Partial Alphabetic TreesOpen Website

2002 (modified: 02 Nov 2022)ESA 2002Readers: Everyone
Abstract: In the partial alphabetic tree problem we are given a multiset of nonnegative weights W = w 1, . . . , w n , partitioned into k ≤ n blocks B 1, . . . , B k. We want to find a binary tree T where the elements of W resides in its leaves such that if we traverse the leaves from left to right then all leaves of B i precede all leaves of B j for every i < j. Furthermore among all such trees, T has to minimize ∑ n i =1 wid(wi), where d(wi) is the depth of wiin T. The partial alphabetic tree problem generalizes the problem of finding a Huffman tree over T (there is only one block) and the problem of finding a minimum cost alphabetic tree over W ( each block consists of a single item). This fundamental problem arises when we want to find an optimal search tree over a set of items which may have equal keys and when we want to find an optimal binary code for a set of items with known frequencies, such that we have a lexicographic restriction for some of the codewords. Our main result is a pseudo-polynomial time algorithm that finds the optimal tree. Our algorithm runs in O(W sum/W min)2α log(W sum/W min)n 2) time where W sum= ∑ n i =1 wi, W min= mini wi, and α = 1/log ∅ 1.44 1. I n particular the running time is polynomial in case the weights are bounded by a polynomial of n. To bound the running time of our algorithm we prove an upper bound of ⌊α log(W sum/W min) + 1⌋ on the depth of the optimal tree. Our algorithm relies on a solution to what we call the layered Hu.- man forest problem which is of independent interest. In the layered Huffman forest problem we are given an unordered multiset of weights W = w 1, . . . , w n, and a multiset of integers D = d 1, . . . , d m. We look for a forest F with m trees, T 1, . . . , T m, where the weights in W correspond to the leaves of F, that minimizes ∑ n i =1 widF(wi) where dF(wi) is the depth of wiin its tree plus dj if wi ∈ Tj. Our algorithm for this problem runs in O(kn 2) time.
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