Question

Which of the following statement(s) is/are CORRECT in the context of image enhancement?

• A. Histogram equalization carries out a contrast stretch such that output values are displayed on the basis of their frequency of occurrence
• B. Compared to a linear contrast stretch, histogram equalization is computationally more expensive
• C. Both histogram equalization and linear contrast stretch are neighborhood operators
• D. Histogram equalization and linear contrast stretch will produce identical results if the histogram of the input image is uniform

Hint:

Neighborhood (spatial) operators mix values from a {window} (e.g., mean, median, Sobel). Histogram equalization and linear stretch are intensity remappings—no spatial averaging.

Answer 1

The Correct Option is A

(A) True. Histogram equalization maps input intensity $r$ to $s=T(r)=\text{CDF}(r)$, where CDF is the cumulative distribution of the image histogram. Hence output gray levels are assigned according to frequency of occurrence (densest ranges get stretched most).
(B) True. Linear contrast stretch uses a simple affine mapping $s=\alpha r+\beta$ based on global min–max (or chosen percentiles), requiring only a few operations per pixel. Histogram equalization needs computing the histogram and cumulative sum and then remapping—more steps and therefore higher computational cost.
(C) False. Both operations are point/global (no spatial window). Histogram equalization is global (depends on the whole-image histogram), and linear stretch is a pointwise affine transform using global parameters; neither is a neighborhood (spatial) operator.
(D) True. If the input histogram is perfectly uniform, its CDF is (approximately) linear; histogram equalization reduces to an (almost) linear mapping, matching a linear contrast stretch up to quantization.