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"spin-off" site
http://www.calcudoku.org/forum/viewtopic.php?f=5&t=832
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Author:  pnm  [ Thu Oct 06, 2016 9:07 am ]
Post subject:  "spin-off" site

Just started in Japan, a Calcudoku "spin-off" site:

Image

As you can see, the 4x4 difficult there is quite tricky [scared]

The site is only available as an "app" with an "au KDDI Smartpass" subscription in Japan.

Author:  marblevolcano  [ Sun Oct 09, 2016 1:33 am ]
Post subject:  Re: "spin-off" site

Nice! Now we have a JapaDutch website creator :-)

That is quite a tough puzzle as well, it seems. Well, everything is better in Japan [thumbsup]

Author:  bram  [ Sun Oct 09, 2016 2:05 am ]
Post subject:  Re: "spin-off" site

pnm wrote:
As you can see, the 4x4 difficult there is quite tricky [scared]

marblevolcano wrote:
That is quite a tough puzzle as well, it seems. Well, everything is better in Japan [thumbsup]

It's not that hard if you check sums in the right and left halves of the puzzle: [2,4] in the right "2–" in the lower row would mean 5 in "17+"'s leftmost cell (its "outie" to the left half), which is of course impossible. It must be [1,3] in the right "2–", 3 in the outie and, given [2,4] in the left "2–", 6 as the sum in "0–". Since the highest number in a "0–" is half the sum of all the numbers, it must contain [1,2,3] and the rest follows automatically.

Or you could start by checking the sum in the three upper rows to find that the sum in "0–" must be 3(10) – 2 – 17 – 5 = 6 and go on from there.

Anyway, what made this approach seem so obvious to me is probably the hefty dose of sum-only puzzles I'm currently getting from the Colossal Killer Sudoku Book :-) (much recommended, by the way) [thumbup]

Author:  paulv66  [ Sun Oct 09, 2016 12:56 pm ]
Post subject:  Re: "spin-off" site

I though it was quite an easy puzzle. As Bram says, b2 must be 3 and a1a2b1 must be 123 and it more or less fills itself in from there. I think it took less than a minute in total and I'm by no means a speed merchant.

Author:  marblevolcano  [ Sun Oct 09, 2016 2:26 pm ]
Post subject:  Re: "spin-off" site

I figured it out by starting with the 3rd row - the sum of c3+d3 must be 5, which means the rest of the values of the 17+ cage with the two has to be 14. Therefore the 0- cage must hold {1,2,3}, of which the 2 would go in a2. This also means c1 must hold the 4 for the first row. Then 3+2+1+5=11, which means that the leftmost 2- cage has to be {2,4} (lf it was {1,3} then like Bram said it would mean a 5 in b2, which is of course impossible). Then b2 must be 3, therefore b1 is 1 and a1 is 3. It also means that the 5+ cage must hold {1,4}.

I love these kinds of 4x4 puzzles! [woot]

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