2 0 2
A = 4 2 3
3 0 3
By iterating A^n = A A^{n-1} where matrix multiplication is done in min-plus
algebra, we get
4 2 3 6 3 5 7 5 6
A^2 = 6 3 5 A^3 = 7 5 6 A^4 = 9 6 8
4 2 3 6 3 5 7 5 6
Note A^4 = 3 A^2 (using "scalar addition") allowing us to conclude that
A^n= 3 A^(n-2) for all n >= 4. It can be shown that we can always find a
recursion with this procedure.
How long does it take to find a recursion? In many applications, the recursion is found with only a few matrix multiplications (sublinear in the matrix size). However when coefficients of a 2x2 matrix are chosen from a uniform distribution, the expected time to find a recursion is infinite.
This talk will start with an introduction to Min-Plus Algebra (with several applications) and conclude with calculations verifying the title result.