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 Post subject: NC Kenken 1 &2
PostPosted: Wed Jul 03, 2019 10:27 pm 
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Grand Master
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Location: Saudi Arabia
NC Kenken 1&2
A long time since I posted a Kenken so here are a couple.

They are Latin Squares.
They are non-consecutive horizontally, vertically and diagonally.
The cages without an algebraic symbol are sum cages.


NC Kenken 1
All cages are non-repeat.
There is a single diagonal D\.
easy to get you started


Image

NC Kenken 2
The killer cages are non-repeat.
The twin killer cage in the centre can repeat.
We have overlapping cages. The cage at r6c67 is 5 minus, the cage at r67c6 is 7 minus.
a little harder


Image


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 Post subject: Re: NC Kenken 1 &2
PostPosted: Sun Jul 07, 2019 2:07 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
First time I've come across this type of puzzle. As HATMAN said, it was fairly easy.

Here is my walkthrough for NC Kenken 1:
Normal cages are sums. Minus sign indicates difference between the numbers in those two-cell cages. Non-consecutive including diagonally (just given as NC).

Prelims
a) 14(4) cage at R4C9 = {1238/1247/1256/1346} (cannot be {2345}, NC), no 9
b) No 5 in 5- cages -> no 5 in R12C7, no 5 in R456789C3, no 5 in R456789C7

1a. R3C7 = 5 (hidden single in C7)
1b. 45 rule on C89 2 innies R3C89 = 16 = {79}, locked for R3
1c. R3C67 is 2- -> R3C6 = 3
1d. 45 rule on C12 2 innies R3C12 = 6 = {24}, locked for R3
1e. R3C34 is 2- -> R3C34 = {68}, locked for R3 -> R3C5 = 1, no 5,7,9 in R2C3 (NC)
1f. R1C3 = 5 (hidden single in C3), R12C3 is 4- -> R2C3 = 1, no 2 in R3C2
1g. R3C12 = [24]
1h. No 1 in the 5- cages in C3 -> no 6 in those cages
1i. R3C3 = 6 (hidden single in C3), placed for D\, R3C4 = 8
1j. 14(4) cage at R4C9 = {1238/1247/1256/1346}, 1 locked for C9
NC clean-ups:
R1C3 = 5 -> no 6 in R1C2, no 4,6 in R12C4
R2C3 = 1 -> no 2 in R12C24
R3C1 = 2 -> no 3 in R2C12, no 1,3 in R4C12
R3C2 = 4 -> no 5 in R24C12, no 3 in R4C3
R3C3 = 6 -> no 7 in R2C2 + R4C23, no 5,7 in R24C4
R3C4 = 8 -> no 9 in R2C4 + R4C34, no 7,9 in R24C5
R3C5 = 1 -> no 2 in R2C45 + R4C456
R3C6 = 3 -> no 4 in R24C56, no 2,4 in R24C7
R3C7 = 5 -> no 6 in R24C67, no 4,6 in R24C8
R3C8 = {79} -> no 8 in R24C78, no 6,8 in R24C9

2a. R2C4 = 3
2b. R12C7 is 5-, R2C7 = {79} -> R1C7 = {24}
NC clean-ups:
R2C4 = 3 -> no 2,4 in R1C5
R1C7 = {24} -> no 3 in R1C8
R2C7 = {79} -> no 8 in R1C68 + R2C6

3a. 18(4) cage at R1C1 = {1368} (only remaining combination, cannot be {1467} because R2C2 only contains 8,9) -> R1C12 = {13}, locked for R1, R2C12 = [68], 8 placed for D\
3b. R2C5 = 5
3c. R2C89 = [24] (hidden pair in R2)
3d. R2C89 = [24] = 6 -> R1C89 = 14 = [68], no 7 in R2C7 (NC)
3e. R2C7 = 9 -> R2C6 = 7
3f. R12C7 is 5-, R2C7 = 9 -> R1C7 = 4
3g. R1C6 = 2 (hidden single in R1)

4a. 14(4) cage at R4C9 = {1256} (only remaining combination), locked for C9
4b. 16(4) cage at R8C8 contains 3 and one of 7,9 for C9 = {1357} (only possible combination, couldn’t be {2347} even if 2 hadn’t already been eliminated) -> R89C8 = {15}, locked for C8, R89C9 = {37}, locked for C9 -> R3C89 = [79]
NC clean-ups:
R89C8 = {15} -> no 2,6 in R89C7

5a. R89C7 is 5- = {38} (only remaining combination), locked for C7
5b. Caged X-Wing for 3 in R89C7 and R89C9, no other 3 in R89
NC clean-ups:
R89C7 = {38} -> no 4,9 in R89C6

6. 2 on D\ only in R5C5 + R7C7, CPE no 2 in R5C7
6a. 2 in C7 on in R67C7 which is 5- -> R67C7 = {27}, locked for C6
6b. R45C7 = [16], no 7 in R6C7 (NC) -> R67C7 = [27], 7 placed for D\, R89C9 = [73], 3 placed for D\, R89C7 = [38], R1C12 = [13], 1 placed for D\, R89C8 = [51], 5 placed for D\
6c. R4C4 = 4, placed for D\ -> R6C6 = 9, placed for D\ -> R5C5 = 2
6d. 45 rule on C1 2 remaining innies R89C1 = 14 = [95] -> R9C6 = 6
6e. 45 rule on C2 2 remaining innies R89C2 = 9 = [27]
6f. R89C3 cage is 5- = [49] (only remaining possibility) -> R9C45 = [24], no 1 in R8C4 (NC)
6g. R8C456 = [681]
NC clean-ups:
R4C4 = 4 -> no 3 in R4C5 + R5C3
R5C5 = 2 -> no 1 in R56C4, no 3 in R6C5
R5C7 = 6 -> no 5 in R45C6
R6C6 = 9 -> no 8 in R5C6
R6C7 = 2 -> no 3 in R567C8
R7C7 = 7 -> no 8 in R6C8 + R7C68
R8C1 = 9 -> no 8 in R7C1
R8C2 = 2 -> no 3 in R7C13, no 1 in R7C2
R8C3 = 4 -> no 5 in R7C24
R8C5 = 8 -> no 9 in R7C45
R8C7 = 3 -> no 4 in R7C68
R8C8 = 5 -> no 6 in R7C9

and the rest is naked singles, without using the diagonal or non-consecutive.

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


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 Post subject: Re: NC Kenken 1 &2
PostPosted: Mon Jul 08, 2019 10:37 pm 
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Thanks, HATMAN!

By the way, why do you call that a "twin killer" cage in the center of #2? (Also, on second look, I realized you did explain the overlapper at R6C6 in #2.)

I enjoy KenKens sometimes, and am looking forward to trying these!


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 Post subject: Re: NC Kenken 1 &2
PostPosted: Wed Jul 24, 2019 8:55 pm 
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
azpaull

The twin killer bit comes from way back when it became popular for a while: two killer Sudoku with sparse and/or large cages with the same solution. JC makes them visually different in his software and I have got in the habit of using twin cages for repeats.

Apologies for the overlapping number but I printed direct from JSudoku for convenience.

Maurice


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 Post subject: Re: NC Kenken 1 &2
PostPosted: Wed Jul 31, 2019 10:39 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
HATMAN's statement "a little harder" felt like an understatement; I'd say quite a lot harder but that could be because I'm new to Kenken puzzles, I don't want to put anyone off trying it.

Here is my walkthrough for NC Kenken 2:
Normal cages are sums. Minus sign indicates difference between the numbers in those two-cell cages. Non-consecutive including diagonally (just given as NC). Repeats are allowed in the central cage.

Prelims
a) No 5 in 5- cages -> no 5 in R4C12, no 5 in R6C34, no 5 in R4C6789
b) 7- cages must be {18/29} -> R67C6 = {18/29}, R89C4 = {18/29}

1. R6 contains three 5- cages from {16/27/38/49}
1a. R6C12 is 4- = {15/59} (cannot be {26/37/48} which only allow two 5- cages in R6), 5 locked for R6
1b. R6C12 contains one of 1,9 -> R6C5 = {46} (to allow three 5- cages in R6)
NC clean-ups:
R6C12 contain 5 -> no 4,6 in R57C12
R6C5 = {46} -> no 5 in R5C456 + R7C45

2. R4 contains three 4- cages from {15/26/37/48/59}
2a. R4C12 is 5- = {16/49} (cannot be {27/38} which would leave two of the three 4- cages as {15} and {59})
2b. R4C5 = {16/49} with three 4- cages -> R4C5 = {28}

3. C4 contains three 2- cages from {13/24/35/46/57/68/79}
3a. R89C4 = {18/29}
R89C4 = {18} gives three 2- cages from {24/35/46/57/79} => two must be {35} and {79} with the third one of {24/46} => R5C4 = {26}
R89C4 = {29} gives three 2- cages from {13/35/46/57/68} => two must be {13} and {57} with the third one of {46/68} => R5C4 = {48}
-> R5C4 = {2468}
3b. Similarly from R67C6 -> R5C6 = {2468}
3c. 30(5) cage at R4C5 contains even numbers in R46C5 and R5C46 -> R5C5 must also be even = {2468}
3d. 30(5) cage contains repeats but since it’s in R5 and C5 no number can be repeated more than twice = {26688/44688} (can only make total of 30 with 8 repeated) -> R4C5 = 8, also 8 in R5C46, locked for R5
3e. The non-repeated number must be in R5C5 -> R5C5 = {26} -> no 2 in R5C46
3f. 6 of {26688} must be in one of R5C46 and in R6C5, 6 of {44688} must be in R5C5) -> 6 in R5C456, locked for R5, 6 in R56C5, locked for C5
3g. R4C5 = 8 -> no 4 in 4- cages in R4 -> 4 in R4 only in R4C12 = {49}, 9 locked for R4
NC clean-ups:
R4C5 = 8 -> no 7,9 in R3C456, no 7 in R4C46
Naked pair {49} in R4C12 -> no 3,5,8 in R3C12, no 3,5 in R5C12

Overlapping cages
4a. R4C4 = {12356} -> no 2,6 in R3C4, no 3 in R4C3
4b. R4C6 = {12356} -> no 2,6 in R3C6, no 3 in R4C7
4c. R6C6 = {1289} -> R6C7 = {3467}

5. Consider combinations for R67C6 = {18/29}
R67C6 = {18}, 8 locked for C6
or R67C6 = [29] => R6C67 = [27] (5- cage) => no 8 in R5C6 (NC)
or R67C6 = [92] => no 8 in R5C6 (NC)
-> R5C6 = {46}
5a. 30(5) cage at R4C5 must contain 8 twice -> R5C4 = 8, no 1 in R89C4 (7- cage)
5b. Naked pair {29} in R89C4, locked for C4
5c. 2- cage at R67C4 = {13/46}, no 7
5d. 7 in C4 only in 2- cage at R12C4 = {57}, 5 locked for C4
NC clean-ups:
R5C4 = 8 -> no 7 in R4C3, no 7,9 in R56C3
R1C4 = {57} -> no 6 in R12C3
R2C4 = {57} -> no 6 in R3C3
R5C6 = {46} -> no 5 in R4C67 + R5C7
Naked pair {57} in R12C4 -> no 8 in R12C3
Naked pair {29} in R89C4 -> no 1,3,8 in R89C3, no 1,3 in R89C5

Overlapping cages
6a. No 7 in R4C3 -> no 3 in R4C4, no 1 in R3C4
6b. No 2,5 in R4C4 -> no 1,6 in R4C3
6c. No 5 in R4C6 -> no 3 in R3C6
6d. No 5 in R4C67 -> no 1 in R4C67
6e. No 7,8 in R6C4 -> no 2,3 in R6C3
6f. No 9 in R6C34 -> no 4 in R6C34 -> no 6 in R7C4 (2- cage)
Also
6g. 7 in R4 only in R4C789 -> no 6 in R4C8 (NC) -> no 2 in R4C9 (2- cage)
6h. 6 in C4 only in R46C4 -> no 5 in R5C3 (NC)

7. 30(5) cage at R4C5 (step 3d) = {26688/44688}, must have the same numbers in R5C6 and R6C5
7a. Consider permutations for R34C4 = [13/46]
R34C4 = [13] => R6C4 = 6 => R6C5 = 4
or R34C4 = [46] => R3C34 = [26] (2- cage) => R4C67 = [37] (2- cage) => no 6 in R5C6 (NC) => R5C6 = 4
-> 30(5) cage = {44688} -> R5C5 = 6, R5C6 = 4, R6C5 = 4, no 3 in R67C4, no 3 in R6C7 (NC)
[Cracked. A lot easier now.]
7b. R3C4 = 3 (hidden single in C4) -> R4C4 = 1 (2- cage), R67C4 = [64], R4C3 = 5 (4- cage), no 4 in R4C2 (NC)
7c. R6C7 = 7 -> R6C6 = 2 (5- cage) -> R7C6 = 9 (7- cage), no 1 in R5C7, no 8 in R6C8 (NC)
7d. R6C4 = 6 -> R6C3 = 1 (5- cage), no 2 in R5C3 (NC)
7e. R4C12 = [49], R6C12 = [95], R5C3 = 3
7f. R4C67 = [62] (4- cage) -> R3C6 = 8 (2- cage)
7g. R6C89 = [38], R4C89 = [73]
7h. R5C7 = 9
NC clean-ups:
R3C4 = 3 -> no 2,4 in R23C3, no 2 in R23C5
R3C6 = 8 -> no 7,9 in R2C5, no 7 in R2C6
R4C3 = 5 -> no 4,6 in R3C2
R4C6 = 6 -> no 5 in R3C57
R4C7 = 2 -> no 1 in R3C78 + R5C8
R4C8 = 7 -> no 6 in R3C789
R4C9 = 3 -> no 2,4 in R3C89, no 2 in R5C89
R5C3 = 3 -> no 2 in R5C2
R6C1 = 9 -> no 8 in R7C12
R6C2 = 5 -> no 6 in R7C3
R6C3 = 1 -> no 2 in R7C23
R6C4 = 6 -> no 7 in R7C35
R6C6 = 2 -> no 1,3 in R7C57
R6C7 = 7 -> no 6,8 in R7C78
R6C8 = 3 -> no 2 in R7C89
R6C9 = 8 -> no 7 in R7C9

8a. R3C7 = 4
8b. R357C8 = [951]
8c. R23C3 = [97]
8d. R3C1259 = [6215]
8e. R7C3579 = [8256]
8f. R5C1 = 2 (hidden single in R5), no 1 in R5C2 (NC)
8g. R5C29 = [71] -> R7C12 = [73], no 2,4 in R8C3 (NC)
8h. R8C3 = 6
NC clean-ups:
R2C3 = 9 -> no 8 in R12C2
R3C2 = 2 -> no 1,3 in R2C1, no 1 in R2C2
R3C3 = 7 -> no 6 in R2C2
R2C2 = 4 -> no 3,5 in R1C1, no 5 in R2C1
R3C7 = 4 -> no 3,5 in R2C6, no 3 in R2C7
R3C8 = 9 -> no 8 in R2C78
R7C1 = 7 -> no 8 in R8C2
R7C5 = 2 -> no 1,3 in R8C6
R7C7 = 5 -> no 4 in R8C8
R7C8 = 1 -> no 2 in R8C89
R7C9 = 6 -> no 7 in R8C9

[Now to use the regular cages. I had realised earlier, before I found the power of step 7, that they cannot contain more than two consecutive numbers since each cells is adjacent to at least three others; that isn’t needed now.]
9a. R2C123 = [849], R3C2 = 2 -> R1C2 = 6 (cage sum)
9b. R789C2 = [318] -> R8C1 = 5
9c. R2C6 = 1 -> R2C789 = [627], R3C8 = 9 -> R1C8 = 4 (cage sum)
9d. R789C8 = [186], R8C7 = 3 -> R8C9 = 4 (cage sum)
9e. R2C6 = 1 -> R1C6 = 3 (2- cage)

and the rest is naked singles.

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


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 Post subject: Re: NC Kenken 1 &2
PostPosted: Thu Aug 01, 2019 6:36 am 
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Posts: 693
Location: Saudi Arabia
Andrew when you have created a puzzle it is often difficult to judge how hard it is.


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