SudokuSolver Forum

A forum for Sudoku enthusiasts to share puzzles, techniques and software
It is currently Tue Mar 19, 2024 11:18 am

All times are UTC




Post new topic Reply to topic  [ 3 posts ] 
Author Message
 Post subject: Assassin 341 X
PostPosted: Thu Dec 15, 2016 4:05 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 30, 2008 9:45 pm
Posts: 692
Location: Saudi Arabia
Assassin 341 X

It is X.

I went for a restrained puzzle - it is unique with the four 5-cages and five 3-cages.

SS gives it 1.5 and JS uses two medium fishes.


Image


JS Code:
3x3:d:k:22:23:5121:5121:3080:5122:5122:3847:3847:24:5121:25:26:3080:27:28:5122:3847:5121:4880:2316:29:3080:30:3853:3349:5122:5121:4880:4880:2316:2316:3853:3349:3349:5122:4618:4618:4618:4111:31:3853:4617:4617:4617:5124:3857:3857:4111:4878:4878:5908:5908:5123:5124:3857:4111:4882:4619:3347:4878:5908:5123:2309:5124:4882:4882:4619:3347:3347:5123:2310:2309:2309:5124:5124:4619:5123:5123:2310:2310:

Solution:
982634157
645871923
731529468
197268345
864153279
253947681
576312894
419786532
328495716


Top
 Profile  
Reply with quote  
 Post subject: Re: Assassin 341 X
PostPosted: Sun Dec 25, 2016 5:50 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for an interesting and challenging puzzle!

Here is my walkthrough for Assassin 341 X:
Prelims

a) 19(3) cage at R3C2 = {289/379/469/478/568}, no 1
b) 9(3) cage at R3C3 = {126/135/234}, no 7,8,9
c) 19(3) cage at R6C5 = {289/379/469/478/568}, no 1
d) 23(3) cage at R6C7 = {689}
e) 19(3) cage at R7C4 = {289/379/469/478/568}, no 1
f) 9(3) cage at R8C1 = {126/135/234}, no 7,8,9
g) 9(3) cage at R8C9 = {126/135/234}, no 7,8,9

1. 45 rule on R6789 1 outie R5C4 = 1 -> R6C4 + R7C3 = 15 = {69/78}
1a. 45 rule on R5 2 innies R5C56 = 8 = {26/35}
1b. Min R4C45 = 6 (cannot be {23} which clashes with R5C56) -> max R3C3 = 3
1c. 45 rule on C5 3 innies R456C5 = 15, R45C5 cannot total 8 (which would clash with R5C56 CCC) -> no 7 in R6C5
1d. R456C5 = 15 = {249/258/348/456}
1e. 8,9 of {249/258/348} must be in R6C5 -> no 2,3 in R6C5
1f. R45C6 cannot total 8 (which would clash with R5C56 CCC) -> no 7 in R3C7
1g. 23(3) cage at R6C7 = {689}, CPE no 6,8,9 in R45C8

2. 45 rule on N7 3 innies R7C23 + R8C3 = 2 outies R6C1 + R9C4 + 16
2a. Min R6C1 + R9C4 = 3 -> min R7C23 + R8C3 = 19, no 1 in R7C2
2b. Max R7C23 + R8C3 = 24 -> max R6C1 + R9C4 = 8, no 7,8,9 in R6C1, no 8,9 in R9C4

3. 45 rule on N9 3 innies R7C78 + R8C7 = 2 outies R6C9 + R9C6 + 16
3a. Min R6C9 + R9C6 = 3 -> min R7C78 + R8C7 = 19, no 1 in R8C7
3b. Max R7C78 + R8C7 = 24 -> max R6C9 + R9C6 = 8, no 8,9 in R6C9 + R9C6

4. Hidden killer pair 8,9 in 18(3) cage at R5C1 and 18(3) cage at R5C7 for R5, 18(3) cage cannot contain both of 8,9 (because no 1) -> each 18(3) cage must contain one of 8,9 = {279/369/378/459/468}
4a. 6 of 23(3) cage at R6C7 must be in R6C78 (R6C78 cannot be {89} which clashes with 18(3) cage at R5C7), 6 locked for R6, N6 and 23(3) cage, no 6 in R7C8, clean-up: no 9 in R7C3 (step 1)
4b. 18(3) cage at R5C7 = {279/378/459}
4c. Killer pair 8,9 in 18(3) cage and 23(3) cage, locked for N6
4d. Max R4C78 = 9 (cannot be {47/57} which clash with 18(3) cage, cannot be {37} because 13(3) cage at R3C8 cannot be 3{37}) -> min R3C8 = 4
4e. 19(3) cage at R6C5 = {289/379/469/478/568}
4f. 6 of {568} must be in R7C7 -> no 5 in R7C7
4g. 2 of {289} must be in R6C6 (R6C56 cannot be {89} which clashes with 23(3) cage), no 2 in R7C7
4h. 3 of {379} must be in R6C6 (R6C56 cannot be {79} which clashes with R6C4 + 23(3) cage), no 3 in R7C7
4i. 6 of {469} must be in R7C7, R6C56 cannot be {78} which clashes with R6C4 + 23(3) cage), no 4 in R7C7
4j. 19(3) cage must contain one of 2,3,4,5 in R6C56 -> must contain one of 7,8,9 in R6C56
4k. Hidden killer triple 7,8,9 in R4C6, R6C4 and R6C56 for N5 -> R4C6 = {789}
4l. Min R45C6 = 9 -> max R3C7 = 6 but R3C7 + R4C6 cannot be [67] which clashes with R6C4 + R7C3 -> max R3C7 = 5

[I saw this after the next step, but it simplifies things to do this step first.]
5. R5C56 (step 1a) = {26/35}
5a. 19(3) cage at R6C5 = {469/478/568} (cannot be {289/379} because R5C56 + R6C6 = {26}3/{35}2 clash with 9(3) cage at R3C3), no 2,3
5b. 19(3) cage = {478/568} (cannot be {469} = {49}6 because R4C6 + R6C4 = {78} clashes with 16(3) cage at R5C4 = 1{78}), no 9
5c. 9 in N5 only in R4C6 + R6C4, locked for D/
[With hindsight I ought to have done step 7b next.]

[I was slow in spotting the design feature of this puzzle; the previous step also seems to be an important design feature.]
6. R37C37 + R5C5 form a 5-cell cage because of the diagonals
6a. 45 rule on N5 4 outies R37C37 = 1 innie R5C5 + 14
6b. R5C5 = {2356} -> R37C37 = 16,17,19,20 -> R37C37 + R5C5 = 18,20,24,26
6c. Min R7C37 = 13, min R3C37 + R5C5 = 6 -> min R37C37 + R5C5 = 20 -> no 2 in R5C5, clean-up: no 6 in R5C6 (step 1a)
[I continued with step 6d, but feel free to jump forward to step 7.]
6d. Possible combinations for R37C37 + R5C5, taking account that 20(5) must contain 3 for R5C5, 24(5) must contain 5 for R5C5 and 26(5) must contain 6 for R5C5 and must contain two of 6,7,8 in R7C37
20(5) = {12368}
24(5) = {13578/14568/23568/24567}
26(5) = {14678/23678}
-> R3C37 = {12/13/14/23/24}, no 5 in R3C7, R7C37 = {67/68/78}

7. R5C56 (step 6c) = [35/53/62]
7a. 19(3) cage at R6C5 (step 5b) = {478} (only remaining combination, cannot be {568} = {58}6 which clashes with R5C56 using D\), 4 locked for R6 and N5, 7 locked for D\, CPE no 8 in R6C7, no 7 in R7C6
7b. Caged X-Wing for 8 in 19(3) cage and 23(3) cage at R6C7, no other 8 in R67 -> 16(3) cage at R5C4 = [196], 6 placed for D/, clean-up: no 2 in R5C6 (step 1a)
7c. R6C78 = [68] -> 19(3) cage = [478], 8 placed for D\, R7C8 = 9
7d. Naked pair {35} in R5C56, locked for R5 and N5
7e. R4C6 = 8, placed for D/ -> R3C7 + R5C6 = 7 = [25/43]
7f. Naked pair {26} in R4C45, locked for R4, R3C3 = 1 (cage sum), placed for D\
7g. 18(3) cage at R5C7 = {279} (only remaining combination), locked for R5 and N6
7h. Naked triple {468} in 18(3) cage at R5C1, locked for N4

8. 15(3) cage at R6C2 = {357} (only remaining combination) -> R6C23 = {35}, locked for R6 and N4, R7C2 = 7, R4C23 = [97], R3C2 = 3 (cage sum), R46C1 = [12], R6C23 = [53], R6C9 = 1
8a. 13(3) cage at R3C8 = {346} (only possible combination) -> R3C8 = 6, R4C78 = {34}, R4C9 = 5
8b. R9C8 = 1 (hidden single in N9) -> R89C9 = 8 = {26}, locked for C9 and N9
8c. 9(3) cage at R8C1 = {234} (only remaining combination) -> R9C2 = 2, R89C1 = {34}, locked for C1 and N7 -> R7C1 = 5, R8C2 = 1 (hidden single in N7), R9C9 = 6, placed for D\, R8C9 = 2, R4C4 = 2, R4C5 = 6, R1C1 = 9, R2C2 = 4, placed for D\
8d. Naked pair {89} in R89C3, locked for C3 -> R5C3 = 4
8e. 20(5) cage at R6C9 = {13457} -> R9C7 = 7
8f. Naked pair {34} in R9C14, locked for R9 -> R9C6 = 5, R5C56 = [53], R3C7 = 4 (cage sum), placed for D/, R8C8 = 3, R7C9 = 4, R9C1 = 3, placed for D/
8g. R8C7 = 5 -> R78C6 = 8 = [26]
8h. R7C4 = 3 -> R8C34 = 16 = [97]
8i. R1C9 + R2C8 = [72] -> R12C3 = [25], R1C8 = 5
8j. R1C3 + R2C2 + R4C1 = [241] = 7 -> R1C4 + R3C1 = 13 = [67]

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

Thanks Ed for detail corrections to my walkthrough; I've also added a few more of my own corrections, spotted while checking through Ed's changes.

Rating Comment:
After checking my walkthrough and making minor changes, I'll rate it not more than Hard 1.25; even that may be a bit high.


Happy Christmas, everybody! :santa:


Top
 Profile  
Reply with quote  
 Post subject: Re: Assassin 341 X
PostPosted: Tue Dec 19, 2017 7:08 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1039
Location: Sydney, Australia
And Merry Christmas to you Andrew! :D

As Andrew said, a very interesting puzzle. My solution is quite different. The way Andrew does those myriad blocking moves reminds of Para.

A341 X
start only:
Prelims courtesy of SudokuSolver
Cage 23(3) n69 - cells ={689}
Cage 9(3) n15 - cells do not use 789
Cage 9(3) n9 - cells do not use 789
Cage 9(3) n7 - cells do not use 789
Cage 19(3) n59 - cells do not use 1
Cage 19(3) n78 - cells do not use 1
Cage 19(3) n14 - cells do not use 1

1. "45" on r6789: 1 outie r5c4 = 1
1a. split 15(2)r6c4+r7c3 = 15 = {69/78}

2. Hidden killer triple 7,8,9 in r9 since the two 20(5) cages at r6c19 can have at most one of 7,8,9 each
2a. r9c5 = (789)
2b. the two 20(5) cages must have at least one of 7,8,9 = {12359/12368/12458/12467/13457}
2c. the one 7,8,9 for each of the 20(5) must be in r9 -> no 7,8,9 in r67c19 nor r8c28
2d. the 20(5) cages must both have 1 -> no 1 in r9c19 since each sees all one 20(5)

3. 8,9 in n9 only in r789c7 or r7c8 -> no 8,9 in r6c7 (CPE)
3a. r6c7 = 6
3b. r67c8 = {89}: both locked for c8
3c. 7 in n9 only in c7: locked for c7
3d. no 9 in r7c3 (sp15(2)r6c4+r7c3))

4. "45" on r789: 2 outies r6c19 - 27 = 4 innies r7c2378
4a. ->r6c19 = 3 = {12} only: both locked for r6
4b. r7c2378 = 30 = {6789} only: all locked for r7
4c. 6 in r7 only in n7: 6 locked for n7

5. Hidden quad 6,7,8,9 in n7 -> r7c23+r89c3 from {6789} only

6. 1 in n4 only in c1: locked for c1

7. 9(3)n7 = {135/234}
7a. must have 3 -> 3 locked for n7

8. r6c1 = {12} and sees all 1 & 2 in n7 apart from r9c2 -> r9c2 = {12}
(Alternatively: r67c1+r8c2 = three of {1245} -> 9(3)n7: {24}[3]/{23}[4] blocked -> r9c2 = {12} only)(personally I find the clone way much easy to find so chuffed I found both!)
8a. 3 in n7 only in c1: locked for c1

9. 9(3)n9 = {126/135/234}
9a. 6 blocked from r8c9 since {12} in r9c89 clashes with r9c2
9b. 6 blocked from r9c8 since {12} in r89c9 clashes with r6c9

10. 6 in n9 only in r8c8, r9c9 only on D\: 6 locked for D\ and no 6 in r9c6 (CPE)

Have a step in common with Andrew so perhaps is the key one for this puzzle. However, I checked SS and it doesn't use it.
11. "45" on n5: 4 outies r3c37+r7c37 - 14 = 1 innie r5c5
11a. Min. 4 outies = {12}[67] = 16 -> min. r5c5 = 3 (can't be 2 since r3c37 has a 2 and it sees r5c5 through the two diagonals

12. "45" on r5: 2 innies r5c56 = 8 = {35} only valid combination: both locked for r5 and n5, and not in r3c7 (CPE)

13. 18(3)n6 = {279} only valid combination: all locked for r5 and n6

14. r6c89 = [81]

15. "45" on c5: 3 innies r456c5 = 15 = [654]: only valid permutation. 5 placed for both D.

On from there. Lots of naked subsets, hidden singles, last combos or cage sums right till the end.
Cheers
Ed


Top
 Profile  
Reply with quote  
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 3 posts ] 

All times are UTC


Who is online

Users browsing this forum: No registered users and 13 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