UGC NET CSE | December 2014 | Part 3 | Question: 65

Question

Given a simple image of size $10 \times 10$ whose histogram models the symbol probabilities and is given by

$P_{1}$	$P_{2}$	$P_{3}$	$P_{4}$
a	b	c	d

 The first order estimate of image entropy is maximum when 

• A. $a = 0, b = 0, c = 0, d = 1$
• B. $a=\frac{1}{2}, b=\frac{1}{2}, c=0, d=0$
• C. $a=\frac{1}{3}, b=\frac{1}{3}, c=\frac{1}{3}, d=0$
• D. $a=\frac{1}{4}, b=\frac{1}{4}, c=\frac{1}{4}, d=\frac{1}{4}$

Answer 1

D is the ans

Comments

how??????

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Explanation please

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In Information Theory, Entropy is defined as: $\Sigma_i \ p_i \log p_i $

So, here, image entropy (H) = $-a \log a \ -b \log b -c \log c \ -d \log d = – \ (a \log a + b \log b + c \log c + d \log d)$

Here, $a,b,c,d$ are the probabilities and $a,b,c,d \in [0,1]$ if we consider the classical theory of probability which is based on some axioms in which we consider probabilities are between $0$ and $1$ (including both).

Now, check options and see for which $H$ is maximum.

$(A)$  $- \ 1 \log 1 = 0$

$(B)$  $- \ (\frac{1}{2} \log 2^{-1} + \frac{1}{2} \log 2^{-1}) = \log 2$

$(C)$ $- \ (\frac{1}{3} \log 3^{-1} + \frac{1}{3} \log 3^{-1} + \frac{1}{3} \log 3^{-1}) = \log 3$

$(D)$ $- \ (\frac{1}{4} \log 4^{-1} + \frac{1}{4} \log 4^{-1} + \frac{1}{4} \log 4^{-1} + \frac{1}{4} \log 4^{-1}) = \log 4$

Hence, $D$