According to the puzzles coder, any puzzle can be solved in 3 clicks (or less). That means that following my algorithm is likely to be significantly less efficient than an optimum solve.
I am not sure I buy it…yet.
Lets see if we can optimize a solve path for a puzzle.
To simplify the notation, I am going to label the boxes using Hexidecimal notation.
- The top left box is 0,
- the one next to it is 1.
- the one at the end of the row is 3
- the first one in the second row is 4
- The last one is F
A solve can thus be listed as a stream of hexidecimal characters. For example, this sequence C 8 C D C E F B F 8 9 D C can solve the puzzle below.
That is 13 clicks, 10 more than the proposed optimum.
However, If I then go to swap cells 4 and E, I end up clicking C 3 times on every other click. Something tells me this is going to be redundant. The same is true for the bottom right: I end up clicking F Twice in alteration. And I end up back on D C.
I note that this sequence now also solves the puzzle:
8 D E B 8 9 D
7 clicks. Over a 40% improvement.
What about that D? What if we skip that? Yep.
8 E B 8 9 Solves it. That is less than half the number of clicks we started with. Can we go further? What if we drop those 8s?
Yep. E B 9 Solves it as well.
It seems like the pattern is to remove any double clicks. Lets try another puzzle and see.
Puzzle 65535 is all black. Using my algorithm there are no repeated clicks. So that does us no good. But it implies that there must be a way to transition from one solve pattern to another.
The algorithmic solution to this puzzle is
0 2 3 4 0 3 2 6 7
If we drop the repeated 0 2 and 3 we get
4 6 7
Which also is a solution.
The blank puzzle (puzzle 0) can be solved with
0 2 5 4 2 3 7 6 8 9 D C A B F E 0 3 F C
What if we drop all of the duplicates?
It does not work…but it is close…if we add a further 2 3 at the very end IT does. Curious.
The following pattern also solves it: Click the three points around the corner, but not the corner itself. In ANY ORDER.
Does that mean that order is meaningless? Lets go back to our original puzzle # 4079 :
EB9 E9B 9BE 9EB Be9 B9E
Yes. They all solve it.
So it appears that the 3 clicks to solve might have some merit. We need a way to convert from non-overlapping solutions to solution that overlap with redundancy. This means we might well got from an algorithmic solution to a less efficient solution, and then from there to a more efficient solution.
Edit: Rereading what he said, I see now that it was “any daily or random” puzzle. Which are, I am guessing, generated using the algorithm of “start with the solution, and randomly flip three dots.”