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"spin-off" site
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Posted on: Thu Oct 06, 2016 9:07 am

Posts: 2258
Joined: Thu May 12, 2011 11:58 pm
"spin-off" site
Just started in Japan, a Calcudoku "spin-off" site:

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

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

Posted on: Sun Oct 09, 2016 1:33 am

Posts: 237
Joined: Sun May 22, 2016 2:17 pm
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

Posted on: Sun Oct 09, 2016 2:05 am

Posts: 193
Joined: Tue May 24, 2011 4:55 pm
Re: "spin-off" site
pnm wrote:
As you can see, the 4x4 difficult there is quite tricky

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

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)

Posted on: Sun Oct 09, 2016 12:56 pm

Posts: 290
Joined: Tue Mar 01, 2016 10:03 pm
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.

Posted on: Sun Oct 09, 2016 2:26 pm

Posts: 237
Joined: Sun May 22, 2016 2:17 pm
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!

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