Answers to Exercises in .NET

Incoporate QR Code in .NET Answers to Exercises
Appendix B
VS .NET qr barcode decoderfor .net
Using Barcode Control SDK for Visual Studio .NET Control to generate, create, read, scan barcode image in Visual Studio .NET applications.
5 6 7
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Answers to Exercises
QR-Code barcode library for .net
Using Barcode recognizer for Visual Studio .NET Control to read, scan read, scan image in Visual Studio .NET applications.
s.pixelGrid[i].c = color((int)random(255)); s.plot(); }
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2. Memory for the points was not allocated. In other words, after allocating memory for the array of points, we also need to allocate memory for each individual point, as in the following code:
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for(int i=0; i<10; i++) p[i] = new MyPoint();
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3. The problem is the same as with exercise 2, except it is harder to detect because there is no compilation error generated. The assignment of point p as a member of seg should been preceded by the allocation of memory for p as shown in the following code:
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p = new MyPoint(); seg.a = p;
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4
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1. Sample Java classes a. Button class
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Button b = new Button( Click Here ); add(b); b.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent e) { println(b.getLabel()); }}); void mouseDragged(){ int xoff = mouseX - pmouseX; // get the offset int yoff = mouseY - pmouseY; MyPoint ref = new MyPoint(0.,0.); for(int i=0; i<group.numShapes; i++) if(group.shapes[i].isSelected){ ref = group.shapes[i].centroid(); //this can be constrcted if(control.status.equals( Move )) //Move group.shapes[i].move(( float)xoff, ( float)yoff); if(control.status.equals( Rotate )) //Rotate group.shapes[i].rotate(( float)xoff, ref); if(control.status.equals( Scale )) //Scale group.shapes[i].scale((float)mouseX/(float)xfirst, (float)mouseY/(float)yfirst, ref); } }
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Appendix B
.net Framework Crystal msi plessey implementationin .net
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Answers to Exercises
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b. Label class
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Label coordsDisplay; //definition
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//Label setup coordsDisplay = new Label(); //display add(coordsDisplay); add(input);
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c. TextField class
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TextField input; //definition
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//TextField setup input = new TextField( Welcome ); //display add(coordsDisplay); add(input); input.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent e) { println( textfield = + input.getText()); } });
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d. Choice class
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Choice transform; //definition transform = new Choice(); transform.addItem( Move ); transform.addItem( Rotate ); transform.addItem( Scale ); add(transform); transform.addItemListener(new ItemListener() { public void itemStateChanged(ItemEvent e) { status = transform.getSelectedIndex(); } });
2. setLayout(null); is missing. 3. MouseUp() is not supported.
Appendix B
Answers to Exercises
5
N o te The answer to question 2 is not provided.
1. You need to multiply the number of rows and add the number of columns. Since i is the number of rows and j the number of columns, the correct answer is D. 3. The algorithm is:
PImage MyImage; noStroke(); MyImage = loadImage( tree_canopy.jpg ); size(MyImage.width,MyImage.height); image(MyImage,0,0); for(int x=0; x<width; x+=5) for(int y=0; y<height; y+=5){ float b = brightness(get(x,y))/50; fill(255); rect(x,y,5,5); fill(0); ellipse(x,y,5-b,5-b); }
The code can be also exported as dxf to produce a 3D effect, as shown in the following code. The result is shown in the figures following the code. In the first figure, the original face image is shown to the left and its perforated version is superimposed in the image to the right. The second figure shows the modeling of the face s pattern perforation.
import processing.dxf.*; PImage MyImage; noStroke(); MyImage = loadImage( face.jpg ); size(MyImage.width,MyImage.height,P3D); image(MyImage,0,0); beginRaw( DXF, out.dxf ); for(int x=0; x<width; x+=5) for(int y=0; y<height; y+=5){ float b = brightness(get(x,y))/50; fill(0); ellipse(x,y,5-b,5-b); } endRaw();
Appendix B
Answers to Exercises
4. The algorithm is:
color c = get(x,y); if(red(c)==0) set(x,y,color(255,255,255)); else set(x,y,color(255,0,0));
5. The algorithm is:
1 2 3 4 5 6 7 8 9 10 11 int [] xd = {0,1,1, 1, 0,-1,-1,-1,0}; int [] yd = {1,1,0,-1,-1,-1, 0, 1,1}; PImage MyImage; int [][] MyCopy; void setup(){ MyImage = loadImage( stockholm white.jpg ); size(MyImage.width,MyImage.height); MyCopy = new int[width][height]; image(MyImage, 0,0); filter(THRESHOLD); for(int x=0; x<width; x++)
Appendix B
Answers to Exercises
12 for(int y=0; y<height; y++) 13 MyCopy[x][y] = getBinary(x,y); 14 for(int g=0; g<3; g++) skeletonize(); 15 } 16 17 void skeletonize(){ 18 for(int x=1; x<width-2; x++) 19 for(int y=2; y<height-1; y++){ 20 int b=0; 21 int a=0; 22 for(int i=0; i<8; i++){ 23 if(getBinary(x+xd[i],y+yd[i])==1)b++; 24 if(getBinary(x+xd[i],y+yd[i])==0 && 25 getBinary(x+xd[i+1],y+yd[i+1])==1) a++; 26 } 27 int a2=0; 27 for(int i=0; i<8; i++) 29 if(getBinary(x+xd[i],y+1+yd[i])==0 && 30 getBinary(x+xd[i+1],y+1+yd[i+1])==1) a2++; 31 int c2 = getBinary(x,y+1)*getBinary(x+1,y)*getBinary(x-1,y); 32 int a3=0; 33 for(int i=0; i<8; i++) 34 if(getBinary(x+1+xd[i],y+yd[i])==0 && 35 getBinary(x+1+xd[i+1],y+yd[i+1])==1) a3++; 36 int c3=getBinary(x,y+1)*getBinary(x+1,y) *getBinary(x,y-1); 37 if((2<=b && b<=6) && a==1 && 38 (c2==0 || a2!=1) && (c3==0 || a3!=1)) 39 if(getBinary(x,y)==1)MyCopy[x][y]=0; 40 } 41 for(int x=1; x<width-1; x++) 42 for(int y=1; y<height-1; y++) 43 if(MyCopy[x][y]==1) 44 set(x,y,color(0,0,0)); //black 45 else 46 set(x,y,color(255,255,255)); //white 47 } 48 49 int getBinary(int x,int y){ 50 return((brightness(get(x,y))>128) 0 : 1); 51 }
The preceding algorithm is also referred to as Hilditch s algorithm. It is a skeletonization process that progresses in steps. In each step, every pixel in the image is evaluated based on its neighboring pixels for the satisfaction of certain conditions. There are two neighboring conditions for a pixel p1: