The essential info in a puzzle.
Let’s observe this puzzle:
There is no information in the cage b1a2b2, but let’s solve it quickly:
We will use the cage “1:” in d4e4e5 as the “key”. This cage has five possible combinations 122, 133, 144, 155 and 224. The cage “60x” in c4c5d5 has the only combination 534 so 133, 144 and 155 are not valid in “1:”. The cage “60x” (60 = 2 x 2 x 3 x 5) in e1e2d3e3 has two possible combinations, 5322 and 5341, but 5322 is not valid with any of the other two combinations for “1:” (122 or 224).
I have shown in blue the hypothesis 224. Now the 4 in e4 forces the 4 of 5341 to go to d3 and this is the relevant point: looking to columns 3 and 4 we have 12 + 2 + 4 + 10 = 28, leaving a total of 30 (2 columns)  28 = 2 for the sum of c2 and c3, which is impossible.
Once we have determined the 122 (now in black) as the final solution for “1:”, the 1 of 5341 goes to d3 and that means that 30  (12 + 2 + 1 +10) = 5, that is, the sum of c2c3 is 5. As we have a 4 in b4, b3 must be 3 (12  4  5) (shown in green). Since c2c3 have a sum of 5 and 23 is not possible now, they must be 14 so c2 = 1 and c3 = 4 (and c5 = 3, d5 = 4). The rest is very easy (completed in orange).
Then, we have solved the entire puzzle with some analysis but regardless of the info of the cage b1a2b2 (it is actually “2:”, but it could have been “8x”, “1“ or “7+”, it is indifferent). And, certainly, we have not needed any additional info, provided as a help, from the twin puzzle (this is the 5x5 difficult on Wednesday Oct 05, 2011, Puzzle id: 361005):
Now let’s see these puzzles (all of them with a unique solution):
We do not need any other information to solve them, and in fact they take very little time (I provide the solutions at the end but try to solve them, you will only need a few seconds). In the case of the 4x4, 6 cells (among 16) have “information”, that is, the 37,50% of the surface of the puzzle. In the case of the 5x5, 9 cells (among 25) have “information”, that is, the 36,00%, and in the case of the 6x6 (12 out of 36) the info represents the 33,33%. Even a 3x3 puzzle with, i.e., a cage “1“ in a2a3b3 (that would be enough to solve the puzzle), has a 33,33% of “info”. Certainly these are very particular puzzles but the question is: How much is the “essential” information in a puzzle to arrive to the unique solution?. How much information is “superfluous”? Is it always possible to remove information leaving generously “blank” cages (like a donation to the program, like when in chess some initial advantage is given to the opponent donating the whites, a pawn or even a knight)?. How can we build a puzzle in which the 100,00% of the information is necessary having, of course, a reasonable size of the cages, and not a extremely reduced number of them?.
