Which grayscale conversion approach generates a grayscale image by assigning…

Concept. Grayscale conversion maps each colour pixel's three channels - red (R), green (G) and blue (B) - onto a single intensity (grey) value. The conversion methods differ only in the weights given to the three channels. The unweighted average uses gray = (R + G + B) / 3, giving every channel the identical weight of one-third, whereas the perceptual luma method uses gray = 0.299*R + 0.587*G + 0.114*B, giving the channels deliberately unequal weights tuned to human vision (the eye is most sensitive to green and least to blue).

Applying it here. The question asks for the approach that treats all three RGB components with equal importance. Only the average rule gray = (R + G + B) / 3 assigns the same coefficient of one-third to R, G and B, so unweighted averaging of the RGB channels is the method that weights the three components equally.

Why the other approaches do not fit:

• Weighted luminance transformation uses unequal coefficients (0.299, 0.587, 0.114), so the channels are not equally important - it is the opposite of equal weighting.

• Edge-based grayscale transformation derives values from spatial intensity gradients between neighbouring pixels, not from combining a single pixel's R, G and B.

• Alpha-transparency mapping works on the opacity (alpha) channel, which carries no colour-intensity information about R, G and B.

• Contrast-normalized grayscale encoding rescales intensities for contrast after a grey value already exists; it is a normalization step, not a channel-combination rule.

Cross-check. In (R + G + B) / 3 the coefficient of each channel is exactly one-third, so the formula is symmetric in R, G and B - swapping any two channels leaves the result unchanged, the hallmark of equal importance. The luma coefficients 0.299, 0.587 and 0.114 also sum to 1 but are unequal, so that formula is not symmetric and does not give equal importance.