SudokuSolver Forum

A forum for Sudoku enthusiasts to share puzzles, techniques and software
It is currently Thu Mar 28, 2024 9:59 am

All times are UTC




Post new topic Reply to topic  [ 4 posts ] 
Author Message
 Post subject: Assassin 52 V2 Revisit
PostPosted: Tue Mar 16, 2021 7:21 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Attachment:
a52v2.JPG
a52v2.JPG [ 63.61 KiB | Viewed 4541 times ]
Assassin 52 V2 Revisit

The first Revisit from Archive B. Gets a score of 1.70 and uses one 'complex intersection'.

Code: Select, Copy & Paste into solver:
3x3::k:4608:4608:1794:1794:5892:1797:1797:2055:2055:4617:4608:2571:2571:5892:3342:3342:2055:6673:4617:4617:4617:4373:5892:2839:6673:6673:6673:3099:5404:4373:4373:5892:2839:2839:5154:3107:3099:5404:5926:5926:5926:5926:5926:5154:3107:3099:5404:3119:3119:5425:4658:4658:5154:3107:6198:6198:6198:3119:5425:4658:4924:4924:4924:6198:2624:1601:1601:5425:3908:3908:3910:4924:2624:2624:2634:2634:5425:2381:2381:3910:3910:
Solution:
+-------+-------+-------+
| 7 8 5 | 2 9 1 | 6 4 3 |
| 9 3 4 | 6 7 8 | 5 1 2 |
| 1 2 6 | 5 3 4 | 9 7 8 |
+-------+-------+-------+
| 6 7 3 | 9 4 5 | 2 8 1 |
| 2 5 8 | 7 1 3 | 4 9 6 |
| 4 9 1 | 8 6 2 | 7 3 5 |
+-------+-------+-------+
| 8 6 7 | 3 2 9 | 1 5 4 |
| 3 1 2 | 4 5 7 | 8 6 9 |
| 5 4 9 | 1 8 6 | 3 2 7 |
+-------+-------+-------+
Cheers
Ed


Top
 Profile  
Reply with quote  
PostPosted: Sun Mar 21, 2021 10:08 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Nice puzzle. Lots of areas to explore but difficult to find a break. Yet not too complicated when it reveals. This is a different path to the two posted WTs on the archive.

A52v2 start:
Preliminaries courtesy of SudokuSolver
Cage 6(2) n78 - cells only uses 1245
Cage 15(2) n89 - cells only uses 6789
Cage 7(2) n12 - cells do not use 789
Cage 7(2) n23 - cells do not use 789
Cage 13(2) n23 - cells do not use 123
Cage 9(2) n89 - cells do not use 9
Cage 10(2) n78 - cells do not use 5
Cage 10(2) n12 - cells do not use 5
Cage 8(3) n3 - cells do not use 6789
Cage 21(3) n4 - cells do not use 123
Cage 10(3) n7 - cells do not use 89
Cage 20(3) n6 - cells do not use 12
Cage 11(3) n256 - cells do not use 9
Cage 26(4) n3 - cells do not use 1

No clean-up done unless stated. This is a highly optimised solution.
1. "45" on c5: 1 innie r5c5 = 1

2. "45" on n7: 2 outies r89c4 = 5 = {14}/[23]
2a. 2 innies r89c3 = 11 = [29/47/56]

3. hidden triple 789 in r1 -> r1c125 = {789}

4. 6 in r1 only in one of two 7(2) cages -> 1 locked for r1 (Locking cages)

5. 8(3)n3 = {125/134}
5a. -> r2c8 = 1 (the tag WT did this differently, very nicely done)
5b. no 9 in r2c34

6. 18(3)n1 must have two of 789 for r1c12 = {279/378}
6a. -> r2c2 = (23)
6b. must have 7: locked for r1 and n1
6c. no 3 in r2c4

7. "45" on n1: 2 outies r12c4 = 8 = [17]/{26} = 2 or 7
7a. r1c3 = (156)
7b. r2c3 = (348)

8. from steps 2. & 7,: r1289c4 = [1723]/{26}[14]
8a. must have both 1 & 2 (killer pair): locked for c4
8b. note: must have 1 in r1c4 or 4 in r9c4

9. "45" on n3: 2 outies r12c6 = 9
9a. but {27} blocked by h8(2)r12c4
9b. and {36} as [36] only, forces r12c46 = [1376], but this is blocked by 7 also in r2c7 ie, two 7s in r2
9c. -> r12c6 = [18]/{45} = 5 or 8
9d. r1c7 = (236)
9e. r2c7 = (589)

10. 3 in r1 only in n3: locked for n3

11. "45" on n9: 2 outies r89c6 = 13
11a. but {58} blocked by r12c6
11b. = [94]/{67}
11c. r8c7 = (689)
11d. r9c7 = (235)

This took a long time to see
12. combined cage r1289c6 = 22(4)
12a. but [1894] blocks all 1 in c4 (step 8b)
12b. = [18]/{67}/{45}{67}
12c. -> r89c6 = {67}: both locked for c6 and n8
12c. no 6 in r8c7
12d. no 5 in r9c7

13. "45" on c6789: 2 innies r5c67 = 7 = {25/34}(no 6789)

14. "45" on c89: 2 outies r37c7 = 10 = {28/46}/[73/91](no 5, no 79 in r7c7)

15. "45" on n3: 2 innies r12c7 = 11 = [29/38/65]
15a. "45" on n9: 2 innies r89c7 = 11 = [83/92]
15b. note: no eliminations yet

16. hidden killer triple 147 in c7: r37c7 can only have one of them -> r456c7 must have two of 147
16a. and "45" on n6: those 3 innies = 13
16b. = {148/157/247}(no 369)
16c. -> no 4 in r5c6 (h7(2)r5c67)

17. 9 in c6 only in 18(3)r6c6 = {189/279/459}(no 3)

18. 11(3)r3c6 = {128/137/245}
18a. 3 in {137} can only be in r4c6 -> no 3 in r3c6

19. 3 in c6 only in n5: locked for n5

Now to the other side
20. "45" on n7: 2 innies r89c3 = 11 = [29/47/56] = 2 or 4 or 6
20a. "45" on c1234: 2 innies r5c34 = 15 = {69/78}

21. "45" on n4: 3 innies r456c3 = 12
21a. the only combinations with 6 possible are {156}: but this is blocked by r1c3 = (156)
21b. and {246}: blocked by r89c3 (step 20)
21c. -> no 6 in r456c3
21d. no 9 in r5c4 (h15(2))

22. 12(3)r6c3 cannot have 9 since 12 are only in r6c3
22a. -> no 9
22a. min. r6c34 = [14] = 5 -> max. r7c7 = 7 (no 8)

23. 9 in c4 is only in 17(3)r3c4
23a. = {179/269/359}(no 48)
23b. no 9 in r4c3

24. 8 in c4 only in r56c4: locked for n5
24a. 8 in r6c4 -> r6c3 + r7c4 = 4 = [13] only
24b. 8 in r5c4 -> r5c3 = 7
24c. -> h12(3)r456c3 must have 1 or 7
24d. but {147} blocked by 21(3)n4 = 4 or 7
24e. = {129/138/237}(no 4,5)
24f. has one of 789 which must go in r5c3 -> no 78 in r46c3

25. 17(3)r3c4 = {179/269/359}
25a. 3 in {359} must go in r4c3 -> no 3 in r3c4

26. 3 in c4 only in n8: locked for n8

27. 21(4)r6c5 = {2469/2478/2568}
27a. must have 6 or 7 which are only in r6c5 -> r6c5 = (67)
27b. must have 2: locked for n8 & c5

28. r89c4 = {14} only: both locked for c4 and n8
28a. -> 21(4)r6c5 = [6]{258}: 5 & 8 locked for c5 and n8
28a. -> r5c34 = {78} only: both locked for r5

29. r1c5 = 9

30. r1c12 = {78} = 15 -> r2c2 = 3
30a. 8 locked for n1

Much easier now
Cheers
Ed


Top
 Profile  
Reply with quote  
PostPosted: Sun Mar 21, 2021 10:52 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
I finished yesterday evening and checked my WT this afternoon, so Ed just beat me to post first. At a glance my breakthrough is different.

Here's how I solved Assassin 52V2 Revisit:
Prelims

a) R1C34 = {16/25/34}, no 7,8,9
b) R1C67 = {16/25/34}, no 7,8,9
c) R2C34 = {19/28/37/46}, no 5
d) R2C67 = {49/58/67}, no 1,2,3
e) R8C34 = {15/24}
f) R8C67 = {69/78}
g) R9C34 = {19/28/37/46}, no 5
h) R9C67 = {18/27/36/45}, no 9
i) 8(3) cage at R1C8 = {125/134}
j) 11(3) cage at R3C6 = {128/137/146/236/245}, no 9
k) 21(3) cage at R4C2 = {489/579/678}, no 1,2,3
l) 20(3) cage at R4C8 = {389/479/569/578}, no 1,2
m) 10(3) cage at R8C2 = {127/136/145/235}, no 8,9
n) 26(4) at R2C9 = {2789/3689/4589/4679/5678}, no 1

1a. 45 rule on C5 1 innie R5C5 = 1
1b. 45 rule on C1234 2 innies R5C34 = 15 = {69/78}
1c. 45 rule on C6789 2 innies R5C67 = 7 = {25/34}

2a. Hidden triple {789} in R1C125
2b. Killer triple 7,8,9 in R1C2 and 21(3) cage at R4C2, locked for C2
2c. 6 in R1 only in R1C34 = {16} or R1C67 = {16} (locking cages), 1 locked for R1
2d. 8(3) cage at R1C8 = {125/134} -> R2C8 = 1, clean-up: no 6 in R1C6, no 9 in R2C34
2e. 18(3) cage at R1C1 = {279/378}, 7 locked for R1 and N1, clean-up: no 3 in R2C4
2f. 18(3) cage = {279/378} -> R2C2 = {23}
2g. 45 rule on R9 1 innie R9C5 = 2 outies R8C28 + 1
2h. Min R8C28 = 3 -> min R9C5 = 4
2i. Max R8C28 = 8 -> no 8,9 in R8C8

3a. 45 rule on N1 2 outies R12C4 = 8 = [17]/{26}, clean-up: no 2,3,4 in R1C3, no 2,6 in R2C3
3b. 45 rule on N3 2 outies R12C6 = 9 = [18/27/36]/{45}, no 9, clean-up: no 4 in R2C7
3c. 45 rule on N7 2 outies R89C4 = 5 = [14/23/41] -> R9C3 = {679}, clean-up: no 1 in R8C3
3d. 45 rule on N9 2 outies R89C6 = 13 = {67}/[85/94] -> R9C7 = {2345}
3e. 45 rule on C12 2 outies R37C3 = 13 = {49/58}/[67]
3f. 45 rule on C89 2 outies R37C7 = 10 = [28]/{37/46}/[91], no 5, no 9 in R7C7
3g. Killer pair 1,2 in R12C4 and R89C4, locked for C4
3h. Max R67C4 = 11 -> no 9 in R67C4
3i. Min R67C4 = 8 (cannot be {34} which clashes with R89C4) -> max R6C3 = 4

4a. 45 rule on N4 3 innies R456C3 = 12 = {129/138/147/237} (cannot be {156} which clashes with R1C3, cannot be {246} which clashes with R89C3, cannot be {345} because no 3,4,5 in R5C3), no 6, clean-up: no 9 in R5C4 (step 1b)
4b. R5C3 = {789} -> no 7,8,9 in R4C3
4c. Killer triple 7,8,9 in 21(3) cage at R4C2 and R5C3, locked for N4
4d. 9 in C4 only in 17(3) cage at R3C4 = {179/269/359}, no 4,8
4e. 12(3) cage at R4C1 = {156/246/345}
4f. Killer triple 4,5,6 in 12(3) cage and 21(3) cage, locked for N4
4g. Max R4C3 = 3 -> min R34C4 = 14, no 3
[I spent time looking at other interactions in C34 before finding a relatively simple forcing chain …]
4h. R5C34 = 15 (step 1b), R12C4 (step 3a) = 8 = [17]/{26}, R89C4 (step 3c) = 5 = [14/23/41]
4i. Consider placement for 8 in C4
R5C4 = 8 => R5C3 = 7 => R46C3 = {23} => R1C3 = 1 (hidden single in C3) => R12C4 = [62]
or 8 in 12(3) cage at R6C3 = 1{38}, 3 locked for C4
-> R89C4 = {14}, locked for C4 and N8, R12C4 = {26}, locked for C4 and N2, clean-up: no 6 in R1C3, no 3 in R1C6 (step 3b), no 4,5 in R1C7, no 3 in R2C3, no 7 in R2C6 (step 3b), no 6,7 in R2C7, no 9 in R5C3 (step 1b), no 4 in R8C4, no 9 in R8C6 (step 3d), no 6 in R8C7, no 7 in R9C4, no 5 in R9C7
4j. Naked pair {78} in R5C34, locked for R5
4k. Killer pair 4,8 in R2C3 and R2C67, locked for R2
4l. 21(3) cage at R4C2 = {489/579} (cannot be {678} which clashes with R5C3), no 6, 9 locked for C2
4m. 6 in N4 only in 12(3) cage at R4C1, locked for N1
4n. 3 in C3 only in R46C3, locked for N4
4o. 6 in N1 only in R3C23, locked for R3
4p. R37C3 (step 3e) = {49}/[67] (cannot be {58} which clashes with R12C3), no 5,8
4q. R89C6 (step 3d) = {67} (cannot be [85] which clashes with R12C6), locked for C6 and N8, clean-up: no 7 in R8C7, no 4 in R9C7
4r. Hidden killer triple 4,6,7 in 23(4) cage at R1C5 and R6C5 for C5, 23(4) cage cannot contain all of 4,6,7 -> R6C5 = {467}
4s. 12(3) cage at R6C3 = 1{38}/[273], no 5
4t. R34C4 = {59} (hidden pair in C4) -> R4C3 = 3 (cage sum)

[I was then doing the more routine steps after the key step 4i before I spotted this final breakthrough.]
5a. 18(3) cage at R1C1 (step 2e) = {279/378}
5b. R2C34 = [46] (cannot be [82] which clashes with 18(3) cage), R1C4 = 2 -> R1C3 = 5, R8C3 = 2 -> R8C4 = 4, R9C4 = 1 -> R9C3 = 9, R367C3 = [617] -> R5C34 = [87]
5c. Naked pair {34} in R1C89, locked for R1 and N3 -> R1C67 = [16], R2C6 = 8 (step 3b) -> R2C7 = 5, R1C5 = 9, R34C4 = [59]
5d. 21(3) cage at R4C2 = {579} (last remaining combination), 5,7 locked for C2, 5 locked for N4, R1C12 = [78] -> R2C2 = 3 (cage sum)
5e. R12C5 = [97] = 16 -> R34C5 = 7 = [34]
5f. R3C6 = 4 -> R4C67 = 7 = [52], R9C7 = 3 -> R9C6 = 6, R8C6 = 7 -> R8C7 = 8
5g. R5C67 (step 1c) = 7 = [34]
5h. R37C7 (step 3f) = 10 = [91]
5i. R8C58 = [56], R6C7 = 7
5j. R4C8 = 8 -> R56C8 = 12 = [93]

and the rest is naked singles.


Top
 Profile  
Reply with quote  
PostPosted: Fri Mar 26, 2021 9:25 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Turns out this puzzle is quite easy. Just have to see Andrew's neat step 5a&b. Here's how the start looks in optimised form.
A52v2 start:
Preliminaries courtesy of SudokuSolver
Cage 6(2) n78 - cells only uses 1245
Cage 15(2) n89 - cells only uses 6789
Cage 7(2) n12 - cells do not use 789
Cage 7(2) n23 - cells do not use 789
Cage 13(2) n23 - cells do not use 123
Cage 9(2) n89 - cells do not use 9
Cage 10(2) n78 - cells do not use 5
Cage 10(2) n12 - cells do not use 5
Cage 8(3) n3 - cells do not use 6789
Cage 21(3) n4 - cells do not use 123
Cage 10(3) n7 - cells do not use 89
Cage 20(3) n6 - cells do not use 12
Cage 11(3) n256 - cells do not use 9
Cage 26(4) n3 - cells do not use 1

No clean-up done unless stated. This is a highly optimised solution.
1. "45" on c5: 1 innie r5c5 = 1

2. "45" on n7: 2 outies r89c4 = 5 = {14}/[23]
2a. 2 innies r89c3 = 11 = [29/47/56]

3. hidden triple 789 in r1 -> r1c125 = {789}

4. 6 in r1 only in one of two 7(2) cages -> 1 locked for r1 (Locking cages)

5. 8(3)n3 = {125/134}
5a. -> r2c8 = 1 (the tag WT did this differently, very nicely done)
5b. no 9 in r2c34

6. 18(3)n1 must have two of 789 for r1c12 = {279/378}
6a. -> r2c2 = (23)
6b. must have 7: locked for r1 and n1
6c. note: must have 8 in r1 or 2 in r2c2
6d. no 3 in r2c4

7. "45" on n1: 2 outies r12c4 = 8 = [17]/{26} = 2 or 7
7a. r1c3 = (156)
7b. r2c3 = (348)

8. from steps 2. & 7,: r1289c4 = [1723]/{26}[14]
8a. must have both 1 & 2 (killer pair): locked for c4

Now to Andrew's step 5a&b which effectively cracks the puzzle
9. 10(2)r2c3: [82] blocked by 18(3)n1 (step 6c)
9a. = [37/46]
9b. no 6 in r1c4 (h8(2)r12c4)
9c. no 1 in r1c3

10. "45" on c12: 2 outies c37c3 = 13 = {49/58}/[67](no 1,2,3, no 6 in r7c3)

11. "45" on c1234: 2 innies r5c34 = 15 = {69/78}

12. "45" on n4: 3 innies r456c3 = 12 and must have 1 for c3
12a. but {147} blocked by 21(3)n4 = 4 or 7
12b. and {156} blocked by r1c3 = (56)
12c. = {129/138}(no 4,5,6,7)
12d. 8,9 must be in r5c3 -> no 8,9 in r46c3
12e. 1 locked for n4
12f. no 8,9 in r5c4 (h15(2))

13. 12(3)r6c3 can't have 9 since 1&2 are only in r6c3

14. 9 in c4 only in 17(3)r4c3 = {179/269/359}(no 4,8)

15. 8 in c4 only in 12(3)r6c3
15. = [1]{38} only: 3 locked for c4

Much easier from here
Cheers
Ed


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 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