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 Transformers Killer Goto page Previous  1, 2
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Andrew
Grandmaster

Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

Posted: Fri Oct 05, 2007 2:51 am    Post subject:

I only got round to starting Transformers Lite yesterday and finished it today.

An interesting puzzle with a fairly difficult cage pattern and, of course, a neat trick by Para.

 CathyW wrote: "The Trick" is clever - seems obvious once you know!!

Very true!

 Para wrote: ... and a "Lite" version for during the lunch break

That would be a very long lunch break! It took me a lot longer than that.

There is a lot of similarity between Cathy's solving path and mine. Must be a narrow solving path needing to make good progress in the bottom third before expanding to the rest of the grid.

Even with "The Trick" I rate it a solid 1.25. No way it's a 1.0!

Here is my walkthrough for Transformers Lite

1. R12C5 = {19/28/37/46}, no 5

2. R56C5 = {49/58/67}, no 1,2,3

3. 21(3) cage at R1C4 = {489/579/678}, no 1,2,3

4. 20(3) cage at R1C6 = {389/479/569/578}, no 1,2

5. 19(3) cage at R5C3 = {289/379/469/478/568}, no 1

6. 9(3) cage at R5C7 = {126/135/234}, no 7,8,9

7. 10(3) cage at R6C1 = {127/136/145/235}, no 8,9

8. 21(3) cage at R6C9 = {489/579/678}, no 1,2,3

9. 32(7) cage at R7C3 = 123{4589/4679/5678}

[Extremely unusual! Nothing has been locked by the preliminary steps.]

10. Para said "It contains a little trick, that could help you in solving. There's always ways around these tricks but if you notice it all the better. The trick is something new to me but actually not that hard. But it has all to do with the cage pattern."

I suppose it could be called Extended CPE.

11. 45 rule on R8 2 innies R8C37 = 8 = {17/26/35}

12. Using "The Trick" the pair in R8C37 must be in R9C456 = 6{17/35} -> R8C37 = {17/35}, no 2,6
12a. 6 locked in R9C456 for R9 and N8

13. 45 rule on N8 3 innies R7C456 = 15
13a. R7C37 = 32 – 8 – 15 = 9 = {18/27/36/45}, no 9

14. R7C456 = {249/258/348} (cannot be {159/357} which clash with R8C37 within 32(7) cage), no 1,7

15. Hidden killer triple 1,2,3 in 32(7) cage which must contain all of 1,2,3 (step 9) -> R7C37, R7C456 and R8C37 must each contain one of 1,2,3 -> no 4,5 in R7C37 = {18/27/36}

16. R7C456 = {249/258} (cannot be {348} => R7C37 = {27} so clash with R8C37 within 32(7) cage) = 2{49/58}, no 3, 2 locked for R7 and N8, clean-up: no 7 in R7C37 (step 13a)

17. R8C456 = {178/349/358} (cannot be {457} which clashes with R8C37)

18. Killer pair 1,3 in R8C37 and R8C456, locked for R8

19. Killer pair 3,7 in R8C37 and R8C456, locked for R8

20. 45 rule on R789 2 outies R6C19 = 10 = [19/28/37/46/64], no 5, no 7 in R6C1

21. 45 rule on R1234 2 outies R5C46 = 7 = {16/25/34}, no 7,8,9

22. 45 rule on C12 2 outies R19C3 = 7 = {25/34}/[61], no 7,8,9, no 1 in R1C3

23. 45 rule on C89 2 outies R19C7 = 13 = {49/58}/[67], no 1,2,3, no 7 in R1C7

24. 45 rule on N2 3 outies R2C37 + R4C5 = 21, max R2C37 = 17 -> min R4C5 = 4
24a. Min R4C5 -> max R3C456 = 11, no 9

25. 45 rule on N7 2 innies R78C3 – 6 = 1 outie R6C1
25a. R78C3 cannot total 12 -> no 6 in R6C1, clean-up: no 4 in R6C9 (step 20)
25b. No 3 in R6C1 because R7C12 = {16} clashes with R78C3 = [63/81], clean-up: no 7 in R6C9 (step 20)
25c. If R6C1 = 1 => R78C3 = [61] => R7C12 = {45} => R6C9 = 9 (step 20) => R7C89 = {48/57} clashes with R7C12 -> no 1 in R6C1, no 9 in R6C9

26. 10(3) cage at R6C1 (step 7) = {127/145/235} (cannot be {136} because R6C1 only contains 2,4), no 6
26a. R6C1 = {24} -> no 4 in R7C12
26b. R7C12 = {15/17/35}

27. R6C1 = {24} -> R78C3 = 8,10 (step 25) = [17/35/37], no 6,8 in R7C3, no 1,3 in R8C3, clean-up: no 1,3 in R7C7 (step 13a), no 5,7 in R8C7 (step 12)
27a. 6 in N7 locked in R8C12, locked for R8

28. Naked quad {1357} in R7C123 + R8C3, locked for N7, clean-up: no 2,4,6 in R1C3 (step 22)

29. 21(3) cage at R6C9 (step 8) = {489/678} (cannot be {579} because R6C9 only contains 6,8), no 5 = 6{78}/8{49}/8{67}
29a. 6 in N9 locked in R7C789
29b. R7C789 = {469/678}

30. 9(3) cage at R5C7 (step 6) = {126/135/234}
30a. R56C7 cannot be {13} which would clash with R8C7 -> no 5 in R6C6

31. 2,5 in N9 locked in 23(5) cage = {12578/23459}
31a. R9C456 (step 12) = 6{17/35} -> R9C789 must contain {17/35}
31b. {12578} contains {17} in R9C789 => 5 in R8C89, R8C7 = 3 => R8C3 = 5 (step 12) clashes with 5 in R8C89 -> 23(5) cage cannot be {12578}
31c. 23(5) cage = {23459}, locked for N9 -> R8C7 = 1, R8C3 = 7 (step 12), R7C3 = 3 (locked for D/), R7C7 = 6 (step 13a, locked for D\), R1C3 = 5, R9C3 = 2 (step 22), clean-up: no 7 in R6C5
31d. R9C789 (step 31a) must contain {35}, locked for R9 and N9

32. Naked pair {78} in R7C89, locked for R7, R6C9 = 6, R6C1 = 4 (step 20), clean-up: no 7,9 in R5C5, no 5 in R7C456 (step 16)
32a. 4 in C3 locked in R23C3, locked for N1

33. Naked triple {249} in R7C456, locked for N8

34. Naked triple {358} in R8C456, locked for R8

35. 9(3) cage at R5C7 (step 6) = {135/234} = 3{15/24}
35a. CPE no 3 in R6C8
35b. 4 of {234} in R5C7 -> no 2 in R5C7

36. 19(3) cage at R5C3 (step 5) = {289/568}, no 7
36a. 2 of {289} and 5 of {568} in R6C4 -> no 8,9 in R6C4
36b. 8 locked in R56C3, locked for C3 and N4
36c. 7 in N5 locked in R4C456, locked for R4

37. 1 in C3 locked in R34C3, locked for 17(4) cage -> no 1 in R45C4, clean-up: no 6 in R5C6 (step 21)

38. 1 in N5 locked in R456C6, locked for C6

39. 7 in N6 locked in 17(3) cage = 7{19/28}, no 3,4,5

40. 7 in N4 locked in 15(3) cage = 7{26/35}, no 1,9

41. 17(3) cage in N6 contains 8/9 -> R4C789 must contain 8/9
41a. 13(2) cage in N5 contains 8/9 -> R4C456 must contain 8/9
41b. Killer pair 8,9 in R4C456 and R4C789, locked for R4

42. Hidden pair {89} in R56C3, locked for C3 -> R6C4 = 2 (locked for D/)

43. 9(3) cage at R5C7 (step 35) = {135} (only remaining combination), no 4 -> R6C6 = 1 (locked for D\)
43a. Naked pair {35} in R56C7, locked for C7 and N6, clean-up: no 8 in R1C7 (step 23)

44. R4C7 = 2 (hidden single in C7), clean-up: no 8 in 17(3) cage in N6 (step 39)
44a. Naked triple {179} in 17(3) cage in N6, locked for N6
44b. Naked pair {48} in R4C89, locked for R4 and 17(4) cage at R2C9
[Guess I should have done more on the 17(4) cage at this stage! I was just thinking ahead to the next two steps.]

45. Naked pair {49} in R19C7, locked for C7
45a. Naked pair {78} in R23C7, locked for N3

46. R3C3 = 4 (naked single, locked for D\), R2C3 = 6, R4C3 = 1, clean-up: no 4 in R1C5, no 4,5,9 in R12C4 (step 3), no 9 in R6C5

47. Naked pair {78} in R12C4, locked for C4 and N2, clean-up: no 2,3 in R12C5

48. R34C3 = [41] -> R45C4 = 12 = [93] (9 locked for D\) -> R8C8 = 2 (locked for D\), R56C7 = [53], R5C6 = 4

49. R5C5 = 8 (naked single, locked for D/ and D\) -> R6C5 = 5, R8C456 = [538], R56C3 = [98], R6C2 = 7, R6C8 = 9

50. R2C2 = 3 (locked for locked for D\) -> R1C1 = 7, R9C9 = 5, R12C4 = [87], R23C7 = [87]
50a. R9C8 = 3 (hidden single in N9)
50b. R3C9 = 3 (hidden single in C9)
50c. R1C6 = 3 (hidden single in C6), R2C6 = 9, clean-up: no 1 in R12C5 = [64], R4C5 = 7, R4C6 = 6 (locked for D/), R7C456 = [492]

51. R9C1 = 9, R8C2 = 4 (both locked for D/), R1C9 = 1, R2C8 = 5

and the rest is naked singles

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

Not sure whether I'll try the harder version. I may just work through Mike's walkthrough.
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

Posted: Sat Oct 27, 2007 12:59 am    Post subject:

 Andrew wrote: [Extremely unusual! Nothing has been locked by the preliminary steps.]

I know a bit late a reply. But it gary mentioning the only 3 cell cages in a killer got me thinking about it again.
As Andrew noted there were no digits locked by the preliminaries. But i'd like to take this one step further. I have always wondered if it is possible to create a killer where the preliminaries don't make one single elimination. Or to say it in a different way: when you open marks in SumoCue it shows all candidates still available. That would mean of course only cages of size 3-7 and some very crafty use of innies and outies. Also it shouldn't be a zero killer, otherwise you could just leave out all the 9's or 1's for example or create a bunch of zero 24(3) or 6(3) cage within houses.

greetings

Para
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