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Ed
 Post subject: Assassin 315
PostPosted: Thu Apr 02, 2015 10:14 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Like the last Assassin, this has a really short way to crack it. Hope it takes you enough looking to find it. I sure did! JSudoku uses all sorts of interesting chains. Unfortunately, this forum won't allow me to attach the JSudoku solver's order I use but email me if you want. Quite different to the one HATMAN uses - much less common to get chains.
JSudoku's log "Deduce all moves": 
4 Complex Intersection
2 Triple Innies & Outies
2 Double Outies minus Innies
1 Complex Naked Pair
2 Complex Hidden Pair
1 Conflicting Pair
2 Quadruple Innies & Outies
4 Triple Outies minus Innies
1 Odd Combinations
3 Pointing Triplet
1 Locked Cages
1 Remote Pairs
1 Skyscraper
1 Generalized X-Wing
5 Conflicting Partial Pair
1 Pointing Multiple
2 Turbot Fish
2 Empty Rectangle
1 Recursively Solve
SudokuSolver gives it a 1.55

Assassin 315
Image

code: paste into solver:
3x3::k:4096:4096:4865:3842:3842:2819:3844:3844:1797:6150:4096:4865:4865:4865:2819:2819:775:1797:6150:6150:4360:4360:8457:5144:5144:775:2827:6668:6150:6150:4360:8457:8457:5144:5144:2827:6668:6668:6668:6668:8457:7950:7950:7950:7950:783:5642:5642:8457:8457:3345:6418:6418:7950:783:3347:5642:5642:8457:3345:3345:6418:6418:1813:3347:2070:2070:6935:6935:6935:3853:6418:1813:3348:3348:2070:2832:2832:6935:3853:3853:
solution: 
+-------+-------+-------+
| 5 2 3 | 8 7 1 | 9 6 4 |
| 6 9 7 | 4 5 2 | 8 1 3 |
| 8 1 4 | 6 3 9 | 7 2 5 |
+-------+-------+-------+
| 9 4 5 | 7 2 8 | 1 3 6 |
| 7 6 1 | 3 4 5 | 2 8 9 |
| 2 3 8 | 9 1 6 | 4 5 7 |
+-------+-------+-------+
| 1 5 9 | 2 6 4 | 3 7 8 |
| 3 8 2 | 5 9 7 | 6 4 1 |
| 4 7 6 | 1 8 3 | 5 9 2 |
+-------+-------+-------+
Cheers
Ed
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Afmob
 Post subject:
PostPosted: Fri Apr 03, 2015 1:39 pm 
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Joined: Mon Apr 21, 2008 9:44 am
Posts: 310
Location: MV, Germany
This Assassin was quite hard and I probably didn't find the short path that Ed mentioned. But thanks nonetheless! :cheesey:

A315 Walkthrough:
1. R123+N7
a) 3(2) @ N7 = {12} locked for C1
b) 7(2) @ N7 = {34} locked for C1+N7
c) Both 13(2) = {58/67} locked N7
d) Hidden Single: R7C3 = 9 @ N7
e) Both 15(2) = {69/78} locked for R1
f) R1C1 = 5
g) Innies R12 = 7(2) = [61] -> R2C1 = 6, R2C8 = 1
h) R3C8 = 2
i) 7(2) = {34} locked for C9+N3

2. N14 !
a) 24(5): R34C2 <> 9 since R3C1 >= 7
b) 26(5) can only have two of (789) -> R5C234 <> 7,8,9
c) Hidden Single: R2C2 = 9 @ C2 -> R1C2 = 2
d) R4C3 <> 1,4,7,8 since it sees all 1,4,7,8 of N1
e) R4C2 <> 7,8 since it sees all 7,8 of C1
f) ! Hidden Killer triple (789) in 22(4) @ N4 -> 22(4) = 9{148/157/238/247} <> 6; R7C4 <> 7,8
g) 6,9 locked in 26(5) @ N4 = 169{28/37} since R45C1 = (789) -> 1,6 locked for R5

3. C456+R789 !
a) Innies+Outies R89: 4 = R7C2 - R8C9 -> R8C9 = (12), R7C2 = (56)
b) ! Consider placement of R6C1 = (12) -> R6C79 <> 1,2:
- R6C1 = 1/2 -> R7C1 = 2/1 -> R8C3 = 1/2 -> R8C9 = 2/1
c) Hidden Single: R4C7 = 1 @ N6
d) Innies+Outies C1234: 3 = R12C5 - R6C4: R2C5 <> 7,8 since R1C5 >= 6; R6C4 = (56789) since R12C5 >= 8
e) Innies+Outies C6789: -2 = R8C5 - R49C6 -> R4C6 <> 2 (IOU @ N8)
f) 33(7) = 126{3489/3579/4578} -> 2 locked for C5
g) ! Finned X-Wing (2) in C57 for R57 with fins R46C5 -> CPE: R5C46 <> 2
h) ! Consider placement of 2 in C1 -> R5C3 <> 2
- R6C1 = 2
- R7C1 = 2 -> R5C7 = 2 (HS @ C7)

4. N134
a) 26(5) = {13679} -> 7 locked for C1+N4; 3 locked for R5
b) 24(5) = 568{14/23} since R3C1 = 8 -> R3C1 = 8; 5 locked for R4+N4
c) 19(4) = 37{18/45} since R12C3 = (1347) -> 7 locked for R2
d) 11(3) = 2{18/45} -> R2C6 = 2
e) 11(2) = [56/92]
f) Innies N3 = 20(3) = 5{69/78} -> R3C7 = (67)

5. R789
a) 11(2) = {38/47} since (56) is a Killer pair of 13(2) @ R9
b) Naked pair (12) locked in R8C39 for R8
c) 8(3) = {125} since 1{34} blocked by Killer pair (34) of 11(2) -> R8C4 = 5
d) 27(4) = 9{378/468/567} -> CPE: R8C8 <> 9
e) R9C7 <> 9 since it sees all 9 of R8
f) 9 locked in 15(3) @ R9 = {249} since R8C8 <> 1,5,9 -> R8C8 = 4, R9C8 = 9, R9C9 = 2
g) R4C9 = 6 -> R3C9 = 5, R2C7 = 8 -> R1C6 = 1

6. C456
a) 13(3) = {346} -> 4 locked for C6
b) 20(4) = {1379} -> R3C7 = 7, R4C8 = 3, R3C6 = 9
c) 27(4) = {5679} since R89C7 = (356) -> R8C7 = 6, R9C7 = 5, R8C6 = 7, R8C5 = 9
d) 17(3) = 6{38/47} since R3C34 = (1346)

7. Rest is singles.

Rating:
1.5. I used a small forcing chain and a finned X-Wing.


Last edited by Afmob on Sun Apr 05, 2015 6:49 am, edited 1 time in total.

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wellbeback
 Post subject: Re: Assassin 315
PostPosted: Sat Apr 04, 2015 12:05 am 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
Thanks Ed. Happy Easter! Here's how I did it.

Thanks Andrew for your corrections and suggestion!
Hidden Text:
1. 3(2)c1 = {12}
-> 7(2)c1 = {34}
-> 2 * 13(2)n7 = {58} & {67}
-> HS 9 n7 -> r7c3 = 9
Also r8c3 from (12)

2. 2 * 15(2)r1 = {69} & {78}
-> NS r1c1 = 5
3(2)c8 = {12}
-> NP Innies r12 r2c18 = +7 = [61]
-> r3c8 = 2

3. r345c1 = {789}
26(5) cannot have all of (789)
-> Max r5c2 = 6 (also applies to r5c3 and r5c4)
Also Max r3c2+r4c23 = +11 -> No 9
-> HS 9 in c2 -> r2c2 = 9
-> r1c2 = 2
Also 9 in r45c1.

4. Whichever of (78) is in r3c1 can go in n4 only in r6c23
-> 6 in n4 in r5c23

5! Since r2c1 = 6 and r3c1 from (78) -> 24(5)n14 must contain a 1 or 2. (2 only in r4c3).
Also since r6c1 from (12) -> 1 cannot be in r56c2. (There's probably a technical name for this move? :) Andrew suggests "Killer CPE")

6. 26(5)n4 either {98}{612} or {97}{613} -> 1 in r5c34
1 also in c34 in 8(3)n78
-> 1 nowhere else in c34
-> HS 1 in n1 -> r3c2 = 1
-> 24(5)n14 can only be [681{45}]
-> r123c3 = {347}
-> r4c23 = [45]

7. Also HS 1 in r1 -> r1c6 = 1
-> r2c67 = [28] or [37]
-> 15(2)n3 = {69} and 15(2)n2 = {78}
-> HS 8 in n3 -> r2c67 = [28]
-> r3c7 = 7
-> r2c3 = 7
Also 11(2)c9 = [56]

8. r3c3 from (34) and r3c456 = {69(4|3)}
Since 5 already in r4 -> no solution for 17(3)r3c3 with a 9 in it.
-> 9 in r3c56
9 in n8 only in r8c56 or r9c6
-> HS 9 in c4 -> r6c4 = 9
-> r3c6 = 9
-> 20(4) = [9713]
Also -> r8c5 = 9
Since min r89c8 = {45} -> max r9c9 = 6
-> r9c8 = 9
-> 15(2)n3 = [96]
-> r5c9 = 9
-> r45c1 = [97]

9. 17(3)r3c3 cannot contain both {34}
-> r3c4 = 6
-> 17(3) = [467] or [368]
-> HS 2 r4 -> r4c5 = 2
Also [r3c5,r4c6] = [38] or [47]
But since 33(7) already contains a 9 it cannot also contain both of (47)
-> [r3c5,r4c6] = [38]
-> 17(3) = [467]
-> 15(2)n2 = [87]
ALso -> r1c3 = 3
-> 7(2)n3 = [43]
Also HS 8 in n8 -> r9c5 = 8
-> r9c6 = 3
-> r5c4 = 3
-> r5c23 = [61]
-> 22(4)n4 = [3892]
etc.

Rating:
Rating Easy 1.25


Last edited by wellbeback on Sun Apr 05, 2015 4:47 pm, edited 1 time in total.

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Andrew
 Post subject: Re: Assassin 315
PostPosted: Sun Apr 05, 2015 4:04 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks Ed. I found a fairly short way in; don't know whether it was your way.

Here is my walkthrough for Assassin 315:
Thanks Afmob for pointing out simplifications to step 10 and that steps 14a and 14b (now deleted) were unnecessary, and thanks to Ed and wellbeback for spotting typos.
Prelims

a) R1C45 = {69/78}
b) R1C78 = {69/78}
c) R12C9 = {16/25/34}, no 7,8,9
d) R23C8 = {12}
e) R34C9 = {29/38/47/56}, no 1
f) R67C1 = {12}
g) R78C2 = {49/58/67}, no 1,2,3
h) R89C1 = {16/25/34}, no 7,8,9
i) R9C23 = {49/58/67}, no 1,2,3
j) R9C56 = {29/38/47/56}, no 1
k) 11(3) cage at R1C6 = {128/137/146/236/245}, no 9
l) 8(3) cage at R8C3 = {125/134}
m) 27(4) cage at R8C5 = {3789/4689/5679}, no 1,2

Steps resulting directly and indirectly from Prelims
1a. Naked pair {12} in R23C8, locked for C8 and N3, clean-up: no 5,6 in R12C9, no 9 in R4C9
1b. Naked pair {34} in R12C9, locked for C9 and N3, clean-up: no 7,8 in R34C9
1c. Naked pair {12} in R67C1, locked for C1, clean-up: no 5,6 in R89C1
1d. Naked pair {34} in R89C1, locked for C1 and N7, clean-up: no 9 in R78C2, no 9 in R9C23
1e. Naked quad {6789} in R1C45 + R1C78, locked for R1 -> R1C1 = 5
1f. Naked quad {5678} in R78C2 + R9C23, locked for N7
1g. R7C3 = 9 (hidden single in N7)
1h. 27(4) cage at R8C5 = {3789/4689/5679}, CPE no 9 in R8C89
1i. R1C1 = 5 -> R12C2 = 11 = [29/38/47]
1j. R9C56 = {29/38/47} (cannot be {56} which clashes with R9C23), no 5,6 in R9C56

2. 45 rule on R12 2 innies R2C18 = 7 -> R2C1 = 6, R2C8 = 1, R3C8 = 2
2a. Naked triple {789} in R345C1, CPE no 7,8,9 in R4C23
2b. 9 in N4 only in R45C1 + R5C2, locked for 26(5) cage at R4C1, no 9 in R5C4
2c. Min R23C1 = 13 -> max R3C2 + R4C23 = 11, no 9 in R3C2

3. 45 rule on R89 2 innies R8C29 = 9 = [72/81], clean-up: no 7,8 in R7C2
3a. Naked pair {12} in R8C39, locked for R8
3b. 6 in R8 only in R8C5678, CPE no 6 in R9C7

4. 11(3) cage at R1C6 = {128/137/245}
4a. R2C7 = {578} -> no 5,7,8 in R2C6

5. 45 rule on N3 3 innies R2C7 + R3C79 = 20 = {569/578}
5a. 6,9 only in R3C79, 7,8 only in R23C7 -> no 5 in R3C7

6. 26(5) cage at R4C1 cannot contain all of 7,8,9, R45C1 = {789} -> no 7,8,9 in R5C234
6a. 9 in N4 only in R45C1, locked for C1
6b. R2C2 = 9 (hidden single in N1) -> R1C2 = 2 (cage sum)
[Remainder of step 6 deleted, hidden killer pair unnecessary after steps 7 and 8.]

7. 24(5) cage at R2C1 contains 6 = {14568/23568/24567} (cannot be {12678} which clashes with R6C1), 5 locked for R4 and N4, clean-up: no 6 in R3C9
7a. R3C1 = {78} -> no 7,8 in R3C2
7b. 24(5) cage = {14568} (only remaining combination, cannot be {23568/24567} because 1 in R13C3 + R4C3 = 2 clashes with R8C3) -> R3C1 = 8
7c. R3C2 + R4C23 = {145}, CPE no 1,4 in R56C2
7d. R34C2 = {14} (hidden pair in C2) -> R4C3 = 5, clean-up: no 8 in R9C2
7e. 3,7 in N1 only in R123C3, locked for C3, clean-up: no 6 in R9C2

8. Naked pair {79} in R45C1, locked for N4
8a. R45C1 = {79} = 16 -> R5C234 = 10 = {136/235} (cannot be {145} because R5C2 only contains 3,6), no 4, 3 locked for R5
8b. 8 in N4 only in R6C23, locked for R6 and 22(4) cage at R6C2, no 8 in R7C4
8c. 22(4) cage contains 8,9 = {1489/2389}, no 5,6,7
8d. 6 in N4 only in R5C23, locked for R5
8e. R5C234 = {136} (only remaining combination), locked for R5
8f. 2 in N4 only in R6C13, locked for R6

9. R2C7 + R3C79 (step 5) = {569/578}
9a. 8 of {578} must be in R2C7 -> no 7 in R2C7
9b. 11(3) cage at R1C6 (step 4) = {128/245} -> R2C6 = 2, R1C6 = {14}, clean-up: no 9 in R9C5

10. 31(5) cage at R5C6 = {16789/25789/45679}
10a. Hidden killer pair 2,4 in R5C5 and 31(5) cage for R5, 31(5) cage cannot contain both of 2,4 -> R5C5 = {24}, 31(5) cage contains one of 2,4 -> 31(5) cage = {25789/45679}, no 1
10b. 31(5) cage contains both of 7,9, R5C6789 can only contain one of 7,9 (both of 7,9 would clash with R5C1) -> R6C9 = {79}
10c. 8 in R5 only in 31(5) cage = {25789} (only remaining combination), 2 locked for R5 and N6 -> R5C5 = 4, R4C9 = 6 -> R3C9 = 5, R2C7 = 8 -> R1C6 = 1 (cage sum), clean-up: no 7 in R1C78, no 7 in R9C6
10d. Naked pair {69} in R1C78, locked for R1 and N3 -> R3C7 = 7
10e. R2C3 = 7 (hidden single in C3)
10f. 1 in N6 only in R46C7, locked for C7

11. 33(7) cage at R3C5 must contain 2 in R47C5, locked for C5, clean-up: no 9 in R9C6
11a. Killer pair 3,4 in R9C1 and R9C56, locked for R9
11b. Killer pair 7,8 in R9C23 and R9C56, locked for R9
11c. 8(3) cage at R8C3 = {125} (only remaining combination, cannot be {134} because 3,4 only in R8C4) -> R8C4 = 5, R9C4 = {12}
11d. R9C49 = {12} (hidden pair in R9)
11e. Naked pair {12} in R89C9, locked for C9 and N9
11f. Naked quad {1234} in R2579C4, locked for C4
11g. R2C5 = 5 (hidden single in R2)
11h. R4C5 = 2 (hidden single in R4)

12. 33(7) cage at R3C5 = {1234689} (only remaining combination), no 7

13. 13(3) cage at R6C6 = {346} (only remaining combination), 4 locked for R7
13a. 4 in N8 only in R789C6, locked for C6

14. 9 in N8 only in R8C56, locked for 27(4) cage at R8C5 -> R9C7 = 5, R9C2 = 7 -> R9C3 = 6

and the rest is naked singles.

Rating Comment:
I wasn't sure what rating to give step 7b; I'll go for 1.25 so that's the rating for my walkthrough.
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Ed
 Post subject: Re: Assassin 315
PostPosted: Thu Apr 09, 2015 10:05 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Found a different way again (step 5) - very pedestrian compared to the multitude of clever steps in the other walkthroughs. Really enjoyed them! Mine is boring. I got lucky but will take it!

start to A315
11 steps:
Preliminaries
Cage 3(2) n47 - cells ={12}
Cage 3(2) n3 - cells ={12}
Cage 15(2) n3 - cells only uses 6789
Cage 15(2) n2 - cells only uses 6789
Cage 7(2) n7 - cells do not use 789
Cage 7(2) n3 - cells do not use 789
Cage 13(2) n7 - cells do not use 123
Cage 13(2) n7 - cells do not use 123
Cage 11(2) n8 - cells do not use 1
Cage 11(2) n36 - cells do not use 1
Cage 8(3) n78 - cells do not use 6789
Cage 11(3) n23 - cells do not use 9
Cage 27(4) n89 - cells do not use 12

Not doing any clean-up unless stated.
1. 3(2)r6c1 = {12} only: both locked for c1
1a. -> 7(3)n7 = {34} only: both locked for c1 and n7
1b. -> two 13(2) cages in n7 = {58/67}: all locked for n7
1c. -> Hidden single 9 in n7: r7c3 = 9

2. Naked quad 6,7,8,9 in r1c4578: all locked for r1
2a. -> r1c1 = 5

3. "45" on r12: 2 innies r2c18 = 7 = [61] only permutation
3a. -> r3c8 = 2
3b. -> 7(2)n3 = {34} only: both locked for c9 and n3

4. Naked triple 7,8,9 in r345c1: r4c23 sees all of those -> no 7,8,9 in r4c23

The key step
5. "45" on n47: 4 outies r5789c4 - 2 = 2 innies r4c23
5a. Min. 4 outies = 10 -> min. r4c23 = 8 -> no 1,2 in r4c23

6. 24(5)r2c1 = 6{1359/1458}(no 7)
6a. must have 1 -> r3c2 = 1
6b. must have 5 which is only in r4c23: 5 locked for r4 and n4

7. 7 in c1 & 9 in n4 only in 26(5)r4c1 = 79{136/145/235}(no 8)
7a. naked pair {79} in r45c1: both locked for n4 and 9 for c1
7b. -> r3c1 = 8

8. r4c23 = 9 (cage sum) = {45} only: 4 locked for n4 and r4
8a. r4c23 = 9 -> r5789c4 = 11 (step 5) = {1235} only: all locked for c4

9. hidden single 9 in n1 -> r2c2 = 9
9a. -> r1c2 = 2 (cage sum)

10. Hidden single 1 in r1 -> r1c6 = 1
10a. r2c67 = 10 (cage sum) = [28/37]

11. 15(2)n3: {78} blocked by r2c7 = (78) -> = {69} only: both locked for r1 and n3

On from there. Much easier.
Cheers
Ed
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