True/False: Suppose we build a Huffman tree T for a character set C with frequency count f. Let c in C be the character with lowest frequency, that is f(c)

Question

True/False: Suppose we build a Huffman tree T for a character set C with frequency count f. Let c in C be the character with lowest frequency, that is f(c) < f(d) for all d in C-c. Then the depthT(c) ≥ depthT(d) for all d in C.

BlackTom AI · Accepted Answer

In an optimal Huffman code, symbols with smaller frequencies cannot occupy strictly shallower leaves than higher-frequency symbols because that would allow a cost-reducing swap.

1) Relevant knowledge points and concepts
- A Huffman tree minimizes the weighted external path length: cost = Σ_x f(x) · depth(x).
- An optimal prefix code has the property that if a symbol a has frequency ≤ b, there exists an optimal tree with depth(a) ≥ depth(b). The usual proof uses an exchange (swap) argument.
- The Huffman greedy algorithm always merges the two least-frequency nodes first, which tends to push low-frequency symbols to larger depth.

2) Complete reasoning process (step by step)
Assume T is an optimal Huffman tree for frequency function f, and c is the unique least-frequency symbol with f(c) < f(d) for every other d. Suppose, for contradiction, that there exists some d with depth(c) < depth(d) (i.e., c is strictly shallower than d).

3) Calculation and intermediate steps
Consider a new tree T' formed by swapping the positions (leaf locations) of c and d in T. All other leaves keep the same depths. The cost change Δ = cost(T') − cost(T) equals
Δ = f(c)·depth(d) + f(d)·depth(c) − [f(c)·depth(c) + f(d)·depth(d)]
   = (f(c) − f(d)) · (depth(d) − depth(c)).

Because f(c) − f(d) < 0 (f(c) < f(d)) and depth(d) − depth(c) > 0 by assumption, their product Δ is negative. So cost(T') < cost(T), contradicting optimality of T.

4) Why the correct answer is correct
This contradiction shows the assumption depth(c) < depth(d) cannot hold. Therefore depth(c) ≥ depth(d) for every d in C. The result follows directly from the weighted-sum optimality and the exchange argument.

5) Why the other option (False) is incorrect
Claiming False would assert there exists an optimal Huffman tree where the unique least-frequency symbol is strictly shallower than some higher-frequency symbol; the swap argument above demonstrates such a configuration cannot be optimal because swapping reduces total cost.

6) Conclusion
Thus the statement is True, which matches the provided answer.

True/False: Suppose we build a Huffman tree T for a character set C with frequency count f. Let c in C be the character with lowest frequency, that is f(c) < f(d) for all d in C-c. Then the depthT(c) ≥ depthT(d) for all d in C.True/False

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