good analytical solution for a specific 4x4 difficult?
in particular, one of the harder ones of June:
http://www.calcudoku.org/en/2014-06-09/4/3(more than a week old, so that link won't give you the puzzle if you're not a subscriber)
Here it is:
I was thinking along these lines:
- d1d2 can either be 1,2 => not possible, because then (20 - 11 - 3) = 6 is impossible,
or 2,3 => then c1 = 4, or 3,4, then c1 = 2
- either way, there must be a 1 in d3 or d4
- if c1 = 4, then a1b1b2 = 1,2,1, which is impossible because of the 7+ cage
- so c1 = 2
- so d3d4 = 1/2, and d1d2 = 3/4
- a2 + a3 = 7, so a2a3 = 3/4
- so a1=1, a4=2
- if a3=3, then c2 = 1, if a3=4, then c2 = 2, which is impossible, so c2 = 1
- so b1b2 = 3,2
- etc.
Any cleaner paths possible?