Digital Electronic MCQ Set 1
1. Which gray-level transformation increase the dynamic range of gray-level in the image?
a) Power-law transformations
b) Negative transformations
c) Contrast stretching
d) None of the mentioned
Answer
Answer: c [Reason:] Increasing the dynamic range of gray-levels in the image is the basic idea behind contrast stretching.
2. When is the contrast stretching transformation a linear function, for r and s as gray-value of image before and after processing respectively?
a) r1 = s1 and r2 = s2
b) r1 = r2, s1 = 0 and s2 = L – 1, L is the max gray value allowed
c) r1 = 1 and r2 = 0
d) None of the mentioned
Answer
Answer: a [Reason:] If r1 = s1 and r2 = s2 the contrast stretching transformation is a linear function.
3. When is the contrast stretching transformation a thresholding function, for r and s as gray-value of image before and after processing respectively?
a) r1 = s1 and r2 = s2
b) r1 = r2, s1 = 0 and s2 = L – 1, L is the max gray value allowed
c) r1 = 1 and r2 = 0
d) None of the mentioned
Answer
Answer: b [Reason:] If r1 = r2, s1 = 0 and s2 = L – 1, the contrast stretching transformation is a thresholding function.
4. What condition prevents the intensity artifacts to be created while processing with contrast stretching, if r and s are gray-values of image before and after processing respectively?
a) r1 = s1 and r2 = s2
b) r1 = r2, s1 = 0 and s2 = L – 1, L is the max gray value allowed
c) r1 = 1 and r2 = 0
d) r1 ≤ r2 and s1 ≤ s2
Answer
Answer: d [Reason:] While processing through contrast stretching, if r1 ≤ r2 and s1 ≤ s2 is maintained, the function remains single valued and so monotonically increasing. This helps in the prevention of creation of intensity artifacts.
5. A contrast stretching result been obtained by setting (r1, s1) = (rmin, 0) and (r2, s2) = (rmax, L – 1), where, r and s are gray-values of image before and after processing respectively, L is the max gray value allowed and rmax and rmin are maximum and minimum gray-values in image respectively. What should we term the transformation function if r1 = r2 = m, some mean gray-value.
a) Linear function
b) Thresholding function
c) Intermediate function
d) None of the mentioned
Answer
Answer: b [Reason:] From (r1, s1) = (rmin, 0) and (r2, s2) = (rmax, L – 1), we have s1 = 0 and s2 = L – 1 and if r1 = r2 = m is set then the result becomes r1 = r2, s1 = 0 and s2 = L – 1, i.e. a thresholding function.
6. A specific range of gray-levels highlighting is the basic idea of ____
a) Contrast stretching
b) Bit –plane slicing
c) Thresholding
d) Gray-level slicing
Answer
Answer: d [Reason:] gray-level slicing is being done by two approach: One approach is to give all gray level of a specific range high value and a low value to all other gray levels.
Second approach is to brighten the pixels gray-value of interest and preserve the background.
I.e. in both highlighting of a specific range of gray-level is been done.
7. What is/are the approach(s) of the gray-level slicing?
a) To give all gray level of a specific range high value and a low value to all other gray levels
b) To brighten the pixels gray-value of interest and preserve the background
c) All of the mentioned
d) None of the mentioned
Answer
Answer: c [Reason:] There are basically two approach of gray-level slicing:
One approach is to give all gray level of a specific range high value and a low value to all other gray levels.
Second approach is to brighten the pixels gray-value of interest and preserve the background.
8. Which of the following transform produces a binary image after processing?
a) Contrast stretching
b) Gray-level slicing
c) All of the mentioned
d) None of the mentioned
Answer
Answer: c [Reason:] The approach of gray-level slicing “to give all gray level of a specific range high value and a low value to all other gray levels” produces a binary image.
One of the transformation in Contrast stretching darkens the value of r (input image gray-level) below m (some predefined gray-value) and brightens the value of r above m, giving a binary image as result.
9. Specific bit contribution in the image highlighting is the basic idea of ____
a) Contrast stretching
b) Bit –plane slicing
c) Thresholding
d) Gray-level slicing
Answer
Answer: b [Reason:] Bit-plane slicing highlights the contribution of specific bits made to total image, instead of highlighting a specific gray-level range.
10. In bit-plane slicing if an image is represented by 8 bits and is composed of eight 1-bit plane, with plane 0 showing least significant bit and plane 7 showing most significant bit. Then, which plane(s) contain the majority of visually significant data.
a) Plane 4, 5, 6, 7
b) Plane 0, 1, 2, 3
c) Plane 0
d) Plane 2, 3, 4, 5
Answer
Answer: a [Reason:] In bit-plane slicing, for the given data, the higher-ordered bits (mostly top four) contains most of the data visually signified.
11. Which of the following helps to obtain the number of bits to be used to quantize each pixel.
a) Gray-level slicing
b) Contrast stretching
c) Contouring
d) Bit-plane slicing
Answer
Answer: d [Reason:] Bits-plane slicing helps in obtaining the importance played by each bit in the image by separating the image into bit-planes.
Digital Electronic MCQ Set 2
1. What does the total number of pixels in the region defines?
a) Perimeter
b) Area
c) Intensity
d) Brightness
Answer
Answer: b [Reason:] The area of a region is defined by the total number of pixels in the region. The perimeter is given the number of pixels along the length of the boundary of the region.
2. What is the unit of compactness of a region?
a) Meter
b) Meter2
c) No units
d) Meter-1
Answer
Answer: c [Reason:] The compactness of a region is defined as (perimeter)2/area. Thus, the compactness of a region is a dimensionless quantity.
3. For which of the following regions, compactness is minimal?
a) Rectangle
b) Square
c) Irregular
d) Disk
Answer
Answer: d [Reason:] We know that, compactness of a region is defined as (perimeter)2/area. Thus, disk shaped region has a minimal value of this ratio and hence the minimal compactness.
4. Compactness is insensitive to orientation.
a) True
b) False
Answer
Answer: a [Reason:] With the exception of errors introduced by the rotation of the digital image, we can state that compactness of a region is insensitive to the orientation of the image.
5. Which of the following measures are not used to describe a region?
a) Mean and median of grey values
b) Minimum and maximum of grey values
c) Number of pixels alone
d) Number of pixels above and below mean
Answer
Answer: c [Reason:] Some of the measures which are used to describe a region are mean and median of grey values, minimum and maximum of grey values and number of pixels above and below mean. The area of the region, i.e., the total number of pixels in the region cannot alone describe the region.
6. We cannot use normalized area as one of the region descriptor.
a) True
b) False
Answer
Answer: b [Reason:] One of the regional descriptor is normalized area. It can be quite useful to extract information from the image. In satellite images of earth, the data can be refined by normalized it with respect to land mass per region.
7. What is the study of properties of a figure that are unaffected by any deformation?
a) Topology
b) Geography
c) Statistics
d) Deformation
Answer
Answer: a [Reason:] We can define topology as the study of properties of a figure that are unaffected by any deformation, as long as there is no joining or tearing of the figure. We use topological properties in the region description.
8. On which of the following operation of an image, the topology of the region changes?
a) Stretching
b) Rotation
c) Folding
d) Change in distance measure
Answer
Answer: c [Reason:] If a topological descriptor is defined by the number of holes in an image, then the number of holes will not vary if the image is stretched or rotated. The number of holes in the region will change only if the image is torn or folded.
9. Topological properties don’t depend on the distance measures.
a) True
b) False
Answer
Answer: a [Reason:] We know that, as stretching affects distance, topological properties do not depend on the notion of distance or any properties implicitly based on the concept of distance measures.
10. What is the Euler number of the image shown below?
a) 0
b) 1
c) 2
d) -1
Answer
Answer: d [Reason:] The image shown in the question has two holes and one connected components. So, the Euler number E is given as 1-2=-1.
11. What is the Euler number of a region with polygonal network containing V,Q and F as the number of vertices, edges and faces respectively?
a) V+Q+F
b) V-Q+F
c) V+Q-F
d) V-Q-F
Answer
Answer: b [Reason:] It is very important to classify the polygonal network. Let V,Q and F denote the number of vertices, edges and faces respectively. Then,
V-Q+F=C-H
Where C,H represents the number of connected components and number of holes in the region respectively. So, the Euler number E is given by V-Q+F.
12. What is the Euler number of the region shown in the figure below?
a) 1
b) -2
c) -1
d) 2
Answer
Answer: b [Reason:] The polygonal network given in the figure has 7 vertices, 11 edges and 2 faces. Thus the Euler number is given by the formula,
E=V-Q+F=7-11+2=-2.
13. The texture of the region provides measure of which of the following properties?
a) Smoothness alone
b) Coarseness alone
c) Regularity alone
d) Smoothness, coarseness and regularity
Answer
Answer: d [Reason:] One of the important approach to region description is texture content. This helps to provide the measure of some of the important properties of an image like smoothness, coarseness and regularity of the region.
14. Structural techniques deal with the arrangement of image primitives.
a) True
b) False
Answer
Answer: a [Reason:] Structural techniques deal with the arrangement of image primitives, such as the description of the texture based on the regularly spaced parallel lines.
15. Which of the following techniques is based on the Fourier transform?
a) Structural
b) Spectral
c) Statistical
d) Topological
Answer
Answer: b [Reason:] Spectral techniques are based on properties of the Fourier spectrum and are used primarily to detect global periodicity in an image by identifying high energy, narrow peaks in the image.
Digital Electronic MCQ Set 3
1. The technique of Enhancement that has a specified Histogram processed image as result, is called?
a) Histogram Linearization
b) Histogram Equalization
c) Histogram Matching
d) None of the mentioned
Answer
Answer: c [Reason:] 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.
2. In Histogram Matching r and z are gray level of input and output image and p stands for PDF, then, what does pz(z) stands for?
a) Specific probability density function
b) Specified pixel distribution function
c) Specific pixel density function
d) Specified probability density function
Answer
Answer: d [Reason:] In Histogram Matching, pr(r) is estimated from input image while pz(z) is Specified probability density function that output image is supposed to have.
3. Inverse transformation plays an important role in which of the following Histogram processing Techniques?
a) Histogram Linearization
b) Histogram Equalization
c) Histogram Matching
d) None of the mentioned
Answer
Answer: c [Reason:] 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.
4. 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, to confirm the single value and monotonous of G-1 what of the following is/are required?
a) G must be strictly monotonic
b) G must be strictly decreasing
c) All of the mentioned
d) None of the mentioned
Answer
Answer: a [Reason:] 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.
5. Which of the following histogram processing techniques is global?
a) Histogram Linearization
b) Histogram Specification
c) Histogram Matching
d) All of the mentioned
Answer
Answer: d [Reason:] All of the mentioned methods modifies the pixel value by transformations that are based on the gray-level of the whole image.
6. What happens to the output image when global Histogram equalization method is applied on smooth and noisy area of an image?
a) The contrast increases little bit with considerable enhancement of noise
b) The result would have a fine noise texture
c) All of the mentioned
d) None of the mentioned
Answer
Answer: a [Reason:] 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.
7. Let us suppose an image containing a quite small square under a large dark square with both having very close gray level values. If an image contains some of this such that the small squares can’t be visualized and some noise blurred enough to reduce its noise content as shown in fig. below, Which of the following method would be preferred for obtaining the small square clear enough?
Figure: original image.
a) Global histogram equalization
b) Local histogram equalization
c) All of the mentioned
d) None of the mentioned
Answer
Answer: b [Reason:] For global histogram enhancement, the small squares have a very close gray value with larger square and have a very small size to be influenced by global histogram equalization method.
But, local histogram enhancement using a 7*7 neighborhood reveals the small square.
(a) Original image. (b) Result using global histogram equalization. (c) Result using local histogram equalization using 7*7 neighborhood about each pixel.
8. In terms of enhancement, what does mean and variance refers to?
a) Average contrast and average gray level respectively
b) Average gray level and average contrast respectively
c) Average gray level in both
d) Average contrast in both
Answer
Answer: b [Reason:] 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.
9. For a local enhancement using mean and variance, there is one condition: ms(x, y) ≤ k0 MG, where, MG is global mean, k0 a constant and ms(x, y) a measure of gray value as light or dark at point (x, y). Then, which fact is true for k0?
a) It is a negative constant with values less than -1.0
b) It is a positive constant with values less than 1.0
c) It is an integer constant with values between -1.0 and 1.0
d) None of the mentioned
Answer
Answer: b [Reason:] In the condition ms(x, y) ≤ k0 MG, k0 is a positive constants whose value is always less than 1.0.
10. For a local enhancement using mean and variance, there is one condition: σs(x, y) ≤ k2DG, where, MDG is global standard deviation, k2 a positive constant and σs(x, y) a measure of contrast at point (x, y). Then, which fact is true for k2 if its values is less than 1.0?
a) Enhancement is being done on light areas
b) Enhancement is being done on dark areas
c) Enhancement is being done independent of value of k0
d) None of the mentioned
Answer
Answer: b [Reason:] In the condition σs(x, y) ≤ k2DG, k0 is a positive constants that helps in enhancing light areas if value is greater than 1.0 and dark areas if value is less than 1.0.
11. For a local enhancement using mean and variance, there is one condition: σs(x, y) ≤ k2DG, where, MDG is global standard deviation, k2 a positive constant and σs(x, y) a measure of contrast at point (x, y). Then, which fact is true for k2 if its values is greater than 1.0?
a) Enhancement is being done on light areas
b) Enhancement is being done on dark areas
c) Enhancement is being done independent of value of k0
d) None of the mentioned
Answer
Answer: a [Reason:] In the condition σs(x, y) ≤ k2DG, k0 is a positive constants that helps in enhancing light areas if value is greater than 1.0 and dark areas if value is less than 1.0.
12. What is standard deviation value for constant area?
a) 0
b) 1
c) -1
d) None of the mentioned
Answer
Answer: a [Reason:] Standard deviation is given by:
that results 0 for constant areas.
13. For a local enhancement using mean and variance, what happens if the lowest value of contrast is not restricted as per the willingness of acceptance of value?
a) There wouldn’t be any enhancement
b) Enhancement will occur for areas with standard deviation value > 1
c) Enhancement of the constant areas will also be the part of procedure
d) Enhancement will occur for areas with standard deviation value > 0 and < 1
Answer
Answer: c [Reason:] If the lowest value of contrast is not restricted as per the willingness of acceptance of value, the Enhancement of the constant areas will also be the part of procedure, since a constant area has standard deviation value 0.
Digital Electronic MCQ Set 4
1. The principal factor to determine the spatial resolution of an image is _______
a) Quantization
b) Sampling
c) Contrast
d) Dynamic range
Answer
Answer: b [Reason:] The spatial resolution of an image principally determine by Sampling.
2. What causes the effect, imperceptible set of very fine ridge like structures in areas of smooth gray levels?
a) Caused by the use of an insufficient number of gray levels in smooth areas of a digital image
b) Caused by the use of huge number of gray levels in smooth areas of a digital image
c) All of the mentioned
d) None of the mentioned
Answer
Answer: a [Reason:] The set of very fine ridge like structures in area of smooth gray levels generally is quite visible in images displayed using 16 or less uniformly spaced gray levels.
3. What is the name of the effect caused by the use of an insufficient number of gray levels in smooth areas of a digital image?
a) Dynamic range
b) Ridging
c) Graininess
d) False contouring
Answer
Answer: d [Reason:] The effect, caused due to insufficient number of gray levels in smooth areas of a digital image, is called false contouring, so called because the ridges resemble topographic contours in a map.
4. Using rough rule of thumb, and assuming powers of 2 for convenience, what image size are about the smallest images that can be expected to be reasonably free of objectionable sampling checkerboards and false contouring?
a) 512*512pixels and 16 gray levels
b) 256*256pixels and 64 gray levels
c) 64*64pixels and 16 gray levels
d) 32*32pixels and 32 gray levels
Answer
Answer: b [Reason:] An image of 128*128pixels shows a pronounced checkerboard pattern, while for 256*256pixels image a minimum gray level of 64 is required to remove false contouring.
Also the effect is quite visible in images displayed using 16 or less uniformly spaced gray levels.
5. What does a shift up and right in the curves of isopreference curve simply means? Verify in terms of N (number of pixels) and k (L=2k, L is the gray level) values.
a) Smaller values for N and k, implies a better picture quality
b) Larger values for N and k, implies low picture quality
c) Larger values for N and k, implies better picture quality
d) Smaller values for N and k, implies low picture quality
Answer
Answer: c [Reason:] Points lying on an isopreference curve correspond to images of equal subjective quality. It was found that the isopreference curves tended to shift right and upward with the details of the image. So, a shift up and right in the curves simply means larger values for N and k, implying better picture quality.
6. How does the curves behave to the detail in the image in isopreference curve?
a) Curves tend to become more vertical as the detail in the image decreases
b) Curves tend to become less vertical as the detail in the image increases
c) Curves tend to become less vertical as the detail in the image decreases
d) Curves tend to become more vertical as the detail in the image increases
Answer
Answer: d [Reason:] The curves in isopreference curve tend to become more vertical as the detail in the image increases.
The right side graph shows the same, curve for crowd is nearly vertical.
7. For an image with a large amount of detail, if the value of N (number of pixels) is fixed then what is the gray level dependency in the perceived quality of this type of image?
a) Totally independent of the number of gray levels used
b) Nearly independent of the number of gray levels used
c) Highly dependent of the number of gray levels used
d) None of the mentioned
Answer
Answer: b [Reason:] For image with high details of the image only a few gray levels may be needed.
8. What is a band-limited function?
a) A function of limited duration whose highest frequency is finite
b) A function of limited duration whose highest frequency is infinite
c) All of the mentioned
d) None of the mentioned
Answer
Answer: a [Reason:] Functions whose area under the curve is finite can be represented in terms of sines and cosines of various frequencies. The highest frequency is determined by the sine/cosine component is the highest “frequency content” of the function. If this highest frequency is finite and that the function is of unlimited duration, then, these functions are called band-limited functions.
9. For a band-limited function, which Theorem says that “if the function is sampled at a rate equal to or greater than twice its highest frequency, the original function can be recovered from its samples”?
a) Band-limitation theorem
b) Aliasing frequency theorem
c) Shannon sampling theorem
d) None of the mentioned
Answer
Answer: c [Reason:] For a band-limited function, Shannon sampling theorem says that “if the function is sampled at a rate equal to or greater than twice its highest frequency, the original function can be recovered from its samples”.
10. What is the name of the phenomenon that corrupts the sampled image, and how does it happen?
a) Shannon sampling, if the band-limited functions are undersampled
b) Shannon sampling, if the band-limited functions are oversampled
c) Aliasing, if the band-limited functions are undersampled
d) Aliasing, if the band-limited functions are oversampled
Answer
Answer: c [Reason:] If the band-limited functions is undersampled, then a phenomenon called aliasing corrupts the sampled image.
11. How aliasing does corrupts the sampled image?
a) By introducing additional frequency components to the sampled function
b) By removing some frequency components from the sampled function
c) All of the mentioned
d) None of the mentioned
Answer
Answer: a [Reason:] Aliasing corrupts the sampled image by introducing additional frequency components into the sampled function. These added components are called aliased frequencies.
12. How can one reduce the aliasing effect on an image?
a) By reducing the high-frequency components of image by blurring the image
b) By increasing the high-frequency components of image by blurring the image
c) By reducing the high-frequency components of image by clarifying the image
d) By increasing the high-frequency components of image by clarifying the image
Answer
Answer: a [Reason:] Aliasing corrupts the sampled image by introducing additional frequency components to the sampled function. So, the principal approach for reducing the aliasing effects on an image is to reduce its high-frequency components by blurring the image prior to sampling.
Digital Electronic MCQ Set 5
1. A pixel p at coordinates (x, y) has neighbors whose coordinates are given by:
(x+1, y), (x-1, y), (x, y+1), (x, y-1)
This set of pixels is called ______
a) 4-neighbors of p
b) Diagonal neighbors
c) 8-neighbors
d) None of the mentioned
Answer
Answer: a [Reason:] The given set of neighbor pixel are 1 unit distance to right, left, up and below respectively from pixel p(x, y). So, are called 4-neighbors of p.
2. A pixel p at coordinates (x, y) has neighbors whose coordinates are given by:
(x+1, y+1), (x+1, y-1), (x-1, y+1), (x-1, y-1)
This set of pixels is called ______
a) 4-neighbors of p
b) Diagonal neighbors
c) 8-neighbors
d) None of the mentioned
Answer
Answer: b [Reason:] The given set of neighbor pixel are 1 unit distance to right-up diagonal, right-down diagonal, left-up diagonal and left-down diagonal respectively from pixel p(x, y). So, are called Diagonal neighbors of p.
3. What is the set of pixels of 8-neighbors of pixel p at coordinates (x, y)?
a) (x+1, y), (x-1, y), (x, y+1), (x, y-1), (x+2, y), (x-2, y), (x, y+2), (x, y-2)
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)
c) (x+1, y+1), (x+1, y-1), (x-1, y+1), (x-1, y-1), (x+2, y+2), (x+2, y-2), (x-2, y+2), (x-2, y-2)
d) (x+2, y), (x-2, y), (x, y+2), (x, y-2), (x+2, y+2), (x+2, y-2), (x-2, y+2), (x-2, y-2)
Answer
Answer: b [Reason:] 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).
4. Two pixels p and q having gray values from V, the set of gray-level values used to define adjacency, are m-adjacent if:
a) q is in N4(p)
b) q is in ND(p) and the set N4(p) ∩ N4(q) has no pixels whose values are from V
c) Any of the mentioned
d) None of the mentioned
Answer
Answer: c [Reason:] 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.
5. Let S, a subset of pixels in an image, is said to be a connected set if:
a) If for any pixel p in S, the set of pixels that are connected to it in Sis only one
b) If it only has one connected component
c) If S is a region
d) All of the mentioned
Answer
Answer: d [Reason:] 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.
6. Let R be a subset of pixels in an image. How can we define the contour of R?
a) If R is a region, and the set of pixels in R have one or more neighbors that are not in R
b) If R is an entire image, then the set of pixels in the first and last rows and columns of R
c) All of the mentioned
d) None of the mentioned
Answer
Answer: c [Reason:] For a subset of pixels in an image R
The boundary or contour of a region R is the set of pixels in the region that have one or more neighbors that are not in R.
In case R is an entire image, then its boundary is defined as the set of pixels in the first and last rows and columns of the image.
7. For pixels p(x, y), q(s, t), and z(v, w), D is a distance function or metric if:
a) D(p, q) ≥ 0
b) D(p, q) = D(q, p)
c) D(p, z) ≤ D(p, q) + D(q, z)
d) All of the mentioned
Answer
Answer: d [Reason:] 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) = 0 if p=q),
(ii) D(p, q) = D(q, p), and
(iii) D(p, z) ≤ D(p, q) + D(q, z).
8. For pixels p(x, y), q(s, t), the Euclidean distance between p and q is defined as:
a) D(p, q) = [(x – s)2 + (y – t)2]1/2
b) D(p, q) = |x – s| + |y – t|
c) D(p, q) = max (|x – s| + |y – t|)
d) None of the mentioned
Answer
Answer: a [Reason:] The Euclidean distance for pixels p(x, y), q(s, t) is:
D(p, q) = [(x – s)2 + (y – t)2]1/2.
9. For pixels p(x, y), q(s, t), the city-block distance between p and q is defined as:
a) D(p, q) = [(x – s)2 + (y – t)2]1/2
b) D(p, q) = |x – s| + |y – t|
c) D(p, q) = max (|x – s| + |y – t|)
d) None of the mentioned
Answer
Answer: b [Reason:] 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|.
10. For pixels p(x, y), q(s, t), the chessboard distance between p and q is defined as:
a) D(p, q) = [(x – s)2 + (y – t)2]1/2
b) D(p, q) = |x – s| + |y – t|
c) D(p, q) = max (|x – s| + |y – t|)
d) None of the mentioned
Answer
Answer: c [Reason:] The chessboard distance for pixels p(x, y), q(s, t) is the D8 distance given by:
D(p, q) = max (|x – s| + |y – t|).
11. The domain that refers to image plane itself and the domain that refers to Fourier transform of an image is/are :
a) Spatial domain in both
b) Frequency domain in both
c) Spatial domain and Frequency domain respectively
d) Frequency domain and Spatial domain respectively
Answer
Answer: c [Reason:] 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.
12. What is the technique for a gray-level transformation function called, if the transformation would be to produce an image of higher contrast than the original by darkening the levels below some gray-level m and brightening the levels above m in the original image.
a) Contouring
b) Contrast stretching
c) Mask processing
d) Point processing
Answer
Answer: b [Reason:] 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.
13. For Image Enhancement a general-approach is to use a function of values of f (input image) in a predefined neighborhood of (x, y) to determine the value of g (output image) at (x, y). The techniques that uses such approaches are called ________
a) Contouring
b) Contrast stretching
c) Mask processing
d) None of the mentioned
Answer
Answer: c [Reason:] 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.
