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Show Source |    | About   «  12.19. Trees versus Tries   ::   Contents   ::   12.21. Binary Tree Chapter Summary  »

12.20. Proof of Optimality for Huffman Coding

12.20.1. Proof of Optimality for Huffman Coding

Huffman tree building is an example of a greedy algorithm. At each step, the algorithm makes a "greedy" decision to merge the two subtrees with least weight. This makes the algorithm simple, but does it give the desired result? This section concludes with a proof that the Huffman tree indeed gives the most efficient arrangement for the set of letters. The proof requires the following lemma.

Lemma: For any Huffman tree built by function buildHuff containing at least two letters, the two letters with least frequency are stored in sibling nodes whose depth is at least as deep as any other leaf nodes in the tree.

Proof: Call the two letters with least frequency \(l_1\) and \(l_2\). They must be siblings because buildHuff selects them in the first step of the construction process. Assume that \(l_1\) and \(l_2\) are not the deepest nodes in the tree. In this case, the Huffman tree must either look as shown in Figure 12.20.1, or effectively symmetrical to this. For this situation to occur, the parent of \(l_1\) and \(l_2\), labeled \(V\), must have greater weight than the node labeled \(X\). Otherwise, function buildHuff would have selected node \(V\) in place of node \(X\) as the child of node \(U\). However, this is impossible because \(l_1\) and \(l_2\) are the letters with least frequency.

Figure 12.20.1: An impossible Huffman tree, showing the situation where the two nodes with least weight, \(l_1\) and \(l_2\), are not the deepest nodes in the tree. Triangles represent subtrees.

Here is the proof.

Theorem: Function buildHuff builds the Huffman tree with the minimum external path weight for the given set of letters.

Proof: The proof is by induction on \(n\), the number of letters.

  • Base Case: For \(n = 2\), the Huffman tree must have the minimum external path weight because there are only two possible trees, each with identical weighted path lengths for the two leaves.
  • Induction Hypothesis: Assume that any tree created by buildHuff that contains \(n-1\) leaves has minimum external path length.
  • Induction Step: Given a Huffman tree \(\mathbf{T}\) built by buildHuff with \(n\) leaves, \(n \geq 2\), suppose that \(w_1 \leq w_2 \leq ... \leq w_n\) where \(w_1\) to \(w_n\) are the weights of the letters. Call \(V\) the parent of the letters with frequencies \(w_1\) and \(w_2\). From the lemma, we know that the leaf nodes containing the letters with frequencies \(w_1\) and \(w_2\) are as deep as any nodes in \(\mathbf{T}\). If any other leaf nodes in the tree were deeper, we could reduce their weighted path length by swapping them with \(w_1\) or \(w_2\). But the lemma tells us that no such deeper nodes exist. Call \(\mathbf{T}'\) the Huffman tree that is identical to \(\mathbf{T}\) except that node \(V\) is replaced with a leaf node \(V'\) whose weight is \(w_1 + w_2\). By the induction hypothesis, \(\mathbf{T}'\) has minimum external path length. Returning the children to \(V'\) restores tree \(\mathbf{T}\), which must also have minimum external path length.

Thus by mathematical induction, function buildHuff creates the Huffman tree with minimum external path length.

Todo

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