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 Post subject: LC Killer 3
PostPosted: Fri Aug 23, 2019 6:52 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
Low Clue Killer 3

All my own work (with a little help from JSudoku).

It is a proper killer so the cages are symmetrical (around the centre), hence the cage structure at say r8c9 matches that at r2c1.

This one felt hard enough and interesting enough, so I did not mess with the cage totals, at first.


Image


OK: here it is with some cage totals removed. Remember the ? represents a digit.

Image


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 Post subject: Re: LC Killer 3
PostPosted: Tue Aug 27, 2019 9:14 pm 
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Thanks! I'll be trying this shortly - after your previous post, I tracked down some easier ones to get the hang of them. I MIGHT now be ready to try yours!


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 Post subject: Re: LC Killer 3
PostPosted: Thu Sep 05, 2019 1:58 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
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Location: Lethbridge, Alberta, Canada
Thanks HATMAN for this nice puzzle!

To anyone who hasn't tried these Low Clue puzzles, I suggest trying Killer Low Clue 1 and 2 (also in this Killer Puzzles forum). They are easier, even though they have ? and ?? in place of numbers as some clues.

While solving this puzzle, it's important to note that the cage pattern is symmetrical about the centre. I only managed to place six cages before I started using this symmetry.

The main part of this puzzle is placing all the cages; once that was done, solving the puzzle was fairly routine. The second version, with one ?? and three ?, will be harder.

Here is my walkthrough for L C Killer 3:
The usual uncaged killer rules, the number is in the top left corner with top dominating left as usual. There are no single cages and no diagonally-connected cages.
There is also the restriction that the cage pattern is symmetrical about the centre.

Prelims, based just on positions of totals.
a) 19(?) cage at R1C2 must contain at least R1C23 + R2C2
b) R12C1 must be 6(2) cage = {15/24}
c) R12C4 must be 5(2) cage = {14/23}
d) R1C56 must be 9(2) cage (cannot be 9(3) cage because 5(2) and 9(3) cages cannot be in the same nonet) = {18/27/36/45}, no 9
e) R9C23 must be 12(2) cage = {39/48/57}, no 1,2,6
f) R89C9 must be 17(2) cage = {89}
Now to use the symmetry of the cages.
g) There’s a 5(2) cage R12C4 -> R89C6 must be 7(2) cage = {16/25/34}, no 7,8,9
h) R9C45 must be 7(2) cage = {16/25/34}, no 7,8,9
i) There’s a 12(2) cage R9C23 -> R1C78 must be 13(2) cage = {49/58/67}, no 1,2,3
j) R9C7 must be in 11(3) cage R8C8 + R9C78 = {128/137/146/236/245}, no 9
k) There’s a 11(3) cage R8C8 + R9C78 -> R1C23 + R2C2 = 19(3) = {289/379/469/478/568}, no 1
l) 8 cage at R4C9 must contain 2 or 3 cells including R5C9 but not R4C8, 9 cage at R5C1 must contain 2 or 3 cells including R6C1 -> from symmetry 8(2) cage R45C9 = {17/26/35}, no 4,8,9, 9(2) cage R56C1 = {18/27/36/45}, no 9
m) From symmetry there must be 16 cage in R3C89 + R45C8 and possibly one more cell, with 20 cage in R56C2 + R7C12 and possibly one more cell
n) From symmetry there must be 26 cage in R1C9 + R2C789 and possibly one more cell, with 16 cage in R8C123 + R9C1 and possibly one more cell
o) 18 cage at R6C4 cannot be R6C456 because R6C6 must be part of 40 cage at R4C4 from symmetry -> R7C4 must be part of 18 cage
p) 14(2) cage at R67C3 = {59/68}
q) From symmetry 11(2) cage at R34C7 = {29/38/47/56}, no 1
r) From symmetry there must be 16(3) cage in R3C6 + R4C56 and 18(3) cage in R6C45 + R7C4, since 18 requires at least 3 cells
s) 40 cage must have 7 cells (cannot have 8 cells for symmetry) in R4C4 + R5C34567 + R6C6

Steps resulting from Prelims
1a. Naked pair {89} in R89C9, locked for C9 and N9
1b. R56C1 = {18/27/36} (cannot be {45} which clashes with R12C1), no 4,5

2. 45 rule on R89 3 innies R8C457 = 20 must contain at least 3 cells, but not more than 3 cells = {479/569/578} (cannot be {389} which clashes with R8C9), no 1,2,3
2a. Killer pair 8,9 in R8C457 and R8C9, locked for R8

3. 45 rule on C123 1 innie R5C3 = 9 (there’s no other position for any innie to C123), clean-up: no 5 in R67C3, no 3 in R9C3
3a. Therefore there must be 23(4) cage in R2C56 + R3C45 and 24(4) cage in R7C56 + R8C45
3b. 45 rule on C789 1 innie R5C7 = 5, clean-up: no 8 in R1C8, no 6 in R34C7, no 3 in R45C9
3c. 45 rule on N2 1 innie R3C6 = 8 -> R4C56 = 8 = {17/26/35}, no 4,9, clean-up: no 1 in R1C56, no 3 in R4C7
3d. 45 rule on N8 1 innie R7C4 = 7 -> R6C45 = 11 = {29/38/56}, no 1,4
3e. Naked pair {68} in R67C3, locked for C3, clean-up: no 4 in R9C2
3f. 40(7) cage at R4C4 must contain 6,7,8 locked for N5
3g. R4C56 = 8 = {35}, locked for R4, 3 locked for N5
3h. Naked pair {29} in R6C45, locked for R6, 2 locked for N5, clean-up: no 7 in R5C1

[The final cages.]
4. 20 cage at R2C3 must contain more than 3 cells because no 8,9 in R234C3 -> R234C3 + R3C2 form 20(4) cage (R4C2 cannot be part of cell because R6C8 cannot be part of 16 cage at R6C7)
4a. This leaves 10(3) cage in R3C1 + R4C12 = {127/136/145} (cannot be {235} because 3,5 only in R3C1), no 8,9
4b. 10(3) cage in R3C1 + R4C12 = {127/136/145}, CPE no 1 in R56C1, clean-up: no 8 in R56C1
4c. From symmetry there must be 16(4) cage in R678C7 + R7C8 and 12(3) cage in R6C89 + R7C9

5. R79C1 = {89} (hidden pair in C1), locked for N7 -> R67C3 = [86], clean-up: no 3,4 in R9C3
5a. Naked pair {57} in R9C23, locked for R9 and N7, clean-up: no 2 in R8C6, no 2 in R9C45
5b. 45 rule on N7 2 innies R7C12 = 11 = [83/92]
5c. 1,4 in N7 only in R8C123, locked for R8, clean-up: no 3,6 in R9C6
5d. R8C457 (step 2) = {569/578}, 5 locked for R8 and N8, clean-up: no 2 in R9C6
5e. Naked quad {1346} in R89C6 + R9C45, locked for N8
5f. 2 in N8 only in R7C56, locked for R7 -> R7C12 = [83], R9C1 = 9, R89C9 = [98]
5g. R8C45 = {58} (hidden pair in N8) -> R8C7 = 7 (step 5d), clean-up: no 6 in R1C8, no 4 in R34C7
5h. 11(3) cage in R8C8 + R9C78 = {236} (hidden triple in N9)
5h. Naked triple {124} in R8C123, 2 locked for R8

6. R8C7 = 7 -> 16(4) cage in R678C7 + R7C8 = {1357} (only possible combination) -> R6C7 = 3, R7C78 = [15], R7C9 = 4 -> R6C89 = 8 = {17}, locked for R6 and N7, R6C1 = 6 -> R5C1 = 3, R6C6 = 4, R9C6 = 1 -> R8C6 = 6, R8C8 = 3, clean-up: no 3,5 in R1C5, no 8 in R1C7, no 8 in R4C7
6a. Naked pair {26} in R45C9, locked for C9 and N6 -> R34C7 = [29], R9C78 = [62], R1C7 = 4 -> R1C8 = 9, clean-up: no 5 in R1C6, no 2 in R2C1, no 1 in R2C4
6b. R45C8 = {48} = 12 -> R3C89 = 4 = [13]
6c. Naked pair {57} in R12C9, 7 locked for C9 and N3 -> R2C78 = [86]
6d. R7C12 = [83], R6C2 = 5 -> R5C2 = 4 (cage sum), R5C8 = 8, R4C4 = 8 (hidden single in N5)

7. 7 in C1 only in 10(3) cage in R3C1 + R4C12 = {127} -> R3C1 = 7, R4C12 = {12}
7a. 5 in C1 only in R12C1 = {15}, locked for C1 and N1 -> R4C12 = [21]
7b. R34C3 = [47] = 11 -> R2C3 + R3C2 = 9 = [36], R1C3 = 2, clean-up: no 7 in R1C56
7c. R1C56 = [63], R1C4 = 1 -> R2C4 = 4

and the rest is naked singles.

Solution, showing the cages:
Attachment:
L C Killer 3.jpg
L C Killer 3.jpg [ 60.49 KiB | Viewed 6963 times ]


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 Post subject: Re: LC Killer 3
PostPosted: Sat Sep 14, 2019 8:36 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
And now I've found time to try the harder version. It took a bit longer but not really harder. Some of my steps have probably been expressed a bit differently; I started again and didn't do any copy and paste from the earlier version.

Here is my walkthrough for the harder version:
The usual uncaged killer rules, the number is in the top left corner with top dominating left as usual. There are no single cages and no diagonally-connected cages. Note each question mark represents a digit so, for example, a double cage with a single ? does not contain a 9.
There is also the restriction that the cage pattern is symmetrical about the centre.

Prelims, based just on positions of totals.
a) 19(?) cage at R1C2 must contain at least R1C23 + R2C2
b) R12C1 must be 6(2) cage = {15/24}
c) R9C23 must be 12(2) cage = {39/48/57}, no 1,2,6
d) 11(?) cage at R8C8 must contain at least R9C8
e) R89C9 must be ??(2) cage
Now to use the symmetry of the cages.
f) There’s a 12(2) cage R9C23 -> R1C78 must be 13(2) cage = {49/58/67}, no 1,2,3
g) ?(?) cage at R1C4 must contain R2C4 but not R1C3, ?(?) cage at R8C6 must therefore contain R9C6 but not R8C7 or R9C5 -> ?(2) cage at R1C4 and ?(2) cage at R8C6, no 9 in R12C4, no 9 in R89C6
h) ?(2) cage in R9C45, no 9 in R9C45
i) ?(2) cage in R9C45 -> 9(2) cage in R1C56 = {18/27/36/45}, no 9
j) 8(?) cage at R4C9 must contain R5C9 but not R4C8, 9(?) cage at R5C1 must contain R6C1 but not R4C1 -> 8(2) cage in R45C9 = {17/26/35}, no 4,8,9, 9(2) cage in R56C1 = {18/27/36/45}, no 9
k) 19(?) cage at R1C2 must include at least R1C3 + R2C2, 11(?) at R8C8 must include R9C8 but not R7C8 -> 19(3) cage in R1C23 + R2C2 = {289/379/469/478/568}, no 1 and 11(3) cage in R8C8 + R9C78 = {128/137/146/236/245}, no 9
l) 26(?) cage at R1C9 must contain at least 4 cells -> 16(?) cage at R8C1 must contain at least R8C123 + R9C1 but not R7C1 -> 26(?) cage must contain at least R1C9 + R2C789 but not R3C9
m) 11(?) cage at R3C7 must include at least R4C7 but not R3C8 or R4C6, 14(?) cage at R6C3 must include R7C3 but not R5C3 -> 11(2) cage R34C7 = {29/38/47/56}, no 1, 14(2) cage = R67C3 = {59/68}
n) 16(?) cage at R3C8 must contain at least R3C89 + R45C8
o) 40(?) cage at R4C4 must be 40(7) cage from symmetry since R5C5 cannot be a separate single cell -> 40(7) cage must contain R6C6 which must be linked to R4C4 through R5C5 with R46C5 and/or R5C5
p) 18(?) cage at R6C4 must contain at least 3 cells including R7C4, 16(?) cage at R3C6 must contain R4C6 but not R2C6 -> 18(3) cage in R6C45 + R7C4, 16(3) cage in R3C6 + R4C56
q) Therefore 40(7) cage must be R4C4 + R5C34567 + R6C6 = {1456789/2356789}
r) 16(?) cage at R6C7 must contain at least R78C6 -> 20(?) cage at R2C3 must contain at least R34C3
s) 10(?) cage at R3C1 and 12(?) cage at R6C8 must have symmetry with each other -> 10(?) cage must contain at least R3C1 + R4C12, 12(?) cage must contain at least R6C89 + R7C9
t) There must therefore be 16(4) cage in R3C89 + R45C8 and 20(4) cage in R56C2 + R7C12

1. 45 rule on C123 1 innie R5C3 (the only available innie) = 9, clean-up: no 5 in R67C3, no 3 in R9C2
1a. Naked pair {68} in R67C3, locked for C3, clean-up: no 4 in R9C2

[Completing the remaining cages]
2a. There are no outies from C123 -> 16(4) cage in R8C123 + R9C1
2b. From symmetry no outies from C789 -> 26(4) cage in R1C9 + R2C789 = {2789/3689/4589/4679/5678}, no 1
2c. Max R234C3 = {457} = 18 -> cage at R2C3 must be 20(4) cage in R234C3 + R3C2
2d. From symmetry 16(4) cage in R678C7 + R7C8
2e. Therefore 10(3) cage in R3C1 + R4C12 = {127/136/145/235}, no 8,9 and 12(3) cage in R6C89 + R7C9
2f. Finally there are 23(4) cage in R2C56 + R3C45 and 24(4) cage in R7C56 + R8C45

3. 45 rule on N3 3 innies R3C789 = 6 = {123}, locked for R3, 2,3 locked for N3
3a. R34C7 = [29/38]
3b. 1 in R3 only in R3C89, locked for 16(4) cage at R3C8

4. 45 rule on C789 3(1+2) innies R5C7 + R89C9 = 22
4a. Max R89C9 = 17 -> min R5C7 = 5
4b. Max R5C7 = 8 -> min R89C9 = 14, no 1,2,3,4

5. 9 in C1 only in R789C1, locked for N7, clean-up: no 3 in R9C3
5a. 45 rule on N7 3 innies R7C123 = 17 = {269/368} (cannot be {179/359} because R7C3 only contains 6,8, cannot be {278/458/467} which clash with R9C23), no 1,4,5,7, 6 locked for R7 and N7
5b. 9 of {269} must be in R7C1 -> no 2 in R7C1
5c. R56C1 = {18/27/36} (cannot be {45} which clashes with R12C1), no 4,5

6. 45 rule on N1 1 innie R3C1 = R4C7 = {457}
6a. 10(3) cage at R3C1 = {127/145/235} (cannot be {136} because R3C1 only contains 4,5,7), no 6
6b. 7 of {127} must be in R3C1 -> no 7 in R4C12
6c. Naked triple {457} in R349C3, locked for C3

7. 45 rule on N14 3 innies R5C2 + R6C23 = 17 = {269/278/368/458/467} (cannot be {179/359} because R6C3 only contains 6,8), no 1

8. 19(3) cage at R1C2 = {289/379} (cannot be {469/478/568} because R1C3 only contains 2,3)
8a. R1C3 = {23} -> R12C2 = {79/89}, 9 locked for N1
8b. R3C2 = 6 (hidden single in N1)
8c. R12C2 = {89} (hidden pair in N1), 8 locked for C2, clean-up: no 4 in R9C3
8d. R12C2 = {89} -> R1C3 = 2, clean-up: no 4 in R12C1, no 7 in R1C56
8e. Naked pair {15} in R12C1, locked for C1 and N1 -> R2C3 = 3, R8C3 = 1, clean-up: no 5 in R4C3 (step 6), no 8 in R56C1
8f. R4C2 = 1 (hidden single in N4) -> R34C1 = 9 = [72], R34C3 [47], R9C23 = [75], clean-up: no 1,6,7 in R5C9
8g. Naked pair {36} in R56C1, locked for C1 and N4 -> R67C3 = [86]
8h. Naked pair {45} in R56C2, 4 locked for C2
8i. Naked triple {589} in R3C456, locked for N2, clean-up: no 1,4 in R1C56
8j. Naked pair {36} in R1C56, locked for N2, 6 locked for R1, clean-up: no 7 in R1C78
8k. Killer pair 8,9 in R1C2 and R1C78, locked for R1
8l. 23(4) cage at R2C5 contains 9 = {24}{89}/{27}{59}, no 1, 2 locked for N2
8m. 1 in N2 only in R12C4, locked for C4

9. Hidden killer pair 1,2 in 40(7) cage at R4C4 and 18(3) cage at R6C4 for N5, 40(7) cage must contain one of 1,2 in N5 -> 18(3) cage must contain one of 1,2 = {189/279}, no 3,4,5,6
9a. 1,2 of {189/279} must be in R6C45 -> no 2 in R7C4
9b. Hidden killer pair 3,4 in 40(7) cage at R4C4 and 16(3) cage at R3C6 for N5, 40(7) cage must contain one of 3,4 in N5 -> 16(3) cage must contain one of 3,4
9c. 16(3) cage at R3C6 = {358} (only possible combination, cannot be {349} = 9{34} which clashes with 40(7) cage), 3 locked for R4 and N5, clean-up: no 5 in R5C9
9d. 40(7) cage = {1456789}, no 2, 1 locked for N5
9e. 18(3) cage = {279}, no 8, 2 locked for R6
9f. 16(3) cage = {358}, CPE no 5,8 in R56C6
9g. 18(3) cage = {279}, CPE no 7 in R5C4
9h. 9 in R4 only in R4C78, locked for N6
9i. 2 in R5 only in R5C89, CPE no 2 in R3C9
9j. 3 in C4 only in R89C4, locked for N8

10. 12(3) cage at R6C8 = {138/147/246/345} (cannot be {129} because 2,9 only in R7C9, cannot be {156} which clashes with R4C9, cannot be {237} which clashes with R5C9), no 9

11. 1 in R3 only in R3C89 -> max R3C89 = 4 -> min R45C8 = 12, no 2,3 in R5C8 (because 16(4) cage at R3C8 cannot be {13}[93])
11a. R5C9 = 2 (hidden single in R5) -> R4C9 = 6
11b. R5C1 = 3 (hidden single in R5) -> R6C1 = 6

12. R6C45 = {29} (hidden pair in R6) -> R7C4 = 7 (cage sum)
12a. Naked pair {14} in R12C4, 4 locked for C4 and N2
12b. Naked pair {27} in R2C56, 7 locked for R2
12c. R2C56 = {27} = 9 -> R3C45 = 14 = {59}, 5 locked for R3
12d. R3C6 = 8 -> R4C56 = {35}, 5 locked for R4 and N5
12e. R45C4 = [86], R4C7 = 9 -> R3C7 = 2, R4C8 = 4, R5C8 = 8 (hidden single in R5, or cage sum), clean-up: no 5 in R1C7
12f. R1C9 = 7 (hidden single in R1)

13. 40(7) cage at R4C4 (step 9d) = {1456789} -> R5C7 = 5, R5C2 = 4, R6C6 = 4 (hidden single in R6)
13a. R5C7 + R89C9 = 22 (step 4), R5C7 = 5 -> R89C9 = 17 = {89}, locked for C9 and N9
13b, Naked pair {13} in R36C9, locked for C9

14. R7C6 = 9 (hidden single in C6), R7C1 = 8, R56C2 = [45] = 9 -> R7C2 = 3 (cage sum), R8C2 = 2
14a. ?(2) cage at R9C4 cannot total more than 9, R9C4 = {23} -> no 8 in R9C5
14b. R8C5 = 8 (hidden single in C5) -> R89C9 = [98], R89C1 = [49]
14c. R7C6 = 9, R8C5 = 8 -> R7C5 + R8C4 = 7 = [25/43]
14d. 1,5 in R7 only in R7C789, locked for R7
14e. 11(3) cage at R8C8 = {236} (only remaining combination) -> R9C8 = 2, R9C4 = 3, R9C7 = 6, R8C8 = 3, R37C8 = [15], R1C8 = 9 -> R1C7 = 4

and the rest is naked singles.
Same solution.


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