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Correct OPTION is (a) 0

Explanation: STANDARD deviation is given by: that RESULTS 0 for constant AREAS.

The correct option is (b) Average gray level and average contrast respectively

To ELABORATE: In terms of enhancement, mean refers to average gray level and variance to average contrast.

Given by, mean as: m = ∑_(i=0)^(L-1) ri p(ri ) and variance as: σ2(r) = ∑_(i=0)^(L-1) (ri –m)2 p(ri ).

Where, ri is histogram COMPONENT of ITH value of r, p(ri) is PROBABILITY occurrence of gray level ri and L is the max gray value allowed.

The CORRECT option is (a) G MUST be strictly monotonic

For EXPLANATION: G being strictly monotonic, confirms that the values of specified histogram pz(zi) can’t be zero. That is G^-1 is also SINGLE valued and monotonic.

The correct choice is (a) The contrast increases little bit with considerable enhancement of NOISE

Easy explanation: To an image’s smooth and noisy area, when global histogram method is applied the contrast increases little bit with considerable enhancement of noise, while for local method the result has a fine noise texture.

(a) Original image. (B) Result using global histogram equalization. (C) Result using local histogram equalization using 7*7 NEIGHBORHOOD about each PIXEL.

The correct OPTION is (d) All of the MENTIONED

To explain I would say: All of the mentioned methods modifies the PIXEL value by transformations that are based on the gray-level of the WHOLE image.

The correct choice is (c) Histogram Matching

Easy explanation: In Histogram Matching or SPECIFICATION, Z = G^-1[T(r)], r and z are gray level of INPUT and output image and T & G are transformations.

In Histogram Linearization or Equalization s = T(r), r and s are gray level of input and output image and T is the only transformations.

Correct option is (B) They LOOK visually very similar to one another

For EXPLANATION I would say: This is because the contents of all IMAGES is same. The difference is just the CONTRAST.

The histogram equalization increases the contrast and make the gray-level difference of output image visually indistinguishable.

The CORRECT answer is (c) Histogram Matching

To explain: Histogram SPECIFICATION method uses a SPECIFIED Histogram, i.e. the shape of histogram can be specified by self, to generate a processed image.

And the same is also KNOWN as Histogram Matching.

The correct answer is (C) All of the mentioned

Best EXPLANATION: For the given transformation, the PDF being positive and the integral providing area under the FUNCTION, the transformation function is SINGLE valued as well as monotonically increasing.

Correct answer is (d) Probability density function

To elaborate: For a RANDOM variable, a PDF, probability density function, is ONE of the most fundamental DESCRIPTOR.

Correct option is (c) Cumulative DISTRIBUTION function

To EXPLAIN: CDF of random variable R, gray value of input image, is cumulative distribution function.

Correct option is (b) It is NEEDED to restrict producing of some inverted gray LEVELS in OUTPUT

For explanation I would say: A T(r) which is not monotonically increasing, could RESULT in an output containing at least a section of inverted intensity range. The T(r) is monotonically increasing in interval 0 ≤ r ≤ 1, is needed to restrict producing of some inverted gray levels in output.

Right CHOICE is (a) It GUARANTEES the existence of inverse transformation

Explanation: The T(r) is SINGLE valued in interval 0 ≤ r ≤ 1, guarantees the existence of inverse transformation.

Right option is (b) The histogram whose component are biased toward high side of GRAY SCALE

To elaborate: The histogram plot is nk VERSUS rk. So, the histogram of a low contrast image will be narrow and centered toward the middle of gray scale.

A dark image will have the histogram that are concentrated on the dark side of gray scale.

A bright image will have the histogram whose component are biased toward high side of gray scale.

A high contrast image will have the histogram that covers wide range of gray scale and the distribution of pixel is approximately UNIFORM.

Right choice is (c) The histogram that is narrow and centered toward the middle of gray scale

Easiest explanation: The histogram plot is nk VERSUS rk. So, the histogram of a low contrast image will be narrow and centered toward the middle of gray scale.

A dark image will have the histogram that are CONCENTRATED on the dark side of gray scale.

A bright image will have the histogram WHOSE COMPONENT are BIASED toward high side of gray scale.

A high contrast image will have the histogram that covers wide range of gray scale and the distribution of pixel is approximately uniform.

Right answer is (a) 1

Explanation: A normalized histogram. P(rk) = nk / n

Where, n is total NUMBER of pixels in image, rk the kth GRAY level and nk total pixels with gray level rk.

Here, p(rk) gives the PROBABILITY of occurrence of rk.

Correct answer is (d) NONE of the mentioned

Easy explanation: All the mentioned devices uses GAMMA CORRECTION and so power-law transformation is generally of use in such CASE.

The correct CHOICE is (d) If each value of histogram is divided by total number of pixels in IMAGE, say n, P(rk)=nk / n

The best explanation: To normalize a histogram each of its value is divided by total number of pixels in image, say n. p(rk) = nk / n.

Right choice is (c) The CONTRAST increases and the detail decreases

Easy explanation: In power-law transformation as gamma decreases is increase in IMAGE DETAILS HOWEVER, the contrast reduces.

Right choice is (b) Power-law transformations

The explanation: The CRT devices has a power function relation between INTENSITY and volt response.

In such devices OUTPUT appears DARKER than input. So, gamma CORRECTION is a MUST in this case.

The correct answer is (d) A process to correct power-law transformation response phenomena

Best EXPLANATION: The exponent used in power-law transformation is CALLED gamma. So, using the ᵞ value, either ᵞ < 1 or ᵞ> 1, various RESPONSES are obtained.

Right answer is (a) The log TRANSFORMATION COMPRESSES the dynamic RANGE of images

Explanation: For compressing gray-levels in an image, power-law transformation is more versatile than log transformation, but log transformation has an IMPORTANT characteristics of compressing dynamic ranges of pixels having a large variation of values.

The CORRECT OPTION is (a) Log transformations

Explanation: Log transformation derives a narrow range of gray-level values in INPUT image to WIDER range of gray-levels in the OUTPUT image, and does performs the above given transformation.

The inverse-log is applied for the opposite.

Correct answer is (C) Negative transformations

Explanation: Negative TRANSFORMATION reverses the intensity LEVELS in the image and produces an equivalent photographic negative. So, WELL suited for the above GIVEN condition.

The CORRECT choice is (b) s = crᵞ, c and ᵞ are POSITIVE constants

To explain: The EXPRESSION for power-law TRANSFORMATION is given as: s = crᵞ, c and ᵞ are positive constants.

The correct ANSWER is (c) s = c LOG (1 + r), c is a constant and r ≥ 0

The EXPLANATION is: The EXPRESSION for log transformation is GIVEN as: s = c log (1 + r), c is a constant and r ≥ 0.

The correct ANSWER is (a) s = L – 1 – R

Explanation: The expression for negative TRANSFORMATION is given as: s = L – 1 – r.

Right answer is (b) negative and IDENTITY transformations

For explanation I WOULD SAY: For IMAGE Enhancement gray-level transformation shows three basic function that are:

LINEARITY for negative and identity transformation

Logarithmic for log and inverse-log transformation, and

Power-law for nth and nthroot transformations.

The CORRECT choice is (a) Log and inverse-log transformations

Easiest EXPLANATION: For Image Enhancement gray-level TRANSFORMATION shows THREE basic function that are:

Linearity for negative and identity transformation

Logarithmic for log and inverse-log transformation, and

Power-law for nth and nthroot transformations.

The correct option is (b) Contrast stretching

Best explanation: For a gray-level transformation FUNCTION “s=T(r)”, where r and s are the gray-level of f(x, y) (input image) and g(x, y) (OUTPUT image) respectively at any point (x, y).

Then the TECHNIQUE, contrast stretching compresses the value of r below m by transformation function into a narrow range of s, towards BLACK and BRIGHTENS the value of r above m.

Right choice is (B) NEGATIVE and identity transformations

The EXPLANATION is: For Image ENHANCEMENT gray-level TRANSFORMATION shows three basic function that are:

Linearity for negative and identity transformation

Logarithmic for log and inverse-log transformation, and

Power-law for nth and nthroot transformations.

Correct CHOICE is (c) Mask PROCESSING

For explanation: The above mentioned approach is BASED on the use of masks. A mask is a small m*n 2-D array in which the values of mask coefficients determine the nature of the process and IMAGE Enhancement on such is called Mask Processing or Filtering.

The CORRECT ANSWER is (c) Spatial domain and FREQUENCY domain respectively

To elaborate: Spatial domain itself refers to the image plane, and approaches in this category are based on direct manipulation of pixels in an image.

Techniques based on Frequency domain PROCESSING are based on modifying the Fourier TRANSFORM of an image.

Right choice is (b) D(p, Q) = |X – s| + |y – t|

For EXPLANATION: The city-block distance for PIXELS p(x, y), q(s, t) is the D4 distance given by:

D(p, q) = |x – s| + |y – t|.

Right choice is (d) All of the mentioned

The EXPLANATION is: For pixels P(x, y), Q(s, t), and z(v, W), D is a distance function or METRIC if:

(i) D(p, q) ≥ 0,(D(p, q) = 0if p=q),

(ii) D(p, q) = D(q, p), and

(iii) D(p, z) ≤ D(p, q) + D(q, z).

Right answer is (a) D(p, q) = [(x – s)^2 + (y – t)^2]^1/2

To ELABORATE: The EUCLIDEAN DISTANCE for pixels p(x, y), q(s, t) is:

D(p, q) = [(x – s)^2 + (y – t)^2]^1/2.

Right choice is (d) All of the mentioned

The EXPLANATION is: For a SUBSET of pixels in an image S

For any pixel P in S, the set of pixels is called a connected component of S if connected to p in S. The set S is called a connected set if it only has ONE connected component.

S, is a region of the image if S is a connected set.

The correct option is (C) Any of the mentioned

The explanation: Mixed adjacency is a modified form of 8-adjacency.

The above CONDITIONED Two pixels P and q are m-adjacent if:

q is in N4(p), or

q is in ND(p) and the set N4(p) ∩ N4(q) has no pixels WHOSE values are from V.

Correct choice is (a) “In general-purpose for a digital image of zooming and shrinking, where Bilinear INTERPOLATION generally is the method of choice over nearest neighbor Interpolation”.

The EXPLANATION: For CASE 32*32 to 1024*1024, the DATA is rather lost in nearest neighbor Interpolation, but the result of Bilinear Interpolation remains REASONABLY good for the same.

Right choice is (b) (X+1, y), (x-1, y), (x, y+1), (x, y-1), (x+1, y+1), (x+1, y-1), (x-1, y+1), (x-1, y-1)

To elaborate: The set of PIXELS of 4-neighbors of p and Diagonal neighbors of p TOGETHER are called as 8-neighbors of PIXEL p(x, y).

Correct option is (a) Aliasing effect

To EXPLAIN: Although Image Shrinking USES the grid analogy of nearest neighbor interpolation, but that we now expand the grid to fit over the original image, do gray-level nearest neighbor or bilinear interpolation, causing the possible aliasing effect, and then shrink the grid BACK to its original specified size.

The correct OPTION is (c) Assign gray level to the new pixel using its FOUR nearest neighbors

Best explanation: Bilinear INTERPOLATION uses the four nearest neighbors of the new pixel. Let (x’, y’) is the coordinates of a POINT in the zoomed IMAGE and the gray level assigned to the point is v(x, y’).

For bilinear interpolation, the assigned gray level is given by

 v(x’, y’) = ax’ + by’ + cx’y’ + d

Here, a, b, c and d are determined from the four equations in four unknowns that can be written using the four nearest neighbors of point (x’, y’).