SudoCue Users

Perfect level for me - thanks Ruud! In other words, I managed to solve it - took about 30 minutes after using a few higher level standard sudoku techniques in addition to the killer ones.

Hi Cathy,

Thanks for the flowers.
Here's a version 1.5:

3x3::k:4608:5633:5633:563333333333:4616:4608:460833323332:4111:4616:4616:4626:46085653:4118:5655:4111:46164626:46265653:4118:5655:4111309838765653:4118:5655:5930:5930:593036303630:4118:4402:4402:4402:593031263640287515973903390331305189:5189:51891352:7242:7242:7242:7242:7242:3151



Ruud

Here's a walkthrough for version 1. I used a couple fancy techniques, but not many. I'll tackle 1.5 in a bit.

Note: This walkthrough should work without any preliminary steps.
Note 2: Numbers between {} are unordered; numbers between [] are left to right first, then top to bottom.
i.e. theoretically a cage shaped would be [975638245].
(There's been some confusion on that at least once in the past, so I just thought I should make it part of a standard walkthrough header.)
Note 3: () surround my comments, | means or, ! means not (most commonly != means not equal)
Note 4. .. indicates a range of values
Everything else is pretty standard.

1. sole combination for 3/2 in n9 = {12} pair -> elim 1|2 from rest of n9 & r9
2. sole combination for 8/2 in n9 = {35} pair (not 17 or 26 due to step 0a) -> elim 3|5 from rest of n9 & r7
3. sole permutation for 4/2 in n8 = [13]
4. sole permutation 11/3 in r34 = [245]
5. sole combination for 21/3 in n9 = {678} triple -> elim 6|7|8 from rest of r8 and n9
6. combinations for 12/3 in n7 = {129} triple -> elim 1|2|9 from rest of n7 and r8 -> r8c4 = 4 (naked single)
7. r7c4 = 9 (hidden single) -> r7c3 = 6
8. sole combination for 9/2 in n7 = {45} pair -> elim 4|5 from rest of n7/r9
9. r9c3 = 3 and r9c7 = 9 (hidden singles)
10. sole combination for 23/3 in r6 = {689} triple -> elim 6|8|9 from rest of r6
11. sole combination for 19/3 in c4 = {568} triple -> elim 5|6|8 from rest of c4 -> r9c4 = 7 (naked single)
12. innies of r1234 = r2c4 = 1
13. combinations for 15/3 in n7 = {258|357} (cannot be 168 or 267 due to r68c7; these would cause four of {678} in the column, which can't happen) -> no 1 or 6; elim 5 from rest of c7; forms killer triple on {678} with r68c7 -> elim 6|7|8 from r15c7
14. r1269c6 = naked quad {6789} in c6 -> elim from rest of c6
15. 7 of c6 locked in n2 -> elim from rest of n2
16. combinations for 9/3 in r1 = {126|135|234} -> no 7..9; forms killer triple {123} with r1c7 -> no 1|2|3 in r1c15689
17. combinations for 9/3 in r6 = {135|234} (not 126 due to 23/3 in r6 eliming 6) -> no 6..9; elim 3 from rest of r6
18. combinations for 18/3 in c3 = {189|279} (cannot be 459 as then r168c3 all set to {12} -- can't have 3 choices from two values in one column) -> no 4|5; elim 9 from rest of c3; forms killer pair {12} with r8c3 -> elim 1|2 from r156c3
19. combinations for 27/4 in n4 = {3789|4689|5679} -> no 1|2; elim 9 from rest of n4
20. 9 of c3 locked in n1 -> elim from rest of n1
21. outies-innies of c89 = r7c58-r8c16 = -7; r7c58 >= 7 -> r8c16 >= 14 && r8c6 <= 9 -> r8c1 >= 5 -> r8c1 != 4
22. 12/4 in n6 = {1236|1245} -> no 7..9; elim 1|2 from rest of n6
23. r16c3 = {45} naked pair -> elim 4|5 from r5c3
24. By combinations, only one of {456} in 9/3 in r1; this is in r1c3; thus no 4|5|6 in r1c2
25. By combinations, only one of {456} in 9/3 in r6; this is in r6c3; thus no 4|5|6 in r6c2
26. combinations for 21/3 in r34 = {489|579|678} -> no 1..3
27. outies of n36 = r16c6 = 15/2 = [69|78|96]
28. Outies-innies of n6 = r3c9+r6c6-r4c7 = 11; r6c6+r3c9 <= 17 -> r4c7 <= 6 -> r4c7 = {35} -> r6c6+r3c9 = 14|16 - r3c9+r6c6 = [59|68|79|86] -> r3c9 = {5678}
29. innies of c6789 = r29c6 = 15/2 = [78|96]
30. combinations for 21/4 in n2 = {1479|1569|1578} -> no 2
31. combinations for 9/3 in r1 = {135|234} -> elim 3 from rest of r1
32. r16c24 = x-wing on 3 -> elim from rest of c2
33. 4 of n3 locked in 18/4 in n3 -> 18/4 in n3 = {1458|1467|2349|3456} (cannot be 2457 as this shares two digits with all combinations for 15/3 in c7, and can see all but one digit of said cage)
34. 12/4 in n6 forms killer pair {35} with r4c7 -> elim 5 from r4c89
35. 7 of n6 locked in r4; elim from rest of r4
36. 7 of n6 locked in 21/3 of r34; elim from r3c9; 21/3 of r34 must be {579|678} -> no 4
37. 4 of n6 locked in 12/4 -> 12/4 in n6 = {1245} quad -> elim from rest of n6 -> r4c7 = 3 (naked single)
38. sole combination for 15/3 in c7 = {357} -> elim 5|7 from rest of c7 and n3
39. sole combination for 18/3 in n3 = {2349} quad -> elim {2349} from rest of n3 -> r1c7 = 1 -> r5c7 = 2 (naked singles)
40. sole combination for 21/3 in r34 = {678} -> no 9
41. r6c8 = 9 (hidden single)
42. r4c5 = 9 (hidden single)
43. r3c3 = 9 (hidden single)
44. r16c24 = Unique Rectangle {23} -> r6c2 = 1 -> a jillion naked singles and last-digit-in-cage moves
45. r1c6 = 7 (hidden single) -> about a dozen naked singles and LDIC moves
46. r5c5 = 7 (hidden single) -> naked singles and last digits solve it.
634257189
287169543
159843726
462591378
398674251
715328694
876912435
921435867
543786912


Much easier than last week's, which I never was able to finish (I think a lot of other people couldn't either judging by the lack of forum activity!)

Managed to do 1.5 as well - amazing!! Didn't keep a record of steps but my last key move was Outies - Innies on N3 and it was all singles from there.

PsyMar wrote:Note: This walkthrough should work without any preliminary steps.
Note 2: Numbers between {} are unordered; numbers between [] are left to right first, then top to bottom.
i.e. theoretically a cage shaped

Code: Select all

   9
7 563
824 5

would be [975638245].
(There's been some confusion on that at least once in the past, so I just thought I should make it part of a standard walkthrough header.)
Note 3: () surround my comments, | means or, ! means not (most commonly != means not equal)
Note 4. .. indicates a range of values
Everything else is pretty standard.

It's good to have standard notation while also having variation in the way that different people write their walkthroughs. It would be boring if all walkthroughs were written the same way.

A few comments on the above notes.

I agree with the first note for cases where a walkthrough can be written without preliminary steps, which is the case for this walkthrough. [Edited after solving the puzzle and then looking at Psymar's walkthrough.]

In more general terms I feel that preliminary steps ought to be given when later moves work forward from them. The exception to this is for any puzzles where an initial population diagram, after basic preliminary steps but not consequential eliminations, has been posted with the puzzle.

Note 2. I definitely agree with the order of numbers when [] are used. In some other cases where combinations are being discussed it may be more appropriate to use a slightly different order. An example of this was step 19 in my Assassin 45 walkthrough where I listed R4C8 before R34C9 which was more logical and easier to read.

Note 3. I assume that <> is an acceptable alternative to ! When using <> I prefer it to be only used for single numbers. When referring to eliminated combinations I'll continue to say "cannot be". It doesn't seem right to use <> for eliminated combinations although ! seems OK for them.

I am not currently able to use the symbol | in my walkthroughs because I have a British computer with a newer Canadian keyboard and there is a compatibility issue because the computer doesn't have the Canadian/US keyboard codes. I'll continue to use / which I also find easier to read than | in lists of alternatives.

Note 3 mentioned comments. One comment that I find particularly useful is "(only remaining combination)" in cases where this isn't immediately obvious. It saves having to spend time thinking "why did the walkthrough make this step?".

Nice puzzle Ruud. About the same level as Assassin 46 Light.

I'm impressed that Cathy solved it in about 30 minutes. She said this was "after using a few higher level standard sudoku techniques in addition to the killer ones". Guess I'd better learn them. Must find time to study Andrew Stuart's book, I've only glanced at bits of it so far.

I managed to solve it by standard killer techniques although it did need a bit more thought at one stage.

1. R7C12 = {69/78}

2. R78C5 = {13}, locked for C5 and N8

3. R7C89 = {17/26/35}, no 4,8,9

4. R9C12 = {18/27/36/45}, no 9

5. R9C89 = {12}, locked for R9 and N9, clean-up: no 6,7 in R7C89, no 7,8 in R9C12
5a. R7C89 = {35}, locked for R7 and N9
5b. R78C5 = [13] (naked singles)

6. R1C234 = {126/135/234}, no 7,8,9

7. 11(3) cage in N14 = {128/137/146/236/245}, no 9

8. R345C4 = {289/379/469/478/568}, no 1

9. R345C6 = 1{25/34}, 1 locked for C6

10. 21(3) cage in N36 = {489/579/678}, no 1,2,3

11. R6C234 = {126/135/234}, no 7,8,9

12. R6C678 = {689}, locked for R6
12a. R6C234 = {135/234} = 3{15/24}, 3 locked for R6

13. 7 in R6 locked in R6C159
13a. 45 rule on R6 3 innies R6C159 = 13 = 7{15/24}

14. 11(3) cage in N89 = {245} (only remaining combination) -> R8C6 = 5, R7C67 = [24]

15. R345C6 = {134}, locked for C6

16. R8C789 = {678} (only remaining combination), locked for R8 and N9 -> R9C7 = 9

17. R9C34567 = 789{36/45}, 3,5 only in R9C3 -> no 4,6,7,8 in R9C3

18. R8C123 = {129} (only remaining combination), locked for R8 and N7 -> R8C4 = 4, clean-up: no 6 in R7C12

19. R7C12 = {78}, locked for R7 and N7 -> R7C34 = [69], clean-up: no 3 in R9C12

20. R9C12 = {45}, locked for R9 -> R9C3 = 3

21. 27(4) cage in N4 must contain 9 = 9{378/468/567}, no 1,2, 9 locked in R5C123 for R5 and N4
21a. R6C159 = 7{15/24} (step 13a), 1 only in R6C9 -> no 5 in R6C9

22. 12(4) cage in N6 must contain 1,2 = 12{36/45}, no 7,8,9, 1,2 locked for N6

23. 7 in N6 locked in R4C789, locked for R4

24. 7 in N4 locked in 27(4) cage = 79{38/56}, no 4

25. 45 rule on R1 3 innies R1C159 = 20 = {389/479/569/578}, no 1,2

26. 45 rule on C123 2 remaining outies R16C4 = 5 = {23}, locked for C4

27. R2C4 = 1 (hidden single in C4)

28. R345C4 = {568} (only remaining combination), locked for C4 -> R9C4 = 7

29. 7 in N5 locked in R56C5, locked for C5

30. 21(4) cage in N2 = 1{479/569/578}, no 2
30a. 7 only in R2C6 -> no 8 in R2C6

31. 45 rule on C789 2 remaining outies R16C6 = 15 = {69}/[78], no 8 in R1C6

32. 2,7 in C5 locked in R3456C5 = 27{49/58}, no 6

33. 45 rule on C5 1 remaining outie R2C6 – 1 = 1 innie R9C5 -> R2C6 = {79}

34. 45 rule on N6 2 outies R3C9 + R6C6 – 11 = 1 innie R4C7, max R3C9 + R6C6 = 18 (they can both be 9) -> no 8 in R4C7, min R4C7 = 3 -> min R3C9 + R6C6 = 14 -> no 4 in R3C9

35. 45 rule on N6 3 innies R4C789 – 10 = 1 outie R6C6, R6C6 = {689} -> R4C789 = 16, 18 or 19 and must contain 7 (step 23)
35a. 12(4) cage in N6 = 12{36/45} (step 22) and R6C78 = {689} -> R4C789 must contain either {457} (with 4 in R4C89) or {37} and one of 6,8,9
35b. R4C789 cannot be {457} (with 4 in R4C89) because this doesn’t provide any valid combinations for 21(3) cage in N36 -> R4C789 = 3{67/78/79} -> R4C7 = 3, R4C89 = {67/78/79}, no 4,5


36. 12(4) cage in N6 = {1245}, no 6
36a. 5 in N6 locked in R5C789, locked for R5

37. 21(4) cage in N36 = {579/678} = 7{59/68}
37a. 7 locked in R4C89 -> no 7 in R3C9
37b. 5 only in R3C9 -> no 9 in R3C9

38. R4C7 = 3 -> R23C7 = 12 = {57}, locked for C7 and N3

39. Naked pair {68} in R68C7, locked for C7

40. R3C9 + R6C6 – 11 = R4C7 (step 34), R4C7 = 3 -> R3C9 + R6C6 = 14 -> no 9 in R6C6
40a. R6C67 = {68} -> R6C8 = 9

41. R4C5 = 9 (hidden single in R4)

42. R1C234 = {135/234} (cannot be {126} which clashes with R1C7), = 3{15/24}, no 6, 3 locked for R1
[PsyMar also spotted killer pair {12} in R1C234 and R1C7, locked for R1. I should have seen that but my next step produced the same result.]

43. R1C678 = [628/718/916] -> R1C8 = {68}

44. Naked triple {678} in R148C8, locked for C8

45. Naked pair {68} in R1C8 + R3C9, locked for N3

46. 9 in N3 locked in 18(4) cage = {2349}, locked for N3 -> R1C7 = 1, R5C7 = 2

47. R1C234 = {234}, no 5, {234} locked for R1, 4 locked in R1C23 for N1 -> R1C9 = 9

48. R1C159 = 20 (step 20), R1C9 = 9 -> R1C15 = 11 = {56}, locked for R1 -> R1C68 = [78], R2C6 = 9, R3C9 = 6, R4C89 = [78]

49. 21(4) cage in N2, R2C4 = 1, R2C6 = 9 -> R12C5 = 11 = {56}, locked for C5 and N2 -> R345C4 = [856], R9C56 = [86], R6C67 = [86], R8C789 = [867]

50. 27(4) cage in N4 (step 24) = {3789}, no 5 -> R6C1 = 7, R7C12 = [87]
50a. R6C159 (step 13a) = 7{15/24} -> R6C59 = [24], R35C5 = [47], R3C6 = 3, R6C4 = 3, R1C234 = [342]

51. 6 in R4 locked in R4C12, 11(3) cage in N14 = {146} (only remaining combination) -> R3C1 = 1, R4C12 = {46}, locked for R4 -> R45C6 = [14], R4C3 = 2
51a. R23C3 = 16 = [79]

52. R2C2 = 8 (hidden single in N1)

and the rest is naked singles

Now to try v1.5

Hi folks,

Here's my walkthrough for the V1.5. It took a couple of strong pushes right at the start to get the proverbial ball rolling, but once it was in motion there was no stopping it...

After that, this puzzle proved to be very much routine, with no real surprises, and often a good selection of choices as to how to progress. This is in contrast to the most difficult Assassins, where one has to be thankful for small mercies in being able to find any way at all of shaving off the next candidate.

Maybe Ruud is just giving us a short break and this is the simply the calm before the storm?

Anyway, enough talk for now, here's my V1.5 walkthrough:

Walkthrough - Assassin 47 V1.5 (http://www.sudocue.net/forum/viewtopic.php?t=663)

1. 22/3 at R1C7 = {(58|67)9} -> no 9 elsewhere in R1

2. 13/4 at R1C5 = {1(237|246|345)} (no 8,9) -> no 1 elsewhere in N2

3. 11/3 at R2C3: no 9

4. 22/3 at R3C4 = {(58|67)9} -> no 9 elsewhere in C4

5. 9 in R1 now locked in R1C23 -> no 9 elsewhere in N1

6. 9 in R2 now locked in R2C789 -> no 9 elsewhere in N3

7. 22/3 at R3C6 = {(58|67)9} -> no 9 elsewhere in C6

8. 9 in N8 now locked in R789C5 -> no 9 elsewhere in C5

9. 12/2 at R7C1: no 1,2,6

10. 12/2 at R7C5: no 1,2,6

11. 11/3 at R7C6: no 9 in R7C7

12. 6/2 at R7C8 = {15|24}

13. 20/3 at R8C7: no 1,2

14. 5/2 at R9C1 = {14|23}

15. 12/2 at R9C8: no 1,2,6

16. Innies N7: R79C3 = 13/2 -> no 1,2,3

17. Innies N9: R79C7 = 7/2 -> no 7,8,9

18a. Innies R8: R8C456 = 10/3 = {127|136|145|235} (no 8,9)
18b. Cleanup: no 3,4 in R7C5

19a. 9 in R8 cannot be in 15/3 cage
({249} blocked due to 5/2 at R9C1, {159} blocked due to h10/3 at R8C456 (step 18))
-> 9 locked in 20/3 at R8C7 = {(38|47|56)9} -> no 9 elsewhere in R8 or N9
19b. 12/2 at R9C8 = {48|57} (no 3)

20. 20/3 at R8C7 must contain exactly two of {6789}
12/2 at R9C8 must contain exactly one of {6789}
6/2 ar R7C8 cannot contain any of {6789}
-> Innies N9 (see step 17) must contain exactly one of {6789}
-> R79C7 = {16} -> no 1,6 elsewhere in C7 or N9

21. 6/2 at R7C8 = {24} (1 no longer available) -> no 2,4 elsewhere in R7 or N9

22. 12/2 at R9C8 = {57} (4 no longer available) -> no 5,7 elsewhere in R9 or N9

23. 20/3 at R8C7 = {389} -> no 3,8 elsewhere in R8

24. Cleanup from previous steps:
24a. 12/2 at R7C1 = {39|57} (no 8)
24b. 12/2 at R7C5 = {48|57} (no 9)

25. Hidden Single (HS) in C5 at R9C5 = 9

26. 8 in N7 locked in h13/2 at R79C3 = [58]

27. 12/2 at R7C1 = {39} (5 unavailable) -> no 3 elsewhere in R7 or N7

28. 5/2 at R9C1 = {14} (3 unavailable) -> no 1,4 elsewhere in R9 or N7

29. Naked Single (NS) at R9C7 = 6

30. NS at R7C7 = 1

31. HS in R8 at R8C4 = 1 -> R7C4 = 8 (last digit in cage)

32. NS at R7C5 = 7 -> R8C5 = 5

33. NS at R7C6 = 6 -> R8C6 = 4 (last digit in cage)

34. Only two possible combinations for 22/3 ({(58|67)9}).
Clearly, 22/3 at R3C4 and 22/3 at R3C6 would clash with each other if both had same

combination
-> One of these cages must be {589} and the other {679}
-> 22/3 at R3C6 = {589} (6 unavailable) -> no 5,8 elsewhere in C4
-> 22/3 at R3C4 = {679} -> no 6,7 elsewhere in C4

35. NS at R1C4 = 5 -> R1C23 = [89]

36. Outie N14: R6C4 = 3 -> R9C46 = [23]

37. NS at R2C4 = 4

38. 16/4 at R3C5 = {1348} (no 2,6) (5,7,9 unavailable) -> no 1,3 in R12C5

39a. HS in C5 at R3C5 = 3
39b. 1 in C5 now locked in N5 -> no 1 in R6C6

40. HS in N2 at R3C6 = 8

41a. HS in R3 at R3C4 = 9
41b. 7 in C4 now locked in N5 -> no 7 in R6C6 -> R6C6 = 2

42. NS at R1C6 = 7 -> R1C78 = {24} -> no 2,4 elsewhere in R1 or N3

43. NS at R2C6 = 1

44. NS at R1C5 = 6 -> R2C5 = 2

45. Naked Pair (NP) on {24} in C8 at R17C8 -> no 2,4 elsewhere in C8

46. 18/4 at R1C1 = {6(147|237|345)} (8,9 unavailable, must have one of {13} due to R1C1)
-> no 6 elsewhere in N1

47. 18/4 at R1C1 and R2C3 at R2C3 form killer pair on {37} -> no 7 in R3C13

48. 18/4 at R1C9 = {13(59|68)} (no 7) (2,4 unavailable) -> no 1,3 elsewhere in N3

49. 18/4 at R1C9 and R3C79 form killer triple on {567} -> no 5,7 in R2C7

50. 16/3 at R2C7 = [853|952] -> R3C7 = 5

51. HS in N3 at R3C9 = 7 -> R4C89 = 5/2 -> R4C8 = {13}, R4C9 = {24}

52. R9C89 = [75]

53. Naked Pair (NP) on {24} in C9 at R47C9 -> no 2,4 elsewhere in C9

54. Split 15/2 cage at R6C78 = [78|96]

55. HS in C8 at R5C8 = 5

56. R45C6 = [59]

57. Innies R1234: R4C45 = 14/2 = [68]

58. NS at R5C4 = 7

59. 8 in C1 locked in 15/4 at R5C1 = {1248} -> no 1,2,4 elsewhere in N4; no 2 in R5C7

60. HS in C1 at R2C1 = 5

61. HS in C1 at R8C1 = 6

62. HS in C1 at R4C1 = 7

--- and it's all naked singles from now on... ---

mhparker wrote:Maybe Ruud is just giving us a short break and this is the simply the calm before the storm?

You may be right, Mike. Ruud is probably planning something special for Assassin 50. Then next year at this time there will be Assassin 100!

CathyW wrote:Perfect level for me - thanks Ruud! In other words, I managed to solve it - took about 30 minutes after using a few higher level standard sudoku techniques in addition to the killer ones.

Hi Cathy

I was just wondering what advanced standard sudoku techniques you used?
Should i be thinking x-wing/xy-wing, turbot fish or ALS-XZ/xy-chain variety?

greetings

Para

Where do advanced techniques start? Of the ones listed by Para, I use X-wing and have once used Swordfish. They don't seem to me to be advanced techniques. However the reason they are often included in that category is presumably because they are closely related to some of them.

I know what XY-wing is but doubt that I would be able to spot one in a puzzle. The other techniques listed are still things that I need to learn when I can find the time.

Multi colouring actually - colouring is one of my favourite techniques and I still use it to help identify strong links in nice loops/AICs. Can't remember now what else I used - I'll have to do it again and keep track of the steps! I've never knowingly used ALS, finned anything or turbots - haven't got that far in Andrew Stuart's book yet .

Hi

I never really use multi-colours. But this is a single digit technique, right?
Cause that was i was mostly wondering about. It is about the same as what SumoCue uses when it gives as hint pattern check, right?

greetings

Para

OK - here's my walkthrough in 30 steps. It took a lot longer the 2nd time with keeping track of what I'd done!

1. 3(2) in r9c89 -> 1,2 not elsewhere in r9, N9
-> 8(2) in N9 = {35} -> r7c5=1, r8c5=3.
-> 9(2) in r9c12 = {36/45}

2. 23(3) in r6c678 -> 6,8,9 not elsewhere in r6
-> 9(3) in r6c234 = {135/234} -> r6c19 <>3

3. 8(3) in c4 must have 1, 1 not elsewhere in c4

4. Innies N7: r79c3=9: 27/45/63

5. Innies N9: r79c7=13: 49/67/76
-> 21(3) in r8c789 must have 8, 8 not elsewhere in r8

6. Innies r6: r6c159=13 -> r6c9 <>5

7. Innies r1: r1c159=20 -> r1c159 <>1 or 2

8. Innies r8: r8c46=9: {27/45} -> Killer combination with 21(3) -> r8c123 can't be 4 or 7.

9. Outies N1+N4=5 -> r16c4 = {14/23}

10. Outies N3+N7=15 -> r16c7 = 69/96/78

11. Innies c1234: r2c4+r9c34=11 -> r2c4={123}, r9c3={35}, r9c4={4567}

12. Innies c6789: r2c6+r9c67=24 -> r2c6={789}, r9c6={6789}, r9c7={79} -> r7c7<>7

13. r7c7=4 (can't be 6 as 11(3) must be {245}) -> r9c7=9, r7c6=2, r8c6=5, r7c3=6, r9c3=3, r8c4=4, r7c4=9

14. Naked Triple {123} in c4. Since r16c4=5, must be {23} -> r2c4=1 -> r9c4=7 -> 19(3) in c4 = {568}

15. 12(4) in N6 must have 1 and 2 -> r4c7<>1 or 2.

16. 27(4) in N4 must have 9 -> r4c3, r5c5 <>9

17. r16c6=15 (from step 10) -> r29c6=15 -> r2c6<>8

18. 7 locked to r12c6 -> r123c5<>7

19. Simple colouring on 3s -> r3c2<>3

20. 7 locked to r4c789 -> r4c1235<>7

21. 18(3) in r234c3 = {189/279/459} -> 9 not elsewhere in N1, c3.
-> From possible combinations r23c3<>2

22. 27(4) in N4 = {3789/4689/5679}
-> If 4689, r6c1=4 -> r5c123<>4
-> If 3789, r5c12 = {39}; If 4689, r5c12 = {69}; If 5679, r5c12 = {69} -> r5c12 = {369}

23. (Perhaps could have done this earlier) Innies c3: r1568c3 = 18 -> r1c3={45}, r5c3={78}, r6c3={45}, r8c3={12} -> r16c3 naked pair, 45 not elsewhere in c3 -> 18(3) = {189/279}

24. 9(3) in r1 and r6 = {135/234} -> r1c2<>6, r1c1789<>3; r16c2<>4 or 5. x-wing on 3 in r16c24 -> r245c2<>3.

25. Innies c7: r1568c7 = 17 = {1268/1367} -> r1c7={12}, r5c7={123}, r6c7={68}, r8c7={678} -> r1c8 = {56789}; r3c7<>1.

26. 15(3) in c7 must have 5 -> {258/357}, cannot have 6.

27. Innies c5: r129c5 = 19 = {469/568} -> r345c5<>6

28. Multi-colouring on 1s -> r1c7=1

-> Singles: r1c3=4, r6c3=5, r6c4=3, r6c2=1, r1c4=2, r1c2=3, r3c6=3
-> Naked Pair (14) in N5 -> r456c5<>4
-> Hidden single in r4 -> r4c6=1, r5c6=4.

29. 21(4) in N1 = {2568}, cannot have 7 -> r2c3={79}, r3c1={17}, r3c3={179}; -> 8 locked to r45c3 -> r4c12<>8.

30. 11(3) in r3c1, r4c12 = {146} -> r3c1=1 ...

Straightforward from here with naked and hidden singles and locked candidates.

Multi-colouring is indeed single digit technique linking conjugate pairs in two chains - in the above scenario blue coloured cells can 'see' both pink and amber coloured cells therefore blue cannot be true and the green coloured cell must be true. I haven't got to grips with using Sumocue (I mostly use JSudoku if not solving on paper) so not sure about the pattern hints comparison - the chains of conjugate pairs don't necessarily form a pattern.

Edit: Sorry if some of the eliminations I made are not clear enough in this walkthrough. I'll try to be more explicit next time.

Hi

That move is(/can be duplicated by) a skyscraper/sashimi x-wing on R1C27 + R6C29 which eliminates 1 from R5C7 and then gives you a hidden single 1 in R1C7.
I use patterns(like skyscraper/finned fish) in sudoku just not in Killer Sudoku. I don't use multicolouring for these eliminations.
Maybe i should. But i guess i generally find it more fun to work combinations and 45-tests to bother looking for regular sudoku techniques.

greetings

Para

More than one way ... as they say.

I guess I should study the 'patterns' more - the skyscraper seems to crop up fairly often.