Given a simple image of size 10 × 10 whose histogram models the symbol probabilities and is given by The first order estimate of image entropy is maximum whenp1 p2 p3 p4 a b c d
a = \(\frac{1}{4}\), b = \(\frac{1}{4}\), c = \(\frac{1}{4}\), d = \(\frac{1}{4}\)
The first order entropy estimate uses the histogram probabilities directly: \(H = -\sum_k p_k \log_2 p_k\) bits/pixel (a term with \(p_k = 0\) contributes 0). The image size 10 × 10 does not affect it.
a = 0, b = 0, c = 0, d = 1: \(H = -1\log_2 1 = 0\) bits (no uncertainty at all).
Two symbols with \(\frac{1}{2}\) each: \(H = 2 \times \frac{1}{2}\log_2 2 = 1\) bit.
Three symbols with \(\frac{1}{3}\) each: \(H = \log_2 3 \approx 1.585\) bits.
Four symbols with \(\frac{1}{4}\) each: \(H = 4 \times \frac{1}{4}\log_2 4 = 2\) bits. This is the largest value; in general entropy of K symbols is maximum, \(\log_2 K\), when all are equally likely.
Hence, the answer is a = \(\frac{1}{4}\), b = \(\frac{1}{4}\), c = \(\frac{1}{4}\), d = \(\frac{1}{4}\).
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