A new one

[nd]

The latest creation. It's dedicated to James Hirschfeld, the trombonist.

Code:
3x3::k3073:5378:53786149:5126:5126:51263073:4875:53786149:6149:5126:6161:4875:4875:487523256149:5126:61612331:4875:8990:8990:8990:8990:8990:61618485:8485:8990:8485:8485:8746:8746:6161:7469:7469:8485:8485:8485:4914:8746:8746:8746:7469:746931284914:8746:4413:4413:5439:5439:54393128:4914:4914:4413:4423:5439:5439365036604423:4423:



Not sure how tough people will find this one. If it's especially hard then it could be solved "tag"-style.

[just made a minor revision to make it a little more elegant--broke a cage in two]

[nd]

Note the revision above. However, if people are looking for a REAAAL challenge, then you can "undo" the revision to make a V2--the only difference between the two puzzles is that the 12(2) and 21(2) cages in N12 are joined to make a 33(5).

The new one is solvable via pure logic; the 33(5) version requires some T&E, I believe.

[Nasenbaer]

This looks like a fantastically difficult one, thanks nd.

I've talked to sudokued and we agreed to do this nice killer in a joined effort. We invite all of you to participate. I'll start the whole thing with the obvious things and as explicite as I can.

Walkthrough nd10 "easy"

1. N1 : 12(2) = {39|48|57} -> no 1,2,6
2. N1 : 11(2) = {29|38|47|56} -> no 1
3. N12 : 21(3) = {489|579|678} -> no 1,2,3
4. N2 : 11(2) = {29|38|47|56} -> no 1
5. N2 : 9(3) = {126|135|234} -> no 7,8,9
6. N4 : 9(3) = {126|135|234} -> no 7,8,9
7. N47 : 29(4) = {5789} -> no 1,2,3,4,6
8. N78 : 12(4) = 12{36|45} -> no 7,8,9 -> 1,2 locked -> no 1,2 possible in r7c6

This is just a starter, I'll add more to the post.

Peter

[nd]

Good luck! -- Quick note (to save confusion)--I'd referred to the harder one as "V2" above, but your "V2" walkthrough is for the easier version. -- Forget about the harder one for now--maybe I'll post it again once the "definitive" version here is solved. It's plenty hard as is!

[Nasenbaer]

OK, I renamed it to nd10 "easy"

Here are some more steps. It' all going veeeeerrrry slow, not much of a progress.


9. N45 : 33(7) = 126{3489|3579|4578}
10. 45 on N9 (3 innies) : r789c7 = 11(3) = {128|137|146|236|245} -> no 9
11. N14 : 19(5) : must have 1, locked in N1 -> no 1 in r4c3
12. 45 on N1 (1 innie, 1 outie) : r4c3 + 3 = r1c3
12a. r1c3 : min: 5, max: 9
12b. r4c3 : min: 2, max: 6
13. 45 on N2 (2 innies, 1 outie) : r1c3 + 4 = r12c6
13a. r1c3 : min: 5, max: 9
13b. r12c6 : min: 9, max: 13
14. 45 on r12 : r2c39 + 9 = r3c78
14a. r2c39 : min: 3, max: 8 -> no 8,9
14b. r3c78 : min: 12, max: 17 -> no 1,2
15. 45 on r123 (1 outie, 2 innies) : r4c3 + 8 = r23c9
15a. r4c3 : min: 2, max: 6
15b. r23c9 : min: 10, max: 14 -> max. r2c9 = 7 -> min r3c9 = 3 -> no 1,2 in r3c9 (edited, thanks Ed)
16. 45 on r1234 : r5c149 = 15(3)

Edit 1: next addition:

17. N69 : 34(6) : must have 9 -> locked in N6 for 34(6) (see also step 10)
18. N56 : 35(6) = 89{1467|2367|2457|3456} -> 9 locked in r4c456 for N5
19. 45 on r123 (3 outies) : r4c3 + r45c9 = 16 (doubles possible!)
19a. r4c3 : min: 2, max: 6
19b. r45c9 : min: 10, max: 14 -> no 1

Edit 2: one more:

20. 45 on r89 (4 innies) : r8c5678 = 24(4)
21. 45 on r789 (3 outies, 1 innie) : r7c7 + 8 = r6c126
21a. r7c7 : min: 5, max: 8 -> no 1,2,3,4
21b. r6c126 : min: 13, max: 16 -> no 5,6,7,8 in r6c6

Edit 3: I think I found some good moves:

22. n9 : (step 10) 11(3) has only one of {5678} -> locked in r7c7 -> r89c7 = {1234}
23. N589 : 19(4) : r6c6 and r8c7 both have {1234} -> r78c6 = {56789} -> no 1,2,3,4 -> combination {1567} for 19(4) not possible
24. r7 : 5,6,7,8,9 are in r7c1267 plus one of them has to be in r7c89 for 17(3) -> no 5,6 in r7c345 possible -> r8c5 = {56} -> no 1,2 in r7c89 (step 8) -> a 3 or a 4 is in r7c89
25. N9 : 17(3) in r78 has at most one of {56789} in r7c89 -> r8c8 = {56789}
26. r8c5 = {56} -> 11(2) in c5 can't have 5,6
27. r8c5 = {56} -> 14(3) in N89 : combination {356} not possible -> one of {789} is in r9c56
28. 45 on N7 (2 innies, 2 outies) : r79c3 + 5 = r6c12
28a. r6c12 : min: 12, max: 17
28b. r79c3 : min: 7, max: 12 -> no 1,2 in r9c3
29. 45 on c12 (3 innies, 1 outie) : r8c3 + 8 = r3c12 + r5c2 (doubles possible!)
29a. r8c3 : min: 1, max: 9
29b. r3c12 + r5c2 : min: 9, max: 17
30. 45 on c12 (1 innie, 4 outies) : r5c2 + 11 = r2348c3
30a. r5c2 : min: 1, max: 9
30b. r2348c3 : min: 12, max: 20

Edit 4: numbering corrected, thanks Andrew.

In r3 might be something going on with 19(4) an 9(3) and the numbers 1,2,3 but now I'm too tired to think because it's 1.30

So, this is how far I got (still not very much):

Code: Select all

.-----------.-----------.-----------------------.-----------.-----------.-----------------------------------.
| 23456789  | 345789    | 56789       456789    | 234789    | 123456789 | 123456789   123456789   123456789 |
|           |           :-----------.           |           |           '-----------.           .-----------:
| 23456789  | 345789    | 1234567   | 456789    | 234789    | 123456789   123456789 | 123456789 | 1234567   |
:-----------'-----------'           :-----------'-----------'-----------.           |           |           |
| 123456789   123456789   123456789 | 123456      123456      123456    | 3456789   | 3456789   | 123456789 |
:-----------------------.           :-----------------------------------'-----------'-----------:           |
| 123456      123456    | 23456     | 123456789   123456789   123456789   12345678    12345678  | 2345678   |
|           .-----------'-----------:           .-----------------------.-----------------------:           |
| 123456    | 123456789   123456789 | 12345678  | 12345678    12345678  | 123456789   123456789 | 2345678   |
:-----------'-----------.           '-----------'           .-----------:                       '-----------:
| 5789        5789      | 123456789   12345678    12345678  | 1234      | 123456789   123456789   123456789 |
|                       :-----------------------------------:           |           .-----------------------:
| 5789        5789      | 1234        1234        1234      | 56789     | 5678      | 3456789     3456789   |
:-----------------------'-----------.-----------.           |           '-----------:           .-----------:
| 123456789   123456789   123456789 | 123456789 | 56        | 56789       1234      | 56789     | 123456789 |
|                       .-----------'           :-----------'-----------------------+-----------'           |
| 123456789   123456789 | 3456789     123456789 | 123456789   123456789   1234      | 123456789   123456789 |
'-----------------------'-----------------------'-----------------------------------'-----------------------'

BTW, SumoCue just crashed on me during this copy operation, luckily I still got something from a previous try, but without cage sums, sorry.

Good night, see you in the morning.

Peter

[sudokuEd]

yow - this one is a typical nd Killer. Havn't found the hidden key(s) yet. Thanks for your efforts too Peter. Unfortunately, I can't add too much at this point. But its been fun trying things while listening to James Hirschfeld's tromboning. Hopefully a good sleep will soften up the brain clog.

Just a few more steps to add.

15. 45 on r123 (1 outie, 2 innies) : r4c3 + 8 = r23c9
15a. r4c3 : min: 2, max: 6
15b. r23c9 : min: 10, max: 14

add to 15b. max. r2c9 = 7 -> min r3c9 = 3

31. "45" r789 -> r6c6 + 21 = r7c127.
a. max r7c127 = 24 -> max r6c6 = 3 (no 4)
-> r7c127 = 22-24
= 22 = {89[5]/79[6]/59[8]}
= 23 = {89[6]}
= 24 = {89[7]}
->9 locked in r7c12 n7, r7 and also no 9 in r6c12

32. 9 in n4 locked in 33(7) = 12369{48/57}

Not much at all. Tomorrow will have a look at r1234 outies combined with the 33(7):n456. Here is the marks only pic for this spot which can be pasted into Sudocue.

Code: Select all

.-----------.-----------.-----------------------.-----------.-----------.-----------------------------------.
| 23456789  | 345789    | 56789       456789    | 234789    | 123456789 | 123456789   123456789   123456789 |
|           |           :-----------.           |           |           '-----------.           .-----------:
| 23456789  | 345789    | 1234567   | 456789    | 234789    | 123456789   123456789 | 123456789 | 1234567   |
:-----------'-----------'           :-----------'-----------'-----------.           |           |           |
| 123456789   123456789   123456789 | 123456      123456      123456    | 3456789   | 3456789   | 3456789   |
:-----------------------.           :-----------------------------------'-----------'-----------:           |
| 123456      123456    | 23456     | 123456789   123456789   123456789   12345678    12345678  | 2345678   |
|           .-----------'-----------:           .-----------------------.-----------------------:           |
| 123456    | 123456789   123456789 | 12345678  | 12345678    12345678  | 123456789   123456789 | 2345678   |
:-----------'-----------.           '-----------'           .-----------:                       '-----------:
| 578         578       | 123456789   12345678    12345678  | 123       | 123456789   123456789   123456789 |
|                       :-----------------------------------:           |           .-----------------------:
| 5789        5789      | 1234        1234        1234      | 5678      | 5678      | 345678      345678    |
:-----------------------'-----------.-----------.           |           '-----------:           .-----------:
| 12345678    12345678    12345678  | 123456789 | 56        | 56789       1234      | 56789     | 123456789 |
|                       .-----------'           :-----------'-----------------------+-----------'           |
| 12345678    12345678  | 345678      123456789 | 123456789   123456789   1234      | 123456789   123456789 |
'-----------------------'-----------------------'-----------------------------------'-----------------------'

[Andrew]

Just one step for now; not as useful as I had first thought because of the doubles possibility but still provides a elimination. I'll look at this puzzle again later and try to find some more.

33. 45 on r12 (5 outies, 1 innies) r2c3 + 33 = r3c789 + r45c9, min r3c789 + r45c9 = 34 (doubles possible, for example 7,8 in r3c78 and in r45c9 so doesn't necessarily contain 9)
33a. Max r3c789 + r45c9 = 39 {78789} -> max r2c3 = 6, no 7
33b. Min r3c789 + r45c9 = 34, max r3c789 = 24, min r45c9 = 10
33c. Min r3c789 + r45c9 = 34, max r345c9 = 24, min r3c78 = 10

[Nasenbaer]

I think we will need every bit of information, so I do this:

34. step 31 effects step 28: r6c12 : max: 15 -> r79c3 : max: 10
35. (step 20) r8 : 24(4) : {3489} not possible
36. (step 15) r5 : 15(3) : {159|249} not possible
37. 45 on N2 (3 outies) : r1c3 + r23c7 = 20(3) (doubles possible) -> no 1 in r2c7

Edit 1: Veeerrrry slow going. So I'll show every thought I have, even those with no immediate progress.

38. 45 on c9 (1 outie, 3 innies) : r9c8 + 4 = r167c9
38a. r167c9 : min: 6, max: 13
38b. r9c8 : min: 2, max: 9 -> no 1
39. 45 on c9 (3 outies, 2 innies) : r16c9 + 13 = r789c8
39a. r16c9 : min: 3, max: 10
39b. r789c8 : min: 16, max: 23
The following step 40 is actually unnecessary because step 41 is more precise but I invested too much thought in it so it will stay
40. 45 on N78 (2 innies, 3 outies) : r6c6 + r89c7 + 10 = r7c12 (doubles possible)
40a. r6c6 + r89c7 : min: 4, max: 7
40b. r7c12 : min: 14, max: 17
41. 45 on N78 (5 outies) : r6c126 + r89c7 = 19
41a. r6c12 = 12 {57}, 13 {58} or 15 {78}
41b. r6c6 + r89c7 = 4, 6 or 7 (doubles possible)
= 4 = [121]
= 6 = [141]|[1]{23}|[2]{13}|[3]{12}
= 7 = [1]{24}|[232]|[2]{14}|[313]
42. 45 on N78 (4 innies, 1 outie) : r9c7 + 29 = r8c6 + r7c126
42a. r9c7 : min: 1, max: 4
42b. r8c6 + r7c126 : min: 30, max: 33 -> max. r7c127 = 24 -> min. r8c6 = 6

Edit 2: Found a good one, how could I have overlooked it?!? First cell is filled! Breakthrough!?!

43. 45 on c89 (4 outies, 1 innie) : r4c8 + 22 = r1567c7
43a. r4c8 : min: 1, max: 8
43b. r1567c7 : min: 23, max: 30
44. N2 : 9(3) : {234} not possible, one of them is needed in 11(2) -> no 4 in 9(3) -> 1 locked in 9(3) for N2 and r3 -> single in N1 : r2c3 = 1

Peter

Edit: Typos in 39 and 41 corrected, thanks Andrew.

[sudokuEd]

Beat me by one minute Peter. Unbelievable that we found the same move at the same time - after how many hours???!

45. 1 in n3 locked in 20(5) -> {23456} lost
45a. 1 in n7 locked in 21(5) -> {23457} lost
45b. 1 required in 33(7) n45 either in n5 or r5c2 -> no 1 in r5c4

[Nasenbaer]

One more before I hit the sack (it's already 2 am, the training for tonight isn't going too well )

Addition to 45: must have exactly one of {789} (might be useful); also must also have two of {234} and one of {56} (thanks Andrew)

46. N8 : 1 locked in r7c45 for r7 and N8

Edit: Step 47 was missing, sent it to Ed to check it tonight before I went to bed, wasn't posted yet.

47. 45 on N4578 (4 outies, 1 innie) : r4c3 + 6 = r489c7 + r4c8 (doubles possible)
47a. r4c3 : min: 2, max: 6
47b: r489c7 + r4c8 : min: 8, max: 12 -> min. r489c7 = 6 -> max. r4c8 = 6 -> no 7,8 in r4c8

[sudokuEd]

Finally worked out how to crack this nut. Used my favourite blunt instrument - contradictions between nonets. Anyone find a subtle way to progress?

Thanks again nd! And really enjoyed working with Peter, Andrew and James H.

Now, back to Assassin 31 - totally defeated at the first attempt

48. 4 in r4c3 -> r3c123 = {239/257/356} but these are all blocked by 9(3) in r3 -> no 4 in r4c3 -> no 7 in r1c3
[edit-deleted invalid 48a]

49. "45" n1:r4c3 + 3 = r1c3
2 in r4c3 -> r1c3 = 5 -> r3c123 = {349} ({358/457} blocked by r1c3;{367} blocked by 9(3):r3)
3 in r4c3 -> r1c3 = 6 -> r3c123 = {249} ({267/456} blocked by r1c3; {258/456} blocked by 9(3):r3)
5 in r4c3 -> r1c3 = 8 -> r3c123 = {247} ({238} blocked by r1c3; {346} blocked by 9(3):r3)
6 in r4c3 -> r1c3 = 9 -> r3c123 = {345} ({237} blocked by 9(3):r3)

50 . In summary r3c123 = {349/249/247/345} (no 6,8) = 4{39/29/27/35}
50a. 4 locked for n1, r3

51.12(2)n1 = {39/57} = [3/5, 7/9...]
-> from step 50, {345} is blocked from r3c123 -> no 6 in r4c3 (step 49), no 9 in r1c3, no 5 in r3c123

52. r3c123 now 4{39/29/27} = [7/9..] -> Killer pair with 12(2) -> 7 and 9 locked for n1
52a. 11(2)n1 = {38/56}

53. 2 locked in n1 in 19(5) -> 2 locked for r3 and no 2 in r4c3, no 5 in r1c3

54. 9(3)r3 now {135} only: locked for n2, r3

55.11(2):n2 = {29/47} (no 8) = [4/9..]

56. r1c3 + 4 = r12c6 -> r12c6 = 10,12
r1c3 = 6 -> r12c4 = {78} -> r12c6 = 10 = [46] [edit: note, [64] not possible since 6 already in r1c3 in this hypothetical]
r1c3 = 8 -> r12c4 = {67} ({49} blocked by 11(2) step 55) -> r12c6 = 12 = [48] [edit: note, [84] not possible since 8 already in r1c3 in this hypothetical]
-> r1c6 = 4, r2c6 = {68}, r12c4 = {78/67} = 7{6/8} -> 7 locked for n2, c4,11(2) = {29}:locked for c5

57. rest of 24{4}:n23 now 20(3) = [659/839/857] -> r23c7 = [59/39/57], r2c7 = 35, r3c7 = {79}

58. Killer pair between r3c123 & r3c7 for {79}:locked for r3

59. r3c89 = {68} locked for r3, n3

and on till the end! Yippee.

[nd]

You got it! Glad you did so, though I think you took the long way round! The key here is really in the bottom N789, which require a "new" (well, maybe not, but not explicitly codified) technique to crack; then some application of the properties of the extra-large cages & you can nail this one handily. Let me write out a walkthrough & I'll post it...

[nd]

I'm typing this quickly as I have to get to bed (up tomorrow for a plane flight) so hopefully it's not got any errors--I'll doublecheck once I get where I'm going (Halifax) & have access to a computer.

1. 29(4) cage in N47 = {5789}. 45 rule on N789 => R7C127 = R6C6 + 21 => R6C6 = {123}, R7C127 = {5..9}, with 9 locked in those cells in R7. 45 rule on N9 => R789C7 = 11(3) => R7C7 = {5678}, R89C7 = {1234}, and the 9 is locked in R7C12 within N7, R7 and the 29(4) cage.

2. R78C6 must contain {5..9} (because of the {1..4} in R6C6 and R8C7). R7C89 must contain one cell of value 5 or greater (because otherwise the max of R7C89 = 3 + 4 which is impossible in a 17(3) cage). So in R7 we have a hidden quint on {56789} in R7C1267 + one of R7C89. Therefore R7C345 = {1234}. Since 12(4) must contain {56}, R8C5 = {56} and {12} are locked in R7C345 within R7.

3. 11(2) cage in N2 = {29|38|47} => 9(3) cage in N2 cannot be {234} => it is {1(26|35)}. R2C3 = 1 (only spot for it now in N1). 45 rule on R3 => R4C3 = {2..6}, R3C789 = 20..24.

4. In N4, 9 must be in the three cells of the 33(7) cage => the 33(7) cage = {12369(48|57)}, i.e. it must contain {123}. Note that the {123} in R6C6 forces one candidate from {123} to be in the three cells R5C23+R6C3 within N4. The 9(3) cage in N4 must also have two candidates from {123}, so in conjunction with the 3 cells of the 33(7) there is a hidden triplet on {123} in N4! => R4C3 = {456}.

5. This in turn means (by 45 rule on R3) that R3C789 = 22..24, i.e. {5..9}. Therefore the 4 in R3 is locked in R3C123 => R4C3 = {56}, R12C2 = {39|57}. 45 rule on N1 => R1C3 = {89}.

6. If R4C3 = 5 then R3C123 = {247} (only possible combo--{346} would conflict with the 9(3) cage in N2). If R4C3 = 6 then R3C123 = {345} (only possible combo). => R3C123 must contain either 5 or 7 => R12C2 = {39}, R1C3 = 8, R12C1 = {56}, R1C123 = {247}, R4C3 = 5, R3C456 = {135}, R3C789 = {689}, R6C12 = {78}, R7C12 = [95].

There's still more to do but it's straightforward mop-up from here on out so I'll leave the rest to you...... -- JC Godart's just posted a walkthrough on DJApe's site which is substantially the same, though he uses a slightly different version of the key hidden-subset move (he uses the split-off 24(4) cage in R8 to narrow things down to a hidden pair of {34} in R7).

[Nasenbaer]

I don't see how r1c6 could be set to 4, 4 was never eliminated from r2c6, so for me it's this: r12c6 = {46|48} = 4{6|8} -> 4 locked for N2 and c6

nd, I haven't looked at your walkthrough yet, and I won't do it until I completed your nice puzzle, so there might come some unnecessary steps.

Please note step 47 which was included above. I repeat it here:

47. 45 on N4578 (4 outies, 1 innie) : r4c3 + 6 = r489c7 + r4c8 (doubles possible)
47a. r4c3 : min: 2, max: 6
47b: r489c7 + r4c8 : min: 8, max: 12 -> min. r489c7 = 6 -> max. r4c8 = 6 -> no 7,8 in r4c8

More to come...

BTW, Happy New Year, Australia!

[Nasenbaer]

Here it is...

60. N4 : r4c3 = {35} -> 9(3) can't be {135} -> no 5 in 9(3) -> 2 locked in 9(3) -> 2 locked in 33(7) in N5
61. c9 : r3c9 = {68} -> 24(4) = 78{36|45} -> no 2 -> 7,8 locked for c9
62. N3 : 20(5) : {12359} und {13457} blocked by r2c7 = {35} -> no 9 -> single in N3 : r3c7 = 9
63. N3 : 20(5) = 124{58|67} ({12368} blocked by r3c9 = {68}) -> 4 locked in 20(5) -> no 4 in r2c9 -> no 3 in 20(5)
64. -> 3 locked in r2c79 for r2 -> no 8 in r1c1, no 9 in r1c2 -> single in r1 : r1c5 = 9, r2c5 = 2 -> single in r2 : r2c2 = 9, r1c2 = 3
65. -> N1 : 11(2) = {56} -> 5,6 locked for N1 and c1 -> r1c3 = 8 -> r4c3 = 5
Now I can place your 4, Ed.
66. r12c4 = {67} -> 6,7 locked for N2 and c4 -> r1c6 = 4, r2c6 = 8, r2c7 = 3

Edit: Here is the rest:

67. single in N3 : r2c8 = 4
68. 45 on N3 : r23c9 = 13(2) = [58]|[76] -> r45c9 = 11(2) = {47}|{38} ({56} not possible, only one of them allowed in 24(4)) -> no 5,6 in r45c9
69. singles in 29(4) in N47 : r7c1 = 9, r7c2 = 5
70. 45 on N7 : r79c3 = 10(2) = [37]|[46] -> single in N8 : r7c4 = 2, r7c5 = 1
71. 45 on r789 : r6c6 + 7 = r7c7 -> r6c6 = 1, r7c7 = 8
72. N6 : 8 locked in r45c9 for N6 and c9 -> r3c9 = 6, r3c8 = 8, r2c9 = 7 (step 68) -> r45c9 = {38} -> 3,8 locked for N6 and c9
73. -> r7c9 = 4 (only candidate left) -> r78c8 = {67} -> 6,7 locked for N9 and c8
74. -> r8c7 = 2 (only candidate left) -> r9c7 = 1

The rest seems to be standard clean-up.

Peter