SudokuSolver Forum

A forum for Sudoku enthusiasts to share puzzles, techniques and software
It is currently Sat Apr 27, 2024 2:04 am

View unanswered posts | View active topics


Board index » Puzzles and Solutions » Killer Puzzles

All times are UTC




Post new topic Reply to topic  Page 1 of 1
  [ 5 posts ] 
  Print view Previous topic | Next topic 
Author Message
Ed
 Post subject: Assassin 313
PostPosted: Thu Mar 19, 2015 10:22 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
sudokuSolver makes a hash of this but JSudoku hardly blinks which is why I tried it. Enjoyed it. Fairly short way to crack it.

Assassin 313

Image
code: paste into solver:
3x3::k:2304:2304:6657:6657:5378:6147:6147:2564:2564:3077:3077:6657:5378:5378:5378:6147:2566:2566:3077:3079:6657:3336:3352:2570:6147:1547:2566:7692:3079:3079:3336:3352:2570:1547:1547:7693:7692:7692:7692:3610:3610:3610:7693:7693:7693:7692:3342:3342:2063:2585:1808:5137:5137:7693:4114:3342:4883:2063:2585:1808:5652:5137:3093:4114:4114:4883:4886:4886:4886:5652:3093:3093:2071:2071:4883:4883:4886:5652:5652:2825:2825:
solution: 
+-------+-------+-------+
| 4 5 2 | 8 9 1 | 6 3 7 |
| 8 3 7 | 4 2 6 | 9 5 1 |
| 1 6 9 | 7 5 3 | 8 2 4 |
+-------+-------+-------+
| 9 2 4 | 6 8 7 | 3 1 5 |
| 3 7 5 | 9 1 4 | 2 6 8 |
| 6 8 1 | 5 3 2 | 4 7 9 |
+-------+-------+-------+
| 2 4 8 | 3 7 5 | 1 9 6 |
| 5 9 3 | 1 6 8 | 7 4 2 |
| 7 1 6 | 2 4 9 | 5 8 3 |
+-------+-------+-------+
Cheers
Ed
Top
 Profile  
Reply with quote  
Afmob
 Post subject:
PostPosted: Fri Mar 20, 2015 6:22 pm 
Offline
Grand Master
Grand Master

Joined: Mon Apr 21, 2008 9:44 am
Posts: 310
Location: MV, Germany
Thanks for the new Assassin, Ed! It took me some time solve it and it's probably not the (technical) easiest way to solve it but at least my walkthrough isn't that long.

A313 Walkthrough:
1. C6789 !
a) Outies N6 = 11(2) = [29/38]
b) Innies+Outies N3: -1 = R1C6 - R3C8 = [12]/[23] -> CPE: R1C789+R2C6+R23C7 <> 2
c) 24(4) = 9{168/258/267} <> 3,4 since R1C6 = (12) -> 9 locked for C7+N3; R123C7 <> 1
d) 10(2) = {37/46}
e) 8,9 locked in 24(4) @ N3 = 89{16/25} locked for C7
f) Outies C89 = 9(3) <> 7
g) ! Using Outies N6 = 11(2): Innies+Outies C89: -6 = R5C7 - R46C8 -> R6C8 <> 6 (IOU @ N6)
h) ! Innies N36 = 34(4+1): R6C7 <> 5,6 since R123C7 = (5689) and R6C8 <> 6

2. C789
a) 20(3) = 9{38/47} since R6C7 = (34) -> 9 locked for C8; R6C8 <> 3,4
b) Outies C89 = 9(3) = 3{15/24} since R6C7 = (34) -> 3 locked for C7+N6
c) 6(3) = {123} -> 1 locked for R4+N6
d) 20(3): R6C8 <> 8 since [389] blocked by Killer pair (38) of 30(5)
e) 5,6,8 locked in 30(5) @ N6 = 568{29/47}

3. R456+C123 !
a) Innies R6789 = 15(2) = {69/78}
b) Innies R1234 = 14(2) = {59/68}
c) Killer pair (79) locked in Innies R6789 + R6C8 for R6
d) Outies N4 = 10(2) <> 5
e) Innies+Outies N4: -3 = R3C2 - R6C23 -> R6C3 <> 3 (IOU @ C2)
f) Innies+Outies N4: -2 = R7C2 - R4C23 -> R4C3 <> 2 (IOU @ C2)
g) Outies C12 = 10(3) <> 8,9: R5C3 <> 7 since R4C3 >= 3
h) ! Hidden Killer pair (13) in 14(3) and 30(5) @ R5 since each can only have one of them -> 30(5) = 9{1578/3468/3567} <> 2 -> 9 locked for N4
i) Outies C123 = 10(2) <> 5; R9C4 <> 9
j) Innies+Outies N1: 2 = R1C4 - R3C2 -> R1C4 <> 2,7; R3C2 <> 3,8,9

4. C123+R4 !
a) ! 2 locked in Innies N4 @ N4 = 15(4) = 24{18/36} since {1257} leaves no combo for 12(3) as R4C23 <> 1 and R3C2 <> 3 -> 4 locked for N4
b) 12(3) = {246} since 7{23} blocked by R4C78 = (123), 1{38} blocked by Killer pair (38) of 30(5) and R4C23 <> 1,7 -> 2 locked for C2; CPE: R56C2 <> 4,6
c) Outies C12 = 10(3) = 1{36/45} since R4C3 = (46) -> R5C3 = (35), R6C3 = 1
d) 13(3) = 1{39/48} -> R7C2 = (49)
e) Outies N4 = 10(2) = [64] -> R7C2 = 4, R3C2 = 6
f) Cage sum: R6C2 = 8
g) R4C2 = 2, R4C3 = 4, R4C8 = 1, R4C7 = 3, R3C8 = 2
h) Outie N14 = R1C4 = 8
i) Outie N47 = R9C4 = 2
j) 19(4) = {2368} since {2359} blocked by R5C3 = (35) -> 3,6,8 locked for C3+N7
k) 8(2) = {17} locked for R9+N7

5. C789
a) Outie N6 = R7C8 = 9
b) Outie N36 = R1C6 = 1
c) Outie N69 = R9C6 = 9
d) 24(4) = {1689} -> 6 locked for C7+N3
e) 1,7 locked in 22(4) @ C7 = {1579} for N7; R9C7 = 5
f) 11(2) = {38} locked for R9+N9

6. R456
a) 5 locked in R6C46 @ R6 for N5
b) 7 locked in R4C456 @ R4 for N5
c) 1 locked in 14(3) @ N5 = {149} locked for R5+N5 -> R5C6 = 4
d) 13(2) @ C4 = {67} -> R3C4 = 7, R4C4 = 6
e) 8(2) = {35} locked for C4

7. Rest is singles.

Rating:
Easy 1.5 - 1.5. I used combo analysis.


Last edited by Afmob on Sat Mar 21, 2015 10:30 pm, edited 1 time in total.

Top
 Profile  
Reply with quote  
Frank
 Post subject: Re: Assassin 313
PostPosted: Fri Mar 20, 2015 11:58 pm 
Offline
Addict
Addict

Joined: Mon Apr 28, 2008 10:58 pm
Posts: 47
Location: Victoria, B.C., Canada
I haven't done an assassin for quite a while. I cracked this with my usual trial and error approach.

walkthrough:
There are a lot of innies and outies to start with. It is easy to show that r1c6 = {12}. I tried 2 and had to go a fair way before getting a contradiction. So r1c6 = 1, and the rest is easy.

Many thanx for this Ed - Cheers - Frank
Top
 Profile  
Reply with quote  
Andrew
 Post subject: Re: Assassin 313
PostPosted: Sat Mar 21, 2015 3:30 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Welcome back, Frank! :D

It's a long time since we've had a symmetrical cage pattern, so welcome stranger. ;)

Ed wrote:
JSudoku hardly blinks
That comment influenced the way I solved this Assassin.

Here is my walkthrough for Assassin 313:
Prelims

a) R1C12 = {18/27/36/45}, no 9
b) R1C89 = {19/28/37/46}, no 5
c) R34C4 = {49/58/67}, no 1,2,3
d) R34C5 = {49/58/67}, no 1,2,3
e) R34C6 = {19/28/37/46}, no 5
f) R67C4 = {17/26/35}, no 4,8,9
g) R67C5 = {19/28/37/46}, no 5
h) R67C6 = {16/25/34}, no 7,8,9
i) R9C12 = {17/26/35}, no 4,8,9
j) R9C89 = {29/38/47/56}, no 1
k) 10(3) cage at R2C8 = {127/136/145/235}, no 8,9
l) 6(3) cage at R3C8 = {123}
m) 20(3) cage at R6C7 = {389/479/569/578}, no 1,2
n) 26(4) cage at R1C3 = {2789/3689/4589/4679/5678}, no 1

1. 45 rule on R1234 2 innies R4C19 = 14 = {59/68}

2. 45 rule on R6789 2 innies R6C19 = 15 = {69/78}

3. 6(3) cage at R3C8 = {123}, CPE no 1,2,3 in R56C8
3a. 45 rule on N6 2 outies R37C8 = 11 = [29/38]
3b. 6(3) cage = {123}, 1 locked for R4 and N6, clean-up: no 9 in R3C6

4. 45 rule on N3 1 innie R3C8 = 1 outie R1C6 + 1, R3C8 = {23} -> R1C6 = {12}
4a. R1C6 + R3C8 = [12/23], CPE no 2 in R1C789 + R23C7 + R3C6, clean-up: no 8 in R1C89, no 8 in R4C6
4b. Max R1C6 = 2 -> min R123C7 = 22, no 1,3,4
4c. 10(3) cage at R2C8 = {127/136/145} (cannot be {235} which clashes with R3C8), 1 locked for N3, clean-up: no 9 in R1C89
4d. 10(3) cage = {127/145} (cannot be {136} which clashes with R1C89), no 3,6
4e. Killer pair 4,7 in R1C89 and 10(3) cage, locked for N3
4f. 8,9 in N3 only in R123C7, locked for C7
4g. R1C12 = {18/27/45} (cannot be {36} which clashes with R1C89), no 3,6 in R1C12

5. 45 rule on C789 2 outies R19C6 = 10 = [19/28]

6. 45 rule on C89 3 outies R456C7 = 9 = {126/135/234}, no 7
6a. 6 of {126} must be in R6C7 -> no 6 in R5C7
6b. 7 in C7 only in R789C7, locked for N9, clean-up: no 4 in R9C89

7. 45 rule on C12 3 outies R456C3 = 10 = {127/136/145/235}, no 8,9

8. 45 rule on C6789 3 innies R258C6 = 18 = {369/378/459/468/567} (cannot be {189} which clashes with R9C6, cannot be {279} which clashes with R19C6), no 1,2

[Since Ed said that JSudoku solved this easily, I’ll try a short forcing chain …]
9. R456C7 (step 6) = {126/135/234}
9a. 45 rule on N3 4 innies R123C7 + R3C8 = 25 = {589}3/{689}2
9b. Consider combinations for R123C7 + R3C8
R123C7 + R3C8 = {589}3, 5 locked for C7 => R4C78 = {12} => R456C7 = {234} (cannot be {126} which clashes with R4C78, CCC)
or R123C7 + R3C8 = {689}2, 6 locked for C7 => R4C78 = {13} => R456C7 = {234} (cannot be {135} which clashes with R4C78, CCC)
-> R456C7 = {234}, locked for C7 and N6
-> R4C8 = 1

10. 4 in N9 only in 12(3) cage at R7C9 = {246/345}, no 1,8,9
10a. R9C89 = {29/38} (cannot be {56} which clashes with 12(3) cage), no 5,6 in R9C89
10b. Killer pair 8,9 in R9C6 and R9C89, locked for R9

11. R6C7 = {34} -> 20(3) cage at R6C7 = {389/479}, no 5,6, 9 locked for C8, clean-up: no 2 in R9C9
11a. 8 of {389} must be in R7C8 (because of interactions between 6(3) cage at R3C8 and R37C8 = 11, step 3a), no 8 in R6C8
[Afmob saw this as [389] clashes with 30(5) cage at R4C9, which is technically simpler.]
11b. 5,6,8 in N6 only in 30(5) cage = {25689/45678}, no 3
11c. Killer pair 7,9 in R6C19 and R6C8, locked for R6, clean-up: no 1 in R7C4, no 1,3 in R7C5

12. 45 rule on C123 2 outies R19C4 = 10 = {37/46}/[82/91], no 5, no 2 in R1C4

13. 45 rule on N1 1 outie R1C4 = 1 innie R3C2 + 2, no 3,8,9 in R3C2
13a. 12(3) cage at R3C2 cannot be 7{23} which clashes with R4C7, no 7 in R3C2, clean-up: no 9 in R1C4, no 1 in R9C4 (step 12)

14. 45 rule on N7 1 innie R7C2 = 1 outie R9C4 + 2, no 1,2,3,7 in R7C2

15. 45 rule on N4 2 outies R37C2 = 10 = [19/28]/{46}, no 5, clean-up: no 7 in R1C4 (step 13), no 3 in R9C4 (step 14)

16. 26(4) cage at R1C3 = {2789/3689/4589/4679} (cannot be {5678} which clashes with R1C12), 9 locked for C3 and N1
[I first saw this combo elimination using hidden killer pair 3,9 for N1, but the clash is simpler.]

17. R1C4 = R3C2 + 2 (step 13) -> R1C4 + R3C2 = [31/42/64/86]
17a. R1C89 = {37/46}, R3C8 = {23}, 3 in N3 only in R1C89 + R3C8 -> combined half cage R1C89 + R3C8 = {37}2/{46}3
17b. R1C4 + R3C2 = [31/64/86] (cannot be [42] which clashes with R1C89 + R3C8, IOD clash), no 4 in R1C4, no 2 in R3C2, clean-up: no 8 in R7C2 (step 15), no 6 in R9C4 (step 12)
[I originally wrote steps 17a and 17b as
Consider placement for 3 in N3
3 in R1C89 = {37} => R3C8 = 2
or R3C8 = 3 => R1C89 = {46}, locked for R1
-> R1C4 + R3C2 = [31/64/86] (cannot be [42], locking out IOD) …]

18. R1C4 + R3C2 (step 17a) = [31/64/86]
18a. 26(4) cage at R1C3 (step 16) = {2789/3689/4589} (cannot be {4679} which clashes with R1C4 + R3C2 = [64], IOD clash)
18b. 26(4) cage = {2789/3689/4589}, CPE no 8 in R1C12, clean-up: no 1 in R1C12
18c. Killer pair 4,7 in R1C12 and R1C89, locked for R1
18d. 1 in R1 only in R1C56, locked for N2, clean-up: no 9 in R4C6

19. R1C12 = {27/45}, R1C89 = {37/46} -> combined cage R1C1289 = {27}{46}/{45}{37}
19a. R1C4 + R3C2 (step 17b) = [31/86] (cannot be [64] which clashes with R1C1289, IOD clash), no 6 in R1C4, no 4 in R3C2, clean-up: no 6 in R7C2 (step 15), no 4 in R9C4 (step 12)
19a. Killer triple 2,3,7 in R9C12, R9C4 and R9C89, locked for R9

20. 12(3) cage at R2C1 = {138/237/345} (cannot be {147/246} which clash with R1C12, cannot be {156} which clashes with R3C2), no 6, 3 locked for N1

21. R1C1289 (step 19) = {27}{46}/{45}{37}, R19C4 (step 12) = [37/82], R19C6 (step 5) = [19/28]
21a. R1C46 = [31/81] (cannot be [32] which clashes with R1C1289, cannot be [82] because R9C46 = [28] clashes with R9C89) -> R1C6 = 1
21b. R1C6 = 1 -> R3C8 = 2 (step 4), R4C7 = 3, R6C7 = 4 -> R67C8 = 16 = [79], R5C7 = 2, R7C2 = 4, R3C2 = 6 (step 15)
21c. 3 in N3 only in R1C89 = [37], R1C4 = 8, R9C4 = 2 (step 12), R9C89 = [83], R9C6 = 9
21d. R1C12 = [45] (only remaining permutation)
21e. Naked triple {279} in R123C3, locked for C3 and N1
[Routine clean-ups omitted from here]

22. R3C2 = 6 -> R4C23 = 6 = [24]
22a. R6C19 (step 2) = {69} (only remaining combination), locked for R6
22b. R7C2 = 4 -> R6C23 = 9 = [81]
22c. R456C3 = 10 (step 7) -> R5C3 = 5, R5C8 = 6, R456C9 = [589], R46C1 = [96]

23. R9C12 = {17} (only remaining combination), locked for R9 and N7, R9C3 = 6, R9C7 = 5, R8C8 = 4
23a. R7C7 = 1 (hidden single in R7), R8C7 = 7 (hidden single in C7)

24. R9C5 = 4, 1 in N8 only in 19(4) cage at R8C4 = {168}4, 6,8 locked for R8 and N8, R7C5 = 7 -> R6C5 = 3

25. R34C4 = [76] (only remaining permutation)

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A313 at 1.5. I used a couple of short chains, combined cages and innie-outie clashes. Then with hindsight I reworked step 17, so only one forcing chain.
Top
 Profile  
Reply with quote  
Ed
 Post subject: Re: Assassin 313
PostPosted: Wed Mar 25, 2015 10:30 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Welcome back Frank! Glad to give you a good one to get back to.

A fourth completely different solution below though Andrew got my key elimination (his 17b. my 2.). Afmob's way is quite similiar to JSudoku which used a different 5 innies to get started. I might have led Andrew a bit astray in my intro. It doesn't do chains quickly.

Start to A313
10 steps:
Prelims courtesy of SudokuSolver
Preliminaries
Cage 7(2) n58 - cells do not use 789
Cage 8(2) n7 - cells do not use 489
Cage 8(2) n58 - cells do not use 489
Cage 13(2) n25 - cells do not use 123
Cage 13(2) n25 - cells do not use 123
Cage 9(2) n1 - cells do not use 9
Cage 10(2) n25 - cells do not use 5
Cage 10(2) n3 - cells do not use 5
Cage 10(2) n58 - cells do not use 5
Cage 11(2) n9 - cells do not use 1
Cage 6(3) n36 - cells ={123}
Cage 10(3) n3 - cells do not use 89
Cage 20(3) n69 - cells do not use 12
Cage 26(4) n12 - cells do not use 1

1. "45" on n6: 2 outies r37c8 = 11 = [29/38] = [2] or [8]

2. "45" on n4: 2 outies r37c2 = 10 (no 5)
2a. but [28] blocked by r37c8 (step 1)
2b. no 2 in r3c2, no 8 in 7c2

3. "45" on n3: 1 innie r3c8 - 1 = 1 outie r1c6 = [12/23]
3a. must have 2 -> no 2 in common peers in r3c6 nor r1c89 nor r123c7
3b. no 8 in r4c6, r1c89

4. 24(4)r1c6 must have 1 or 2 for r1c6 = {1689/2589/2679)(no 3,4)
4a. must have 9: 9 locked for c7 and n3
4b. can only have one of 1 or 2 -> no 1 in r123c7
4c. no 1 in r1c89

5. 10(2)n3 = {37/46} = 3 or 6
5a. 10(3)n3 must have 1 for n3 = {127/136/145}
5b. but {136} blocked by 10(2)
5c. 10(3) = {127/145}(no 3,6)

6. [23] in r1c6+r3c8 (step 2) or 3 for n3 in 10(2) = {37}
6a. ie, must have 2 or 7 in one of r1c6 or r1c89 because of 3's in n3 (Blocking cages)
6b. ->{27} blocked from 9(2)n1: no 2,7

7. 9(2)n1: {36} blocked by 10(2)n3 (step 5.)
7a. 9(2) = {18/45}(no 3,6): note, must have 5 or 8

8. 26(4)r1c3 sees all of 9(2)n1 -> {4589/5678} blocked (step 7b)
8a. = {2789/3689/4679}(no 5)

9. "45" on n1: 1 outie r1c4 - 2 = 1 innie r3c2
9a. no 2,4,7 in r1c4; no 3,8,9 in r3c2
9b. no 1,2,7 in r7c2 (h10(2)r37c2)

10. from step 9. 6 or 9 in r1c4 -> 4 or 7 in r3c2
10a. 26(4)r1c3 = {2789/3689/4679}: but {4679} blocked from 26(4)r1c3 by IOD n3 = -2 (IOD block)
10b. = {2789/3689}(no 4)
10c. must have 8 -> no 8 in common peers in r1c12
10d. -> 9(2)n1 = {45} only: both locked for n1 and r1
10e. -> 10(2)n3 = {37} only: both locked for r1 and n3

cracked.
Cheers
Ed
Top
 Profile  
Reply with quote  
Display posts from previous:  Sort by  
Post new topic Reply to topic  Page 1 of 1
  [ 5 posts ] 

Board index » Puzzles and Solutions » Killer Puzzles

All times are UTC


Who is online

Users browsing this forum: Bing [Bot] and 9 guests

You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB® Forum Software © phpBB Group