SampuZ4 Killer + V2

[sudokuEd]

Tried very hard to make a V2 for Special Killer X 4: couldn't find one hard enough.

So instead, here's a really nice hard Killer - another one inspired by the Killer Samurai from Flower-sudoku with lots of diagonal cages. Found it a terrible struggle to solve - but suspect I missed something.

So, as insurance, have included a V2 that slams shut a couple of front doors . Havn't tried to solve V2 -but would love to do it 'tag' style (especially on the weekend) if someone is interested to have a crack. Havn't included a pic of V2 yet (spoiler) - if you need one, just send a PM.

SampuZ4 - (note: this is not a diagonals puzzle)
3x3::k:4608:46082051:56364870:4359:4359:46085636:563633334870:4359:4608:5636:788:788:4630:4630:46304870207523342848:48974870:541236134136:41364897541223503888:4136:48974897:5412:77355945:594536432622:7735:7735:77355945:5444544477353650:59451868:54445444:

Text for SampuZ4V2 (same solution, different cages,not a diagonals puzzle)
3x3::k:4608:46082051:56363846:4359:4359:46085636:563633333846:4359:4608:5636:788:788360633432075361323344897:4898:4898:541236134136:41364897:4898:541223503888:4136:48974897:5412:77355945:594536432622:7735:7735:77355945:5444544477353650:59451868:5444:2622:5444:


Here is a puzzle pic for SampuZ4- and an initial population pic.

[Para]

Ok was just looking over this puzzle quickly and found an interesting elimination. Just as a headstart for everyone.

Check out how the 9 is locked in N7 for 30(5) and R7C1. So no 9's anywhere else in N7.
Explanation 30(5) = 9{....}/{87654}(either a 9 in 30(5) or {87654} in 30(5) and {87654} -->> R7C1 = 9
No clue how useful it is, but it just struck me as an interesting move.

Para

[edit]
p.s.
Ok not useful. Eliminations are to be made on easier way

[rcbroughton]

Ok Ed,

I'll get you started on the V2. I like the look of it but have a feeling it's going to be a long one . . .

Some simple steps to begin:

1. Naked pair {12} r3c34

2. {69} must be in either 14(2) or 13(2) in n2 - nowhere else in that nonet

3. {4689} must be in the 14(2), 13(2) or 8(2) in n2 - nowhere else in that nonet

4. 4 now locked in column 6 in N2 in 13(2) - must be {49}. {49} locked for n2 and c6

5. 8(2) in n2 can now only be {17}/{35} - no 2

6. {68} now locked in r3 of n2 - 14(2) must be {68}

7. 13(3) in n3 can now only be {139}/{157}/{247}/{346} no 345789 in r2c7

8 22(4) in n12 - no valid combination with 2/3 at r2c3 or r3 at r3c2
8a. {2479} r2c3/r3c2 must be {79}
8b. {2569} r2c3/r3c2 must be [69]
8c. {2389} r2c3/r3c2 must be [89]
8d. {2578} r2c3/r3c2 must be [85]/[87]
8e. {3478} r2c3/r3c2 must be [84]

9. 8(3) in n4 must use a 1 - nowhere else in that nonet

10. Must use {123} in combination of 9(2) & 8(3) in n4 - nowhere else in that nonet

11. 14(3) in n45 - r5c4 can now only be 1/2/3

12. 9(2) in n4 can now only be {27}/{36}

13. Similarly - must use {789} in combination of 19(3) & 19(4) in n6

14. 11(2) in n56 can now only be {56}/{47} or [83]

Over to someone else?

Richard

[sudokuEd]

First, here is a pic for SampuZ4 V2: and a population pic




Thanks for getting us started Richard. I still have trouble working out how you do things. So, hope you don't mind if I restate your steps using the "45" rule.

I have one disagreement with your output too - step 14.

1. Naked pair {12} r3c34

2. {69} must be in either 14(2) or 13(2) in n2 - nowhere else in that nonet

just to explain this:
2a. 14(2)n2 = {59/68} = [5/8,6/9..]
2b. 13(2)n2 = {49/67} ({58}Blocked step 2a) = [6/9..]

3. {4689} must be in the 14(2), 13(2) or 8(2) in n2 - nowhere else in that nonet

3a. In n2:8(2) = {17} -> 13(2) = {49} -> 14(2) = {68}
3b. In n2:8(2) = {35} -> 14(2) = {68} -> 13(2) = {49}

4. 4 now locked in column 6 in N2 in 13(2) - must be {49}. {49} locked for n2 and c6

5. 8(2)n2 = {17/35} (no 2)

6. {68} now locked in r3 of n2 - 14(2) must be {68}

7. 13(3) in n3 can now only be {139}/{157}/{247}/{346} no 345789 in r2c7

7a. r2c7 = {126}

8 22(4) in n12 - no valid combination with 2/3 at r2c3 or r3 at r3c2

8f. "45"n1 -> r3c3 + r3c2 & r2c3 = 15
8g. r3c3 = {12} -> other 2 = 13/14 -> min 4 in those two cells

9. 8(3) in n4 must use a 1 - nowhere else in that nonet

9a. = 1{25/34} = [2/3..]

10. Must use {123} in combination of 9(2) & 8(3) in n4 - nowhere else in that nonet

10a. from step 9a, 8(3) = [4/5] -> {45} blocked from 9(2)
10b. 9(2) = {27/36} = [2/3..]
10c. [2/3..]locked in 8(3), 9(2) for n4

11. 14(3) in n45 - r5c4 can now only be 1/2/3

12. 9(2) in n4 can now only be {27}/{36}

13. Similarly - must use {789} in combination of 19(3) & 19(4) in n6

13a. "45" n6 -> r5c7 + r6c8 = 7 = [34/43/52/61]
13b. r5c7 = {3456}, r6c8 = {1234}

14. 11(2) in n56 can now only be {56}/{47} or [83]

14a. I think this should be {56}/[74/83]
14b. r4c6 = {5678}
14c. r7c8 = {6789}

Now some extra ones
15. "45" n789 -> r7c138 = 16
15a. min r7c18 = {46} = 10 -> max r7c3 = 6 but 2 6's in r7
15b. min r7c18 = {47} = 11 -> max r7c3 = 5
15c. r7c13 = 7..10

16. "45" n89 -> r8c4 + r7c8 = 15 = {69/78}
16a. r8c4 = {6789}
16b. min r8c4 = 6 -> max r9c23 = 8 -> (no 89)
I like para's way better! Nice move.

17. "45" c6789 -> r359c5 = 19
17a. Max r39c5 = [86] = 14 -> min r5c5 = 5
17b. Max r35c5 = [89] = 17 -> min r9c5 = 2
17c. no 6 r9c6

Now a sum/marks pic:

Code: Select all

.-----------------------.-----------.-----------.-----------.-----------.-----------.-----------------------.
|(18)                   |(12)       |(8)        |(22)       |(13)       |(15)       |(17)                   |
| 123456789   123456789 | 345789    | 1357      | 12357     | 49        | 123456789 | 123456789   123456789 |
|           .-----------+-----------'-----------+-----------:           :-----------+-----------.           |
|           |           |                       |           |           |(13)       |           |           |
| 123456789 | 345789    | 456789      12357     | 1357      | 49        | 126       | 123456789 | 123456789 |
|           :-----------+-----------------------+-----------'-----------:           '-----------+-----------:
|           |           |(3)                    |(14)                   |                       |           |
| 34579     | 4579      | 12          12        | 68          68        | 34579       34579     | 34579     |
:-----------'-----------+-----------.-----------'-----------.-----------+-----------.-----------'-----------:
|(8)                    |(14)       |(9)                    |(11)       |(19)       |(19)                   |
| 12345       12345     | 456789    | 12345678    12345678  | 5678      | 123456789 | 23456789    23456789  |
:-----------.           |           '-----------.-----------'-----------+-----------+-----------.           |
|(21)       |           |                       |(16)                   |           |           |           |
| 456789    | 12345     | 456789      123       | 56789       1235678   | 3456      | 123456789 | 23456789  |
|           :-----------'-----------.-----------'-----------.           :-----------+-----------+-----------:
|           |(9)                    |(15)                   |           |           |(10)       |           |
| 456789    | 2367        2367      | 123456789   123456789 | 1235678   | 123456789 | 1234      | 123456789 |
|           :-----------.-----------+-----------------------+-----------'-----------:           :-----------:
|           |(30)       |           |(23)                   |(14)                   |           |(10)       |
| 456789    | 123456789 | 12345     | 123456789   123456789 | 1235678     123456789 | 6789      | 1234567   |
:-----------'           '-----------+-----------.           :-----------.           :-----------:           |
|                                   |(14)       |           |(21)       |           |           |           |
| 123456789   123456789   123456789 | 6789      | 123456789 | 1235678   | 123456789 | 123456789 | 1234567   |
|           .-----------------------+-----------+-----------'-----------+-----------+-----------+-----------:
|           |                       |           |(7)                    |           |           |           |
| 123456789 | 1234567     1234567   | 123456789 | 23456       1235      | 123456789 | 1234567   | 123456789 |
'-----------'-----------------------'-----------'-----------------------'-----------'-----------'-----------'

[rcbroughton]

Ed

thanks for expanding on my steps. Just to explain how I come to step 2 - I do it slightly differently.

2. {69} must be in either 14(2) or 13(2) in n2 - nowhere else in that nonet

I take 14(2) and 13(2) and say that since they are all in the same nonet, you can treat it as a single cage 27(4), so we know between them they can only be {3789}, {4689} or {5679}, but {3789} doesn't split into two pairs for 14 & 13, so only possiblitlies are {69}{48/57}. Same result as you point out, I just look at it differently.

I use the same type of analysis in step 3, step 10 and step 13.

14a. I think this should be {56}/[74/83]

Your observation at step 14 is spot on - [74]

I like the point you make on 18f - having pointed out the 45 for me (I ssometimes struggle to see these at the best of times - these non-contiguous cages make it even harder to spot sometimes) there's an extra step:

8h. since r3c3 ={1/2} r2c3+r3c2=13 or 14. If 14, can't be 7/7, if 13 can only be [67] so no 7 at r2c3

now an extra one I should have spotted after step 4

18. 7(2) in n8 can't now have a 3 at r9c5

[sudokuEd]

Actually, lets just get rid of that [67] Richard.
19. from step 8h, [76] = 13 in r3c2 + r2c3 with 2 in r3c3
19b. -> rest of 22(4) cage in n2 = 9 = {18} only
19c. but 1 is already in r3c4 -> no [76] in 22(4) in n1

20. 6 in n1 only in 18(4) = 6{138/147/237}(no 5,9) ({1269} blocked by r3c3): {3456} blocked by 22(4) in n1)

21. from step 17: r359c5 = 19
21a. = [685/694/856/865/874/892]
21b. [892] combo -> r3569c6 = [6165/6255]: Not good
21c. ->no 2 r9c5, no 7 r9c6

That's me for the day - Happy solving!

[rcbroughton]

Ed

god spot!!

now, since you've reminded me about the 45 rule . . .

22. 45 rule on n4 - r5c4 r7c1 total 7
22a. no 7/8/9 at r7c1
22b. 21(3) at n45 - now has no 4/5/6 in r56c1

23. 45 rule on n7 - r7c1 plus r7c3 minus r8c4 equals 1, so r7c1 plus r7c3 equals 7, 8, 9 or 10
23a. [43]/[52]/[61] ok for 7
23b. [53]/[62] ok for 8
23c. [63] ok for 9 (can't have {54} otherwise no options left for 30(5))
23d. [64] ok for 10
23e. no option with 5 in r7c3

24. No 4 in r1c1 or r2c1 - only combo for 18(4) with 4 is {1467}
24a. 21(3) at n45 uses 4 or 7 - so cannot have 4&7 in 18(4) c1
24b. r3c1=4 -> r1c2 =7
24c. r3c1 or r3c2 = 4 -> r3c1=7 - not allowed

and now a real tricky one!!

25. 45 rule on c1 - r123489c1 equal 24
25a only combo for 24(6) continaining 9 is {123459}
25a r3c1= 3 -> {459} in r489 -> r12 = {12} not possible because of r3c3
25b r3c1 = 4 -> r12 = {12}, {13} or {23} - none possilble because can't make 18(4) total
So no 9 in r89c1

26. 9 now locaked in r56 for c1 in n4 and 21(3)
26a no 9 in 14(3)
26b no 6 in 21(3) and no 1 in r5c4 since these total 7 (step 22)

27. 9 now locked in n6 for r4 - nowhere else in that nonet

28. 45 rule on N5 - r5c4 plus r4c6 minus cells r7c3 equals 5
28a minimum in r7c3 is 2 - no 1 in r7c3
28b maximum of r5c4+r4c6 is 9. r4c6 can't be 8

29. 45 rule on N6. r4c6 minus r6c8 equals 4
29a r6c8 cant be 4

30 45 rule again on n6 r5c7 + r6c8 equal 7
30a r5c7 can't be 3

[sudokuEd]

OK - now we're getting somewhere: must be blessed .

23d. [64] ok for 10

This one blocked: from step 15, r7c138 = 16 -> [646]
23e. this is a very important extra step-> no 4 r7c3, no 9 r8c4

24. No 4 in r1c1 or r2c1 - only combo for 18(4) with 4 is {1467}

very clever

25. ..So no 9 in r89c1..
26. 9 now locaked in r56 for c1 in n4 and 21(3)

Another way to see this - by ignoring "45" and looking at combo's!!.
25c. 21(3)c1:{678} blocked by 18(4)n1
Explanation: from step 20, each combo in 18(4)n1 ({1368/1467/2367}) shares two common digits with {678}, but only 1 of the common digits can be in r1c2 -> {678}blocked -> 9 locked in 21(3)c1

Some more
31. from step 23e, 28a, r7c3 now = {23}
31a. 15(3)n57: must use one of 2,3 but not bot -> no 123 in r6c45

32. "45" n5: r5c4 + r4c6 - r7c3 = 5
32a. r7c3 = {23} -> 2 innies = 7/8 -> no 7 r7c6

33. 11(2)n56 = {56} only
33a. no 5 or 6 r5c56
33b. no 5 or 6 r4c789

34. "45"c6789 -> r359c5 = 19 = [685/694/874]
34a. r9c5 = {45}, r9c6 = {23}

35.from step 23ab
r7c13 = [43/52/53] = 7/8
35a. -> r8c4 = {67}
35b. -> r7c8 = {89} (step 15)
35c. r6c8 = {12}

36. 7(2)r9 = [43/52] = [3/5..]
36a. -> 14(3)n78 {356} blocked by 7(2)
36b. 14(3) = {167/257/347} = 7{16/25/34}
36c. no 7 r9c4
36d. no 7 r8c123

37."45"n8 -> r8c4 + r78c6 = 15 = h15(3)
37a.r8c4 = 6 -> r78c6 = 9 = [18]/{27} ([81]blocked by 8 in r7c8 step35b})
...............= 7 -> r78c6 = 8 = {26} ({35} blocked by [53] in r7c13 step 35)
37b. r7c6 = {1267}
37c. r8c6 = {2678}
37d. In summary h15(3)n8 = {168/267} = 6{18/27}
37e.6 locked for n8


38.9 in n8 only in 23(4) = 9{158/257/347} ({2489}blocked by 7(2))

Marks pic

Code: Select all

.-----------------------.-----------.-----------.-----------.-----------.-----------.-----------------------.
| 123678      1234678   | 345789    | 1357      | 12357     | 49        | 123456789 | 123456789   123456789 |
|           .-----------+-----------'-----------+-----------:           :-----------+-----------.           |
| 123678    | 345789    | 4589        12357     | 1357      | 49        | 126       | 123456789 | 123456789 |
|           :-----------+-----------------------+-----------'-----------:           '-----------+-----------:
| 347       | 459       | 12          12        | 68          68        | 34579       34579     | 34579     |
:-----------'-----------+-----------.-----------'-----------.-----------+-----------.-----------'-----------:
| 12345       12345     | 45678     | 12345678    12345678  | 56        | 1234789   | 234789      234789    |
:-----------.           |           '-----------.-----------'-----------+-----------+-----------.           |
| 789       | 12345     | 45678       23        | 789         12378     | 56        | 123456789 | 23456789  |
|           :-----------'-----------.-----------'-----------.           :-----------+-----------+-----------:
| 789       | 2367        2367      | 456789      456789    | 1235678   | 123456789 | 12        | 123456789 |
|           :-----------.-----------+-----------------------+-----------'-----------:           :-----------:
| 45        | 123456789 | 23        | 12345789    12345789  | 1267        123456789 | 89        | 1234567   |
:-----------'           '-----------+-----------.           :-----------.           :-----------:           |
| 1234568     12345689    12345689  | 67        | 12345789  | 2678      | 123456789 | 123456789 | 1234567   |
|           .-----------------------+-----------+-----------'-----------+-----------+-----------+-----------:
| 12345678  | 1234567     1234567   | 1234589   | 45          23        | 123456789 | 1234567   | 123456789 |
'-----------'-----------------------'-----------'-----------------------'-----------'-----------'-----------'

[rcbroughton]

Ed

Ok, made a lot more headway. I get the feeling it's fairly close after this, but it's about time I put down my pencil for the afternoon.

Quite a lot of steps and quite a few placements, but not quite the major collapse I'd been hoping for. A Couple of moves you might find interesting.

39. 19(3) in n6 - {568} no longer possible so no 5 at r5c9

40. 5 locked in n5 for column 6 - nowhere else in that nonet

41. 4 locked in row6 of n5 - nowhere else in the row
41.a 4 locked also locked in 15(3) n57 so {267} no longer valid - remove {67} from r6c45

42. 9 locked in n6 for row 4 - nowhere else in that nonet

43. 45 rule on n1 r3c4+r1c5+r2c4 total 10
43a. 1{17} not OK
43b. 1{27} Ok
43c. 2{17} Ok
43d. 2{35} not OK - would need {97} in other two cells of 22(4) and no possible combination
43e. 8(2) in n2 now {35}

Now for the first placement - a bit convoluted but stick with me

44. 45 rule on n5 r5c4 plus r4c6 minus r7c3 equals 5 - r5c4={23} r4c6={56} r7c3={23}
44a. [252] - ok
44b. [353] - ok
44c. [263] - not ok because {26} would remove all possiblilities for 16(3) and 9(2) in n5 - reasoning 16(3)+9(2) = 25(5) -> {13678}/{23569}/{23578} (4 locked outside, any other combos don't break down to 16 + 9)
44d. r4c6<>6 -> r4c6=5

45. 11(2) n56 now [56]

46. 5 now locked in row 5 in n4 - no 5 in r5c8

47. must use 1 in 15(3) or 13(3) in n3 - not in 17(3)
47a. only combo for 28(6)=15(3)+13(3) without 1 is {234568}->{348}+{256} or {456}+{238} but no 6 or 8 in 13(3)

48. must use 2 in 19(3) or 19(4) in n6
48a only combo for 38(6)=19(3)+19(4) without 2 is {1346789} -> but 6 is outside at r5c7
48b. r6c8<>2 -> r6c8=1
48c. 10(2) n69=[19]

49. hidden single 9 at r9c4 for the row

50. 1 locked in col 7 in n3 - nowhere else in the col

51. 5 locked in n7 for col 1 - nowhere else in the nonet

52. 45 rule on n89 -> r8c4=6
52a. 14(3) n87 must now be 6{17}
52b. naked pair {17} in r9c23 for r9 and n7
52c. 9(2) in n5 can't now be [63] - no 3 in r4c5

53. 6 locked in n4 for column 3 nowhere else in the nonet
53a 9(2) in n4 can't be [63] - no 3 in r6c3

54. 45 rule on n7 - r56c1 plus r6c45 = 29 with r56c1={79} or {89} and r6c45=[49] or {48}
54a. r6c45=[49] -> r56c1=[97]
54b. r6c45={48} -> r56c1=[89]
54c. no 7 in r5c1, no 8 in r6c1

55. 45 rule on n8 r9c23 plus r789c7 plus r8c8 plus r9c9 total 34 - but r9c23 total 8, so rest total 26
55a. 26(5) (excluding 9 and 1) = {23678}/{34568}(6 locked in r9c9) or {24578} no 3
55b. no 3 in r9c9

56. 45 rule on r3 - r3c12789 total 28
56a. only combo is {34579} with r3c78 total 11 {47} or 12 [93]/{75}
56b. no 3 in r3c7

57. r4c9 <> 3 and <> 2
57a. if r4c9=3 then r4c8+r5c9 = 16 in 19(3) but no 9
57b. if r4c9=2 then r4c8+r5c9 = 17 in 19(3) but no 9

58. 16(3) in n5 - no 7,8 in r5c6 no 2,3 in r6c6
58a. r5c5=7 -> r56c6=9 = [18]/[36]
58b r5c5=8 -> r56c9=8 = [17]/[26]
58c. r5c5=9 -> r56c9=7 = [16]

59. 45 16(3) in n5 - no 2 at r5c6
59a. if r5c6=2 it forces r5c5=8, r5c4=3, r6c6=6 -> no candidates left for 9(2) in n5

60. 2 now locked in c6 in n8 - nowhere else in the nonet

61. 13(3) in n3 is {139}/{247}/{157} - so 15(3) cannot be {357}, only other option with 7 for 15(3) is {267} - 6 only in r8c8, no 2 in r3c9 so r1c7<>7 and r2c8<>7

62. 8(3) in n4 only option with 2 is {12}5 - no 2 in r5c2

63. 10(3) in n9 - only option with 2 is {17}2 - so no 2 in r78c9

64. 21(4) in n89 - only options with 5 are {2568}/{3567} both use 6 which must be at r9c9 - so no 5 at r9c9

Now another convoluted one

65. 45 rule on r6. r6c1679 minus r7c3 equals 20.
65a. r7c3=2 - r6c45=[49]->r6c1=7 -> r6c679=15={258} so 8 must be in r6c6
65b. r7c3=3 - r6c45={48}->r6c1679=23=97{25}/96{35}
65c. no 7,8 in r6c79

66. 8 now locked in n5 for r6 - nowhere else in n5
66a. no 1 in 9(2) in n5
66b. r5c6=1

67. hidden single 3 at r9c6 for col 6
67b 7(2) in n8 = [43]

68. hidden single 4 at r6c4 for row 6

69. hidden single 8 at r7c4 for col 4
69a. 10(3) in n9 only options remaining are {13}6 {14}5 {17}2
69b. no 5/6 in r7c9
69c. no 6 at r8c9

70. hidden single 5 at r1c4 for col 4
70a. 8(3) in n2=[53]
70b. 12(2) in n1-> no 7 in r2c2, no 9 in r1c3

71. 1 locked in column 5 of N8

72. 7 locked in column 6 of N8

73. 7 locked in n4 for row 6
73a. 19(4) in n6 - no option with 2/3 in r4c7

74. Follows from 69 - 6 hidden single at r7c2 for r7

which should leave it here for you to pick up.

Code: Select all

.-----------------------.---------------.------------------------.
| 123678  123478  3478  | 5   27   49   | 123489  234678 2346789 |
| 12678   4589    4589  | 127 3    49   | 12      24568  2456789 |
| 347     459     12    | 12  68   68   | 4579    3457   34579   |
:-----------------------+---------------+------------------------:
| 1234    1234    468   | 237 267  5    | 4789    23478  4789    |
| 89      345     458   | 23  79   1    | 6       23478  23478   |
| 79      237     267   | 4   89   68   | 235     1      235     |
:-----------------------+---------------+------------------------:
| 45      6       23    | 8   15   27   | 23457   9      1347    |
| 23458   23489   23489 | 6   15   27   | 234578  234578 1347    |
| 258     17      17    | 9   4    3    | 258     256    268     |
'-----------------------.---------------.------------------------'

happy hunting!!

Rgds
Richard

[Para]

Now another convoluted one

65. 45 rule on r6. r6c1679 minus r7c3 equals 20.
65a. r7c3=2 - r6c45=[49]->r6c1=7 -> r6c679=15={258} so 8 must be in r6c6
65b. r7c3=3 - r6c45={48}->r6c1679=23=97{25}/96{35}
65c. no 7,8 in r6c79

This is actually a fun step. We heard of the killer pair. Well this move is actually a Killer Quad. 7/9 in R6C1, 6/7 in R6C23, 8/9 in R6C45 and 6/7/8 in R6C6. So a Killer Quad {6789} in R6C123456.

[Para]

I am sorry i missed everything you did. I have a way to finish this but i'll put it in small text for if you 2 want to finish it. It seems i only get to do the final steps of these killer tags. If my input is appreciated, i'll enlarge it.

75. R7C3 = 3; R5C4 = 3; R7C1 = 4
75a. 45 on N7 : R7C13 = 7; 45 on N4: R5C4 + R7C1 = 7 -->> R7C3 = R5C4
75b. 2 in R3C4 = 2 -->> R5C4 = 3 -->> R7C3 = 3
75c. 2 in R3C3 = 2 -->> R7C3 = 3
75d. R5C4 = 3; R7C1 = 4

76. R6C5 = 8; R6C6 = 6; R5C5 = 9

77. R3C56 = [68]

78. R56C1 = [89]

79. Hidden single 6 in R4C3 -->> R5C3 = 5 -->> R5C2 = 4

80. Naked pair {72} in R5C89 for N6

81. Naked pair {53} in R6C79 for N6 and R6

82. Naked pair {72} in R6C23 for N4

83. Naked pair {13} in R4C12 for R4

84. 8 locked in 19(3) in N6, no 8 in R4C7

85. Naked pair {25} in R89C1 for C1 and N7

86. Naked pair {89} in R8C23 for R8

87. 7 locked in N1 and 18(4) for C1 no 7 anywhere else in N1
87a. 18(4) = {2367} -->> locked for N1
87b. R1C2 = 2 -->> R3C34 = [12]
87c. Hidden single 1 in R4C1 -->> R4C2 = 3

88. 12(2) in N1 = {48} -->> R1C3 = 4; R 2C2 = 8

89. R2C3 = 9; R3C2 = 5

90. Hidden single 1 in R1C7

91. R2C8 + R3C9 in 15(3) = [59]

And the rest is singles (naked and hidden) or cage sums

greetings

Para

[Nasenbaer]

It seems that I'm too late. But here are a few comments:

28. 45 rule on N5 - r5c4 plus r4c6 minus cells r7c3 equals 5
28a minimum in r7c3 is 2 - no 1 in r7c3

I think here is a mistake, 1 was never excluded from r7c3!

So the following steps are also faulty:

31. from step 23e, 28a, r7c3 now = {23}
31a. 15(3)n57: must use one of 2,3 but not bot -> no 123 in r6c45

... must use one of 1,2,3 ...

32. "45" n5: r5c4 + r4c6 - r7c3 = 5
32a. r7c3 = {23} -> 2 innies = 7/8 -> no 7 r7c6

I don't understand the second half of this step. Do you mean no 7 r4c6? But that would also be a mistake, [17] for the innies would still be possible even with 1 excluded from r7c3

33. 11(2)n56 = {56} only
33a. no 5 or 6 r5c56
33b. no 5 or 6 r4c789

From this step it seems you meant no 7 r4c6 in step 32. But, as said before, I think this is not valid.

I'll stop here with the walkthrough, maybe you could explain these steps to me.

Peter

[Para]

It seems that I'm too late. But here are a few comments:

28. 45 rule on N5 - r5c4 plus r4c6 minus cells r7c3 equals 5
28a minimum in r7c3 is 2 - no 1 in r7c3

I think here is a mistake, 1 was never excluded from r7c3!

You read the step wrong. It explains why there isn't a 1 in R7C3. minimum R5C4 + R4C6 = 7 -->> Minimum R7C3 = 2 -->> eliminate 1 from r7c3

same goes for 28b. maximum R7C3 = 4 -->> Maximum R5C4 +R4C6 = 9 -->> eliminate 8 from R4C6

I hope this makes it more clear. and now all other steps work too.

Para

[Nasenbaer]

Thanks to Ed and Para for explaining it to me.

26b no 6 in 21(3) and no 1 in r5c4 since these total 7 (step 22)

I totally missed this step which removed the 1 from r5c4 and the 6 from r7c1. So my thoughts went the wrong way. I shouldn't do killers and watch Superbowl at the same time.

Peter

[sudokuEd]

Here's a condensed walk-through for SampuZ4 V2 - though many of the key steps are needed for V1 also.

All the steps have been reordered and renumbered, and in some cases reworked to make the shortest possible solution with clarity. If there are any suggestions for improvements, please let me know.

There are many contradiction/ short chain moves - with 2 "money" moves that are very clever. These make this a very, very difficult puzzle. So, it was really great to work as a team. Thanks a lot.

SampuZ4 V2
1. Naked pair {12} r3c34:locked for r3

2. 14(2)n2 = {59/68} = [5/8,6/9..]

3. 13(2)n2 = {49/67} ({58}Blocked by 14(2)) = [6/9..]

4. Killer pair {69} in 14(2) and 13(2): locked for n2

5. 8(2)n2 = {17/35}(no 2)
5a. In n2:8(2) = {17} -> 13(2) = {49} -> 14(2) = {68}
5b. In n2:8(2) = {35} -> 14(2) = {68} -> 13(2) = {49}

6. ->13(2) = {49} locked for n2 and c6
6a. no 2 or 7 r5c7
6b. no 3 r9c5

7. 14(2)n2 = {68}:locked for n2,r3

8. "45" n1 -> r3c3 + {r3c2 & r2c3} = 15 = h15(3)
8a. r3c3 = {12} -> other 2 = 13/14 -> min 4 in those two cells
8b. h15(3) = {159/249/258}(no 6,7) ({168} blocked:6,8 only in r2c3;
.........................................................{267} blocked:no {18/45} in rest of 22(4) in n2)

9. "45" n1 -> [r3c4] + {r2c4 + r1c5} = 10 = h10(3) = [2]{17}/[1]{27} = {127} ({235} blocked: rest of 22(4) in n1 can only be {68}:but no 6 available)
9a. {127} locked for n2
9b. 8(2)n2 = {35}
9c. 22(4)n21 = 7{159/249/258}

10. 6 in n1 only in 18(4) = 6{138/147/237}(no 5,9) ({1269} blocked by r3c3): {3456} blocked by h15(3)(step 8b)

11. 8(3)n4 must use a 1 = 1{25/34} = [2/3,4/5..]
11a. 1 locked for n4
11b. 8(3) = [4/5] -> {45} blocked from 9(2)
11c. 9(2)n4 = {27/36}(no 8) = [2/3..]
11d. Killer pair {2/3} in 8(3), 9(2) for n4

12. 14(3) in n45: r5c4 can now only be 1/2/3

13. "45" n4: r5c4 + r7c1 = 7
13a. r7c1 = {456}
13b. -> no 4/5/6 in r56c1

This is one of the "money" moves to crack this puzzle
14. 21(3)c1:{678} blocked by 18(4)n1
14a.Explanation: from step 10, each combo in 18(4)n1 ({1368/1467/2367}) shares two common digits with {678}, but only 1 of the common digits can be in r1c2
-> {678} blocked
14b.21(3)c1 = 9{48/57} (no 6)
14c. 9 locked in r56 for c1 and n4
14d. no 6 in r7c1 -> no 1 in r5c4 (step 13)

15. "45" n6 -> r5c7 + r6c8 = 7 = [34/43/52/61]
15a. r5c7 = {3456}, r6c8 = {1234}

16. r4c6 = {5678}
16a. r7c8 = {6789}

17. "45" n789 -> r7c138 = h16(3)
17a. min r7c18 = {46} = 10 -> max r7c3 = 6 but 2 6's in r7
17b. min r7c18 = {47} = 11 -> max r7c3 = 5
17c. h16(3) = 16 = [439/529/538] ([457/547]Blocked: 30(5)n7 needs [4/5])
17d. r7c3 = {23}, r7c8 = {89}
17e. r6c8 = {12} -> r4c6 = {56} (step 13) -> r4c6 = {56}

18. 11(2)n56 = {56} only
18a. no 5 or 6 r5c56
18b. no 5 or 6 r4c789

19. "45" c6789 -> r359c5 = 19
19a. Max r39c5 = [86] = 14 -> min r5c5 = 5
19b. Max r35c5 = [89] = 17 -> min r9c5 = 2
19c. no 6 r9c6

20. 16(3)n5 = {169/178/259/367} ({268} -> r4c6 = 5 but {2568} is blocked by 9(2)n5;{358}-> r5c4 = 2 but{2358} blocked by 9(2)n5)

21. 15(4)n57: must use one of 2,3 but not both -> no 123 in r6c45
21a. r6c45 = {49/58/48}(no 6,7) ({67} blocked by 9(2)r6;{57} ->16(3) = {169} but {5..6.} blocked by r4c6)
21b. r6c45 = {4589}

22. "45"n5 -> 4 innies = 20
22a. r6c45(step 21a) + r5c4 + r4c6 = 20 = [{49}25/{48}35] ([{58}25] 2 5's;[{48}26] blocked by 9(2)n5)
22b. -> r4c6 = 5
22c. r6c45 = {48/49} = 4{8/9}:4 locked for n5,r6

23. r5c7 = 6, r6c8 = 1, r7c8 = 9

24. "45" n89 ->r8c4 = 6

Now the second "money" move
25. "45" n7 -> r7c13 = 7; 45 n4 -> r5c4 + r7c1 = 7
25a. -> r7c3 = r5c4
25b. only 2 2's in r3, r3c34
.... 2 in r3c4 -> r5c4 = 3 -> r7c3 = 3
.... 2 in r3c3 -> r7c3 = 3 -> r5c4 = 3

26. -> r7c3 = 3; r5c4 = 3; r7c1 = 4

27. r6c45 = {48}:locked for n5, r6

28. r56c1 = [89]

29. 9(2)n5 = {27}:locked for n5. r4

30. r6c6 = 6; r5c56 = [91]

31. r3c56 = [68]

32. Hidden single 6 r4c3, r5c3 = 5 -> 9(2)n4 = {27}:locked for n4, r6

33. r5c2 = 4, r4c12 = {13}:locked for r4

34. 8 locked in 19(3) in n6 (all remaining combinations have 8), no 8 in r4c7

35. r9c23 = {17}(only remaining combination):locked for n7, r9

36. 7 in c1 only in n1 and 18(4):7 locked for n1
36a. 18(4) must have 6 and 7 = {2367}: locked for N1
36b. r3c34 = [12]

37. 12(2) in n1 = {48} -> r1c3 = 4; r2c2 = 8

38. r2c3 = 9; r3c2 = 5

And the rest are singles (naked and hidden) or cage sums
Thanks Andrew for some corrections & clarifications