As you have identified (tagged) correctly, this indeed is very similar to the matrix multiplication problem (in what order do I multiply matrixes in order to do it quickly).
This can be solved polynomially using a dynamic algorithm.
I'm going to instead solve a similar, more classic (and identical) problem, given a formula with numbers, addition and multiplications, what way of parenthesizing it gives the maximal value, for example
6+1 * 2 becomes (6+1)*2 which is more than 6+(1*2).
Let us denote our input a1 to an real numbers and o(1),...o(n-1) either * or +. Our approach will work as follows, we will observe the subproblem F(i,j) which represents the maximal formula (after parenthasizing) for a1,...aj. We will create a table of such subproblems and observe that F(1,n) is exactly the result we were looking for.
Define
F(i,j)
- If i>j return 0 //no sub-formula of negative length
- If i=j return ai // the maximal formula for one number is the number
- If i<j return the maximal value for all m between i (including) and j (not included) of:
F(i,m) (o(m)) F(m+1,j) //check all places for possible parenthasis insertion
This goes through all possible options. TProof of correctness is done by induction on the size n=j-i and is pretty trivial.
Lets go through runtime analysis:
If we do not save the values dynamically for smaller subproblems this runs pretty slow, however we can make this algorithm perform relatively fast in O(n^3)
We create a n*n table T in which the cell at index i,j contains F(i,j) filling F(i,i) and F(i,j) for j smaller than i is done in O(1) for each cell since we can calculate these values directly, then we go diagonally and fill F(i+1,i+1) (which we can do quickly since we already know all the previous values in the recursive formula), we repeat this n times for n diagonals (all the diagonals in the table really) and filling each cell takes (O(n)), since each cell has O(n) cells we fill each diagonals in O(n^2) meaning we fill all the table in O(n^3). After filling the table we obviously know F(1,n) which is the solution to your problem.
Now back to your problem
If you translate the polygon into n different formulas (one for starting at each vertex) and run the algorithm for formula values on it, you get exactly the value you want.
