Previous determination of cages in a 6x6 difficult. (Oriented to beginners / medium players)
This is the 6x6 “difficult” (solver rating 44.3) (saturday May 19, 2012). Though many puzzlers have solved it, the idea of opening this post is to show a different point of view on the analysis that can decrease the time for the solution and can avoid many “trial and error” (TAE). The main idea is determining the “0” cages arithmetically. We will solve this puzzle in three steps:
First: We define which will be the cages (the only possible combinations) (graphic 1).
Second: We assign the combinations found to the proper positions in the grid (Graphic 2).
Third: We easily complete the rest of the puzzle (Graphic 3).
In order to better understand the first step we must remind that, in a 6x6, always, the numbers in the inner area (the 4x4 inner box) are the corners plus twice the numbers 1 thru 6; it’s an easy demonstration, very intuitive, which is anyway explained in the post “Why in a 4x4 numbers in corners are those in inner area (2)” in the section “Solving strategies and tips”. (Comment: This rule is valid for puzzles of any size, i.e., in a 7x7, the 25 numbers in the inner area would be the corners plus thrice the numbers 1 thru 7, ... ).
Graphic 2: If we assign [6,4,2] to s3 >>> b6 = 6. Now, since the columns b and c must sum 42, being s1 = [5,3,2] (and consequently c2 = 2), we are left with a sum of 7 for the three cells in gray colour and that’s impossible, what means that [5,3,2] must go to s3 and [6,4,2] to s1.
Graphic 3: We complete the puzzle. We start (since [5,3,2] goes to s3) with d5 = 3 >>> e4e5 = [2,5] then e6 = 6, e1 = 4 >>> d6 = 4, d1 = 5. Additionally d5 = 3 >>> b4 = 3 >>> c1 = 3 >>> b1 = 1 >>> c5 = 1 >>> b5 = 4 >>> c2 = 4. The rest of numbers in blue colour are very easy to follow. And finally we place the numbers in brown colour.
The “official solution”:
