locked for R5
6b. R5C34 = R4C12 + 6 (step 5), R5C34 = {68} = 14 -> R4C12 = 8 contains 3 (step 5b) = [35], R5C2 = 7 (cage sum), R6C2 = 4, R7C2 = 1 (cage sum), clean-up: no 4 in R4C8 (step 2q), no 6 in R6C8 (step 2r)
6c. Naked pair {19} in R56C1, locked for C1 -> R7C1 = 6 (cage sum)
6d. R2C2 = 9 (hidden single in N1), placed for D\, clean-up: no 4 in R8C9
6e. R9C23 = [37], R12C3 = {35} = 8 -> R1C12 = 14 = [86], 8 placed for D\, R3C2 = 2, R8C2 = 8, placed for D/, clean-up: no 1,3 in R1C56, no 3 in R3C8, no 5 in R89C9, no 1 in R9C4
6f. Naked pair {27} in R1C56, locked for R1 and N2, clean-up: no 4 in R12C4, no 5 in R2C56
6g. Naked triple {149} in R1C789, locked for R1 and N3 -> R12C4 = [51], R8C4 = 7 -> R9C4 = 2, R8C9 = 9 -> R9C9 = 4, placed for D\, R1C9 = 1, placed for D/, R8C3 = 4, R7C3 = 9, placed for D/, R4C4 = 6, placed for D\, R9C1 = 5, placed for D/, R5C5 = 2, placed for D\, clean-up: no 7 in R23C8
6h. R23C8 = [65], R3C7 = 7, placed for D/
6i. R4C8 = 2 -> R4C9 = 7 (step 2q), R6C8 = 8 (step 2r)
6j. Naked pair {36} in R3C56, locked for R3 -> R23C9 = [38], R567C9 = [562]
6k. R8C8 = 3, R7C7 = 5, placed for D\
6l. R6C5 = 5 (hidden single in N5)

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

Rating Comment:
I'll rate my reworked WT for A352 at 1.5. However Ed's WT is in the 1.25 range.


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 Post subject: Re: Assassin 352
PostPosted: Mon Jul 09, 2018 8:38 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 793
Location: Sydney, Australia
Some wonderful steps in these WTs.

Show Andrew's first since it was in his original posted WT.
Attachment:
a352AndrewStep5a.JPG
a352AndrewStep5a.JPG [ 26.31 KiB | Viewed 1093 times ]
Quote:
5. ...... 3 remaining innies R4C123 = 1 outie R5C4 + 8
5a. R4C12 are both odd, R4C3 is even -> R4C123 must be even -> R5C4 must be even = {468}, no 5,7
Technically, need to add that the IOD is even to make this work. What an elegant way to make eliminations!

wellbeback's is great vision.
Attachment:
a352Wstep12.JPG
a352Wstep12.JPG [ 93.22 KiB | Viewed 1093 times ]
Quote:
12! r4c8 and r7c9 are both 1 or both 2
But in disjoint cage 8(3)r4c1 - 1 must be in r4c1 or r7c2
-> r4c8 = r7c9 = 2
Thanks!


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