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?