This is Exercise 15.5-4 of Introduction to Algorithms, 3rd edition, which is about Knuth’s improvement to the DP approach to Optimal Binary Search Tree.
The DP algorithm of Optimal Binary Search Tree is:
OPTIMAL_BST(p, q, n) let e[1..n+1, 0..n], w[1..n+1, 0..n], and root[1..n, 1..n] be new tables for i = 1 to n+1 e[i, i - 1] = q[i - 1]; w[i, i - 1] = q[i - 1]; for l = 1 to n for i = 1 to n - l + 1 j = i + l - 1 e[i, j] = INFINITY w[i, j] = w[i, j - 1] + p[j] + q[j] for r = i to j t = e[i, r - 1] + e[r + 1, j] + w[i, j] if t < e[i, j] e[i, j] = t root[i, j] = r return e and root
The complexity is O(n3). Knuth had observed that
root[i, j - 1] <= root[i, j] <= root[i + 1, j], so Exercise 15.5-4 asks to implement an O(n2) algorithm by doing some modification to the original algorithm.
Well after some effort I have figured this out: in the innermost loop, replace the line
for r = i to j
for r = r[i, j - 1] to r[i + 1, j]
This has been proved by this link: Optimal binary search trees
However, I’m not sure this is really O(n2): since during each innermost loop, distance from r[i, j - 1] to r[i + 1, j] is not constant, I suspect it is still O(n3).
So my question is: can you please explain to me why the improvement to DP algorithm yields O(n2) complexity?
PS: Maybe I might have read Knuth’s paper first, but really I searched the web but found no free access to the paper.
You’re correct that the distance from
r[i, j - 1] to
r[i + 1, j] is not constant in the worst case, but it is constant on average, which suffices to imply a quadratic running time. The total number of iterations for
therefore the average is S / (n - l + 1), which is a constant
by simplifying the telescoping sum.