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 Post subject: Sudokakuro Posted: Fri Sep 14, 2018 11:00 am Grand Master Joined: Wed Apr 30, 2008 9:45 pm
Posts: 550
Mathimagics has published a new variant:

http://forum.enjoysudoku.com/sudokakuro-t35041.html

It is essentially an anti-king zero twin killer so you will probably enjoy them.

Top Post subject: Re: Sudokakuro Posted: Fri Sep 14, 2018 1:06 pm Grand Master Joined: Wed Apr 30, 2008 9:45 pm
Posts: 550
SK1

Here is one I put together
It is X

It is Anti-King

Not too difficult - remember the anti-king.

Each number is the sum of horizontally and vertically adjacent cells. Hence a central number is the sum of four cells, an edge number is the sum of three cells and a corner is the sum of two cells.
For example the 14 at r1c1 is the sum of r1c2 and r2c1, the 24 at r1c3 is the sum of r1c2, r1c4 and r2c3 and the 28 at r2c2 is the sum of r1c2, r2c1, r2c3 and r3c2

explanation corrected Last edited by HATMAN on Wed Mar 27, 2019 2:07 pm, edited 2 times in total.

Top Post subject: Re: Sudokakuro Posted: Thu Sep 20, 2018 8:07 am Grand Master Joined: Wed Apr 30, 2008 9:45 pm
Posts: 550
correct Azpauli

clarified above

Top Post subject: Re: Sudokakuro Posted: Wed Sep 26, 2018 10:37 pm Addict Joined: Mon Apr 21, 2008 11:35 pm
Posts: 39
Hatman, the clues don't mean that there isn't a 1-->9 in those cells, correct? For example, the cell A1 still should have a 1-->9 in it, even though you've inserted the clue "14"? And, the "14" refers to the total of A2 and B1? And, the anti-King means that A2 and B1 can't be 7's?

Thanks!

Top Post subject: Re: Sudokakuro Posted: Wed May 01, 2019 2:00 am Grand Master Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1647
Location: Lethbridge, Alberta, Canada
Thanks Mathimagics for publishing the new variant and HATMAN for creating this nice puzzle.

As HATMAN said "Not too difficult - remember the anti-King". I was struggling until I realised that I'd missed some AKs, some of which proved to be key although not obviously so when they first became available. I've reworked my walkthrough, putting those AKs in their correct places which made my solving path a lot easier.

Here is my reworked walkthrough for SK1:
Each number is the sum of horizontally and vertically adjacent cells (but not the cell containing the number). Hence a central number is the sum of four cells, an edge number is the sum of three cells and a corner is the sum of two cells.
For example the 14 at R1C1 is the sum of R1C2 and R2C1, the 24 at R1C3 is the sum of R1C2, R1C4 and R2C3, and the 28 at R2C2 is the sum of R1C1, R2C1, R2C3 and R3C2
X - no repeats on diagonals, Anti-King (AK)

Prelims
Because of AK all the numbers surrounding a total must form a cage and the cell containing the total cannot contain any of the numbers in the cage, nor can any cell pointed to by the cage.
a) Total in R1C1 = 14 -> R1C2 + R2C1 = {59/68}
b) Total in R1C9 = 14 -> R1C8 + R2C9 = {59/68}
c) Total in R9C1 = 10 -> R8C1 + R9C2 = {19/28/37/46}, no 5
d) Total in R9C9 = 7 -> R8C9 + R9C8 = {16/25/34}, no 7,8,9
e) Total in R1C3 = 24 -> R1C24 + R2C3 = {789}, no 7,8,9 in R1C3
f) Total in R3C1 = 19 -> R24C1 + R3C2 = {289/379/469/478/568}, no 1
g) Total in R3C9 = 20 -> R24C9 + R3C8 = {389/479/569/578}, no 1,2
h) Total in R2C2 = 28 -> R13C2 + R2C13 = {4789/5689}, no 1,2,3, no 8,9 in R2C2
i) Total in R3C5 = 13 -> R24C5 + R3C46 = {1237/1246/1345}, no 8,9, no 1 in R13C5
j) Total in R4C4 = 13 -> R35C4 + R4C35 = {1237/1246/1345}, no 8,9, no 1 in R4C4
k) Total in R5C3 = 13 -> R46C3 + R5C24 = {1237/1246/1345}, no 8,9, no 1 in R5C13

1a. R1C2 = {89} -> R2C1 = {56}
1b. R13C2 + R2C13 = {5689} (only remaining combination), locked for N1 -> R2C3 = {89}, R3C2 = {56}
1c. R1C24 + R2C3 = {789} -> R1C4 = 7
1d. Naked pair {56} in R2C1 + R3C2, total 19 in R3C1 -> R4C1 = 8
1e. 7 in N1 only in R2C2 + R3C13 -> no 7 in R4C2 (AK)

[Now to what is clearly the key concept of this type of puzzle, which I was a bit slow in realising.
Cascading from known totals of diagonal pairs to totals of adjacent diagonal pairs. I’m listing each cascade separately for clarity.]
2a. Total in R3C3 = 20, R2C3 + R3C2 = 14 -> R3C4 + R4C3 = 6 = {15/24} (cannot be , AK)
2b. Total in R4C4 = 13, R3C4 + R4C3 = 6 -> R4C5 + R5C4 = 7 = {16/34} (cannot be {25} which clashes with R3C4 + R4C3), no 2,5,7
2c. Total in R5C5 = 22, R4C5 + R5C4 = 7 -> R5C6 + R6C5 = 15 = {69/78}
2d. Total in R6C6 = 23, R5C6 + R6C5 = 15 -> R6C7 + R7C6 = 8 = {17/26/35} (cannot be , AK), no 4,8,9
2e. Total in R7C7 = 19, R6C7 + R7C6 = 8 -> R7C8 + R8C7 = 11 = {29/38/47/56}, no 1
[Which matches up with R8C9 + R9C8 = 7. Now for cascade along the other diagonal.]
2f. Total in R2C8 = 24, R1C8 + R2C9 = 14 -> R2C7 + R3C8 = 10 = [19/28]/{37/46} (cannot be , AK), no 5, no 8,9 in R2C8
2g. Total in R3C7 = 20, R2C8 + R3C8 = 10 -> R3C6 + R4C7 = 10 = [19/37]/{46} (cannot be , AK), no 2,5, no 1,3 in R4C7
2h. Total at R4C6 = 21, R3C6 + R4C7 = 10 -> R4C5 + R5C6 = 11 = [38/47], no 1,6 in R4C5, no 6,9 in R5C6
2i. Total at R5C5 = 22, R4C5 + R5C6 = 11 -> R5C4 + R6C5 = 11 = [38/47], no 1,6 in R5C4, no 6,9 in R6C5
2j. Total at R6C4 = 22, R5C4 + R6C5 = 11 -> R6C3 + R7C4 = 11 = [29/38/56/65/74], no 1,4 in R6C3, no 1,2,3 in R7C4
2k. Total at R7C3 = 24, R6C3 + R7C4 = 11 -> R7C2 + R8C3 = 13 = {49/58/67}, no 1,2,3
[Which matches up with R8C1 + R9C2 = 10.]
2l. Naked pair {34} in R4C5 + R5C4, locked for N5 and R35C4 + R4C35
2m. R3C4 + R4C3 = 6 = {15}, no 1 in R3C3, no 5 in R4C4
2n. Naked pair {15} in R3C4 + R4C3, locked for R24C3 + R3C24
2o. R3C2 = 6 -> R2C3 = 8, R1C2 = 9, R2C1 = 5
2p. Naked pair {78} in R5C6 + R6C5, locked for N5 and R57C6 + R6C57
2q. R6C7 + R7C6 = 8 = {26/35}, no 1
2r. 8 in C4 only in R789C4, locked for N8
2s. 8 on D\ only in R7C7 + R8C8 + R9C9, locked for N9
2t. R7C8 + R8C7 = {29/47} (cannot be {56} which clashes with R6C7 + R7C6), no 3,5,6
2u. R4C5 + R5C6 = [38/47] -> R3C6 + R4C7 = [19/46] (cannot be  which clashes with R4C5 + R5C6), no 3 in R3C6, no 4,7 in R4C7
2v. 7 in R4 only in R4C89, locked for N6
Clean-ups: no 6 in R1C8, no 4 in R2C7, no 5 in R7C2, no 1,4 in R8C1, no 4,7 in R8C3, no 2 in R9C2

3a. R24C5 + R3C46 = 13 = {1345} (only remaining combination, cannot be {1246} because 2,6 only in R2C5), 1 locked for N2, 3 locked for C5, no 4,5 in R13C5
3b. R24C5 + R3C46 = {1345} -> R3C4 = 5, R4C3 = 1 (step 2m)
3c. 1 in N2 only in R2C5 + R3C6 -> no 1 in R2C7 (AK)
3d. 1 in N5 only in R5C5 + R6C46 -> no 1 in R7C5 (AK)
3e. 5 in N5 only R46C6 + R5C5 -> no 5 in R5C7 (AK)
Clean-ups: no 9 in R3C8, no 6 in R6C3

4. R46C3 + R5C24 = {1237/1345}, no 3 in R5C13
4a. R4C5 + R5C6 (step 2h) = [38/47] -> R5C46 = [37/48]
4b. R46C3 + R5C24 = {1237} can only be  (R5C24 cannot be  which clashes with R5C46 = ) -> no 7 in R5C2, no 2 in R6C3
Clean-up: no 9 in R7C4

5. R24C9 + R3C8 = {389/479} (cannot be {569} because R3C8 only contains 3,4,7,8, cannot be {578} because R2C9 only contains 6,9), no 5,6
5a. R2C9 = 9 -> R1C8 = 5
5b. 8 of {389} must be in R3C8 -> no 3 in R3C8
5c. R3C5 = 9 (hidden single in N2) -> no 9 in R4C46
5d. 8 in N2 only in R1C56, locked for R1
5e. 9 in R4 only in R4C78, locked for N6
5f. 9 in N5 only in R6C46, locked for R6
Clean-ups: no 7 in R2C7, no 2 in R8C9

6. 7 in N3 only in R2C8 + R3C789 -> no 7 in R4C8 (AK)
6a. R4C9 = 7 (hidden single in R4) -> R3C8 = 4 (step 5) -> R2C7 = 6 (step 2f)
6b. R3C6 = 1, R4C7 = 9
6c. Naked pair {34} in R24C5, locked for C5
6d. Naked triple {234} in R2C456, locked for R2 and N2
6e. 1 in N5 only in R5C5 + R6C4, locked for D/
6f. R2C8 = 7, placed for D/
6g. R2C2 = 1, placed for D\
6h. R6C46 =  (hidden pair in N5), 9 placed for D\
6i. R7C1 = 1 (hidden single in N7)
6j. R7C8 = 9 (hidden single in N9) -> R8C7 = 2 (step 2e), no 2 in R7C6 (AK)
6k. R6C7 + R7C6 (step 2d) = {35} -> no 3,5 in R7C7
6l. 2 in N3 only in R13C9, locked for C9
6m. 5 in N5 only in R4C6 + R5C5, locked for D/
6n. 9 on D/ only in R7C3 + R9C1, locked for N7
Clean-ups: no 4 in R7C2, no 3,5 in R8C9, no 8 in R9C2

7a. R7C2 + R8C3 (step 2k) = 13 =  (cannot be  which clashes with R8C1 + R9C2 = {37}/)
7b. R6C3 + R7C4 (step 2j) = 11 = 
7c. R46C3 + R5C24 = 13, R46C3 =  = 8 -> R5C24 = 5 = 
7d. R1246C5 =  -> R5C5 = 5, placed for D\
7e. Naked pair {26} in R4C46, locked for R4 -> R4C28 = 
7f. R8C9 + R9C8 = 7 = {16}, locked for N9 -> R8C8 = 8
7g. Naked triple {234} in R1C1 + R3C3 + R9C9, locked for D\, 2 also locked for N1
7h. R4C46 = , 2 placed for D/
7i. Naked pair {34} in R68C2, locked for C2 -> R9C2 = 7, R8C1 = 3
7j. R6C7 = 5 -> R7C6 = 3 (step 6k)

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

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

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