SudokuSolver Forum

A forum for Sudoku enthusiasts to share puzzles, techniques and software
It is currently Sat Apr 27, 2024 6:28 pm

All times are UTC




Post new topic Reply to topic  [ 4 posts ] 
Author Message
PostPosted: Mon Aug 20, 2012 7:15 am 
Offline
Expert
Expert

Joined: Sat Jul 28, 2012 11:05 pm
Posts: 92
13th August 2012 Solution
Hidden Text:
Image


20th August 2012
Image

SS Score: 1.65
Jsudoku Code:
3x3::k:3345:3345:2578:2578:2323:2323:7440:7440:7440:3345:11520:11520:11520:11520:3348:2574:2574:7440:5142:11520:4367:11520:11520:3348:1547:1547:1547:5142:11520:4367:4367:11520:4360:4360:4360:1287:5142:2581:3085:3085:2826:2825:11521:4360:1287:2581:2581:3852:11522:2826:2825:11521:11521:3846:2581:11522:3852:11522:11522:4357:11521:3846:3846:11522:11522:11522:11522:4357:4357:11521:11521:11521:11522:2307:2307:5380:5380:5380:5380:11521:11521:



Moderator edit: changed formatting of corrected code; changed SS score to default scoring. Ed


Top
 Profile  
Reply with quote  
PostPosted: Mon Aug 20, 2012 11:23 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Wow! That's the first time I've seen a cage pattern with 3 45(9) cages!


Top
 Profile  
Reply with quote  
PostPosted: Fri Aug 24, 2012 9:42 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Another superb killer Pinata! I found this one easier than #3 & 4 since step 6 popped out quickly and it kept helping.

Start to PKS#5 20th August 2012
20 steps:
Let me know if anything is not correct or could be clearer.[Thanks Andrew!]

1. "45" on r1: 2 outies r2c19 = 16 = {79} only: both locked for r2

2. 6(3)n3 = {123} only: all locked for r3 and n3

3. 10(2)n3 = {46} only: both locked for r2 an n3

4. 13(2)n2 = {58} only: both locked for c6 and n2

5. "45" on n69: 1 outie r4c6+8 = 1 innie r9c7
5a. -> r4c6 = 1, r9c7 = 9

6. r7c45 sees all n7 apart from r9c23 -> must be equal (and contain the same pair of numbers), r9c23 = 9 ->r7c45 = h9(2)(no 9)

7. "45" on n8: remembering h9(2) at r7c45 -> 1 innie r8c4 = 7
7a. no 2 in r7c45 (h9(2))
7b. no 2,7 in r9c23 (step 6)

8. 10(3)r5c2 = {1234} only

9. r7c3 = 7 (hsingle n7)
9a. r6c3 = 8

10. "45" on n78: 1 outie r6c4 = 1 innie r7c1: both = (234)

11. 12(2)r5c3 = {39}[48](no 1,2,5,6; no 4 in r5c4)

12. 11(2)r5c6 = {29/47}(no 3,6) = [2/4,2/7..]

13. 5&6 in n4 in r4c123+r5c1: ie, must have at least one of 5/6 in r4 (no eliminations yet)
[edit: Andrew's step 17a uses a direct way to make this powerful compared to my step 14&15]

14. "45" on n5: 4 remaining innies r4c45+r56c4 = 22
14a. but {2479/2578} blocked by 11(2)r5c6 (step 12)
14b. {2569/3568/4567} all blocked by step 13 since 5&6 only in r4c45
14c. = {2389/3469/3478}(no 5)

15. 5 in n5 only in 11(2)r5c5 = {56} only: both locked for c5 and 6 for n5
15a. no 3,4 in r7c4 (h9(2)r7c45)

16. 9(2)r1c5 = {27}[36](no 1,4,9: no 3 in r1c6)

17. 3 in c6 only in n8: 3 locked for n8
17a. no 6 in r7c4 (h9(2)r7c45)(step 6)
17b. no 3,6 in 9(2)n7 (step 6)

18. 5 in n8 in h9(2)r7c45 = [54] or in split 12(3)r9c456 -> 4 in r9c456 must also have 5 or there would be no 5 for n8 (Locking-out cages)
18a. ->r9c456: {246} blocked
18b. h9(2)r7c45 = {18}[54] = [1/5..]
18c. -> {156} blocked from r9c456
18d. -> r9c456 = {138/345}(no 2,6)
18e. must have 3 -> r9c6 = 3

19. 6 in n8 only in r78c6: 6 locked for c6
19a. 17(3)r7c6 must have 6 = {269} only

20. 9(2)n2 = {27} only: both locked for r1 and n2

easier now. The final spot to check after lots of easier placements is the 15(3)r6c9. Some permutations don't work.
Cheerio
Ed


Top
 Profile  
Reply with quote  
PostPosted: Wed Aug 29, 2012 2:11 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks Pinata for a very enjoyable puzzle! :D

I deliberately waited until I'd finished Assassin 243 before I started this puzzle (but I then did this puzzle before checking and posting my walkthrough for A243).

My breakthrough steps were similar to Ed's:
I used "sees all except" steps leading to "clones". They made it easier than some recent puzzles; without them I think it would be a difficult one.

Here is my walkthrough for Pinata #5:
Prelims

a) R1C34 = {19/28/37/46}, no 5
b) R1C56 = {18/27/36/45}, no 9
c) R23C6 = {49/58/67}, no 1,2,3
d) R2C78 = {19/28/37/46}, no 5
e) R45C9 = {14/23}
f) R5C34 = {39/48/57}, no 1,2,6
g) R56C5 = {29/38/47/56}, no 1
h) R56C6 = {29/38/47/56}, no 1
i) R67C3 = {69/78}
j) R9C23 = {18/27/36/45}, no 9
k) 6(3) cage at R3C7 = {123}
l) 20(3) cage at R3C1 = {389/479/569/578}, no 1,2
m) 29(4) cage at R1C7 = {5789}
n) 10(4) cage at R5C2 = {1234}
o) and, of course, all three 45(9) cages = {123456789}

Steps resulting from Prelims
1a. Naked triple {123} in 6(3) cage at R3C7, locked for R3 and N3
1b. Naked quad {5789} in 29(4) cage at R1C7, locked for N3
1c. Naked quad {1234} in 10(4) cage at R5C2, CPE no 3,4 in R45C1
1d. Naked pair {46} in R2C78, locked for R2, clean-up: no 7,9 in R3C6

2. Max R3C9 + R45C9 = 8(3) must contain 1, locked for C9

3. 45 rule on N69 1 innie R9C7 = 1 outie R4C6 + 8 -> R4C6 = 1, R9C7 = 9, clean-up: no 8 in R1C5, no 4 in R5C9
3a. R6C8 = 9 (only place for 9 in 45(9) cage at R5C7), clean-up: no 2 in R5C5, no 2 in R5C6, no 6 in R7C3
3b. 1 in 45(9) cage at R2C2 only in R2C2345, locked for R2
3c. 1 in N4 only in R5C2 + R6C12, locked for 10(4) cage at R5C2, no 1 in R7C1

4. 45 rule on R1 2 outies R2C19 = 16 = {79}, locked for R2, clean-up: no 4,6 in R3C6
4a. 5,8 in N3 only in R1C789, locked for R1, clean-up: no 2 in R1C34, no 1,4 in R1C5, no 4 in R1C6
4b. R1C34 = {19/46} (cannot be {37} which clashes with R1C56), no 3,7 in R1C34

5. Naked pair {58} in R23C6, locked for C6 and N2, clean-up: no 3,6 in R56C6
5a. Killer triple 1,2,3 in R1C56 and R2C45, locked for N2, 1 also locked for R2, clean-up: no 9 in R1C3
5b. 2,3 in R2 only in R2C2345, locked for 45(9) cage at R2C2, no 2,3 in R4C25

6. 13(3) cage at R1C1 = {139/247}, no 6
6a. R2C1 = {79} -> no 7,9 in R1C12

7. 20(3) cage at R3C1 = {569/578} (cannot be {479} which clashes with R2C1), no 4, 5 locked for C1
7a. Killer pair 7,9 in R2C1 and 20(3) cage, locked for C1

8. 17(4) cage at R4C6 contains 1 = {1268/1358/1367/1457}
8a. 45 rule on N6 3 remaining innies R56C7 + R6C9 = 15 = {168/258/267/357} (cannot be {348} which clashes with R45C9, cannot be {456} which clashes with 17(4) cage), no 4
8b. 4 in 45(9) cage at R5C7 only in R7C7 + R8C789 + R9C89, locked for N9

9. 45 rule on N78 2 remaining outies R6C34 = 1 innie R7C1 + 8
9a. Max R7C1 = 4 -> max R6C34 = 12 -> max R6C4 = 5 (because R6C34 cannot be [66])

10. R7C3 “sees” all 7,8,9 in 45(9) cage at R6C4 except for R8C4 -> R7C3 = R8C4 = {789}
10a. 9 in N7 only in R78C23, CPE no 9 in R7C45

11. 17(3) cage at R7C6 = {269/359/368/467} (cannot be {179/278} which clash with R56C6, cannot be {458} because 5,8 only in R8C5), no 1
11a. 5,8 of {359/368} must be in R8C5 -> no 3 in R8C5

12. R7C1 “sees” all cells of 45(9) cage at R6C4 except for R68C4, R7C3 = R8C4 = {789} (step 10) cannot equal R7C1 -> R7C1 = R6C4 = {234}
12a. R7C1 = R6C4, R7C3 = R8C4 -> R7C45 = R9C23 (the same pair of numbers totalling 9, Law of Leftovers)
[Ed saw this more directly. R7C45 “see” all of N7 except for R9C23 …]
12b. 45 rule on N8 3 remaining innies R7C45 + R8C4 = 16, R7C45 = 9 -> R8C4 = 7, R7C3 = 7, R6C3 = 8, clean-up: no 5 in R5C3, no 4,5 in R5C4, no 3 in R5C5, no 1,2 in R9C2, no 2 in R9C3
12c. 7 in R9 only in R9C89, locked for 45(9) cage at R5C7, no 7 in R56C7

13. 7 in R9 only in R9C89
13a. 45 rule on R9 3 innies R9C189 = 15 = {267/357}, no 1,4,8
13b. R9C23 = {45}/[81] (cannot be {36} which clashes with R9C189), no 3,6 in R9C23
13c. R9C23 = {45}/[81] -> R7C45 = {18/45} (step 12a), no 2,3,6

14. 15(3) cage at R6C9 = {168/258/267/357}
14a. 7 of {357} must be in R6C9 -> no 3 in R6C9
14a. 5 of {258} must be in R6C9 (R7C89 cannot be {58} which clashes with R7C45), 7 of {267} must be in R6C9 -> no 2 in R6C9

15. 17(3) cage at R3C3 = {269/359/368/458}
15a. 8 of {458} must be in R4C4 -> no 4 in R4C4

16. 20(3) cage at R3C1 (step 7) = {569/578}
16a. 8 of {578} must be in R3C1 -> no 7 in R3C1
16b. 7 in R3 only in R3C25, locked for 45(9) cage at R2C2, no 7 in R4C25

17. 5,6 in N4 only in R4C123 + R5C1 -> R4C123 must contain at least one of 5,6
17a. 5,6 in N5 only in R4C45 or R56C5 (locking cages because R56C5 is a 11(2) cage), but R4C45 cannot be {56} (which clashes with R4C123) -> R56C5 = {56}, locked for C5 and N5, clean-up: no 3 in R1C6, no 4 in R7C4 (step 13c)
17b. 7 in N5 only in R56C6 = {47}, locked for C6 and N5, clean-up: no 2 in R1C5, no 4 in R7C1 (step 12)

18. 10(4) cage at R5C2 = {1234}, 4 locked for N4, clean-up: no 8 in R5C4
18a. Naked pair {39} in R5C34, locked for R5, clean-up: no 2 in R4C9

19. 9 in C6 only in R78C6, locked for N8
19a. 17(3) cage at R7C6 (step 11) contains 9 = {269} (only remaining combination) -> R8C5 = 2, R78C6 = {69}, locked for C6 and N8 -> R1C6 = 2, R1C5 = 7, R9C6 = 3

20. R12C9 = [97] (hidden pair in N3), R2C1 = 9, R9C8 = 7 (hidden single in R9), R4C7 = 7 (hidden single in N6), R5C1 = 7 (hidden single in C1), R56C6 = [47]

21. 20(3) cage at R3C1 (step 7) = {578} (only remaining combination) -> R34C1 = [85], R23C6 = [85]
21a. R4C5 = 8 (only remaining place for 8 in 45(9) cage at R2C2), R3C5 = 9 (hidden single in C5), R4C2 = 6, R3C4 = 4, R1C4 = 6, R1C3 = 4, clean-up: no 1 in R7C4 (step 13c), no 5 in R9C2

22. R2C5 = 3 (hidden single in C5), R2C4 = 1

23. R3C3 = 6, R4C34 = 11 = {29}, locked for R4

24. Naked pair {34} in R4C89, locked for N6
24a. R4C67 = [17] = 8 -> R45C8 = 9 = [36/45]
24b. Naked pair {56} in R5C8 + R6C9, locked for N6

25. R5C7 = 8 (hidden single in N6), R1C78 = [58]

26. 15(3) cage at R6C9 (step 14) = {168/258} -> R7C9 = 8, R7C4 = 5, R7C5 = 4 (step 13c), R9C45 = [81], R9C23 = [45], R2C23 = [52], R4C34 = [92], R5C34 = [39], R8C3 = 1, R6C4 = 3, R7C1 = 3 (step 12)

27. R89C1 = [62], R9C9 = 6, R6C9 = 5, R5C8 = 6, R2C78 = [64], R4C8 = 3, R4C9 = 4, R5C9 = 1

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for Pinata #5 at 1.5. I used "sees all except" steps leading to "clones" and Law of Leftovers. My solving path felt easier than 1.5, but that's the lowest rating I can give when using "clones".

Sudoku Solver:
v3.3.1 (I'm still using this earlier version for Archive purposes until I finish Archive Part G) says that it used "Simple T&E". Don't know why it needed to; none of my steps were T&E.

Now to try Pinata #6.


Top
 Profile  
Reply with quote  
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 4 posts ] 

All times are UTC


Who is online

Users browsing this forum: No registered users and 55 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:  
Powered by phpBB® Forum Software © phpBB Group