Dynamic Programming: Why Knuth's improvement to Optimal Binary Search Tree O(n^2)?

Question

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(n ³ ). 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(n ² ) 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

with

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(n ² ): 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(n ³ ).

So my question is: can you please explain to me why the improvement to DP algorithm yields O(n ² ) 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.

Answer 1

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 l is

  S = sum_{i = 1}^{n - l + 1} (r[i + 1, j] + 1 - r[i, j - 1]),  j = i + l - 1
    = sum_{i = 1}^{n - l + 1} (r[i + 1, i + l - 1] + 1 - r[i, i + l - 2])
    = r[n - l + 2, n] + n - l + 1 - r[1, l - 1]

therefore the average is S / (n - l + 1), which is a constant

by simplifying the telescoping sum.

Answer 2

You can find the exact running time analysis with a google search or just start to write your own analysis wrt for loops. But just note that in all of them sum in total is calculated by telescopic sum, I mean may be one of them is big but in each iteration for first loop takes O(n), and totally takes O(n ² ).