Maverick 1

[mhparker]

Hi folks,

Welcome to the first puzzle in the Maverick series, a generic name for any alternative Killer Sudokus with custom cage patterns that are not based on any official Assassins. Note that they do not necessarily have to come from me. Anyone who wants to publish his or her own custom puzzle is free to add to this series. Furthermore, there are no rules about frequency of appearance, level of difficulty, or type of Killer puzzle (Killer-X, Center Dot Killer, Zero Killer, etc.). As with the official Assassins, variants may also be posted.

So, without further ado, let's get down to business:


Maverick 1 (M1) (Est. rating: 1.5)

This puzzle offers a wide variety of different moves, and should provide plenty of opportunity for creativity and individualism, as its name impies.



3x3::k:3840358643563333154333384356:4356:435630872066309164223352386651483093:64223352:51545148:51482342:64223625:5154:51545148:439964222099:51543885:4399:439964222099210133925186:5186:51863141:442315992634:51863149:4423:4423:

Have fun!

[gary w]

Hi Mike et al,

I'm finding this one tough going!! I've made some progress,I think but it hasn't led to much yet.,



I've placed r3c7=6 and have r1c7={79} r5c7={89} and r9c7={78} and it's pretty easy to see that r1c7=9 rather quickly leads to a contradiction but I haven't seen a nice way to prove this yet.or am I barking up the wrong tree entirely I wonder?So..simple to complete from here using this initial hypo?/t&e but the proof is eluding me so far.



Will we be getting an assassin and a maverick each week now ?Ay caramba!

Regards

Gary

[gary w]

Mike,

Finally cracked it.

Hopefully get a chance to put a reasonably detailed wt together but in the meantime ..here's just an indication..


Innies n6/9 =25.Innies N369 -> R159C7=24={789}.This together with some combo work on cages at r159c67 -> r1c7={79} r5c7={89} r9c7={78} r8c7<>9

r4c7=1/2/3 only possibilities to complete 13(3) cage N3
Therefore in N6/9 6 is in C89.The X wing on 6 -> r3c7=6

O-I r1 -> r1c5<>1/2/3 -> r1 a 1,2 or 3 must be at c12

I next considered the placement of 4s in c1 and c9.
In c1 if 4 is at r1 -> 7 at r1c7 if 4 at r2c1 -> 29 or 38 in r1c12 -> r1c7=7
If 4 is at r5c1 (only other place) in c9 4 must be at r89 -> 12(2) cage N9=39

Now combos conflict in (c7r159,innies on N6/9) if 9 at r1c7

Thus in all cases r1c7<>9 so r3c9=9

Mop up now.



A great work out Mike..many thanks

Regards

Gary

[mhparker]

Will we be getting an assassin and a maverick each week now ?

No, definitely not! The idea is just to publish them "on demand". For example, if the weekly Assassin has been dealt with very quickly and a V2 has neither been announced, nor is expected, as was the case this week.

Also, not all of my puzzles will be labelled as Mavericks. Some will have a theme name, like the Vortex or Concentric Squares Killers.

BTW, Gary, thanks for giving it a go and finally cracking it! Anyone else want to try it, and maybe provide a full WT?

[Afmob]

I'm trying to solve it but I seem to be stuck. There are quite some interesting moves like XY-Wings, Hidden Triples, Hidden Killer triples so I hope I can post my walkthrough soon.

[gary w]

I should have been more explicit re one point....



I said that outies-innies on r1 -> r1c5<>1,2 or 3.This is true but only in combination with the fact that r1c5=1,2 or 3,because of the 6(3) cage in N3, -> r2c1=3.But this of course is impossible because of the other 6(3) cage in N7.A 3 at r2c1 -> r89c1={12} and means the 8(2) cage in c1 cannot be completed.After that wt goes thro' as above.



Regards

Gary

[Afmob]

Definitely a fun assassin with lots of cool moves. My original/unpolished walkthrough had 3 XY-Wings so sometimes I felt I was solving a normal Sudoku. So it was no surprise that the "finishing move" (step 11a) was another Sudoku technique though it took me quite a while to find it.

M1 Walkthrough:

1. C1234
a) 6(3) = {123} locked for N7
b) 10(2): R9C4 <> 7,8,9
c) 1,2,3 locked in 8(2) + R89C1 for C1
d) Innies C1234 = 7(2) = {16/25/34}

2. C6789
a) 6(3) = {123} locked for N3
b) 12(2) @ N3 = {48/57}
c) 13(3) must have 1,2 or 3 -> only possible @ R4C7 -> R4C7 = (123)
d) 13(3) <> 9 because R3C78 <> 1,3
e) 1,2,3 locked in 8(2) + R12C9 for C9
f) Innies C789 = 24(3) = {789} locked for C7
g) Both 12(2) @ N9 = [39/48/57]
h) Hidden Single: R3C7 = 6 @ C7
i) Innies C6789 = 10(2) -> no 5

3. C67
a) 13(2) = [49/58/67]
b) 14(2) = [59/68]
c) Outies C789 = 15(3) = {456} -> locked for C6
d) 11(2) = {29/38}
e) 9(2) = {18/27}
f) Killer pair (28) locked in 11(2) + 9(2) for C6
g) 12(2) = [48/57]

4. R12
a) Killer pair (58) of 14(2) @ R1 blocks {58} of 13(2)
b) Killer pair (69) locked in 14(2) + 13(2) for R1
c) Killer pair (58) of 12(2) blocks {58} of 13(2) @ R2
d) Killer pair (47) locked in 13(2) + 12(2) for R2
e) Killer pair (46) of 13(2) @ N1 blocks {456} of 15(3)
f) 15(3) must have 1,2 or 3 -> only possible @ R1C2 -> R1C2 = (123)
g) Naked triple (123) locked in R1C289 for R1
h) 12(2) = [48/57]

5. N36
a) 13(3) = 6[43/52] -> R3C8 = (45), R4C7 = (23)
b) 15(2) = [78/87/96]

6. N9
a) 7,8,9 locked in R8C8 + R9C7 + 17(3) -> R7C9 <> 7
b) 8(2): R6C9 <> 1
c) 17(3) must have 7,8 xor 9 -> 17(3) = {269/359/368/458/467} -> no 1
d) 17(3): R9C8 <> 9 since R89C9 >= 9

7. R789
a) 1 locked in R7C789 for R7
b) Both 9(2): R6C46 <> 8
c) Innies+Outies R9: R8C19 = R9C5
-> R9C5 <> 1,2,3,4 since R8C19 >= 5
-> R8C9 <> 9 because R9C5 <= 9

8. C12
a) Innies+Outies C1: -1 = R19C2 - R5C1
-> R5C1 = (456) because R19C2 = 3/4/5
b) 8(2): R4C1 <> 2
c) 1,2,3 locked in R19C2 + 20(4) @ R456C2 for C2

9. N1
a) Hidden triple (123) in R1C2 + R3C13 -> no other candidates
b) 8(2) = [17/26/35]

10. N369
a) Consider both candidates of R1C7 = (79)
-> R1C7 = 7 -> R2C8 = 8 -> R3C9 = 9 -> 15(2) = [96]
-> R1C7 = 9 -> R5C7 = 8 -> 15(2) = [87]
-> 15(2) = [87/96]

11. C1 !
a) ! Coloring 4 in C1:
- i) If R1C1 = 4 -> R2C7 = 4 (Hidden Single @ R2) -> R78C7 <> 4
- ii) If R5C1 = 4 -> 4 locked in R89C9 for C9 -> R78C7 <> 4
-> R78C7 <> 4

12. C789 !
a) 12(2) @ R8 = [39/57]
b) ! XY-Wing: R2C8 = (78) and R3C9 = (89), R8C8 = (79) -> R9C9 <> 9
c) Hidden Single: R8C8 = 9 @ N9 -> R8C7 = 3
d) 13(3) = {256} -> R3C8 = 5, R4C7 = 2
e) R2C7 = 4 -> 12(2) = [48] -> R2C8 = 8
f) R3C9 = 9, R1C7 = 7, R9C7 = 8, R5C7 = 9

13. Rest is clean-up (might involve simple cage combo analysis) and singles.

Thanks for posting this Killer Sudoku mhparker!

[mhparker]

Hi folks,

It was interesting to see that both Gary and Afmob had similar breakthrough moves for solving this puzzle, namely...

...using the 4's in C19. However, Afmob made more eliminations than mentioned in Gary's solving outline, and (in particular) had already removed the 4 from r2c1, allowing a "purer" breakthrough combination to be used.

In particular, I would like to present the excellent combination used by Afmob to crack this puzzle to a wider audience, by elaborating on it below. It would be a shame to leave it buried at the end of a walkthrough that may only be read by a few. Also, I would like to express the first part of the combination in a somewhat different form to the way Afmob expressed it in hs WT.

To begin with, let's look at the grid state after Afmob's step 10:

Code: Select all

.-----------------------.-----------------------.-----------.-----------------------.-----------------------.
| 4578        123       | 5689        5689      | 4578      | 46          79        | 123         123       |
|           .-----------'-----------.-----------'           '-----------.-----------'-----------.           |
| 5689      | 4679        4679      | 12356       1235689     139       | 45          78        | 123       |
:-----------+-----------------------+-----------.-----------.-----------+-----------------------+-----------:
| 123       | 45789       123       | 345789    | 12345789  | 2389      | 6           45        | 89        |
|           :-----------.           |           |           |           |           .-----------:           |
| 567       | 123456789 | 123456789 | 345789    | 123456789 | 2389      | 23        | 123456789 | 67        |
:-----------'           :-----------'-----------:           :-----------'-----------:           '-----------:
| 456         123456789 | 12345678    12345678  | 123456789 | 56          89        | 123456789   456789    |
:-----------.           :-----------.-----------:           :-----------.-----------:           .-----------:
| 6789      | 123456789 | 123456789 | 1234567   | 123456789 | 127       | 12345     | 123456789 | 23567     |
|           :-----------'           |           |           |           |           '-----------:           |
| 6789      | 456789      456789    | 2345678   | 23456789  | 278       | 12345       12345     | 12356     |
:-----------+-----------------------+-----------'-----------'-----------+-----------------------+-----------:
| 123       | 456789      456789    | 123456      123456789   1379      | 345         789       | 45678     |
|           '-----------.-----------'-----------.           .-----------'-----------.-----------'           |
| 123         123       | 46789       12346     | 56789     | 45          78        | 2345678     456789    |
'-----------------------'-----------------------'-----------'-----------------------'-----------------------'

From this position, Afmob essentially applied the following breakthrough combination:

11. Grouped X-Cycle with 5 links on 4 as follows:

(4)r89c9=r5c9-r5c1=r1c1-r2c23=r2c7 => r78c7<>4

This can be explained in verbose form as follows:
(i) Either r89c9 contains a 4 or
(ii) r89c9 does not contain a 4
(iii) => r5c9=4 (strong link, c9)
(iv) -> r5c1<>4 (weak link, r5)
(v) => r1c1=4 (strong link, c1)
(vi) -> r2c23<>4 (weak link, n1)
(vii) => r2c7=4 (strong link, r2)
Conclusion: r89c9 contains a 4 and/or r2c7 contains a 4
=> 4 can be eliminated from all common peers of r89c9 and r2c7
=> no 4 in r78c7; cleanup: no 8 in r8c8

12. r8c8 now {79}
=> XY-Wing on 9 with pivot at r2c8 and pincers at r3c9 and r8c8
=> no 9 in r9c9

This can be explained in verbose form as follows:
(i) Either r2c8 contains a 7, implying r8c8=9, OR...
(ii) ...r2c8 contains an 8, implying r3c9=9
=> At least one of r3c9 and r8c8 must contain a 9
=> no 9 in r9c9 (common peer of r3c9 and r8c8)

This left a hidden single in n9 at r8c8=9, cracking the puzzle.

Now that's what I call Sudoku!

[gary w]

Hi Mike and Afmob et al,

I don't know if we're the only ones who tackled this puzzle??I hope not because it was excellent and as Afmob's wt and Mike's detailed explication makes clear it had a rather lovely solving path (which I "sort of" used myself but could never have expressed it as well as the other two did).

Regards and many thanks to Mike for the killer

Gary

[Andrew]

Hi Mike and Afmob et al,

I don't know if we're the only ones who tackled this puzzle??I hope not because it was excellent ...

I'm still working/struggling on this puzzle. Out of curiosity I compared my current grid state with that posted in Mike's message and was somewhat surprised to find that I've made extra eliminations in almost a quarter of the cells and don't seem to have missed any.

It's a bit surprising that the diagram hadn't done simple eliminations such as

1 locked in R67C7 -> no 1 in R7C8, and

45 rule on C123 -> 3 outies R159C4 = 17 giving two eliminations in R5C4

Maybe they weren't needed, or spotted, before Afmob made his breakthrough move.

I must admit I've no idea at the moment what that move was. I'll continue trying to solve the puzzle myself. I haven't yet done any steps that can be described as serious combo crunching.

My current feeling, from the amount of time that I've spent on this puzzle, and the difficulty I'm having in making progress, is that the rating can't be only 1.5.

In case anyone is interested, here is how far I've got

458 - 123 - 5689 - 5689 - 4578 - 46 - 79 - 123 - 123
589 - 4679 - 4679 - 1356 - 123569 - 139 - 45 - 78 - 123
123 - 45789 - 123 - 345789 - 12345789 - 2389 - 6 - 45 - 89
567 - 123456789 - 1234567 - 345789 - 123456789 - 2389 - 23 - 123456789 - 67
456 - 123456789 - 12347 - 25678 - 123456789 - 56 - 89 - 123456789 - 456789
789 - 123456789 - 12345678 - 123567 - 123456789 - 127 - 12345 - 123456789 - 23567
678 - 456789 - 456789 - 234678 - 2346789 - 278 - 12345 - 245 - 12356
123 - 456789 - 456789 - 1246 - 12346789 - 179 - 345 - 789 - 4567
123 - 123 - 46789 - 12346 - 56789 - 45 - 78 - 234678 - 46789

[sudokuEd]

here is how far I've got

In friendlier format.

Code: Select all

.-------------------------------.-------------------------------.-------------------------------.
| 458       123       5689      | 5689      4578      46        | 79        123       123       |
| 589       4679      4679      | 1356      123569    139       | 45        78        123       |
| 123       45789     123       | 345789    12345789  2389      | 6         45        89        |
:-------------------------------+-------------------------------+-------------------------------:
| 567       123456789 1234567   | 345789    123456789 2389      | 23        123456789 67        |
| 456       123456789 12347     | 25678     123456789 56        | 89        123456789 456789    |
| 789       123456789 12345678  | 123567    123456789 127       | 12345     123456789 23567     |
:-------------------------------+-------------------------------+-------------------------------:
| 678       456789    456789    | 234678    2346789   278       | 12345     245       12356     |
| 123       456789    456789    | 1246      12346789  179       | 345       789       4567      |
| 123       123       46789     | 12346     56789     45        | 78        234678    46789     |
'-------------------------------.-------------------------------.-------------------------------'

[Andrew]

Thanks Ed

In friendlier format.

I suppose it is friendlier for those who use a solver as a worksheet. If anyone else uses an Excel worksheet like I do, or even pencil and paper, then I think my format is a bit friendlier. In any case it's the best that I can produce.

Having said that, I hope someone finds Ed's version of my diagram useful.

Edit. I've deleted the rest of this message, now that I've finished this puzzle. My walkthrough is posted in a later message.

[mhparker]

Hi folks,

I note that, from Andrew's current position (Ed's marks pic), there's an interesting variation of Afmob's breakthrough move that can be applied (after a preliminary cleanup-like step that Andrew appears to have overlooked ), which also cracks the puzzle. It's not often that something like this crops up in Killers, so I wouldn't want to let it go without a mention:

1. 8(3)n69 = {125/134}
1a. {13} only in r67c7
1b. -> no 4 in r67c7

Note: the 4 in n6 is now locked within the 20(4) cage, making the following move possible:

2. Grouped X-Cycle with 7 links on 4 as follows:

(4)r2c7=r3c8-r456c8=r5c9-r5c1=r1c1-r2c23=r2c7 => r2c7=4

This can be explained in verbose form as follows:
(i) Either r2c7 contains a 4 or...
(ii) ...r2c7 does not contain a 4
(iii) => r3c8=4 (strong link, n3)
(iv) -> r456c8<>4 (weak link, c8)
(v) => r5c9=4 (strong link, n6)
(vi) -> r5c1<>4 (weak link, r5)
(vii) => r1c1=4 (strong link, c1)
(viii) -> r2c23<>4 (weak link, n1)
(ix) => r2c7=4 (strong link, r2)
Conclusion: r2c7=4

This placement breaks the deadlock.

I thought this was worth mentioning because, although grouped X-Cycles with three links (aka. "grouped Turbot fishes") are relatively common in Killers, grouped X-Cycles with 5 links are much less common, and ones with 7 links (as in the above example) are rarer still.

P.S. Now off to "meet up" with Afmob at the Brick Wall...

[Para]

Hi all

I see all this talk about x-wing and x-cycles going on. When i solved this i used our basic killer techniques. There's no need really to go looking for these techniques in this puzzle really, but they are a nice shortcut. It is also solvable by just sticking to the basic killer techniques as 45-test cage combos and Killer Subsets.
I'll give my walk-through from Andrew's position, just to show how I solved the final bit of the puzzle. Some steps can be left out, but i just solved it again and didn't try to keep it short. Maybe later on I'll add the beginning steps if Andrew doesn't post his.

Complete Walk-Through Maverick 1:

1. R1C34 and R5C67 = {59/68} = {58..}: no 1,2,3,4,7

2. R1C67 = {49/67}: {58} blocked by R1C34: no 1,2,3,5,8

3. 6(3) at R1C8 = {123} -> locked for N3

4. R2C78 = {48/57} = {58..}: no 6,9

5. R2C23 = {49/67}: {58} blocked by R2C78: no 1,2,3,5,8

6. R34C1 and R67C9 = {17/26/35}: no 4,8,9

7. R34C4, R8C78 and R9C67 = {39/48/57}: no 1,2,6

8. R34C6 = {29/38/47/56}: no 1

9. R34C9 and R67C1 = {69/78}: no 1,2,3,4,5

10. R5C34, R67C4 and R67C6 = {18/27/36/45}: no 9

11. 8(3) at R6C7 = {125/134}: no 6,7,8,9

12. R8C23 = {49/58/67}: no 1,2,3

13. R9C34 = {19/28/37/46}: no 5

14. 6(3) at R8C1 = {123} -> locked for N7
14a. Clean up: R9C4: no 7,8,9

15. Killer Pair {69} in R1C34 + R1C67 -> locked for R1

16. Killer Pair {47} in R2C23 + R2C78 -> locked for R2

17. Killer Triple {123} in R34C1 + R89C1 -> locked for C1

18. Killer Triple {123} in R12C9 + R67C9 -> locked for C9

19. 15(3) at R1C1 = {159/168/249/258/267/348/357}: {456} blocked by R2C23: Needs one of {123} only place is R1C2 -> R1C2 = {123}
19a. Naked Triple {123} in R1C2 + R1C89 -> locked for R1

20. 45 on C789: 3 innies: R159C7 = {789} -> locked for C7
20a. Clean up: R1C6: no 7,9; R2C8: no 4,5; R5C6: no 8,9; R8C8: no 3,4,5; R9C6: no 7,8,9

21. 6 in N3 locked for R3

22. 45 on C789: 3 outies R159C6 = 15 = {456} -> locked for C6
22a. Clean up: R9C7: no 9; R34C6: no 7; R67C6: no 3

23. Killer Pair {28} in R34C6 + R67C6 -> locked for C6

24. 13(3) at R3C7 = [481/571/472]/{56}[2]/{46}[3] -> R3C8: no 9; R4C7 = {123}
24a. R3C7 = 6(hidden)
24b. Clean up: R3C8: no 7,8; R4C7: no 1; R4C9: no 9

25. 1 in C7 locked within 8(3) cage at R6C7 -> R7C8: no 1

26. 17(3) at R8C9 = {269/359/368/458/467} = {789..}: {179/278} blocked by R8C8 + R9C7: no 1
26a. Killer Triple {789} in R8C8 + R9C7 + 17(3) at R8C9 -> locked for N9
26b. Clean up: R6C9: no 1

27. 1 in N9 locked for R7
27a. Clean up: R6C46: no 8

28. 45 on N69: 4 innies: R4C79 + R59C7 = 25 = [2698/3787]: [2797/3697] blocked by R1C7: R4C9: no 8
28a. Clean up: R3C9: no 7

29. 8(3) at R6C7 = {12}[5]/{13}[4]/{15}[2]: {14}[3] blocked by R28C7: R67C7: no 4; R7C8: no 3

30. 4 in C7 locked within 12(2) cage in R2C78 + R8C78 -> 4 in C8 locked within R28C8: locked for C8

31. 8 in N6 locked in R5C79 for R5
31a. Clean up: R5C34: no 1

32. Clearly R3C9 = R5C7(check combos for {789} in C7 and N3) -> 8 in locked in R5C79; 8 locked in R35C9: locked for C9

33. 4 in N6 locked in 20(4) at R4C8 = {1469/147[8]/2459/246[8]/345[8]}(only place for 8 in R5C9): {3467} blocked by R4C9
33a. 20(4) can't have 2 of {245} in R456C8 because of R37C8: {2459/246[8]/345[8]} blocked
33b. 20(4) = {1469/147[8]} = {67..}: no 2,3,5; R5C9: no 7; 1 locked for N6 and C8(only place for 1 in R456C8);

34. R7C7 = 1(hidden single)
34a. Clean up: R6C9: no 7

35. 5 in N6 locked for R6
35a. Clean up: R7C4: no 4

36. 2 in C7 locked for N6
36. Clean up: R7C9: no 6

37. Killer Pair {67} in R4C9 + 20(4) in R4C8 -> locked for N6
37a. Clean up: R7C9: no 2

38. Naked Pair {35} in R67C9 -> locked for C9
38a. R1C8 = 3(hidden single)

39. 3 in N1 loked for R3
39a. Clean up: R4C4: no 9; R4c6: no 8

40. 45 on C1234: 2 innies: R28C4 = {16/25}/[34]: no 7,8,9; R8C4: no 3

41. 9 in C4 locked for N2
41a. Clean up: R4C6: no 2

42. 45 on C6789: 2 innies: R28C6 = 10 = [19/37]: R8C6 = {79}

43. Killer Triple {789} in R8C23 + R8C6 + R8C8 -> locked for R8

44. 45 on R9: 2 outies + 1 innie: R8C19 = R9C5: Min R8C19 = 5 -> Min R9C5 = 5: no 1,2,3,4

45. 17(3) at R8C9 = [629]/[4]{67}: [6]{47} blocked by R9C67

46. 45 on R9: 3 innies and 1 outie: R8C1 + 17 = R9C589 = [1]-[5]{67}/[1]-[7][29]/[2]-[8][29]: R8C1: no 3; R9C5: no 6,9
46a. R9C589 = [5]{67}/[729]/[829] = {78..}

47. Killer Pair {78} in R9C589 + R9C7 -> locked for R9
47a. Clean up: R9C4: no 2,3

48. 9 in R2 locked for N1
48a. Clean up: R1C4: no 5

49. 45 on C1: 2 outies and 1 innie: R19C2 + 1 = R5C1: R19C2 = 3/4/5 -> R5C1 = {456}

50. R5C34 = {27/36}: {45} blocked by R5C16

51. 45 on C123: 3 innies: R159C3 = 16 = [529/826/574]: R1C3: no 6; R5C3: no 3,6
51a. Clean up: R1C4: no 8; R5C4: no 3,6

52. Naked Pair {27} in R5C34 -> locked for R5

53. 15(3) at R1C1 = [519/429]: [518] blocked by R1C3; [618/627] blocked by R67C1: R1C1 = {45}; R2C1 = 9
53a. Clean up: R2C23: no 4; R67C1: no 6

53. Naked Pair {67} in R2C23 -> locked for R2 and N1

And now it breaks down to naked and hidden singles.

greetings

Para

[Andrew]

Interesting to see that Mike and Para have both posted continuations from my position. It was clearly worth Ed making the effort to convert my diagram. Thanks again Ed.

[Edit. Most of this message now deleted. I've posted my full walkthrough later in this thread.