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"spinoff" 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:  "spinoff" site 
Just started in Japan, a Calcudoku "spinoff" 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. 
Author:  marblevolcano [ Sun Oct 09, 2016 1:33 am ] 
Post subject:  Re: "spinoff" 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 
Author:  bram [ Sun Oct 09, 2016 2:05 am ] 
Post subject:  Re: "spinoff" 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 sumonly puzzles I'm currently getting from the Colossal Killer Sudoku Book (much recommended, by the way) 
Author:  paulv66 [ Sun Oct 09, 2016 12:56 pm ] 
Post subject:  Re: "spinoff" 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: "spinoff" 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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