Re: 22/7/2014 6x6 difficult help on twitter

clm wrote:

Hi, patrick, the strategy is the usual, then the sum of the inner 4x4 box is 57 and "3-" and "2-" cages sum 11, etc., however no further analysis is necessary in this case, it's immediate from the beginning that "3-" = [2,5] since [1,4] is not possible due to the two 1's already present in "5-" and"4+" but at the same time [3,6] is not possible because it would produce two 1's in the same row, in b3 and c3.

clm beat me to it. This is the method I used to solve it. Mass addition is generally farther down the queue of FAS techniques for me, though, I'll be honest.

Doing the above gives you the location for each of the cells in the 2d [4,2] and 3d [2,5] because of the 5 in 4d [5,6]. Doing so puts a 2 and a 5 in the same row as the 4e 2-, leaving only [1,3] and [4,6] as options. [4,6] is out because it, along with the 4d [5,6] would block all 6s in the two 9+ cages, leaving those as both [4,5]s, which doesn't work with the existing 4s and 5s in the 4d 11+ and 4e 2- cages. So that leaves 4e as [1,3].

It's probably just snobbery on my part to resist the more brute force mass cage addition approach if I can find a purely logical solution, however.