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. The entropy of a source is a measure of its average information content per symbol. For a source with M possible symbols, the entropy H is given by: \[ H = -\sum_{i=1}^{M} p_i \log_2(p_i) \] where \(p_i\) is the probability of the i-th symbol.
Step 2: Understand the condition for maximum entropy. The entropy is maximized when all symbols are equally likely. In this case, \(p_i = 1/M\) for all i. The maximum entropy is then: \[ H_{max} = -\sum_{i=1}^{M} \frac{1}{M} \log_2\left(\frac{1}{M}\right) = -M \left(\frac{1}{M} \log_2\left(\frac{1}{M}\right)\right) = -\log_2(M^{-1}) = \log_2(M) \]
Step 3: Apply the formula to the given problem. The question asks for the entropy associated with \textit{one image pixel}. A single pixel is the "symbol". The number of possible symbols (distinguishable intensity levels) is M = 8. The maximum entropy occurs when each of the 8 levels is equally probable. \[ H_{max} = \log_2(M) = \log_2(8) \] Since \(2^3 = 8\), \[ H_{max} = 3 \] The unit of entropy is bits per symbol (in this case, bits per pixel). The total number of pixels (\(512 \times 512\)) is irrelevant for calculating the entropy of a single pixel.