计算机图形学答案,第七章
《计算机图形学》习题与解答
《计算机图形学》习题与解答第一章概述1. 试描述你所熟悉的计算机图形系统的硬软件环境。
计算机图形系统是计算机硬件、图形输入输出设备、计算机系统软件和图形软件的集合。
例如：计算机硬件采用PC、操作系统采用windows2000，图形输入设备有键盘、鼠标、光笔、触摸屏等，图形输出设备有CRT、LCD等，安装3D MAX图形软件。
2. 计算机图形系统与一般的计算机系统最主要的差别是什么？3. 图形硬件设备主要包括哪些？请按类别举出典型的物理设备?图形输入设备：鼠标、光笔、触摸屏和坐标数字化仪，以及图形扫描仪等。
图形显示设备：CRT、液晶显示器（LCD）等。
图形绘制设备：打印机、绘图仪等。
图形处理器：GPU（图形处理单元）、图形加速卡等等。
4. 为什么要制定图形软件标准？可分为哪两类？为了提高计算机图形软件、计算机图形的应用软件以及相关软件的编程人员在不同计算机和图形设备之间的可移植性。
图形软件标准通常是指图形系统及其相关应用系统中各界面之间进行数据传送和通信的接口标准，另外还有供图形应用程序调用的子程序功能及其格式标准。
5. 请列举出当前已成为国际标准的几种图形软件标准，并简述其主要功能。
（1）CGI(Computer Graphics Interface)，它所提供的主要功能集包括控制功能集、独立于设备的图形对象输出功能集、图段功能集、输入和应答功能集以及产生、修改、检索和显示以像素数据形式存储的光栅功能集。
（2）GKS(Graphcis Kernel System)，提供了应用程序和图形输入输出设备之间的接口，包括一系列交互和非交互式图形设备的全部图形处理功能。
主要功能如下：控制功能、输入输出功能、变换功能、图段功能、询问功能等。
6. 试列举计算机图形学的三个应用实例。
（1）CAD/CAM（2）VISC（3）VR.第二章光栅图形学1. 在图形设备上如何输出一个点？为输出一条任意斜率的直线，一般受到哪些因素影响？若图形设备是光栅图形显示器，光栅图形显示器可以看作是一个像素的矩阵，光栅图形显示器上的点是像素点的集合。
计算机图形学总复习
第一章：（蓝色字体为部分答案）●计算机图形学的定义？计算机图形学是研究通过计算机将数据转换为图形，并在专门显示设备上显示的原理、方法和技术的学科。
●计算机图形学常见的应用领域有哪些？（应用领域的标题）●计算机图形学的相关学科有哪些？和计算机图形学互逆的学科是？●CRT中为什么需要刷新？刷新频率是什么？由于荧光物质存在余晖时间，为了让荧光物质保持一个稳定的亮度值，电子束必须不断的重复描绘出原来的图形，这个过程叫做刷新刷新频率：每秒钟重绘屏幕的次数（次/秒、HZ)●彩色CRT和单色CRT的区别：⏹在荧光屏的内表面安装一个影孔板，用于精确定位像素的位置⏹CRT屏幕内部涂有很多组呈三角形的荧光粉，每一组由三个荧光点，三色荧光点由红、绿、蓝三基色组成（一组荧光点对应一个像素）⏹三支电子枪, 分别与三基色相对应●光栅扫描显示器中帧缓存是什么？位面是什么？⏹存储用于刷新的图像信息。
也就是存储屏幕上像素的颜色值。
⏹帧缓存的单位是位面。
⏹光栅扫描显示器屏幕上有多少个像素，该显示器的帧缓存的每个位面就有多少个一位存储器●1024×1024像素组成的24位真彩色光栅扫描显示器所需要的最小帧缓存是多少？第二章●什么是CDC？在微软基类库MFC中，CDC类是定义设备上下文对象的基类，所有绘图函数都在CDC基类中定义。
⏹简述CDC的4个派生类的名称，以及作用CClientDC类：显示器客户区设备上下文类CClientDC只能在窗口的客户区（不包括边框、标题栏、菜单栏以及状态栏的空白区域）进行绘图CMetaFileDCCMetaFileDC封装了在一个Windows图元文件中绘图的方法CPaintDC类该类一般用在响应WM_PAINT消息的成员函数OnPaint()中使用CWindowDC类整个窗口区域的显示器设备上下文类，包括客户区和非客户区（即窗口的边框、标题栏、菜单栏以及状态栏）⏹什么是映射模式？映射模式定义了Windows如何将绘图函数中指定的逻辑坐标映射为设备坐标输出到显示器或者打印机上。
交互式计算机图形学(第五版)1-7章课后题答案
Angel: Interactive Computer Graphics, Fifth Edition Chapter 1 Solutions1.1 The main advantage of the pipeline is that each primitive can be processed independently. Not only does this architecture lead to fast performance, it reduces memory requirements because we need not keep all objects available. The main disadvantage is that we cannot handle most global effects such as shadows, reflections, and blending in a physically correct manner.1.3 We derive this algorithm later in Chapter 6. First, we can form the tetrahedron by finding four equally spaced points on a unit sphere centered at the origin. One approach is to start with one point on the z axis(0, 0, 1). We then can place the other three points in a plane of constant z. One of these three points can be placed on the y axis. To satisfy the requirement that the points be equidistant, the point must be at(0, 2p2/3,−1/3). The other two can be found by symmetry to be at(−p6/3,−p2/3,−1/3) and (p6/3,−p2/3,−1/3).We can subdivide each face of the tetrahedron into four equilateral triangles by bisecting the sides and connecting the bisectors. However, the bisectors of the sides are not on the unit circle so we must push thesepoints out to the unit circle by scaling the values. We can continue this process recursively on each of the triangles created by the bisection process.1.5 In Exercise 1.4, we saw that we could intersect the line of which theline segment is part independently against each of the sides of the window. We could do this process iteratively, each time shortening the line segment if it intersects one side of the window.1.7 In a one–point perspective, two faces of the cube is parallel to the projection plane, while in a two–point perspective only the edges of the cube in one direction are parallel to the projection. In the general case of a three–point perspective there are three vanishing points and none of the edges of the cube are parallel to the projection plane.1.9 Each frame for a 480 x 640 pixel video display contains only about300k pixels whereas the 2000 x 3000 pixel movie frame has 6M pixels, or about 18 times as many as the video display. Thus, it can take 18 times asmuch time to render each frame if there is a lot of pixel-level calculations.1.11 There are single beam CRTs. One scheme is to arrange the phosphors in vertical stripes (red, green, blue, red, green, ....). The major difficulty is that the beam must change very rapidly, approximately three times as fast a each beam in a three beam system. The electronics in such a system the electronic components must also be much faster (and more expensive). Chapter 2 Solutions2.9 We can solve this problem separately in the x and y directions. The transformation is linear, that is xs = ax + b, ys = cy + d. We must maintain proportions, so that xs in the same relative position in the viewport as x is in the window, hencex − xminxmax − xmin=xs − uw,xs = u + wx − xminxmax − xmin.Likewiseys = v + hx − xminymax − ymin.2.11 Most practical tests work on a line by line basis. Usually we use scanlines, each of which corresponds to a row of pixels in the frame buffer. If we compute the intersections of the edges of the polygon with a line passing through it, these intersections can be ordered. The first intersection begins a set of points inside the polygon. The second intersection leaves the polygon, the third reenters and so on.2.13 There are two fundamental approaches: vertex lists and edge lists. With vertex lists we store the vertex locations in an array. The mesh is represented as a list of interior polygons (those polygons with no otherpolygons inside them). Each interior polygon is represented as an array of pointers into the vertex array. To draw the mesh, we traverse the list of interior polygons, drawing each polygon.One disadvantage of the vertex list is that if we wish to draw the edges inthe mesh, by rendering each polygon shared edges are drawn twice. Wecan avoid this problem by forming an edge list or edge array, each elementis a pair of pointers to vertices in the vertex array. Thus, we can draw each edge once by simply traversing the edge list. However, the simple edge list has no information on polygons and thus if we want to render the mesh in some other way such as by filling interior polygons we must add somethingto this data structure that gives information as to which edges form each polygon.A flexible mesh representation would consist of an edge list, a vertex listand a polygon list with pointers so we could know which edges belong to which polygons and which polygons share a given vertex.2.15 The Maxwell triangle corresponds to the triangle that connects thered, green, and blue vertices in the color cube.2.19 Consider the lines defined by the sides of the polygon. We can assigna direction for each of these lines by traversing the vertices in acounter-clockwise order. One very simple test is obtained by noting thatany point inside the object is on the left of each of these lines. Thus, if we substitute the point into the equation for each of the lines (ax+by+c), we should always get the same sign.2.23 There are eight vertices and thus 256 = 28 possible black/white colorings. If we remove symmetries (black/white and rotational) there are14 unique cases. See Angel, Interactive Computer Graphics (Third Edition) or the paper by Lorensen and Kline in the references.Chapter 3 Solutions3.1 The general problem is how to describe a set of characters that might have thickness, curvature, and holes (such as in the letters a and q). Suppose that we consider a simple example where each character can be approximated by a sequence of line segments. One possibility is to use a move/line system where 0 is a move and 1 a line. Then a character can be described by a sequence of the form (x0, y0, b0), (x1, y1, b1), (x2, y2, b2), .....where bi is a 0 or 1. This approach is used in the example in the OpenGL Programming Guide. A more elaborate font can be developed by using polygons instead of line segments.3.11 There are a couple of potential problems. One is that the application program can map different points in object coordinates to the same point in screen coordinates. Second, a given position on the screen when transformed back into object coordinates may lie outside the user’s window.3.19 Each scan is allocated 1/60 second. For a given scan we have to take 10% of the time for the vertical retrace which means that we start to draw scan line n at .9n/(60*1024) seconds from the beginning of the refresh. But allocating 10% of this time for the horizontal retrace we are at pixel m on this line at time .81nm/(60*1024).3.25 When the display is changing, primitives that move or are removed from the display will leave a trace or motion blur on the display as the phosphors persist. Long persistence phosphors have been used in text only displays where motion blur is less of a problem and the long persistence gives a very stable flicker-free image.Chapter 4 Solutions4.1 If the scaling matrix is uniform thenRS = RS(α, α, α) = αR = SRConsider R x(θ), if we multiply and use the standard trigonometric identities for the sine and cosine of the sum of two angles, we findR x(θ)R x(φ) = R x(θ + φ)By simply multiplying the matrices we findT(x1, y1, z1)T(x2, y2, z2) = T(x1 + x2, y1 + y2, z1 + z2)4.5 There are 12 degrees of freedom in the three–dimensional affine transformation. Consider a point p = [x, y, z, 1]T that is transformed top_ = [x_y_, z_, 1]T by the matrix M. Hence we have the relationshipp_ = Mp where M has 12 unknown coefficients but p and p_ are known. Thus we have 3 equations in 12 unknowns (the fourth equation is simplythe identity 1=1). If we have 4 such pairs of points we will have 12equations in 12 unknowns which could be solved for the elements of M.Thus if we know how a quadrilateral is transformed we can determine theaffine transformation.In two dimensions, there are 6 degrees of freedom in M but p and p_ haveonly x and y components. Hence if we know 3 points both before and after transformation, we will have 6 equations in 6 unknowns and thus in two dimensions if we know how a triangle is transformed we can determine theaffine transformation.4.7 It is easy to show by simply multiplying the matrices that theconcatenation of two rotations yields a rotation and that the concatenationof two translations yields a translation. If we look at the product of arotation and a translation, we find that the left three columns of RT arethe left three columns of R and the right column of RT is the rightcolumn of the translation matrix. If we now consider RTR_ where R_ is arotation matrix, the left three columns are exactly the same as the leftthree columns of RR_ and the and right column still has 1 as its bottomelement. Thus, the form is the same as RT with an altered rotation (whichis the concatenation of the two rotations) and an altered translation.Inductively, we can see that any further concatenations with rotations and translations do not alter this form.4.9 If we do a translation by -h we convert the problem to reflection abouta line passing through the origin. From m we can find an angle by whichwe can rotate so the line is aligned with either the x or y axis. Now reflectabout the x or y axis. Finally we undo the rotation and translation so the sequence is of the form T−1R−1SRT.4.11 The most sensible place to put the shear is second so that the instance transformation becomes I = TRHS. We can see that this order makessense if we consider a cube centered at the origin whose sides are alignedwith the axes. The scale gives us the desired size and proportions. Theshear then converts the right parallelepiped to a general parallelepiped.Finally we can orient this parallelepiped with a rotation and place it wheredesired with a translation. Note that the order I = TRSH will work too.4.13R = R z(θz)R y(θy)R x(θx) =⎡⎢⎢⎢⎣cos θy cos θz cos θz sin θx sin θy −cos θx sin θz cos θx cos θz sin θy + sin θx sin θz 0cos θy sin θz cos θx cos θz + sin θx sin θy sin θz −cos θz sin θx + cos θx sin θy sin θz 0 −sin θy cos θy sin θx cos θx cos θy 00 0 0 1⎤⎥⎥⎥⎦4.17 One test is to use the first three vertices to find the equation of theplane ax + by + cz + d = 0. Although there are four coefficients in theequation only three are independent so we can select one arbitrarily ornormalize so that a2 + b2 + c2 = 1. Then we can successively evaluateax + bc + cz + d for the other vertices. A vertex will be on the plane if weevaluate to zero. An equivalent test is to form the matrix⎡⎢⎢⎢⎣1 1 1 1x1 x2 x3 x4y1 y2 y3 y4z1 z2 z3 z4⎤⎥⎥⎥⎦for each i = 4, ... If the determinant of this matrix is zero the ith vertex isin the plane determined by the first three.4.19 Although we will have the same number of degrees of freedom in theobjects we produce, the class of objects will be very different. For exampleif we rotate a square before we apply a nonuniform scale, we will shear the square, something we cannot do if we scale then rotate.4.21 The vector a = u ×v is orthogonal to u and v. The vector b = u ×a is orthogonal to u and a. Hence, u, a and b form an orthogonal coordinatesystem.4.23 Using r = cos θ2+ sin θ2v, with θ = 90 and v = (1, 0, 0), we find forrotation about the x-axisr =√22(1, 1, 0, 0).Likewise, for rotation about the y axisr =√22(1, 0, 1, 0).4.27 Possible reasons include (1) object-oriented systems are slower, (2)users are often comfortable working in world coordinates with higher-level objects and do not need the flexibility offered by a coordinate-free approach, (3) even a system that provides scalars, vectors, and points would have to have an underlying frame to use for the implementation. Chapter 5 Solutions5.1 Eclipses (both solar and lunar) are good examples of the projection of an object (the moon or the earth) onto a nonplanar surface. Any time a shadow is created on curved surface, there is a nonplanar projection. All the maps in an atlas are examples of the use of curved projectors. If the projectors were not curved we could not project the entire surface of a spherical object (the Earth) onto a rectangle.5.3 Suppose that we want the view of the Earth rotating about the sun. Before we draw the earth, we must rotate the Earth which is a rotation about the y axis. Next we translate the Earth away from the origin. Finally we do another rotation about the y axis to position the Earth in its desired location along its orbit. There are a number of interesting variants of this problem such as the view from the Earth of the rest of the solar system.5.5 Yes. Any sequence of rotations is equivalent to a single rotation abouta suitably chosen axis. One way to compute this rotation matrix is to form the matrix by sequence of simple rotations, such asR = RxRyRz.The desired axis is an eigenvector of this matrix.5.7 The result follows from the transformation being affine. We can also take a direct approach. Consider the line determined by the points(x1, y1, z1) and (x2, y2, z2). Any point along can be written parametrically as (_x1 + (1 − _)x2, _y1 + (1 − _)y2, _z1 + (1 − _)z2). Consider the simple projection of this point 1d(_z1+(1−_)z2) (_x1 + (1 − _)x2, _y1 + (1 − _)y2)which is of the form f(_)(_x1 + (1 − _)x2, _y1 + (1 − _)y2). This form describes a line because the slope is constant. Note that the function f(_) implies that we trace out the line at a nonlinear rate as _ increases from 0 to 1.5.9 The specification used in many graphics text is of the angles the projector makes with x,z and y, z planes, i.e the angles defined by the projection of a projector by a top view and a side view.Another approach is to specify the foreshortening of one or two sides of a cube aligned with the axes.5.11 The CORE system used this approach. Retained objects were kept in distorted form. Any transformation to any object that was defined with other than an orthographic view transformed the distorted object and the orthographic projection of the transformed distorted object was incorrect.5.15 If we use _ = _ = 45, we obtain the projection matrixP =266641 0 −1 00 1 −1 00 0 0 00 0 0 1377755.17 All the points on the projection of the point (x.y, z) in the direction dx, dy, dz) are of the form (x + _dx, y + _dy, z + _dz). Thus the shadow of the point (x, y, z) is found by determining the _ for which the line intersects the plane, that isaxs + bys + czs = dSubstituting and solving, we find_ =d − ax − by − czadx + bdy + cdz.However, what we want is a projection matrix, Using this value of _ we findxs = z + _dx =x(bdy + cdx) − dx(d − by − cz)adx + bdy + cdzwith similar equations for ys and zs. These results can be computed by multiplying the homogeneous coordinate point (x, y, z, 1) by the projection matrixM =26664bdy + cdz −bdx −cdx −ddx−ady adx + cdz −cdy −ddy−adz −bdz adx + bdy −ddz0 0 0 adx + bdy + cdz37775.5.21 Suppose that the average of the two eye positions is at (x, y, z) and the viewer is looking at the origin. We could form the images using the LookAt function twice, that isgluLookAt(x-dx/2, y, z, 0, 0, 0, 0, 1, 0);/* draw scene here *//* swap buffers and clear */gluLookAt(x+dx/2, y, z, 0, 0, 0, 0, 1, 0);/* draw scene again *//* swap buffers and clear */Chapter 6 Solutions6.1 Point sources produce a very harsh lighting. Such images are characterized by abrupt transitions between light and dark. The ambient light in a real scene is dependent on both the lights on the scene and the reflectivity properties of the objects in the scene, something that cannot be computed correctly with OpenGL. The Phong reflection term is not physically correct; the reflection term in the modified Phong model is even further from being physically correct.6.3 If we were to take into account a light source being obscured by an object, we would have to have all polygons available so as to test for this condition. Such a global calculation is incompatible with the pipeline model that assumes we can shade each polygon independently of all other polygons as it flows through the pipeline.6.5 Materials absorb light from sources. Thus, a surface that appears red under white light appears so because the surface absorbs all wavelengths of light except in the red range—a subtractive process. To be compatible with such a model, we should use surface absorbtion constants that define the materials for cyan, magenta and yellow, rather than red, green and blue. 6.7 Let ψ be the angle between the normal and the halfway vector, φ be the angle between the viewer and the reflection angle, and θ be the anglebetween the normal and the light source. If all the vectors lie in the same plane, the angle between the light source and the viewer can be computer either as φ + 2θ or as 2(θ + ψ). Setting the two equal, we find φ = 2ψ. Ifthe vectors are not coplanar then φ < 2ψ.6.13 Without loss of generality, we can consider the problem in two dimensions. Suppose that the first material has a velocity of light of v1 andthe second material has a light velocity of v2. Furthermore, assume thatthe axis y = 0 separates the two materials.Place a point light source at (0, h) where h > 0 and a viewer at (x, y)where y < 0. Light will travel in a straight line from the source to a point(t, 0) where it will leave the first material and enter the second. It willthen travel from this point in a straight line to (x, y). We must find the tthat minimizes the time travelled.Using some simple trigonometry, we find the line from the source to (t, 0)has length l1 = √h2 + t2 and the line from there to the viewer has length1l2 = _y2 + (x − t)2. The total time light travels is thus l1v1 + l2v2 .Minimizing over t gives desired result when we note the two desired sinesare sin θ1 = h√h2+t2 and sin θ2 = −y √(y2+(x−t)2 .6.19 Shading requires that when we transform normals and points, we maintain the angle between them or equivalently have the dot productp ·v = p_ ·v_ when p_ = Mp and n_ = Mp. If M T M is an identity matrix angles are preserved. Such a matrix (M−1 = M T ) is called orthogonal. Rotations and translations are orthogonal but scaling and shear are not.6.21 Probably the easiest approach to this problem is to rotate the givenplane to plane z = 0 and rotate the light source and objects in the sameway. Now we have the same problem we have solved and can rotate everything back at the end.6.23 A global rendering approach would generate all shadows correctly. Ina global renderer, as each point is shaded, a calculation is done to seewhich light sources shine on it. The projection approach assumes that wecan project each polygon onto all other polygons. If the shadow of a given polygon projects onto multiple polygons, we could not compute these shadow polygons very easily. In addition, we have not accounted for thedifferent shades we might see if there were intersecting shadows from multiple light sources.Chapter 7 Solutions7.1 First, consider the problem in two dimensions. We are looking for an _ and _ such that both parametric equations yield the same point, that isx(_) = (1 − _)x1 + _x2 = (1 − _)x3 + _x4,y(_) = (1 − _)y1 + _y2 = (1 − _)y3 + _y4.These are two equations in the two unknowns _ and _ and, as long as the line segments are not parallel (a condition that will lead to a division by zero), we can solve for _ _. If both these values are between 0 and 1, the segments intersect.If the equations are in 3D, we can solve two of them for the _ and _ where x and y meet. If when we use these values of the parameters in the two equations for z, the segments intersect if we get the same z from both equations.7.3 If we clip a convex region against a convex region, we produce the intersection of the two regions, that is the set of all points in both regions, which is a convex set and describes a convex region. To see this, consider any two points in the intersection. The line segment connecting them must be in both sets and therefore the intersection is convex.7.5 See Problem 6.22. Nonuniform scaling will not preserve the angle between the normal and other vectors.7.7 Note that we could use OpenGL to, produce a hidden line removed image by using the z buffer and drawing polygons with edges and interiors the same color as the background. But of course, this method was not used in pre–raster systems.Hidden–line removal algorithms work in object space, usually with either polygons or polyhedra. Back–facing polygons can be eliminated. In general, edges are intersected with polygons to determine any visible parts. Good algorithms (see Foley or Rogers) use various coherence strategies to minimize the number of intersections.7.9 The O(k) was based upon computing the intersection of rays with the planes containing the k polygons. We did not consider the cost of filling the polygons, which can be a large part of the rendering time. If we consider a scene which is viewed from a given point there will be some percentage of 1the area of the screen that is filled with polygons. As we move the viewer closer to the objects, fewer polygons will appear on the screen but eachwill occupy a larger area on the screen, thus leaving the area of the screen that is filled approximately the same. Thus the rendering time will be about the same even though there are fewer polygons displayed.7.11 There are a number of ways we can attempt to get O(k log k) performance. One is to use a better sorting algorithm for the depth sort. Other strategies are based on divide and conquer such a binary spatial partitioning.7.13 If we consider a ray tracer that only casts rays to the first intersection and does not compute shadow rays, reflected or transmitted rays, then the image produced using a Phong model at the point of intersection will be the same image as produced by our pipeline renderer. This approach is sometimes called ray casting and is used in volume rendering and CSG. However, the data are processed in a different order from the pipeline renderer. The ray tracer works ray by ray while the pipeline renderer works object by object.7.15 Consider a circle centered at the origin: x2 + y2 = r2. If we know thata point (x, y) is on the curve than, we also know (−x, y), (x,−y),(−x,−y), (y, x), (−y, x), (y,−x), and (−y,−x) are also on the curve. This observation is known as the eight–fold symmetry of the circle. Consequently, we need only generate 1/8 of the circle, a 45 degree wedge, and can obtain the rest by copying this part using the symmetries. If we consider the 45 degree wedge starting at the bottom, the slope of this curve starts at 0 and goes to 1, precisely the conditions used for Bresenham’s line algorithm. The tests are a bit more complex and we have to account for the possibility the slope will be one but the approach is the same as for line generation.7.17 Flood fill should work with arbitrary closed areas. In practice, we can get into trouble at corners if the edges are not clearly defined. Such can be the case with scanned images.7.19 Note that if we fill by scan lines vertical edges are not a problem. Probably the best way to handle the problem is to avoid it completely by never allowing vertices to be on scan lines. OpenGL does this by havingvertices placed halfway between scan lines. Other systems jitter the y value of any vertex where it is an integer.7.21 Although each pixel uses five rays, the total number of rays has only doubled, i.e. consider a second grid that is offset one half pixel in both the x and y directions.7.23 A mathematical answer can be investigated using the notion of reconstruction of a function from its samples (see Chapter 8). However, a very easy to see by simply drawing bitmap characters that small pixels lead to very unreadable characters. A readable character should have some overlap of the pixels.7.25 We want k levels between Imin and Imax that are distributed exponentially. Then I0 = Imin, I1 = Iminr,I2 = Iminr2, ..., Ik−1 = Imax = Iminrk−1. We can solve the last equation for the desired r = ( ImaxImin)1k−17.27 If there are very few levels, we cannot display a gradual change in brightness. Instead the viewer will see steps of intensity. A simple rule of thumb is that we need enough gray levels so that a change of one step is not visible. We can mitigate the problem by adding one bit of random noise to the least significant bit of a pixel. Thus if we have 3 bits (8 levels), the third bit will be noise. The effect of the noise will be to break up regions of almost constant intensity so the user will not be able to see a step because it will be masked by the noise. In a statistical sense the jittered image is a noisy (degraded) version of the original but in a visual sense it appears better.。
《计算机图形学》练习题答案
《计算机图形学》练习题(答案)《计算机图形学》练习题1．直线扫描转换的Bresenham算法(1) 请写出生成其斜率介于0和1之间的直线的Bresenham算法步骤。
(2) 设一直线段的起点和终点坐标分别为(1,1)和(8,5)，请用Bresenham算法生成此直线段，确定所有要绘制象素坐标。
(1)①输入线段的两个端点，并将左端点存储在(x0,y0)中②将(x0,y0)装入帧缓存，画出第一个点③计算常量∆x, ∆y, 2∆y, and 2∆y-2∆x,并得到决策参数的第一个值：p0 = 2∆y - ∆x④从k=0开始，在沿线路径的每个xk处，进行下列检测:如果pk < 0,下一个要绘制的点就是(xk +1,yk) ，并且pk+1 = pk + 2∆y否则下一个要绘制的点就是(xk +1, yk +1)，并且 pk+1 = pk + 2∆y- 2∆x⑤重复步骤4，共∆x-1次(2)m=(5-1)/(8-1)=0.57∆x=7 ∆y=4P0=2∆y-∆x=12∆y=8 2∆y-2∆x=-6 k pk (xk+1,yk+1)0 1 (2,2)1 -5 (3,2)2 3 (4,3)3 -3 (5,3)4 5 (6,4)5 -1 (7,4)6 7 (8,5)2．已知一多边形如图1所示，其顶点为V1、V2、V3、V4、V5、V6，边为E1、E2、E3、E4、E5、E6。
用多边形的扫描填充算法对此多边形进行填充时(扫描线从下到上)要建立边分类表(sorted edge table)并不断更新活化边表(active edge list)。
(1)在表1中填写边分类表中每条扫描线上包含的边(标明边号即可)；(2)在表2中写出边分类表中每条边结构中各成员变量的初始值(3) 指出位于扫描线y=6,7,8,9和10时活化边表中包含那些边，并写出这些边中的x值、ymax值、和斜率的倒数值1/m。
表1边分y1边 x y max 1/m 4 1 1 97 4 60 05 1 9 76 0 0 61 9 6 6 0 0Y 值(Scan Line Number ) 边(Edge Number ) 1 0 2 0 3 0 4 E1 5 E6,E2 6 E6 7 E3 8 E5,E3 9E4 10 01 2 3 4 5 6 7 8 9 1表 2 边的初7 1 18 7 790 1-18 2 7 9 9 1 -19 3 36 9 991-13. 二维变换(1) 记P(xf,yf)为固定点，sx、sy分别为沿x 轴和y轴方向的缩放系数，请用齐次坐标(Homogeneous Coordinate)表示写出二维固定点缩放变换的变换矩阵。
《计算机图形学》试卷与答案
《计算机图形学》试卷与答案1、计算机图形学中的图形是指由点、线、面、体等______________________ 和明暗、灰度（亮度）、色彩等______________________ 构成的，从现实世界中抽象出来的带有灰度、色彩及形状的图或形。
2、一个计算机图形系统至少应具有_______________ 、______________ 、输入、输出、 ______________ 等基本功能。
3、常用的字符描述方法有：点阵式、___________________ 和___________________ 。
4、字符串剪裁的策略包括_________________________ 、____________________ 和笔划/像素精确度__________________________________________________ 。
5、所谓齐次坐标就是用______________ 维向量表示一个n 维向量。
6、投影变换的要素有：投影对象、_____________________ 、_________________ 、投影线和投影。
7、输入设备在逻辑上分成定位设备、描画设备、定值设备、_____________________ 、拾取设备和 _______________________ 。
8人机交互是指用户与计算机系统之间的通信，它是人与计算机之间各种符号和动作的。
9、按照光的方向不同，光源分类为：__________________ ，________________ ，________________________________________________ 。
10、从视觉的角度看，颜色包含3个要素：即___________________ 、 ___________________ 和亮度。
二、单项选择题（每题2分，共30分。
请将正确答案的序号填在题后的括号内）1、在CRT 显示器系统中，（。
【中南大学】计算机图形学习题及答案
计算机图形学作业答案第一章序论第二章图形系统1．什么是图像的分辨率？解答：在水平和垂直方向上每单位长度（如英寸）所包含的像素点的数目。
2．计算在240像素/英寸下640×480图像的大小。
解答：（640/240）×(480/240)或者（8/3）×2英寸。
3．计算有512×512像素的2×2英寸图像的分辨率。
解答：512/2或256像素/英寸。
第三章二维图形生成技术1．一条直线的两个端点是（0，0）和（6，18），计算x从0变到6时y所对应的值，并画出结果。
解答：由于直线的方程没有给出，所以必须找到直线的方程。
下面是寻找直线方程（y ＝mx＋b）的过程。
首先寻找斜率：m ＝⊿y/⊿x ＝（y2－y1）/（x2－x1）＝（18－0）/(6－0) ＝ 3 接着b在y轴的截距可以代入方程y＝3x＋b求出 0＝3（0）＋b。
因此b＝0，所以直线方程为y＝3x。
2．使用斜截式方程画斜率介于0°和45°之间的直线的步骤是什么？解答：（1）计算dx：dx＝x2－x1。
（2）计算dy：dy＝y2－y1。
（3）计算m：m＝dy/dx。
（4）计算b: b＝y1－m×x1（5）设置左下方的端点坐标为（x，y），同时将x end设为x的最大值。
如果dx < 0，则x＝x2、y＝y2和x end＝x1。
如果dx > 0,那么x＝x1、y＝y1和x end＝x2。
（6）测试整条线是否已经画完，如果x > x end就停止。
（7）在当前的（x，y）坐标画一个点。
（8）增加x：x＝x＋1。
（9）根据方程y＝mx＋b计算下一个y值。
（10）转到步骤（6）。
3．请用伪代码程序描述使用斜截式方程画一条斜率介于45°和－45°（即|m|>1）之间的直线所需的步骤。
假设线段的两个端点为（x1，y1）和（x2，y2），且y1<y2int x = x1, y = y1;float x f, m = （y2－y1）/（x2－x1）, b = y1－mx1;setPixel( x, y );/*画一个像素点*/while( y < y2 ) {y++;x f = ( y－b)/m;x = Floor( x f +0.5 );setPixel( x, y );}4．请用伪代码程序描述使用DDA算法扫描转换一条斜率介于－45°和45°（即|m| ≤1）之间的直线所需的步骤。
图形学复习资料 (含部分答案)
计算机图形学复习指导一、考试大纲要求掌握计算机图形学和图形系统所必须的基本原理，其主要内容包括：（一）计算机图形学和图形系统基本知识计算机图形学研究对象及应用领域；图形系统的硬软件及图形标推接口。
(二)二维基本图形生成算法直线和二次曲线生成的常用算法；字符和区域填充的实现方法。
(三)图形的剪裁和几何变换窗口视图变换；二维图形的裁剪的原理与方法；二维和三维图形的各种几何变换及其表示。
(四)三维物体的表示方法与输出显示处理各种不同类型曲面的参数表示；实体的定义、性质及各种几何表示方法；投影变换原理与实现；观察空间的定义和转换；三维裁剪。
(五)常用的光学模型及其算法实现(六)消隐显示和阴影生成等实现真实感图形的常用技术二、复习指南2(一)计算机图形学和图形系统基本知识1．计算机图形学研究对象及应用领域2．图形硬件设备3．图形软件系统4．图形标准接口(二)二维图形生成1．直线的生成算法(1)生成直线的常用算法---逐点比较法、数字微分(DDA)法和Bresenham 算法。
(2)直线属性——线型、线宽和线色。
2．曲线的生成算法(1)二次曲线的生成算法---圆弧的逐点比较插补法、圆/椭圆弧的角度数字微分(DDA)法、Bresenham 画圆算法和参数拟合法。
(2)自由曲线的设计---抛物线参数样条曲线、Hermite 曲线、三次参数样条曲线、Bezier 曲线和B 样条曲线。
3．字符(1)字符编码---ASCII 码和汉字国标码。
(2)矢量字符的存储与显示。
(3)点阵字符的存储与显示。
4．区域填充(1)种子填充算法。
(2)扫描转换填充算法。
(3)区域填充属性---式样、颜色和图案。
(三)图形的剪裁和几何变换1．窗口视图变换窗口区与视图区及其变换。
2．二维图形的裁剪(1)二维图形的裁剪的策略及原理。
(2)二维线段的裁剪方法---矢量裁剪法、编码裁剪法和中点分割裁剪法。
(3)字符的裁剪---矢量裁剪、字符裁剪和字符串裁剪法。
计算机图形学教程课后习题参考答案
计算机图形学教程课后习题参考答案(总26页)--本页仅作为文档封面，使用时请直接删除即可----内页可以根据需求调整合适字体及大小--第一章1、试述计算机图形学研究的基本内容答：见课本P5-6页的节。
2、计算机图形学、图形处理与模式识别本质区别是什么请各举一例说明。
答：计算机图形学是研究根据给定的描述，用计算机生成相应的图形、图像，且所生成的图形、图像可以显示屏幕上、硬拷贝输出或作为数据集存在计算机中的学科。
计算机图形学研究的是从数据描述到图形生成的过程。
例如计算机动画制作。
图形处理是利用计算机对原来存在物体的映像进行分析处理，然后再现图像。
例如工业中的射线探伤。
模式识别是指计算机对图形信息进行识别和分析描述，是从图形（图像）到描述的表达过程。
例如邮件分捡设备扫描信件上手写的邮政编码，并将编码用图像复原成数字。
3、计算机图形学与CAD、CAM技术关系如何答：见课本P4-5页的节。
4、举3个例子说明计算机图形学的应用。
答：①事务管理中的交互绘图应用图形学最多的领域之一是绘制事务管理中的各种图形。
通过从简明的形式呈现出数据的模型和趋势以增加对复杂现象的理解，并促使决策的制定。
②地理信息系统地理信息系统是建立在地理图形基础上的信息管理系统。
利用计算机图形生成技术可以绘制地理的、地质的以及其它自然现象的高精度勘探、测量图形。
③计算机动画用图形学的方法产生动画片，其形象逼真、生动，轻而易举地解决了人工绘图时难以解决的问题，大大提高了工作效率。
5、计算机绘图有哪些特点答：见课本P8页的节。
6、计算机生成图形的方法有哪些答：计算机生成图形的方法有两种：矢量法和描点法。
①矢量法：在显示屏上先给定一系列坐标点，然后控制电子束在屏幕上按一定的顺序扫描，逐个“点亮”临近两点间的短矢量，从而得到一条近似的曲线。
尽管显示器产生的只是一些短直线的线段，但当直线段很短时，连成的曲线看起来还是光滑的。
②描点法：把显示屏幕分成有限个可发亮的离散点，每个离散点叫做一个像素，屏幕上由像素点组成的阵列称为光栅，曲线的绘制过程就是将该曲线在光栅上经过的那些像素点串接起来，使它们发亮，所显示的每一曲线都是由一定大小的像素点组成的。
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习题
2．试证明下述几何变换的矩阵运算具有互换性：
（1）两个连续的旋转变换；（2）两个连续的平移变换；
（3）两个连续的变比例变换；（4）当比例系数相等时的旋转和比例变换；
（1）证明：设第一次的旋转变换为：
cosθ1 sinθ1 0
T1= - sinθ1 cosθ1 0
0 0 1
第二次的旋转变换为：
Cosθ2 s inθ2 0
T2= - sinθ2 cosθ2 0
0 0 1
则因为
T1*T2 = cosθ1 sinθ1 0 cosθ2
sinθ2 0
- sinθ1 cosθ1 0 - sinθ2 cosθ2 0
0 0 1 0 0
1
= cosθ1 cosθ2+sinθ1 sinθ2 cosθ1 sinθ2+ sinθ1 cosθ2 0
- sinθ1 cosθ2- cosθ1 sinθ2 -sinθ1 sinθ1+ cosθ1 cosθ2 0
0 0
1
Cos（θ1+θ2）sin（θ1+θ2） 0
= - sin（θ1+θ2） cos（θ1+θ2） 0
0 0 1
cosθ2 sinθ2 0 cosθ1 sinθ1 0
T2*T1 = - sinθ2 cosθ2 0 - sinθ1
cosθ1 0
0 0 1 0 0
1
cosθ1 cosθ2+ sinθ1 sinθ2 cosθ1 sinθ2+
sinθ1 cosθ2 0
= - sinθ2cosθ1- cosθ2 sinθ1 -sinθ1 sinθ1+
cosθ1 cosθ2 0
0 0 1
Cos（θ1+θ2）sin（θ1+θ2） 0
= - sin（θ1+θ2） cos（θ1+θ2） 0
0 0 1
即T1*T2= T2*T1, 两个连续的旋转变换具有互换性
（2）证明：设第一次的平移变换为：
1 0 0
T1= 0 1 0
Tx1 Ty1 1
第二次的平移变换为：
1 0 0
T2= 0 1 0
Tx2 Ty2 1
则因为
T1*T2 = 1 0 0 1 0 0
0 1 0 0 1 0
Tx1 Ty1 1 Tx2 Ty2 1
1 0 0
= 0 1 0
Tx1+Tx2 Ty1+Ty2 1
而
T2*T1 = 1 0 0 1 0 0
0 1 0 0 1 0
Tx2 Ty2 1 Tx1 Ty1 1
1 0 0
= 0 1 0
Tx1+Tx2 Ty1+Ty2 1
即T1*T2= T2*T1, 两个连续的平移变换具有互换性
（3）证明：设第一次的变比例变换为：
Sx1 0 0
T1= 0 Sy1 0
0 0 1
第二次的变比例变换为：
Sx2 0 0
T2 = 0 Sy2 0
0 0 1
则因为
T1*T2 = Sx1 0 0 Sx2 0 0 0 Sy1 0 0 Sy2 0 0 0 1 0 0 1
Sx1*Sx2 0 0
= 0 Sy1*Sy2 0
0 0 1
而
T2*T1 = Sx2 0 0 Sx1 0 0 0 Sy2 0 0 Sy1 0 0 0 1 0 0 1
Sx1*Sx2 0 0
= 0 Sy1*Sy2 0
0 0 1
即T1*T2= T2*T1, 两个连续的变比例变换具有互换性
（4）证明：设第一次为比例系数相等时的比例变换：
S 0 0
T1= 0 S 0
0 0 1
第二次的为旋转变换：
cosθ sinθ 0
T2= - sinθ cosθ 0
0 0 1
则因为
T1*T2 = S 0 0 cosθ sinθ 0
0 S 0 - sinθ cosθ 0
0 0 1 0 0 1
S cosθ S sinθ 0
= - S sinθ S cosθ2 0
0 0 1
而
T2*T1 = cosθ sinθ 0 S 0 0
- sinθ cosθ 0 0 S 0
0 0 1 0 0 1
S cosθ S sinθ 0
= -S sinθ S cosθ 0
0 0 1
即T1*T2= T2*T1, “当比例系数相等时的旋转和比例“变换具有互换性
3、证明二维点相对x轴作对称，紧跟着相对y=-x直线作对称变换完全等价于该点相对坐标原点作旋转变换。
证明：
(1) 点相对x轴作对称的变换矩阵
1 0 0
T1= 0 -1 0
0 0 1
(2) 相对于y=-x直线作对称变换矩阵
0 -1 0
T2= -1 0 0
0 0 1
1 0 0 0 -1 0 0 -1 0
因为 T1*T2= 0 -1 0 * -1 0 0 = 1 0 0
0 0 1 0 0 1 0 0 1
cos(-90o) sin(-90 o) 0
= - sin(-90 o) cos(-90o) 0
0 0 1
即该点相对坐标原点作顺时针方向转90 o的旋转变换
4、证明
1-t2 2t
1+ t 2 1+t2
T= 完全表示一个旋转变换。
-2t 1-t2
1+t2 1+t2
证明：令t=tg(θ/2)
则：（1-t2）/(1+ t 2)= cosθ
（2t）/(1+ t 2)= sinθ
即
cosθ sinθ
T=
- sinθ cosθ
将T扩充为一个三行齐次坐标的变换矩阵为：
cosθ sinθ 0
T= - sinθ cosθ 0
0 0 1
该矩阵表示为一个旋转变换
5、例：三角形ABC各顶点坐标为A（3，0）B（4，2）C（6，0），其绕原点逆时针旋转90°，再向X方向平移2，Y方向平移-1。
解：因为：θ=90°
变换矩阵为
COS90° SIN90° 0 0 1 0 TR= - SIN90° COS90° 0 = -1 0 0
2 -1 1 2 -1 -1
则
A 3 0 1 0 1 0 2 2 1 A‘
B 4 2 1 -1 0 0 = 0 3 1 B‘
C 6 0 1 2 -1 1 2 5 1 C‘
如果先进行平移变换，再进行旋转变换，
1 0 0 COS90° SIN90° 0 0 1 0 Tr= 0 1 0 - SIN90° COS90° 0 = -1 0 0
2 -1 1 0 0 1 1 2 1
则
A 3 0 1 0 1 0 1 5 1 A‘
B 4 2 1 -1 0 0 = -1 6 1 B‘
C 6 0 1 1 2 1 1 8 1 C‘结论：变换顺序不同，结果也不同。