Complexity of edit distance (Levenshtein distance) recursion top down implementation

问题

I have been working all day with a problem which I can't seem to get a handle on. The task is to show that a recursive implementation of edit distance has the time complexity Ω(2^max(n,m)) where n & m are the length of the words being measured.

The implementation is comparable to this small python example

def lev(a, b):
    if("" == a):
       return len(b)   # returns if a is an empty string
    if("" == b):
        return len(a)   # returns if b is an empty string
    return min(lev(a[:-1], b[:-1])+(a[-1] != b[-1]), lev(a[:-1], b)+1, lev(a, b[:-1])+1)

From: http://www.clear.rice.edu/comp130/12spring/editdist/

I have tried drawing trees of the recursion depth for different short words but I cant find the connection between the tree depth and complexity.

Recursion Formula from my calculation

m = length of word1
n = length of word2
T(m,n) = T(m-1,n-1) + 1 + T(m-1,n) + T(m,n-1)
With the base cases:
T(0,n) = n
T(m,0) = m

But I have no idea on how to proceed since each call leads to 3 new calls as the lengths don't reach 0.

I would be grateful for any tips on how I can proceed to show that the lower bound complexity is Ω(2^max(n,m)).

回答1:

Your recursion formula:

T(m,n) = T(m-1,n-1) + T(m-1,n) + T(m,n-1) + 1
T(0,n) = n
T(m,0) = m

is right.

You can see, that every T(m,n) splits of into three paths. Due to every node runs in O(1) we only have to count the nodes.

A shortest path has the length min(m,n), so the tree has at least 3^min(m,n) nodes. But there are some path that are longer. You get the longest path by alternately reduce the first and the second string. This path will have the length m+n-1, so the whole tree has at most 3^m+n-1 nodes.

Let m = min(m,n). The tree contains also at least