Special KillerX5oct06

[Nasenbaer]

That was a good one. Took me really long to get to the first number, but from then on it unfolded rapidly.

I used the diagonals in the "endgame", so maybe they aren't really necessary, I don't know. Has anyone tried it without them?

Walkthrough KillerX5oct06

1. 14(2) in r1 : {59/68} -> no 1,2,3,4,7
2. 5(2) in r1: {14/23} -> no 5,6,7,8,9
3. 6(2) in c1: {15/24} -> no 3,6,7,8,9
4. 15(2) in c1: {69/78} -> no 1,2,3,4,5
5. 14(2) in c9: {59/68} -> no 1,2,3,4,7
6. 3(2) in c9: {12} -> 1,2 locked in c9 -> nowhere else
7. 17(2) in r9: {89} -> 8,9 locked in r9 -> nowhere else
8. 16(5) in n8: {12346} -> 1,2,3,4,6 locked in n8 -> nowhere else
8a. -> r7c4 = {5789}
8b. -> r7c6 = {5789}
9. 9(2) in r9: using steps 8 and 7: r9c3={24}, r9c4={57}
10. "45" on n7 -> r7c13 + r9c3 = 19 -> {289/469/478} -> r7c3 = {6789}
11. 11(3) in n7 : no 9
12. 18(3) in n9 : at most one of 8 or 9 (one used for step 7)
-> {189} not possible -> no 1
13. "45" on n9 -> r7c79 + r9c7 = 13 -> {148/139/238} -> r7c7 = {34}
14. 18(4) in n4578 : min. r7c34 = 11 = {56}
-> max. r6c34 = 7 = {1234} (5 and 6 not possible)
15. "45" on r89 : r7c258 = 10 = {127/136/145/235} -> no 8,9
16. "45" on n3 : r3c79 + r1c7 = 10 = {127/136/145/235} -> no 8,9
16a. -> {127} not possible because r3c9 = {56} -> no 7
16b. -> r13c7 = {1234} (step 16a) -> no 5,6
17. r4c9 = {89} (because r3c9 = {56})
18. 22(3) in n3 : {589/679} -> 9 locked in 22(3), nowhere else
19. 22(3) in n3 has either 5 or 6, r3c9 = {56} -> 5,6 nowhere else in n3
20. 13(3) in n3 : {148/238/247} -> r1c8 = {12}
21. "45" on c89 -> r258c7 = 18 -> at most one of 8 or 9 (one used for step 7)
-> no 1
22. "45" on n6 : r46c79 = 21 -> because of {12} in r6c9 and {89} in r4c9 it
can have only one of {12} -> no 1,2 in r46c7
23. 1 locked in r13c7 for c7 -> r1c8 = 2
24. 5(2) in r1 = {14} -> 1,4 locked in r1
25. 2 locked in 18(3) in c7 (step 21) -> only {279} possible, 2,7,9 locked
in c7, nowhere else -> r9c7 = 8, r9c6 = 9
26. naked triple {134} in c7 -> 3,4 removed from r46c7 -> r46c7 = {56}
-> 5,6 locked in n6
27. 14(3) in n9 has to have 5 or 6 (from c9) -> no 9 possible
-> 9 locked in 18(3) in n9, locked in r8 -> 9 locked in r7c13 for r7
28. 19(3) in n7 (step 10) has 9 -> {478} not possible -> no 7
28a. 8 removed from r6c1
29. "45" on r6789 : r6c258 = 21 = {489/579/678} -> no 1,2,3

EDIT:
Old 30. 14(3) in n9 (continued from step 27) : {167/257/356} -> no 4
-> no 7 in r9c8
New 30. 14(3) in n9 (continued from step 27) : {167/356} -> no 4
-> no 7 in r9c8 (Thanks, Andrew)

31. 18(3) in n9 : {279/369/459} -> no 7 in r8c78
32. 23(4) in n5689 : max. (r7c67 + rr6c7) = 8+4+6 = 18 -> min. r6c6 = 5
-> no 1,2,3,4 in r6c6 -> 23(4) = {3578/4568}
33. 13(4) in n2356 : {1246/1345} -> {56} in r4c7 -> r34c6 = {1234}
34. "45" on n5 : r46c46 = 14 -> no 9 -> 9 locked in 31(5) in n5
34a. r6c6 = {5678} (step 32) -> no 7,8 in r4c4
35. 8 locked in r7c46 for r7 -> 8 removed from r7c13 -> 6,9 locked for r7 and n7
36. 19(3) in n7 (step 10) has 6 and 9 -> r9c3 = 4, r9c4 = 5
37. 7 locked in r7c46 for r7 -> nowhere else

EDIT: SudokuEd just pointed out that step 38 is flawed because the elimination of 5,6 is not logical. I couldn't find how I did it, so I will insert a new step (n38). I will leave the old version, maybe someone can tell me how I did it.

n38: hidden triple {123} in r6c349 -> 21(3) in r6 (step 29) has to have 4 -> {489} -> r6c1 = 6 -> r6c7 = 5 -> r6c6 = 7 -> r7c6 = 8 (EDIT: last part added, thanks Andrew)


Old version:
38. from step 32: 5,6 removed from r6c6 (locked in r67c7) -> r6c6 = {78}
38a. naked pair {78} in c6 -> nowhere else in c6
38b. 14(4) in n5 (step 34) : {1238/1247} -> no 1,2 in 31(5) in n5

39. "45" on c6789 -> r258c6 = 14 = {356} -> 3,5,6 locked in c6
39a. 2 locked in r34c6 for c6 and 13(4) in n2356 -> 13(4) = {1246}
-> r4c7 = 6 -> r6c7 = 5 -> r7c7 = 3 (step 32)

Now everything unfolds rapidly.

40. 14(3) in n9 : only {167} left -> r9c8 = 1, r89c9 = {67}
41. r7c9 = 2, r6c9 = 1, r3c9 = 5, r4c9 = 9
42. r8c7 = 9, r78c8 = {45}
43. r2c7 = 7, r5c7 = 2, r23c8 = {69}, r12c9 = {38} -> r5c9 = 4 (EDIT: error in r5c7 corrected, thanks Andrew)
44. 2,3 locked in r6c34 for r6 and 18(4) -> r7c3 = 6, r7c4 = 8
45. r2c8 = 9 (through D/), r3c8 = 6
46. r7c1 = 9, r6c1 = 6, r7c6 = 8, r6c6 = 7
47. r9c9 = 6 (through D\), r8c9 = 7, r6c8 = 8
48. 15(3) has to have 7 -> {357} -> r8c1 = 5, r9c12 = {37} -> r9c5 = 2
49. "45" on c12 -> r257c3 = 10 -> no 8,9
50. r7c2 = 1 -> r8c3 = 2, r8c2 = 8

Now everything should be clear.

I hope everything is clear although my use of brackets might not be fully correct (sorry for that).

Have fun!


Peter

[Nasenbaer]

I just tried to solve it without the diagonals. Sorry, but they have to stay, no unique solution without them.

Peter

[sudokuEd]

Yeah. That works now Peter. That hidden triple was great as the new step 38. Actually, the whole rest of the puzzle was just hidden and naked singles from that point. If that is your first ever walk-through Peter, it is very impressive. I remember how cryptic my first attempts were - I'm sure nd does too!!

I just tried to solve it without the diagonals. Sorry, but they have to stay, no unique solution without them.

SumoCue says 4 solutions without the diagonals.

I used the diagonals in the "endgame", so maybe they aren't really necessary

True - in the sense of solving techniques. But there are some really fun tricks with Diagonal Killers - just not as often as I'd like. I'm having trouble making Diagonal Killers that require those tricks.

So PLEASE everyone - tell Ruud how much you enjoyed the Special Killer X so that I can have one of these regularly

[Andrew]

So PLEASE everyone - tell Ruud how much you enjoyed the Special Killer X so that I can have one of these regularly

I'll add my support to that. I came late to SKX1 and only solved it in December. I've only just found time to check through my walkthrough which I'm posting now because it has some interesting moves in r78. I also tried to make use of the diagonals as early as I could although I didn't find any clever moves using them, only eliminations.

Many thanks to Ed for reviewing my walkthrough. I've added his comments to steps 22, 27 and 28 which provide useful insight into quicker routes to a solution and included his addition to step 24.

I'm also posting my walkthrough for SKX3 today.


1. R9C67 = {89}, locked for R9

2. 45 rule on N8 4 innies = 29 = {5789} -> R7C46 = {5789}, R9C34 = [27/45]

3. 16(5) cage in N8 = {12346}

4. 31(5) cage in N5 contains 9, locked for N5

5. 22(3) cage in N3 = 9{58/67}, 9 locked for N3

6. 13(4) cage in N2356 contains 1, min of any 3 cells = 6 so max remaining cell 7, no 8,9

7. R67C9 = {12}, locked for C9

8. 10(3) cage in N1, no 8,9

9. 11(3) cage in N7, no 9

10. 45 rule on N9 3 innies = 13 -> R7C7 = {34} (cannot be 2 which would clash with 2,9 in the other two innies)

11. 45 rule on N7 3 innies = 19 -> R7C13 = 15 or 17 = {69/78} or {89} [8/9], R7C46 = [5/7,8/9 because of R9C4 and R9C6 respectively], killer pair 8/9 in R7C1346 for R7

12. R67C1 = {69/78}

13. R7C34 min 11 -> R6C34 max 7, no 7,8,9 in R6C34, R6C34 min 4 (cannot be {12} because of R6C9) -> R7C34 max 14 -> R7C4 = {578}, 9 in N8 locked in R79C6, locked for C6

14. R34C9 = [59]/{68}, 45 rule on N3 3 innies = 10 -> R34C9 = [59/68], R1C67 = {14/23} -> R3C7 = {1234}

15. 22(3) cage in N3 = 9{58/67}, killer pair 5/6 with R3C9 for N3

16. 13(3) cage in N3 must contain 7 or 8 -> {247} or 8{14/23} -> R1C8 = {12}

17. R13C7 = {13} or {14/23}, R7C7 = {34}, killer pair 3/4 in R137C7 for C7

18. R1C67 = {14/23}, R1C8 = {12}, killer pair 1/2 in R1C678 for R1

19. R34C1 = {15/24}

20. R1C34 = {59/68}

21. 45 rule on N1 2 outies R1C4 + R4C1 – 2 = 1 innie R3C3, min R1C4 + R4C1 = 6 -> min R3C3 = 4

22. 45 rule on N6 4 innies = 21, max R4C79 + R6C9 = 18 -> min R6C7 = 3, min R6C7 actually 5 because 3,4 blocked in C7, max R4C9 + R6C79 = 19 -> min R4C7 = 2 but cannot have two 2s in N6 so min R4C7 = 5
[Ed. FYI - can go further here. combo's = [91{56}/81{57}/82{56}] -> R6C7 = {567} and 5 locked for C7, N6 and the 4 innies = [8/9, 6/7..] -> 24(5) = 34{179/269/278} ({12489/12678}blocked by the 4 innies)]

23. 24(5) cage in N6 contains 1 or 2

24. 13(4) cage in N2356, min R4C7 + any two others = 8, max R3C6 or R4C6 = 5
but this would mean two 5s in 13(4) since min R4C7 = 5 -> max R3C6 or R4C6 = 4 [Thanks Ed. I’d missed that last point]

25. 45 rule on C9 2 outies R19C8 + 1 = 1 innie R5C9 -> min R5C9 = 4

26. 45 rule on R9 2 outies R8C19 – 10 = 1 innie R9C5 -> min R8C19 = 11, no 1 in R8C1

27. 23(4) cage in N5689, max R6C7 + R7C67 = 21 -> min R6C6 = 2
[If Ed’s mods to step 22 had been used, this max = {479} = 20 -> min R6C6 = 3 = {3479/3569/3578/4568} = [8/9..] in C6 -> killer pair with R9C6]

28. R9C7 = {89} so R8C789 must have 8 or 9 -> R8C123 must have 8 or 9 -> R7C13 = {69/78}(can’t be {89} which would clash with R8C123) -> R9C3 = 4 (step 11), R9C4 = 5, R1C34 = [59]/{68}
[Ed. One thing you missed from this is that R7C4 = {78} -> R7C13 {78} is blocked -> = {69} only in R7C13. You pick this up in the next couple of steps so no problem]

29. R7C1346 = naked quad {6789}, 6,7 locked for R7

30. 7 in N8 locked in R7C46, locked for R7 -> R7C13 = {69}, locked for R7 and N7, R6C1 = {69}, locked for C1

31. R7C46 = {78} -> R9C6 = 9, R9C7 = 8

32. No 8 in R8C1 because R9C12 cannot total 7 -> 8 in N7 locked in R8C23 -> 11(3) cage = {128}, R7C2 = {12}, R8C23 = 8{12}, R9C12 = {37}, locked for R9, R8C1 = 5

33. R9C9 = 6 (naked single), locked for D\, R3C9 = 5, R4C9 = 9, R13C7 = {14/23} (from step 14), R34C1 = {24}, locked for C1

34. 22(3) cage in N3 = {679}, locked for N3, 8 in N3 locked in R12C9, locked for C9

35. Only valid combination for 14(3) cage in N9 = [716] -> R67C9 = [12], R9C5 = 2

36. R5C9 = 4, R12C9 = {38}, R1C8 = 2

37. R1C67 = {14}, R3C7 = {14}, locked for R1 and C7

38. R5C7 = 2 (hidden single in C7 and N6)

39. R7C7 = 3 (naked single), R7C8 = 5 (hidden single in R7 and N9) -> R8C78 = [94], 3,4 locked for D\

40. R7C5 = 4 (hidden single in N8) -> R8C456 = {136}, R8C23 = {28}, R7C2 = 1

41. 24(5) cage in N6, 3 remaining cells R456C8 = 18 = {378}(subtraction combo), locked for C8 and N6 -> R23C8 = {69}, R2C7 = 7, R46C7 = {56}

42. 45 rule on N1 3 innies R1C3 + R3C13 = 18, no valid combination with R3C1 = 2 -> R3C1 = 4, R13C3 = [59]/[68], R1C34 = [59]/[68], R4C1 = 2

43. R3C7 = 1, locked for D/ and 13(4) cage, R1C7 = 4, R1C6 = 1

44. R4C4 = 1 (hidden single on D\), R8C5 = 1 (hidden single in R8, C5 and N8)

45. 2 in N1 locked in 10(3) cage = 2{17/35}, no 6

46. 2 on D\ can only be in R2C2 or R6C6 -> no 2 in R2C6

47. R5C5 = 5 (hidden single on D/), locked for D\

48. R2C2 = 2 (naked single), R8C23 = [82], 2 locked for D\, 8 locked for D/

49. R2C8 + R7C3 = {69}, killer pair on D/

50. R9C1 = 7 (hidden single on D/), R9C2 = 3

and the rest is naked and hidden singles and simple elimination remembering to use the diagonals and cage sums