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PostPosted: Fri Aug 06, 2010 11:26 am 
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Expert
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Joined: Sat Jan 17, 2009 8:30 am
Posts: 118
Location: france
Hi,

I wanted to produce a killer puzzle with only two cells cages. Of course, blank cells are necessary, since a grid contains an odd number of cells. Solving this kind of puzzle is really different from classical ones, so I post this one in "other variants" thread. It can be solves without TE or chainy steps, but it took me several hours to produce a walkthrough avoiding them. Enjoy.

Thanks to Ed for helping me to producing a picture : I had some troubles with my own computer.

Image

3x3::k:1536:1:1794:3:2308:5:3846:7:2312:1536:3082:1794:2828:2308:3854:3846:1296:2312:18:3082:20:2828:22:3854:24:1296:26:3355:3355:29:3358:3358:32:1569:1569:35:36:2597:2597:39:1832:1832:42:3115:3115:2349:2349:47:3376:3376:50:2867:2867:53:3126:55:3128:57:3898:59:1852:61:2110:3126:3136:3128:1602:3898:2884:1852:2118:2110:72:3136:74:1602:76:2884:78:2118:80:

Solution :

Hidden Text:
276315984
431789625
598426137
859673241
164852793
723941568
317298456
985164372
642537819


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PostPosted: Sat Aug 07, 2010 6:55 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks manu for another fun puzzle!

It is, of course, be possible to construct cage patterns with 40 2-cell cages, thus giving the maximum possible number of Prelims, but maybe that wouldn't lead to a good puzzle.

I liked the cage pattern for this puzzle, nicely symmetrical and plenty of empty space. That allowed me to colour it using only two colours, the first time I've ever done that.

Image

My solving path was a bit more difficult than it needed to be, as I've explained in the comment in step 13. However I didn't want to rewrite my walkthrough completely because that would have omitted the fairly unusual step 13a.

My key solving techniques were:
locking-out cages and locking cages; there's also the interesting step 13a where I used a hidden cage to obtain an innie-outie difference.

Here is my walkthrough:
Prelims (all cells)

a) R12C1 = {15/24}
b) R12C3 = {16/25/34}, no 7,8,9
c) R12C5 = {18/27/36/45}, no 9
d) R12C7 = {69/78}
e) R12C9 = {18/27/36/45}, no 9
f) R23C2 = {39/48/57}, no 1,2,6
g) R23C4 = {29/38/47/56}, no 1
h) R23C6 = {69/78}
i) R23C8 = {14/23}
j) R4C12 = {49/58/67}, no 1,2,3
k) R4C45 = {49/58/67}, no 1,2,3
l) R4C78 = {15/24}
m) R5C23 = {19/28/37/46}, no 5
n) R5C45 = {16/25/34}, no 7,8,9
o) R5C89 = {39/48/57}, no 1,2,6
p) R6C12 = {18/27/36/45}, no 9
q) R6C45 = {49/58/67}, no 1,2,3
r) R6C78 = {29/38/47/56}, no 1
s) R78C1 = {39/48/57}, no 1,2,6
t) R78C3 = {39/48/57}, no 1,2,6
u) R78C5 = {69/78}
v) R78C7 = {16/25/34}, no 7,8,9
w) R78C9 = {17/26/35}, no 4,8,9
x) R89C2 = {39/48/57}, no 1,2,6
y) R89C4 = {15/24}
z) R89C6 = {29/38/47/56}, no 1
aa) R89C8 = {17/26/35}, no 4,8,9

1. Naked sextet {345789} in R78C1, R78C3 and R89C2, locked for N7

2. 8,9 in N9 only in R7C8 + R9C79
2a. 45 rule on N9 3 innies R7C8 + R9C79 = 22 = {589} (only combination containing both of 8,9), 5 locked for N9, clean-up: no 2 in R78C7, no 3 in R78C9, no 3 in R89C8
2b. R78C7 = {34} (hidden pair in N7), locked for C7, clean-up: no 2 in R4C8, no 7,8 in R6C8
2c. Killer pair 1,2 in R23C8 and R89C8, locked for C8, clean-up: no 5 in R4C7, no 9 in R6C7
2d. R12C9 = {18/36/45} (cannot be {27} which clashes with R78C9), no 2,7

3. Killer triple 4,5,6 in R4C12, R4C45 and R4C8, locked for R4
3a. Killer triple 4,5,6 in R4C45, R5C56 and R6C45, locked for N5

4. 45 rule on N1 3 innies R1C2 + R3C13 = 20 = {389/479/569/578}, no 1,2
[I could have used {479} clashes with R23C2 but that isn’t necessary.]

5. Hidden killer pair 1,2 in R12C1 and R12C3 for N1, R12C1 contains one of 1,2 -> R12C3 must contain one of 1,2 -> R12C3 = {16/25}, no 3,4
5a. R12C1 = {24} (cannot be {15} which clashes with R12C3), locked for C1 and N1, clean-up: no 5 in R12C3, no 8 in R23C2, no 9 in R4C2, no 5,7 in R6C2, no 8 in R78C1
5b. Naked pair {16} in R12C3, locked for C3 and N1 -> R9C3 = 2, clean-up: no 4,8,9 in R5C2, no 4 in R8C4, no 9 in R8C6, no 6 in R8C8

6. 45 rule on N2 3 innies R1C46 + R3C5 = 10 = {127/136/145/235}, no 8,9

7. Hidden killer pair 7,9 in R12C7 and R1C8 + R3C79 for N3, R12C7 contains one of 7,9 -> R1C8 + R3C79 must contain one of 7,9
7a. 45 rule on N3 3 innies R1C8 + R3C79 = 16 = {169/178/259/457} (cannot be {268/358} which don’t contain 7 or 9, cannot be {349} which clashes with R23C8, cannot be {367} which clashes with R12C7), no 3

8. 2 in N4 only in R5C23 = {28} or R6C12 = {27} -> R5C23 cannot be {37}, R6C12 cannot be {18} (locking out cages)
8a. 1 in N4 only in R5C12, locked for R5, clean-up: no 6 in R5C56
8b. 1 in N5 only in R46C6, locked for C6

9. 6 in R4 only in R4C12 or R4C45 -> either R4C12 or R4C45 = {67}, 7 locked for N5
9a. 6 in N5 only in R4C45 or R6C45 -> either R4C45 or R6C45 = {67}, 7 locked for N5

10. 45 rule on N4 3 innies R46C3 + R5C1 = 13 = {139/148/346} (cannot be {157} because R4C3 only contains 3,8,9), no 5,7
10a. 1,6 only in R5C1 -> R5C1 = {16}
10b. 4 of {148} only in R6C3 -> no 8 in R6C3
10c. Naked pair {16} in R59C1, locked for C1, clean-up: no 7 in R4C2, no 3 in R6C2

11. 45 rule on R5 3 innies R5C147 = 16 = {169/178/268/367} (cannot be {259/349/358/457} because R5C1 only contains 1,6), no 5
11a. 6 of {169} must be in R5C7 -> no 9 in R5C7

12. 1 in R6 only in R6C69
12a. 45 rule on R6 3 innies R6C369 = 12 = {129/138/147} (cannot be {156} because R6C3 only contains 3,4,9), no 5,6
12b. R6C3 = {349} -> no 3,4,9 in R6C69

13. 45 rule on C3 4 innies R3456C3 = 24
[With hindsight I could have got more out of this hidden cage, limiting it to {3489} because 5,7 only in R3C3. However at the time I’d overlooked something in an earlier step which I’ve since corrected. Continuing directly with step 14 would now have been simpler but this way is more interesting.]
13a. 45 rule on N4 (using hidden cage R3456C3 = 24), 1 outie R3C3 = 2 innies R5C12 + 1
13b. R5C12 must contain 1 -> R5C12 = {16} (cannot be [12] because no 4 in R3C3), locked for R5 and N4, clean-up: no 7 in R4C1, no 8 in R5C3, no 3 in R6C1
13c. R5C12 = {16} = 7 -> R3C3 = 8, clean-up: no 3 in R2C4, no 7 in R2C6, no 4 in R78C3

14. R78C3 = {57} (hidden pair in C3), locked for N7
14a. Naked pair {39} in R78C1, locked for C1 and N7, clean-up: no 4 in R4C2
14b. Naked pair {48} in R89C2, locked for C2 -> R4C12 = [85], R4C8 = 4, R4C7 = 2, R6C12 = [72], R3C1 = 5, clean-up: no 7 in R23C2, no 6 in R2C4, no 1 in R23C8, no 9 in R4C45, no 8 in R5C89, no 6 in R6C45, no 9 in R6C8
14c. R1C2 = 7 (hidden single in N1), clean-up: no 2 in R2C5, no 8 in R2C7

15. 45 rule on N6 3 innies R46C9 + R5C7 = 16 = {178} (only remaining combination) -> R4C9 = 1, R6C9 = 8, R5C7 = 7, R6C6 = 1, clean-up: no 8 in R1C7, no 5 in R5C89, no 5 in R6C45, no 3 in R6C8, no 7 in R78C9

16. Naked pair {23} in R23C8, locked for C8 and N3 -> R5C89 = [93], R5C3 = 4, R5C2 = 6, R59C1 = [16], R7C2 = 1, clean-up: no 6 in R12C9, no 5 in R8C6

17. Naked pair {69} in R12C7, locked for C7 and N3 -> R3C7 = 1, R6C78 = [56], R9C7 = 8, R7C8 = 5, R9C9 = 9, R1C8 = 8, clean-up: no 1 in R2C5, no 2,3 in R8C6
17a. R3C9 = 7 (hidden single in N3), clean-up: no 4 in R2C4, no 8 in R2C6

18. Naked pair {69} in R2C67, locked for R2 -> R12C3 = [61], R12C7 = [96], R23C6 = [96], R23C2 = [39], R23C8 = [23], R12C1 = [24], R12C9 = [45]

19. R12C5 = {18} (only remaining combination) = [18], R2C4 = 7, R3C4 = 4

and the rest is naked singles.


Last edited by Andrew on Sat Aug 07, 2010 10:15 pm, edited 1 time in total.

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PostPosted: Sat Aug 07, 2010 8:23 pm 
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Addict
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Joined: Fri May 30, 2008 9:16 am
Posts: 27
All you need to do is work out what the 11 in nonet 6 is(which takes about 5 mins) and the puzzle falls over(given the info you can deduce for nonets 1,7 and 9.But is this an acceptable way to proceed?I don't really know.Anyway it's what I did.


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PostPosted: Mon Aug 09, 2010 6:28 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
Goooders

I tried your approach and it works as you say. However most members of the forum would consider my solution T&E - perhaps yours was neater?

Maurice


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