IGNOU MMTE004 Solved Assignment 2024  M.Sc. MACS
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IGNOU MMTE004 Assignment Question Paper 2024
mmte004solvedassignment2024584fdbf367d440959ab50c19a7e9ae3b

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) and$A(1,2)$ $B(10,8)$ $B(10,8)$ 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${y}^{2}=4x$ ${y}^{2}=4x$ y^(2)=4x y^2=4 x in the interval${y}^{2}=4x$ $[0,10]$ $[0,10]$ [0,10] [0,10] .$[0,10]$ 
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) . Prepare a sorted edge list, and then make the active edge list for the scanlines$(15,30)$ $y=5,10,15,20,25,30$ $y=5,10,15,20,25,30$ y=5,10,15,20,25,30 y=5,10,15,20,25,30 .$y=5,10,15,20,25,30$
b) Develop and implement the flood fill algorithm.
c) Prove or disprove: "Multiplication of transformation matrices for two successive rotations is commutative." 
a) Transform the quadrilateral
$ABCD$ $ABCD$ ABCD A B C D with vertices$ABCD$ $A(1,0),B(4,1),C(5,3)$ $A(1,0),B(4,1),C(5,3)$ A(1,0),B(4,1),C(5,3) A(1,0), B(4,1), C(5,3) and$A(1,0),B(4,1),C(5,3)$ $D(1,5)$ $D(1,5)$ D(1,5) D(1,5) under a translation by the point$D(1,5)$ $(4,5)$ $(4,5)$ (4,5) (4,5) followed by a counterclockwise rotation by an angle of$(4,5)$ ${45}^{\circ}$ ${45}^{\circ}$ 45^(@) 45^{\circ} .${45}^{\circ}$
b) If you perform an$x$ $x$ x x direction shear transformation, and then a$x$ $y$ $y$ y y 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.$y$
c) Let$\mathrm{W}$ $\mathrm{W}$ W \mathrm{W} be a window with corners$\mathrm{W}$ $(0,0),(8,0),(8,4)$ $(0,0),(8,0),(8,4)$ (0,0),(8,0),(8,4) (0,0),(8,0),(8,4) and$(0,0),(8,0),(8,4)$ $(0,4)$ $(0,4)$ (0,4) (0,4) . Clip a triangle with vertices$(0,4)$ $(1,1),(10,2)$ $(1,1),(10,2)$ (1,1),(10,2) (1,1),(10,2) and$(1,1),(10,2)$ $(5,9)$ $(5,9)$ (5,9) (5,9) against the window$(5,9)$ $\mathrm{W}$ $\mathrm{W}$ W \mathrm{W} by tracing Liang Barskey line clipping algorithm.$\mathrm{W}$ 
a) Write a boundary fill procedure to fill an 8connected region.
b) Let$\mathrm{W}$ $\mathrm{W}$ W \mathrm{W} be the window having two diagonally opposite corners at$\mathrm{W}$ $(10,2)$ $(10,2)$ (10,2) (10,2) and (30, 15). Trace the CohenSutherland line clipping algorithm for the line segment joining the points$(10,2)$ $(0,0)$ $(0,0)$ (0,0) (0,0) and$(0,0)$ $(15,30)$ $(15,30)$ (15,30) (15,30) .$(15,30)$ 
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=y x=y if it is viewed from the point$x=y$ $(1,2,0)$ $(1,2,0)$ (1,2,0) (1,2,0) ? Justify your answer.$(1,2,0)$
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) . The view plane normal vector is$(1,1,2)$ $(4,2,5)$ $(4,2,5)$ (4,2,5) (4,2,5) and the view up vector is$(4,2,5)$ $(1,4,0)$ $(1,4,0)$ (1,4,0) (1,4,0) .$(1,4,0)$ 
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) and has normal vector$P(4,5,3)$ $(1,2,1)$ $(1,2,1)$ (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 anticlockwise direction. 
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=x y=x is equivalent to reflection along the$y=x$ $y$ $y$ y y axis followed by a counterclockwise rotation by$y$ ${90}^{\circ}$ ${90}^{\circ}$ 90^(@) 90^{\circ} .${90}^{\circ}$
c) Shear a square whose opposite vertices are at$(1,1)$ $(1,1)$ (1,1) (1,1) and$(1,1)$ $(2,2)$ $(2,2)$ (2,2) (2,2) by$(2,2)$
i) 2 units along the$x$ $x$ x x axis and reference line$x$ $y=0$ $y=0$ y=0 y=0 .$y=0$
ii) 4 units along the$y$ $y$ y y axis and reference line$y$ $x=0$ $x=0$ x=0 x=0 .$x=0$
MMTE004 Sample Solution 2024
mmte004solvedassignment2024ss–8e24e61006c94b4384f6a5bf6ef5ab5c
 a) Explain what do you understand by the terms persistence, refresh rate, resolution, aspect ratio, horizontal and vertical retrace.

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.

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 fastpaced video content and gaming.

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.

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).

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.
 Initialize:
$x=1,y=2$ $x=1,y=2$ x=1,y=2 x = 1, y = 2 $x=1,y=2$  For
$x=1$ $x=1$ x=1 x = 1 to$x=1$ $10$ $10$ 10 10 :$10$  Plot the pixel at
$(x,\text{round}(y))$ $(x,\text{round}(y))$ (x,”round”(y)) (x, \text{round}(y)) $(x,\text{round}(y))$  Increment
$x$ $x$ x x by 1$x$  Increment
$y$ $y$ y y by$y$ $m=\frac{2}{3}$ $m=\frac{2}{3}$ m=(2)/(3) m = \frac{2}{3} $m=\frac{2}{3}$
 Plot the pixel at
Step  Rounded 
Pixel Position  

1  1  2  2  (1, 2) 
2  2  3  (2, 3)  
3  3  4  (3, 4)  
4  4  4  (4, 4)  
5  5  5  (5, 5)  
6  6  6  (6, 6)  
7  7  6  (7, 6)  
8  8  7  (8, 7)  
9  9  8  (9, 8)  
10  10  8  (10, 8) 

Initialization:
 Start with the initial point
${P}_{0}(0,0)$ ${P}_{0}(0,0)$ P_(0)(0,0) P_0(0, 0) .${P}_{0}(0,0)$  Determine the region where the curve is steep or shallow. For the curve
${y}^{2}=4x$ ${y}^{2}=4x$ y^(2)=4x y^2 = 4x , the curve is shallow in the interval${y}^{2}=4x$ $[0,10]$ $[0,10]$ [0,10] [0, 10] since$[0,10]$ $dy/dx<1$ $dy/dx<1$ dy//dx < 1 dy/dx < 1 .$dy/dx<1$
 Start with the initial point

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
${y}^{2}=4x$ ${y}^{2}=4x$ y^(2)=4x y^2 = 4x , the decision parameter at any point${y}^{2}=4x$ $(x,y)$ $(x,y)$ (x,y) (x, y) can be defined as:$(x,y)$ $$d={y}^{2}4x$$ $$d={y}^{2}4x$$ d=y^(2)4x d = y^2 – 4x $$d={y}^{2}4x$$  For the initial point
${P}_{0}(0,0)$ ${P}_{0}(0,0)$ P_(0)(0,0) P_0(0, 0) , the decision parameter is${P}_{0}(0,0)$ ${d}_{0}={0}^{2}4\times 0=0$ ${d}_{0}={0}^{2}4\times 0=0$ d_(0)=0^(2)4xx0=0 d_0 = 0^2 – 4 \times 0 = 0 .${d}_{0}={0}^{2}4\times 0=0$
 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

Iteration:
 For each step in the xdirection, we need to decide whether to increment the ycoordinate based on the decision parameter.
 If
$d<0$ $d<0$ d < 0 d < 0 , the next point is$d<0$ $(x+1,y)$ $(x+1,y)$ (x+1,y) (x + 1, y) , and we update the decision parameter as$(x+1,y)$ ${d}_{\text{new}}=d+4y+4$ ${d}_{\text{new}}=d+4y+4$ d_(“new”)=d+4y+4 d_{\text{new}} = d + 4y + 4 .${d}_{\text{new}}=d+4y+4$  If
$d\ge 0$ $d\ge 0$ d >= 0 d \geq 0 , the next point is$d\ge 0$ $(x+1,y+1)$ $(x+1,y+1)$ (x+1,y+1) (x + 1, y + 1) , and we update the decision parameter as$(x+1,y+1)$ ${d}_{\text{new}}=d+4y+8$ ${d}_{\text{new}}=d+4y+8$ d_(“new”)=d+4y+8 d_{\text{new}} = d + 4y + 8 .${d}_{\text{new}}=d+4y+8$  Repeat this process until
$x$ $x$ x x reaches 10.$x$

Symmetry:
 Since the curve is symmetric about the xaxis, for each point
$(x,y)$ $(x,y)$ (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.$(x,y)$
 Since the curve is symmetric about the xaxis, for each point
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