Question

An image uses \(512 \times 512\) picture elements. Each of the picture elements can take any of the 8 distinguishable intensity levels. The maximum entropy associated with one image pixel will be

• A. 3
• B. 8
• C. 512
• D. 786432

Hint:

The maximum entropy of a source with M equally likely outcomes is simply \(\log_2(M)\). This is also the number of bits required to uniquely represent each outcome.

Answer 1

The Correct Option is A

Step 1: Define Entropy.Entropy, denoted as H, quantifies the average information per symbol from a source. For a source emitting M distinct symbols, entropy is calculated as:\[ H = -\sum_{i=1}^{M} p_i \log_2(p_i) \]where \(p_i\) represents the probability of the i-th symbol occurring.
Step 2: Condition for Maximum Entropy.Maximum entropy arises when all symbols have equal probability. In this scenario, \(p_i = 1/M\) for every i. The maximum entropy is then determined by:\[ H_{max} = -\sum_{i=1}^{M} \frac{1}{M} \log_2\left(\frac{1}{M}\right) = -M \left(\frac{1}{M} \log_2\left(\frac{1}{M}\right)\right) = -\log_2(M^{-1}) = \log_2(M) \]
Step 3: Solution.The problem asks for the entropy of a single image pixel.Consider a single pixel as the "symbol".Given 8 distinguishable intensity levels, M = 8.Maximum entropy is achieved when each of the 8 levels has equal probability.\[ H_{max} = \log_2(M) = \log_2(8) \]Since \(2^3 = 8\),\[ H_{max} = 3 \]Therefore, the maximum entropy is 3 bits per pixel. The image dimensions (\(512 \times 512\)) are not relevant to the per-pixel entropy calculation.