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 highest frequency, that is f(c) f(d) for all d in C-c. Then the depthT(c) ≤ depthT(d) for all d in C.

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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 highest 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 Huffman coding, the most frequent character is assigned the shortest codeword on average, because the overall expected code length is minimized by giving shorter codes to more frequent symbols.
Option 1: True. Since f(c) is strictly greater than f(d) for every other character d, c is uniquely the most frequent symbol. In an optimal prefix code produced by Huffman’s algorithm, the symbol with the highest frequency has the smallest possible code length, and no other symbol can have a shorter code than c. Therefore depthT(c) ≤ depthT(d) for all d ≠ c, because a shorter depth would imply a shorter codeword, which would contradict c’s status as the unique most frequent symbol.
Option 2: False. This would claim that, despite c having strictly the highest frequency, some other symbol could have a strictly smaller depth than c. That would violate the fundamental property of Huffman codes, where more frequent symbols receive shorter or equal-length codewords compared to less frequent ones. The unique maximum frequency ensures c cannot be deeper than any other symbol in the code tree.
In summary, the logic relies on the alignment between frequency and code length in optimal prefix codes: higher frequency tends to yield shorter depth, and with a unique highest frequency, there’s no reason for any other symbol to be strictly shorter than c’s depth.

COMP2123 (ND) Quiz 8

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 highest frequency, that is f(c) > f(d) for all d in C-c. Then the depthT(c) ≤ depthT(d) for all d in C.

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