IGNOU MMTE-004 Solved Assignment 2024, M.Sc. MACS

IGNOU MMTE-004 Solved Assignment 2024 | M.Sc. MACS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MMTE-004 Assignment Question Paper 2024

mmte-004-solved-assignment-2024-584fdbf3-67d4-4095-9ab5-0c19a7e9ae3b

mmte-004-solved-assignment-2024-584fdbf3-67d4-4095-9ab5-0c19a7e9ae3b

MMTE-004 Solved Assignment 2024
  1. a) Explain what do you understand by the terms persistence, refresh rate, resolution, aspect ratio, horizontal and vertical retrace.
    b) Compute the pixel positions along the line path of the lien joining the points A ( 1 , 2 ) A ( 1 , 2 ) A(1,2)A(1,2)A(1,2) and B ( 10 , 8 ) B ( 10 , 8 ) B(10,8)B(10,8)B(10,8).
    c) Using the midpoint method and symmetry in account, develop an efficient method for scan converting the curve y 2 = 4 x y 2 = 4 x y^(2)=4xy^2=4 xy2=4x in the interval [ 0 , 10 ] [ 0 , 10 ] [0,10][0,10][0,10].
  2. a) Consider a polygon with vertices at (5, 20), (12, 5), (15, 15), (25, 5), (30, 25), and ( 15 , 30 ) ( 15 , 30 ) (15,30)(15,30)(15,30). Prepare a sorted edge list, and then make the active edge list for the scanlines y = 5 , 10 , 15 , 20 , 25 , 30 y = 5 , 10 , 15 , 20 , 25 , 30 y=5,10,15,20,25,30y=5,10,15,20,25,30y=5,10,15,20,25,30.
    b) Develop and implement the flood fill algorithm.
    c) Prove or disprove: "Multiplication of transformation matrices for two successive rotations is commutative."
  3. a) Transform the quadrilateral A B C D A B C D ABCDA B C DABCD with vertices A ( 1 , 0 ) , B ( 4 , 1 ) , C ( 5 , 3 ) A ( 1 , 0 ) , B ( 4 , 1 ) , C ( 5 , 3 ) A(1,0),B(4,-1),C(5,3)A(1,0), B(4,-1), C(5,3)A(1,0),B(4,1),C(5,3) and D ( 1 , 5 ) D ( 1 , 5 ) D(-1,5)D(-1,5)D(1,5) under a translation by the point ( 4 , 5 ) ( 4 , 5 ) (4,5)(4,5)(4,5) followed by a counter-clockwise rotation by an angle of 45 45 45^(@)45^{\circ}45.
    b) If you perform an x x xxx-direction shear transformation, and then a y y yyy-direction shear transformation, will the result be the same as the one which is obtained when it is simultaneous shear in both the directions? Justify your answer.
    c) Let W W W\mathrm{W}W be a window with corners ( 0 , 0 ) , ( 8 , 0 ) , ( 8 , 4 ) ( 0 , 0 ) , ( 8 , 0 ) , ( 8 , 4 ) (0,0),(8,0),(8,4)(0,0),(8,0),(8,4)(0,0),(8,0),(8,4) and ( 0 , 4 ) ( 0 , 4 ) (0,4)(0,4)(0,4). Clip a triangle with vertices ( 1 , 1 ) , ( 10 , 2 ) ( 1 , 1 ) , ( 10 , 2 ) (1,1),(10,2)(1,1),(10,2)(1,1),(10,2) and ( 5 , 9 ) ( 5 , 9 ) (5,9)(5,9)(5,9) against the window W W W\mathrm{W}W by tracing Liang Barskey line clipping algorithm.
  4. a) Write a boundary fill procedure to fill an 8-connected region.
    b) Let W W W\mathrm{W}W be the window having two diagonally opposite corners at ( 10 , 2 ) ( 10 , 2 ) (10,2)(10,2)(10,2) and (30, 15). Trace the Cohen-Sutherland line clipping algorithm for the line segment joining the points ( 0 , 0 ) ( 0 , 0 ) (0,0)(0,0)(0,0) and ( 15 , 30 ) ( 15 , 30 ) (15,30)(15,30)(15,30).
  5. a) What is the difference between a parallel projection and a perspective projection? Explain with examples.
    b) What will be the perspective projection of a unit cube on the plane x = y x = y x=yx=yx=y if it is viewed from the point ( 1 , 2 , 0 ) ( 1 , 2 , 0 ) (1,2,0)(1,2,0)(1,2,0) ? Justify your answer.
    c) Transform the scene in the world coordinate system to the viewing coordinate system with viewpoint at ( 1 , 1 , 2 ) ( 1 , 1 , 2 ) (1,1,2)(1,1,2)(1,1,2). The view plane normal vector is ( 4 , 2 , 5 ) ( 4 , 2 , 5 ) (-4,2,5)(-4,2,5)(4,2,5) and the view up vector is ( 1 , 4 , 0 ) ( 1 , 4 , 0 ) (1,4,0)(1,4,0)(1,4,0).
  6. a) If the origin is taken as the centre of projection, then what will be the perspective projection when the projection plane passes through the point P ( 4 , 5 , 3 ) P ( 4 , 5 , 3 ) P(4,5,3)P(4,5,3)P(4,5,3) and has normal vector ( 1 , 2 , 1 ) ( 1 , 2 , 1 ) (1,2,-1)(1,2,-1)(1,2,1).
    b) Write a program that produces different views of a cuboid, that is, how the cuboid looks from the top, from the front or from the right.
    c) Write a code to continuously rotate a pentagon about a corner point in the anti-clockwise direction.
  7. a) Devise an efficient algorithm that takes advantage of symmetry properties to display a sine function.
    b) Prove that the reflection along the line y = x y = x y=-xy=-xy=x is equivalent to reflection along the y y yyy-axis followed by a counter-clockwise rotation by 90 90 90^(@)90^{\circ}90.
    c) Shear a square whose opposite vertices are at ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) and ( 2 , 2 ) ( 2 , 2 ) (2,2)(2,2)(2,2) by
    i) 2 units along the x x xxx-axis and reference line y = 0 y = 0 y=0y=0y=0.
    ii) 4 units along the y y yyy-axis and reference line x = 0 x = 0 x=0x=0x=0.
\(c=a\:cos\:B+b\:cos\:A\)

MMTE-004 Sample Solution 2024

mmte-004-solved-assignment-2024-ss–8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

mmte-004-solved-assignment-2024-ss–8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

MMTE-004 Solved Assignment 2024 SS
  1. a) Explain what do you understand by the terms persistence, refresh rate, resolution, aspect ratio, horizontal and vertical retrace.
Answer:
The terms persistence, refresh rate, resolution, aspect ratio, horizontal and vertical retrace are commonly used in the context of display technologies, particularly in computer monitors and television screens. Here’s what each term means:
  1. Persistence:
    • Persistence refers to the duration for which a phosphor coating on a cathode ray tube (CRT) screen or an individual pixel on an LCD/LED screen continues to emit light after being excited by an electron beam or electrical signal. High persistence results in less flicker, but can cause motion blur, while low persistence reduces motion blur but may increase flicker.
  2. Refresh Rate:
    • The refresh rate is the number of times per second that the display updates its image. It is measured in hertz (Hz). A higher refresh rate results in smoother motion and reduces flicker, making it particularly important for fast-paced video content and gaming.
  3. Resolution:
    • Resolution refers to the number of distinct pixels that can be displayed on a screen. It is usually expressed as the width x height, such as 1920×1080. Higher resolution means more pixels and therefore more detail and clarity in the displayed image.
  4. Aspect Ratio:
    • The aspect ratio is the ratio of the width of the display to its height. Common aspect ratios include 4:3 (traditional TV and computer monitors), 16:9 (widescreen displays and HDTVs), and 21:9 (ultrawide monitors).
  5. Horizontal and Vertical Retrace:
    • In CRT displays, horizontal and vertical retrace refer to the process by which the electron beam returns to the starting position after scanning a line (horizontal retrace) or a frame (vertical retrace). During retrace, the beam is turned off (blanked) to avoid drawing visible lines on the screen. In modern LCD/LED displays, these terms are less relevant, as they do not use electron beams for image formation.
These terms are important for understanding the performance and quality of display devices, as well as for optimizing settings for specific applications, such as gaming, graphic design, or video playback.
b) Compute the pixel positions along the line path of the lien joining the points A ( 1 , 2 ) A ( 1 , 2 ) A(1,2)A(1,2)A(1,2) and B ( 10 , 8 ) B ( 10 , 8 ) B(10,8)B(10,8)B(10,8).
Answer:
To compute the pixel positions along the line path joining the points A ( 1 , 2 ) A ( 1 , 2 ) A(1,2)A(1,2)A(1,2) and B ( 10 , 8 ) B ( 10 , 8 ) B(10,8)B(10,8)B(10,8), we can use the digital differential analyzer (DDA) algorithm, which is a simple and commonly used method for rasterizing lines.
First, we calculate the slope of the line:
m = y 2 y 1 x 2 x 1 = 8 2 10 1 = 6 9 = 2 3 m = y 2 y 1 x 2 x 1 = 8 2 10 1 = 6 9 = 2 3 m=(y_(2)-y_(1))/(x_(2)-x_(1))=(8-2)/(10-1)=(6)/(9)=(2)/(3)m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{8 – 2}{10 – 1} = \frac{6}{9} = \frac{2}{3}m=y2y1x2x1=82101=69=23
Since the slope m m mmm is less than 1, we will increment the x x xxx-coordinate in unit steps and compute the corresponding y y yyy-coordinates.
Starting from point A ( 1 , 2 ) A ( 1 , 2 ) A(1,2)A(1,2)A(1,2), we can compute the pixel positions as follows:
  1. Initialize: x = 1 , y = 2 x = 1 , y = 2 x=1,y=2x = 1, y = 2x=1,y=2
  2. For x = 1 x = 1 x=1x = 1x=1 to 10 10 101010:
    • Plot the pixel at ( x , round ( y ) ) ( x , round ( y ) ) (x,”round”(y))(x, \text{round}(y))(x,round(y))
    • Increment x x xxx by 1
    • Increment y y yyy by m = 2 3 m = 2 3 m=(2)/(3)m = \frac{2}{3}m=23
Let’s perform the calculations:
Step x x xxx y y yyy Rounded y y yyy Pixel Position
1 1 2 2 (1, 2)
2 2 2 + 2 3 2 + 2 3 2+(2)/(3)2 + \frac{2}{3}2+23 3 (2, 3)
3 3 2 + 4 3 2 + 4 3 2+(4)/(3)2 + \frac{4}{3}2+43 4 (3, 4)
4 4 2 + 2 2 + 2 2+22 + 22+2 4 (4, 4)
5 5 2 + 8 3 2 + 8 3 2+(8)/(3)2 + \frac{8}{3}2+83 5 (5, 5)
6 6 2 + 10 3 2 + 10 3 2+(10)/(3)2 + \frac{10}{3}2+103 6 (6, 6)
7 7 2 + 4 2 + 4 2+42 + 42+4 6 (7, 6)
8 8 2 + 14 3 2 + 14 3 2+(14)/(3)2 + \frac{14}{3}2+143 7 (8, 7)
9 9 2 + 16 3 2 + 16 3 2+(16)/(3)2 + \frac{16}{3}2+163 8 (9, 8)
10 10 2 + 6 2 + 6 2+62 + 62+6 8 (10, 8)
So, the pixel positions along the line path joining the points A ( 1 , 2 ) A ( 1 , 2 ) A(1,2)A(1,2)A(1,2) and B ( 10 , 8 ) B ( 10 , 8 ) B(10,8)B(10,8)B(10,8) are:
( 1 , 2 ) , ( 2 , 3 ) , ( 3 , 4 ) , ( 4 , 4 ) , ( 5 , 5 ) , ( 6 , 6 ) , ( 7 , 6 ) , ( 8 , 7 ) , ( 9 , 8 ) , ( 10 , 8 ) ( 1 , 2 ) , ( 2 , 3 ) , ( 3 , 4 ) , ( 4 , 4 ) , ( 5 , 5 ) , ( 6 , 6 ) , ( 7 , 6 ) , ( 8 , 7 ) , ( 9 , 8 ) , ( 10 , 8 ) (1,2),(2,3),(3,4),(4,4),(5,5),(6,6),(7,6),(8,7),(9,8),(10,8)(1, 2), (2, 3), (3, 4), (4, 4), (5, 5), (6, 6), (7, 6), (8, 7), (9, 8), (10, 8)(1,2),(2,3),(3,4),(4,4),(5,5),(6,6),(7,6),(8,7),(9,8),(10,8)
c) Using the midpoint method and symmetry in account, develop an efficient method for scan converting the curve y 2 = 4 x y 2 = 4 x y^(2)=4xy^2=4 xy2=4x in the interval [ 0 , 10 ] [ 0 , 10 ] [0,10][0,10][0,10].
Answer:
To scan-convert the curve y 2 = 4 x y 2 = 4 x y^(2)=4xy^2 = 4xy2=4x in the interval [ 0 , 10 ] [ 0 , 10 ] [0,10][0, 10][0,10] using the midpoint method and taking symmetry into account, we can follow these steps:
  1. Initialization:
    • Start with the initial point P 0 ( 0 , 0 ) P 0 ( 0 , 0 ) P_(0)(0,0)P_0(0, 0)P0(0,0).
    • Determine the region where the curve is steep or shallow. For the curve y 2 = 4 x y 2 = 4 x y^(2)=4xy^2 = 4xy2=4x, the curve is shallow in the interval [ 0 , 10 ] [ 0 , 10 ] [0,10][0, 10][0,10] since d y / d x < 1 d y / d x < 1 dy//dx < 1dy/dx < 1dy/dx<1.
  2. Decision Parameter:
    • The decision parameter for the midpoint method is based on the difference between the curve’s equation and the midpoint’s coordinates. For the curve y 2 = 4 x y 2 = 4 x y^(2)=4xy^2 = 4xy2=4x, the decision parameter at any point ( x , y ) ( x , y ) (x,y)(x, y)(x,y) can be defined as: d = y 2 4 x d = y 2 4 x d=y^(2)-4xd = y^2 – 4xd=y24x
    • For the initial point P 0 ( 0 , 0 ) P 0 ( 0 , 0 ) P_(0)(0,0)P_0(0, 0)P0(0,0), the decision parameter is d 0 = 0 2 4 × 0 = 0 d 0 = 0 2 4 × 0 = 0 d_(0)=0^(2)-4xx0=0d_0 = 0^2 – 4 \times 0 = 0d0=024×0=0.
  3. Iteration:
    • For each step in the x-direction, we need to decide whether to increment the y-coordinate based on the decision parameter.
    • If d < 0 d < 0 d < 0d < 0d<0, the next point is ( x + 1 , y ) ( x + 1 , y ) (x+1,y)(x + 1, y)(x+1,y), and we update the decision parameter as d new = d + 4 y + 4 d new = d + 4 y + 4 d_(“new”)=d+4y+4d_{\text{new}} = d + 4y + 4dnew=d+4y+4.
    • If d 0 d 0 d >= 0d \geq 0d0, the next point is ( x + 1 , y + 1 ) ( x + 1 , y + 1 ) (x+1,y+1)(x + 1, y + 1)(x+1,y+1), and we update the decision parameter as d new = d + 4 y + 8 d new = d + 4 y + 8 d_(“new”)=d+4y+8d_{\text{new}} = d + 4y + 8dnew=d+4y+8.
    • Repeat this process until x x xxx reaches 10.
  4. Symmetry:
    • Since the curve is symmetric about the x-axis, for each point ( x , y ) ( x , y ) (x,y)(x, y)(x,y) on the curve, there is a corresponding point ( x , y ) ( x , y ) (x,-y)(x, -y)(x,y). We can plot both points simultaneously to take advantage of this symmetry.
By following these steps, we can efficiently scan-convert the curve y 2 = 4 x y 2 = 4 x y^(2)=4xy^2 = 4xy2=4x in the interval [ 0 , 10 ] [ 0 , 10 ] [0,10][0, 10][0,10] using the midpoint method while taking advantage of the curve’s symmetry.

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\(b^2=c^2+a^2-2ac\:Cos\left(B\right)\)

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