GATE CS 2017 Set 2 Q.50 (CS Q.50): Greedy Algorithms

A message is made up entirely of characters from the set $X = \{P, Q, R, S, T\}$ . The table of probabilities for each of the characters is shown below:

Character	Probability
P	0.22
Q	0.34
R	0.17
S	0.19
T	0.08
Total	1.00

If a message of 100 characters over $X$ is encoded using Huffman coding, then the expected length of the encoded message in bits is

Your answer

Official answer and solution

Official answer: 225
From: Official GATE 2017 answer key

Solution

Construct the Huffman tree by repeatedly merging the two least probable nodes, determine codeword lengths, and calculate expected message length.

Given character probabilities:

$T: 0.08$
$R: 0.17$
$S: 0.19$
$P: 0.22$
$Q: 0.34$

Apply the greedy Huffman tree algorithm:

Combine the two lowest probabilities $T (0.08)$ and $R (0.17)$ :
Node $(TR)$ with probability $0.08 + 0.17 = 0.25$ .
Remaining probabilities: $S (0.19), P (0.22), (TR) (0.25), Q (0.34)$ .
Combine the two lowest probabilities $S (0.19)$ and $P (0.22)$ :
Node $(SP)$ with probability $0.19 + 0.22 = 0.41$ .
Remaining probabilities: $(TR) (0.25), Q (0.34), (SP) (0.41)$ .
Combine the two lowest probabilities $(TR) (0.25)$ and $Q (0.34)$ :
Node $((TR)Q)$ with probability $0.25 + 0.34 = 0.59$ .
Remaining probabilities: $(SP) (0.41), ((TR)Q) (0.59)$ .
Combine $(SP) (0.41)$ and $((TR)Q) (0.59)$ to form the root with probability $1.00$ .

Determine the length $\ell_i$ of each character's codeword (its depth in the tree):

$S$ : depth 2 (in $SP$ ) $\implies \ell(S) = 2$ bits
$P$ : depth 2 (in $SP$ ) $\implies \ell(P) = 2$ bits
$Q$ : depth 2 (in $(TR)Q$ ) $\implies \ell(Q) = 2$ bits
$R$ : depth 3 (in $TR$ inside $(TR)Q$ ) $\implies \ell(R) = 3$ bits
$T$ : depth 3 (in $TR$ inside $(TR)Q$ ) $\implies \ell(T) = 3$ bits

Compute the expected length per character ( $L$ ):

L = \sum p_i \ell_i = (0.19 + 0.22 + 0.34) \times 2 + (0.17 + 0.08) \times 3

L = 0.75 \times 2 + 0.25 \times 3 = 1.50 + 0.75 = 2.25 \text{ bits/character}

For a message of 100 characters:

\text{Expected total length} = 100 \times 2.25 = 225 \text{ bits}

Final answer

225

Common trap

Forgetting to multiply the per-character expected length (2.25 bits) by the message length of 100 characters.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2017 Computer Science and Information Technology (CS) Session 2
Question: 50 of 65, printed as CS Q.50
Section: Computer Science and Information Technology

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