Discussion
You are given four images represented as
\(I_1 = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\)
\(I_2 = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}\)
\(I_3 = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\)
\(I_4 = \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix}\)
The value of entropy is maximum for image
Answer: C
The entropy of a binary image depends on the proportion of 1s and 0s, and is highest when they are equally likely. For \(I_1\) all four pixels are 0, so the entropy is 0. For \(I_2\) one pixel in four is 1, and for \(I_4\) three in four are 1, so both give
\[ H = -\tfrac{1}{4}\log_2\tfrac{1}{4} \]\[ \quad - \tfrac{3}{4}\log_2\tfrac{3}{4} \approx 0.81 \]For \(I_3\) two pixels are 1 and two are 0, so p = 1/2 and the entropy is 1 bit, the maximum. The keyed answer is \(I_3\).
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