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 Post subject: Assassin 2X Revisit
PostPosted: Mon Sep 14, 2020 8:12 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 860
Location: Sydney, Australia
Thought it might be worthwhile to have a go at an old puzzle to see if the new techniques we've been finding give an interesting alternative solution. Just picked the first puzzle that gets a 1.50 rating in the archives. However, it has so many small cages it may not yield anything but could be fun trying.

Assassin 2X
Attachment:
a2x.JPG
a2x.JPG [ 106.09 KiB | Viewed 72 times ]


Archive post here
SSscore = 1.60, JSudoku (not default routines) uses 3 "complex intersections"
triple click code:
3x3:d:k:4352:4352:3841:3841:4610:3587:3587:3332:3332:4352:4613:4613:3841:4610:3587:4358:4358:3332:4615:4613:3848:3848:4610:3081:3081:4358:3338:4615:4615:3848:3595:3595:2572:3081:3338:3338:2573:2573:2573:3595:2572:5390:3599:3599:3599:3344:3344:4113:2572:5390:5390:4370:3347:3347:3344:4116:4113:4113:3093:4370:4370:6166:3347:4375:4116:4116:3864:3093:2585:6166:6166:3354:4375:4375:3864:3864:3093:2585:2585:3354:3354:
solution:
Code:
+-------+-------+-------+
| 8 6 7 | 3 9 2 | 5 1 4 |
| 3 4 9 | 5 1 7 | 6 2 8 |
| 2 5 1 | 6 8 4 | 7 9 3 |
+-------+-------+-------+
| 7 9 8 | 2 5 3 | 1 4 6 |
| 5 1 4 | 7 6 8 | 2 3 9 |
| 6 3 2 | 1 4 9 | 8 5 7 |
+-------+-------+-------+
| 4 2 5 | 9 7 6 | 3 8 1 |
| 1 8 6 | 4 3 5 | 9 7 2 |
| 9 7 3 | 8 2 1 | 4 6 5 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 2X Revisit
PostPosted: Sat Sep 19, 2020 4:22 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1727
Location: Lethbridge, Alberta, Canada
Thanks Ed for reposting this excellent variant by Mike Parker from over a decade ago. If it was posted now it would certainly still qualify as an Assassin.

I haven't looked at how I solved it back then but I think I've almost certainly found a more interesting way this time. My key breakthrough move was a bit like a wellbeback type of step.

Here is my walkthrough for Assassin 2X Revisited:
Prelims

a) 10(3) cage at R4C6 = {127/136/145/235}, no 8,9
b) 10(3) cage at R5C1 = {127/136/145/235}, no 8,9
c) 21(3) cage at R5C6 = {489/579/678}, no 1,2,3
d) 24(3) cage at R7C8 = {789}
e) 10(3) cage at R8C6 = {127/136/145/235}, no 8,9

1a. 45 rule on N1 3 innies R1C3 + R3C13 = 10 = {127/136/145/235}, no 8,9
1b. Naked triple {789} in 24(3) cage at R7C8, locked for N9
1c. 45 rule on N9 3 innies R7C79 + R9C7 = 8 = {125/134}, no 6, 1 locked for N9
1d. 45 rule on R5 3 innies R5C456 = 21 = {489/579/678}, no 1,2,3
1e. 4 of {489} must be in R5C5 -> no 4 in R5C46
1f. 45 rule on R12 3 outies R3C258 = 22 = {589/679}, 9 locked for R3
1g. 45 rule on C12 3 outies R258C3 = 19 = {289/379/469/478/568}, no 1
1h. 2 of {289} must be in R5C3 -> no 2 in R28C3
1i. Max R3C34 = 13 (cannot be {68} which clashes with R3C258) -> min R4C4 = 2

2a. 45 rule on C6789 3 outies R5C5 + R6C45 = 11 = {146/245} (cannot be {128/137/236} because 1,2,3 only in R6C4), no 3,7,8,9
2b. 1,2 only in R6C4 -> R6C4 = {12}
2c. R5C5 + R6C45 = 11 = {146/245}, 4 locked for C5 and N5
2d. 10(3) cage at R4C6 = {136/145/235} (cannot be {127} because no 1,2,7 in R5C5), no 7
2e. 45 rule on C5 3 innies R456C5 = 15 = {456} (only remaining combination), 5,6 locked for C5 and N5
2f. 10(3) cage at R4C6 = {136/235} (cannot be {145} because 4,5 only in R5C5) -> R4C6 = 3, placed for D/
2g. R6C5 = 4 (hidden single in C5) -> R56C6 = 17 = {89}, locked for C6 and N5
2h. R5C4 = 7
2i. Naked pair {12} in R46C4, locked for C4
2j. 10(3) cage at R5C1 = {136/145/235}
2k. Killer pair 5,6 in 10(3) cage and R5C5, locked for R5
2l. R258C3 (step 1g) = {289/379/469/478/568}
2m. 3 of {379} must be in R5C3 -> no 3 in R28C3
2n. Max R7C67 = [75] -> min R6C7 = 6
2o. Min R89C4 = 7 -> max R9C3 = 8

3a. 45 rule on N8 2 innies R7C46 = 2 outies R9C37 + 8
3b. Min R9C37 = 3 -> min R7C46 = 11, no 3 in R7C4, no 1 in R7C6
3c. Max R7C46 = 16 -> max R9C37 = 8, no 8 in R9C3

4a. R5C4 = 7 -> R4C45 = 7 = [16/25]
4b. 45 rule on R5 2 remaining innies R5C56 = 14 = [59/68]
4c. 45 rule on R6789 2 remaining innies R6C46 = 10 = [19/28]
4d. 45 rule on N1 6(3+3) outies R123C4 + R4C123 = 38, max R123C4 = {689} = 23 -> min R4C123 = 15
4e. 14(3) cage at R5C7 = {149/239} (cannot be {248} which forces R5C56 = [59], R6C456 = [248] and R4C123 = {248} = 14 but R4C123 must total at least 15), 9 locked for R5 and N6
4f. R56C6 = [89] -> R5C5 = 6, placed for both diagonals, 9 placed for D\, R4C5 = 5, R4C4 = 2, placed for D\, R6C4 = 1, placed for D/
4g. 10(3) cage at R5C1 = {145/235}, 5 locked for N4
4h. 5 in R6 only in R6C89, locked for 13(3) cage at R6C8
4i. 13(3) cage at R6C8 = {157/256}, no 3,4,8
4j. 3 in N6 only in 14(3) cage = {239}, 2,3 locked for R5, 2 locked for N6
4k. Naked triple {145} in 10(3) cage at R5C1, 1,4 locked for N4
4l. R258C3 (step 1g) = {469/478/568}
4m. R5C3 = {45} -> no 4,5 in R28C3
4n. Max R6C7 + R7C6 = 15 -> no 1 in R7C7
4o. 1 on D\ only in R1C1 + R2C2 + R3C3, locked for N1
4p. Min R4C12 = 13 -> max R3C1 = 5
4q. Consider placement for 9 in R4
9 in R4C12 => 18(3) cage at R3C1 = 2{79}/3{69}
or 9 in R4C3 => R3C34 = 6 = [15]
-> no 5 in R3C1
4r. R1C3 + R3C13 = 10 (step 1a), min R1C3 + R3C1 = 5 -> max R3C3 = 5
4s. Min R2C3 + R3C2 = 11 -> max R2C2 = 7
4t. 8 on D\ only in R1C1 + R8C8, CPE no 8 in R1C8 + R8C1
4u. R7C79 + R9C7 (step 1c) = {125/134}
4v. 5 of {125} must be in R7C7 -> no 5 in R9C7
4w. 12(3) cage at R3C6 = {147/156/246}, no 8

5a. 12(3) cage at R7C5 = {129/138/237}
5b. 45 rule on N47 3 innies R4C123 = 3 outies R789C4 + 3
5c. 9 in R4 only in R4C123, R4C123 = 22, 23, 24 -> R789C4 = 19, 20, 21
5d. R789C4 = 19, 20, 21 = {469/568/569/489} (cannot be {389} = 20 which clashes with 12(3) cage), no 3
5e. 3 in N8 only in 12(3) cage = {138/237}, no 9, 3 locked for C5
5f. 9 in N8 only in R789C4 = {469/569/489}, 9 locked for C4
5g. 15(3) cage at R8C4, min R789C4 = 19 -> R7C4 at least 4 more than R9C3, no 4 in R7C4, no 6,7 in R9C3
5h. R789C4 = 19, 20, 21, 12(3) cage at R7C5 -> R789C6 = 12, 13, 14
5i. 10(3) cage at R8C6, min R789C6 = 12 -> R7C6 at least 2 more than R9C7, no 2 in R7C6
5j. R1C3 + R3C13 (step 1a) = {127/136/145/235}
5k. 4 of {145} must be in R3C1 -> no 4 in R13C3

6a. 6 in N9 only in 13(3) cage at R8C9 = {256/346}
6b. 5 of {256} must be in R9C9 -> no 5 in R8C9 + R9C8
6c. 5 in N9 only in R7C7 + R9C9, locked for D\
6d. 15(3) cage at R3C3 = {159/168/348/357} = [159/168/186/348/357], no 3 in R3C4
6e. 3 in C4 only in R12C4, locked for 15(3) cage at R1C3
6f. 15(3) cage at R1C3 = {357} (only possible combination, cannot be {348} because no 3,4,8 in R1C3) -> R12C4 = {35}, 5 locked for C4 and N2, R1C3 = 7
6g. R789C4 (step 5f) = {469/489}, 4 locked for C4, N8 and 15(3) cage at R8C4
6h. 15(3) cage at R3C3 = [168/186] -> R3C3 = 1, R4C3 = {68}
6i. R258C3 (step 4l) = {469} (only remaining combination, cannot be {568} which clashes with R4C3) -> R5C3 = 4, R28C3 = {69}, locked for C3 -> R4C3 = 8, R3C4 = 6

7a. R1C3 + R3C13 = 10 (step 1a), R1C3 = 7, R3C3 = 1 -> R3C1 = 2 (cage sum)
7b. R3C1 = 2 -> R4C12 = 16 = {79}, 7 locked for R4 and N4
7c. R6C7 = 8 (hidden single on R6) -> R7C67 = 9 = [54/63]
7d. R8C8 = 7 (hidden single on D\) -> R8C7 = 9, R7C8 = 8, R7C4 = 9, R28C3 = [96]
7e. Naked pair {48} in R89C4, 8 locked for N8 -> R9C3 = 3 (cage sum), R67C3 = [25], 5 placed for D/
7f. Naked pair {36} in R6C12, 6 locked for R6 -> R7C1 = 4 (cage sum)
7g. R6C89 = [57] -> R7C9 = 1 (cage sum)
7h. R8C3 = 6 -> R78C2 = 10 = [28], 8 placed for D/, R8C1 = 1
7i. 12(3) cage at R7C5 = {237} (only remaining combination), 2,7 locked for C5 and N8
7j. R789C6 = [651] -> R9C7 = 4 (cage sum), R7C7 = 3, placed for D\
7k. R3C67 = [47], 7 placed for D/, R4C7 = 1 (cage sum)
7l. R3C8 = 9 -> R2C78 = 8 = [62]

and the rest is naked singles, without using the diagonals.
Same rating as first time, which Ed has already posted.


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 Post subject: Re: Assassin 2X Revisit
PostPosted: Fri Sep 25, 2020 7:02 pm 
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Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 860
Location: Sydney, Australia
Wonderful Andrew! Really like 4e and 5g. Both are definitely interesting. Mike would have been really pleased I'm sure. I couldn't solve it and never noticed that "45" behind 4e. However, I did enjoy trying. Loved the 4 13(3) cages in c9. They must all have different combinations since they see 2 cells of each of the others. The top three 13(3) have 1 for both c8 & c9. Those 9 cells sum to 39 so the repeat 1 could be replacing a 7 to get the 9 cells down to 39 (ie 45 - 39 = 6). However, those 3 cages must have 7 for c9 so must have at least two repeats. But couldn't take it any further.

I haven't looked at how it was done previously but plan to.
Cheers
Ed


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