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 Post subject: CNC 50 double X
PostPosted: Thu Oct 10, 2013 1:30 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
CNC 50 Double X

All four diagonals have one common number missing.

r3c6 is in three cages and r7c4 is in three cages.

I hope that you will find it hard.



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 Post subject: Re: CNC 50 double X
PostPosted: Mon Oct 21, 2013 4:05 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
HATMAN wrote:
I hope that you will find it hard.
Yes, it was. After a fairly easy start it proved to be very difficult; I wasn't sure whether I'd manage to finish it.

There's a bit of a shortcut:
See my comment after step 19. If I'd used that it would have made my solving path a bit shorter and slightly easier, but it would still have been a very hard puzzle. I could be wrong, but my feeling is that using the "rotational symmetry" shortcut doesn't eliminate 5 from R2C7 and R8C3; thus more work would still be needed to demonstrate that 5 is the number missing from all the diagonals.

Here is my walkthrough for CNC 50 Double X:
Red border cages contain consecutive numbers, not necessarily in order; black border cages cannot contain consecutive numbers.

1. 12(3) killer cage at R1C4 = {138/147/246} (other combinations contain consecutive numbers), no 5,9
1a. 19(3) killer cage at R2C5 = {379/469} (other combinations contain consecutive numbers) -> R3C5 = 9
1b. 12(3) and 19(3) killer cages overlap -> R2C5 + R3C6 = {46} (only combination possible in both cages), locked for N2, R1C4 = 2 (cage sum)
1c. Naked pair {46} in R2C5 + R3C6, CPE no 4,6 in R2C7 + R4C5 using D/ at R1C8
1d. Remban group at R2C7 = {234/345/567/678} (cannot be {456} because 4,6 only in R3C6, other consecutive groups don’t contain 4 or 6), no 1,9

2. 11(3) killer cage at R7C4 = {137/146} (other combinations contain consecutive numbers), no 2,5,8,9, 1 locked for N8
2a. R7C4 + R8C5 are common cages for 11(3) and 18(3) killer cages at R7C4 -> R9C6 = R7C5 + 7 (difference of cage totals) -> R7C5 = 1, R9C6 = 8
2b. 18(3) killer cage at R7C4 = {468} (cannot be {378} which contains consecutive numbers) -> R7C4 + R8C5 = {46}, locked for N8
2c. Naked pair {46} in R28C5, locked for C5
2d. Naked pair {46} in R7C4 + R8C5, CPE no 4,6 in R8C3 using D/ at R2C9
2e. Remban group at R6C5 = {234/345/567/678} (cannot be {456} because 4,6 only in R7C4, other consecutive groups don’t contain 4 or 6), no 1,9

3. No 1,9 in R5C5 -> one Remban group in R5 must be {1234} and the other Remban group in R5 must be {6789} -> R5C5 = 5
3a. Remban group at R4C6 = {345/456/567}, no 1,2,8,9

4. 12(3) killer cage at R1C9 = {138/147/246} (other combinations contain consecutive numbers), no 5,9

5. 18(3) killer cage at R7C1 = {279/369/468} (other combinations contain consecutive numbers), no 1,5

6. 45 rule on C1 2 innies R56C1 = 7 = {16/34}/[25], no 7,8,9, no 2 in R6C1

7. 45 rule on C9 2 innies R45C9 = 13 = {49/67}/[58], no 1,2,3, no 8 in R4C9

8. 18(3) killer cage at R7C1 (step 5) = {279/369/468}, R56C1 (step 6) = {16/34}/[25] -> combined cage = {279}{16}/{279}{34}/{369}[25]/{468}[25], 2 locked for C1
8a. 20(4) killer cage at R1C1 must contain 1 (cannot be {3579} which clashes with combined cage), locked for C1, clean-up: no 6 in R56C1 (step 6)

9. R5C1 = {234} -> Remban group at R5C1 = {1234}, locked for R5, 2 also locked for N4, clean-up: no 9 in R4C9 (step 7)
9a. Naked quad {6789} in Remban group at R5C6, 8 locked for N6

10. 12(3) killer cage at R1C9 (step 4) = {138/147/246}, R45C9 (step 7) = {49/67}/[58] -> combined cage = {138}{49}/{138}{67}/{147}[58]/{246}[58], 8 locked for C9
10a. 20(4) killer cage at R6C9 must contain 9 (cannot be {1357} which clashes with combined cage), locked for C9, clean-up: no 4 in R4C9 (step 7)

11. Combined cage in C1 (after step 8a) = {279}{34}/{369}[25]/{468}[25] -> 20(4) killer cage at R1C1 = {1379} (cannot be {1478/1568} which contain consecutive numbers), locked for C1 -> R56C1 = [25], 18(3) killer cage at R7C1 = {468}, locked for N7

12. Combined cage in C9 (after step 10a) = {138}{67}/{147}[58]/{246}[58] -> 20(4) killer cage at R6C9 = {1379} (cannot be {2369/2459} which contain consecutive numbers), locked for C9 -> R45C9 = [58], 12(3) killer cage at R1C9 = {246}, locked for N3

[I was stuck after my original step 21 (now step 19), so I had another look at the puzzle diagram and realised that I’d overlooked the killer cages at R1C3 and R7C7 when setting up my worksheet. They are useful for the next two steps, eliminating more Remban group combinations, so I’ve re-worked from here.]

13. Remban group at R1C2 = {345/456/567} (cannot be {123/789} (which clash with 20(4) killer cage at R1C1, ALS block, cannot be {234/678} because killer cage at R1C3 cannot be {245/568}), no 1,2,8,9, 5 locked for C2 and N2
13a. 2,8 in N1 only in killer cage at R1C3, locked for C3
13b. 2,8 in N1 only in killer cage at R1C3 -> no 1,3,7,9 in killer cage
13c. 1,9 in N1 only in 20(4) killer cage at R1C1, locked for C1
13d. Hidden killer pair 3,7 in 20(4) killer cage at R1C1 and Remban group at R1C2 for N1 -> Remban group must contain one of 3,7 -> Remban group = {345/567}

14. Remban group at R7C8 = {345/456/567} (cannot be {123/789} (which clash with 20(4) killer cage at R6C9, ALS block, cannot be {234/678} because killer cage at R7C7 cannot be {245/568}), no 1,2,8,9, 5 locked for C8 and N9
14a. 2,8 in N9 only in killer cage at R7C7, locked for C7
14b. 2,8 in N9 only in killer cage at R7C7 -> no 1,3,7,9 in killer cage
14c. 1,9 in N9 only in 20(4) killer cage at R6C9, locked for C9
14d. Hidden killer pair 3,7 in 20(4) killer cage at R6C9 and Remban group at R7C8 for N9 -> Remban group must contain one of 3,7 -> Remban group = {345/567}

15. Naked triple {134} in R5C234, CPE no 1,3,4 in R4C3 using D\ at R2C1, no 1,3,4 in R6C3 using D/ at R1C8

16. Naked triple {679} in R5C678, CPE no 6,7,9 in R4C7 using D/ at R2C9, no 6,7,9 in R6C7 using D\ at R1C2

17. Naked pair {46} in R28C5, naked pair {46} in R2C5 + R3C6, naked pair {46} in R7C4 + R8C5 -> R3C6 + R7C4 form naked pair {46}
17a. Naked pair {46} in R3C6 + R7C4, CPE no 4 in R5C4 using D/ at R1C8, no 6 in R5C6 using D/ at R2C9
17b. 4 in R5 only in R5C23, locked for N4
17c. 6 in R5 only in R5C78, locked for N6
[Note that R3C6 + R7C4 = {46} means that 4,6 must be on all the short diagonals.]

18. 4 on D/ at R1C8 only in R3C6 => R2C5 = 6, R8C5 = 4 or in R8C1 -> 4 in R8C15, locked for R8

19. 6 on D/ at R2C9 only in R2C9 or in R7C4 => R8C5 = 4, R2C5 = 6 -> 6 in R2C59, locked for R2

[It seems very likely, because of the symmetry of this puzzle, that the value 5 in R5C5 will be the number missing from all the short diagonals. Also the symmetry and the fact that all the placements and cages reduced to single combinations remind me of the “rotational symmetry” of Assassin 123 and its variants where corresponding cell pairs such as R1C4 and R9C6, R3C5 and R7C5, must total 10. This would require the central Remban group at R4C6 to be {456}, not {345} or {567}, with one of the Remban groups at R2C7 and R6C5 {234} and the other one {678}. However I can’t see any way to demonstrate any of this, apart from udosuk’s explanation in the Assassin 123 thread.]

20. If 5 is on all the short diagonals then R2C7 and R8C3 must contain 5 => 5 in N1 in R13C2, 5 in N9 in R79C8, locked for D\ at R1C2 and D\ at R2C9
or 5 not on short diagonals
-> no 5 in R3C4 and R7C6
[Taking this a bit further]
20a. If 5 is on all the short diagonals then R2C7 = 5 and R3C2 = 5
or 5 not on short diagonals => R2C2 = 5 (hidden single in C2), R3C7 = 5 (hidden single in R3)
-> no 5 in R1C27
20b. If 5 is on all the short diagonals then R7C8 = 5 and R8C3 = 5
or 5 not on short diagonals => R7C3 = 5 (hidden single in R7) and R8C8 = 5 (hidden single in C8)
-> no 5 in R9C38
20c. R1C6 = 5 (hidden single in R1)
20d. R9C4 = 5 (hidden single in R9)

21. Remban group at R2C7 (step 1d) = {234/345/567/678}, Remban group at R6C5 (step 2e) = {234/345/567/678}
21a. If 5 is on all the short diagonals then R2C7 and R8C3 must contain 5 => R46C5 = {37} (from Remban groups) => R5C46 = [19] => R5C23 = {34}, R5C67 = {67} => R4C1 = 7, R6C9 = 3
or 5 not on short diagonals => R2C7 = {37}, R8C3 = {37} => no 3,7 in R46C5 (because Remban groups only contain one of 3,7)
-> no 7 in R4C5, no 3 in R6C5

22. Consider placements for R4C1
R4C1 = 3 => 3 in N1 only in Remban group at R1C2 = {345}, locked for N1, naked triple {268} in killer group at R1C3, locked for C3 => naked pair {79} in R46C3, locked for N4
or R4C1 = 7
-> no 7 in R46C2

23. Consider placements for R6C9
R6C9 = 3
or R6C9 = 7 => 7 in N9 only in Remban group at R7C8 = {567}, locked for N9, naked triple {248} in killer group at R7C7, locked for C7 => naked pair {13} in R46C7, locked for N6
-> no 3 in R46C8

24. Remban group at R6C5 (step 2e) = {234/567/678} (cannot be {345} because 3,5 only in R8C3)
24a. Consider combinations for Remban group at R2C7 (step 1d) = {234/345/678} (cannot be {567} because 5,7 only in R2C7)
Remban group = {234/345} => 3 in R2C7 + R4C5
or Remban group = {678}, R3C6 = 6, R2C5 = 4, R8C5 = 6, R7C4 = 4 => R8C3 = 3
-> 3 in R2C7 or R4C5 or R8C3, CPE no 3 in R4C7 using D/ at R2C9, no 3 in R7C2 using D/ at R1C8
24b. 3 in N6 only in R6C79, locked for R6
24c. 3 in N6 only in R6C79, CPE no 3 in R8C9 using D\ at R1C2

25. Remban group at R2C7 (step 24a) = {234/345/678}
25a. Consider combinations for Remban group at R6C5 (step 24) = {234/567/678}
Remban group = {234}, R7C4 = 4, R8C5 = 6, R2C5 = 4, R3C6 = 6 => R2C7 = 7
or Remban group = {567/678} => 7 in R6C5 + R8C3
-> 7 in R2C7 or R6C5 or R8C3, CPE no 7 in R6C3 using D/ at R1C8, no 7 in R3C8 using D/ at R2C9
25b. 7 in N4 only in R4C13, locked for R4
25c. 7 in N4 only in R4C13, CPE no 7 in R2C1 using D\ at R2C1

[Note than steps 24a and 25a mean that 3,7 are also on all the short diagonals.]

26. Remban group at R4C6 (step 3a) = {345/456/567}
26a. Consider placements for R4C6 = {346}
R4C6 = 3 => R6C4 = 4 (Remban group)
or R4C6 = {46}, naked pair {46} in R34C6, locked for C6
-> no 4 in R6C6
26b. Consider placements for R6C4 = {467}
R6C4 = {46}, naked pair {46} in R67C4, locked for C4
or R6C4 = 7 => R4C6 = 6 (Remban group)
-> no 6 in R4C4

27. Remban group at R1C2 (step 13d) = {345/567}
27a. Consider combinations for killer cage at R1C3 (step 13b) = {248/268}
Killer cage = {248} => Remban group = {567}, locked for N1 => R4C1 = 7 (hidden single in C1) => R46C3 = {69}
or Killer cage = {268} => R6C3 = 9, R4C3 = 7
-> R46C3 = {69}/[79], 9 locked for C3 and N4

28. Remban group at R7C8 (step 14d) = {345/567}
28a. Consider combinations for killer cage at R7C7 (step 14b) = {248/268}
Killer cage = {248} => R46C7 = [13]
or Killer cage = {268} => Remban group = {345}, locked for N9 => R6C9 = 3 (hidden single in C9) => R46C7 = {14}
-> R46C7 = [13]/{14}, 1 locked for C7 and N6

29. Remban group at R1C2 (step 13d) = {345/567}, Remban group at R2C7 (step 24a) = {234/345/678}
29a. Consider combinations for killer cage at R1C3 (step 13b) = {248/268}
Killer cage = {248} => Remban group at R1C2 = {567}
or Killer cage = {268} => Remban group at R1C2 = {345}, locked for C2 => R5C2 = 1, R5C4 = 3 => Remban group at R2C7 = {678} => R3C6 = 6
-> no 6 in R3C3, no 3 in R3C2 using D\ at R2C1
29b. Killer cage = {248} => Remban group at R1C2 = {567}
or Killer cage = {268} => R1C3 = 6, R1C9 = 4
-> no 4 in R1C2
29c. 3 of Remban group at R1C2 = {345} must be in R1C2 -> no 3 in R2C2

30. Remban group at R6C5 (step 24) = {234/567/678}, Remban group at R7C8 (step 14d) = {345/567}
30a. Consider combinations for killer cage at R7C7 (step 14b) = {248/268}
Killer cage = {248} => Remban group at R7C8 = {567}, locked for C8 => R5C8 = 9, R5C6 = 7 => Remban group at R6C5 = {234} => R7C4 = 4
or Killer cage = {268} => Remban group at R7C8 = {345}
-> no 4 in R7C7, no 7 in R7C8 using D\ at R1C2
30b. Killer cage = {248} => R9C7 = 4, R9C1 = 6
or Killer cage = {268} => Remban group at R7C8 = {345}
-> no 6 in R9C8
30c. 7 of Remban group at R7C8 = {567} must be in R9C8 -> no 7 in R8C8

[Without invoking the comments I made after step 19, the only way forward I can see from here is to use contradiction moves.]

31. Remban group at R4C6 (step 3a) = {345/456/567}, Remban group at R2C7 (step 24a) = {234/345/678}, Remban group at R6C5 (step 24) = {234/567/678}
31a. Remban group at R4C6 = {345} => R4C6 = 3, R6C4 = 4, R36C6 = [46] (hidden singles in C6) => Remban group at R2C7 = {234} => R2C7 = 3, R4C5 = 2, R46C7 = [41], R4C8 = 9, R5C6 = 9 (hidden single in R5), R6C5 = 7 (hidden single in N5), R7C4 = 6 => R8C3 = 5 (Remban group at R6C5) => 5 on D/ at R2C9 but no 5 on D/ at R1C8 (all short diagonals but include the same 8 numbers)
-> no 3 in R4C6
31b. Remban group at R4C6 = {567} => R4C6 = 6, R6C4 = 7, R47C4 = [46] (hidden singles in C4) => Remban group at R6C5 = {678} => R6C5 = 8, R8C3 = 7, R46C3 = [96], R6C2 = 1, R5C4 = 1 (hidden single in R5), R4C5 = 3 (hidden single in N5), R3C6 = 4 => R2C7 = 5 (Remban group at R2C7) => 5 of D/ at R1C8 but no 5 on D/ at R2C9 (all short diagonals but include the same 8 numbers)
-> no 7 in R6C4
31c. Naked pair {46} in R4C6 + R6C4, locked for N5

32. Remban group at R2C7 (step 24a) = {234/345/678}, Remban group at R6C5 (step 24) = {234/567/678}
32a. Remban group = {345} => R2C7 = 5, R3C6 = 4, R4C5 = 3, placed for D\ at R1C2, R4C6 = 6, R6C4 = 4, R6C7 = 1, placed for D\ at R1C2, R7C4 = 6, R8C3 = 5 (for diagonal because 5 in R2C7), R6C5 = 7(Remban group), R6C2 = 8 (hidden single in R6), R4C4 = 8 (hidden single in R4) => R3C4 = 7, R5C6 = 9 and R6C7 = 1 clash with R8C9 on D\ at R1C2
-> Remban group = {234/678}, no 5 in R2C7, no 3 in R4C5
32b. No 5 in D/ at R1C8 -> no 5 in the other diagonals -> no 5 in R3C2 + R7C8 + R8C3
32c. R2C2 = 5, R3C7 = 5, R7C3 = 5, R8C8 = 5 (hidden singles in R2, R3, R7 and R8)

33. Remban group at R6C5 (step 24) = {234/678}
33a. R8C3 = {37} -> no 7 in R6C5
33b. Naked pair {28} in R46C5, locked for C5 and N5
33c. 3 in N5 only in R45C4, locked for C4
33d. 7 in N5 only in R56C6, locked for C6
33e. 8 in C4 only in R23C4, CPE no 8 in R2C3 using D\ at R1C1
33f. 2 in C6 only in R78C6, CPE no 2 in R7C8 using D\ at R2C1

[Even after eliminating 5 from all the diagonals and showing symmetry in the Remban group at R4C6, there still seems quite a lot to do.]

34. Consider combinations for Remban group at R2C7 (step 32a) = {234/678}
Remban group = {234} => R2C7 = 3, R5C4 = 1
or Remban group = {678} => R3C6 = 6, R4C5 = 8, R6C3 = 9, R6C5 = 2, R4C67 = [41], R4C8 = 2 (hidden single in R4), R6C2 = 8 (hidden single in R6), 1 in N4 only in R5C23, locked for R5 => R5C4 = 3, 9 in N6 only in R5C78, locked for R5 => R5C6 = 7 => R3C4 = 1
-> 1 in R35C4, locked for C4, 3 in R2C7 + R5C4, locked for D/ at R1C8, 3 in R2C7 + R5C4, CPE no 3 in R2C1 using D\ at R2C1

35. Consider combinations for Remban group at R6C5 (step 33) = {234/678}
Remban group = {234} => R7C4 = 4, R6C5 = 2, R4C7 = 1, R4C5 = 8, R6C34 = [96], R6C2 = 8 (hidden single in R6), R4C8 = 2 (hidden single in R4), 1 in N4 only in R5C23, locked for R5 => R5C4 = 3 => R7C6 = 9, 9 in N6 only in R5C78, locked for R5 => R5C6 = 7
or Remban group = {678} => R8C3 = 7, R5C6 = 9
-> 9 in R57C6, locked for C6, 7 in R5C6 + R8C3, locked for D/ at R2C9, 7 in R5C6 + R8C3, CPE no 7 in R8C9 using D\ at R1C2

36. Remban group at R2C7 (step 32a) = {234/678}
36a. Consider placement for 3 in R4
3 in R4C12 => R4C4 = 9, R5C4 = 3 (hidden single in C4), placed for D/ at R1C8 => Remban group at R2C7 = {678}
or R4C4 = 3 => R5C5 = 1, R6C6 = 7, R6C9 = 3, R5C6 = 9, placed for D\ at R1C2 => R8C9 = 1, placed for D\ at R1C2 => R6C7 = 4, R6C4 = 6, R4C6 = 4, R3C6 = 6 => Remban group at R2C7 = {678}
-> Remban group at R2C7 = {678} -> R2C7 = 7, R3C6 = 6, R4C5 = 8, all placed for D/ at R1C8, 8 also placed for D\ at R1C2, R4C6 = 4, R6C4 = 6
[Cracked, at last.]
36b. R6C5 = 2, R7C4 = 4 -> Remban group at R6C5 (step 33) = {234} -> R8C3 = 3, 2,3,4 placed for D/ at R2C9, 2 also placed for D\ at R2C1
36c. R4C7 = 1, placed for D/ at R2C9, R6C3 = 9, placed for D/ at R1C8, R1C8 = 1, placed for D/ at R1C8, R9C2 = 9, placed for D/ at R2C9, R5C4 = 3, placed for D\ at R2C1, R5C6 = 7, placed for D\ at R1C2, R3C4 = 1, placed for D\ at R1C2, R7C6 = 9, placed for D\ at R2C1

37. R1C1 = 9 (hidden single in C1)

and the rest is naked singles, without using the diagonals.

Solution:
9 6 8 2 7 5 3 1 4
1 5 2 8 4 3 7 9 6
3 7 4 1 9 6 5 8 2
7 3 6 9 8 4 1 2 5
2 4 1 3 5 7 9 6 8
5 8 9 6 2 1 4 7 3
8 2 5 4 1 9 6 3 7
4 1 3 7 6 2 8 5 9
6 9 7 5 3 8 2 4 1

Rating Comment:
I'll rate my walkthrough for CNC 50 Double X at least 1.75. I used a lot of forcing chains, plus some contradiction moves; it felt a lot harder than CNC 27 which I also rated at least 1.75.


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