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**Kamla Nehru College - RTMNU**- Computer Science Engineering
- B.Tech
**6 Views**- Uploaded 8 months ago

11/24/2018 Computer Graphics Line Generation Algorithm LINE GENERATION ALGORITHM https://www.tutorialspoint.com/computer_graphics/line_generation_algorithm.htm Copyright © tutorialspoint.com Advertisements A line connects two points. It is a basic element in graphics. To draw a line, you need two points between which you can draw a line. In the following three algorithms, we refer the one point of line as X , Y and the second point of line as X , Y . 0 1 0 1 DDA Algorithm Digital Differential Analyzer DDA algorithm is the simple line generation algorithm which is explained step by step here. Step 1 − Get the input of two end points (X 0, Y0 ) and (X 1, Y1 ) . Step 2 − Calculate the difference between two end points. dx = X1 - X0 dy = Y1 - Y0 Step 3 − Based on the calculated difference in step-2, you need to identify the number of steps to put pixel. If dx > dy, then you need more steps in x coordinate; otherwise in y coordinate. if (absolute(dx) > absolute(dy)) Steps = absolute(dx); else Steps = absolute(dy); Step 4 − Calculate the increment in x coordinate and y coordinate. Xincrement = dx / (float) steps; Yincrement = dy / (float) steps; Step 5 − Put the pixel by successfully incrementing x and y coordinates accordingly and complete the drawing of the line. for(int v=0; v < Steps; v++) { x = x + Xincrement; y = y + Yincrement; putpixel(Round(x), Round(y)); } Bresenham’s Line Generation The Bresenham algorithm is another incremental scan conversion algorithm. The big advantage of this algorithm is that, it uses only integer calculations. Moving across the x axis in unit intervals and at each step choose between two different y coordinates. https://www.tutorialspoint.com/cgi-bin/printpage.cgi 1/5

11/24/2018 Computer Graphics Line Generation Algorithm For example, as shown in the following illustration, from position 2, 3 you need to choose between 3, 3 and 3, 4 . You would like the point that is closer to the original line. At sample position X k + 1, the vertical separations from the mathematical line are labelled as d From the above illustration, the y coordinate on the mathematical line at x k Y = m$X So, d upper and d lower k$ + 1 + 1 upper and d lower . is − +b are given as follows − https://www.tutorialspoint.com/cgi-bin/printpage.cgi 2/5

11/24/2018 Computer Graphics Line Generation Algorithm d lower = y − yk = m(X k + 1) + b − Y k and d upper = (yk + 1) − y = Y k + 1 − m(X k + 1) − b You can use these to make a simple decision about which pixel is closer to the mathematical line. This simple decision is based on the difference between the two pixel positions. d lower − d upper = 2m(xk + 1) − 2yk + 2b − 1 Let us substitute m with dy/dx where dx and dy are the differences between the end-points. dy dx(d lower − d upper ) = dx(2 dx (xk + 1) − 2yk + 2b − 1) = 2dy. xk − 2dx. yk + 2dy + 2dx(2b − 1) = 2dy. xk − 2dx. yk + C So, a decision parameter P for the kth step along a line is given by − k pk = dx(d lower − d upper ) = 2dy. xk − 2dx. yk + C The sign of the decision parameter P is the same as that of d k lower − d upper . If p is negative, then choose the lower pixel, otherwise choose the upper pixel. k Remember, the coordinate changes occur along the x axis in unit steps, so you can do everything with integer calculations. At step k+1, the decision parameter is given as − pk+1 = 2dy. xk+1 − 2dx. yk+1 + C Subtracting p from this we get − k pk+1 − pk = 2dy(xk+1 − xk ) − 2dx(yk+1 − yk ) But, x k+1 is the same as (xk) + 1 . So − pk+1 = pk + 2dy − 2dx(yk+1 − yk ) Where, Y k+1 – Yk is either 0 or 1 depending on the sign of P . k The first decision parameter p is evaluated at (x 0 0, y0 ) is given as − p0 = 2dy − dx Now, keeping in mind all the above points and calculations, here is the Bresenham algorithm for slope m < 1 − https://www.tutorialspoint.com/cgi-bin/printpage.cgi 3/5

11/24/2018 Computer Graphics Line Generation Algorithm Step 1 − Input the two end-points of line, storing the left end-point in (x 0, Step 2 − Plot the point (x 0, y0 ) y0 ) . . Step 3 − Calculate the constants dx, dy, 2dy, and 2dy– 2dx and get the first value for the decision parameter as − p0 = 2dy − dx Step 4 − At each X along the line, starting at k = 0, perform the following test − k If p < 0, the next point to plot is (x k k + 1, yk ) and pk+1 = pk + 2dy Otherwise, (xk , yk + 1) pk+1 = pk + 2dy − 2dx Step 5 − Repeat step 4 dx– 1 times. For m > 1, find out whether you need to increment x while incrementing y each time. After solving, the equation for decision parameter P will be very similar, just the x and y in the equation gets interchanged. k Mid-Point Algorithm Mid-point algorithm is due to Bresenham which was modified by Pitteway and Van Aken. Assume that you have already put the point P at x, y coordinate and the slope of the line is 0 ≤ k ≤ 1 as shown in the following illustration. Now you need to decide whether to put the next point at E or N. This can be chosen by identifying the intersection point Q closest to the point N or E. If the intersection point Q is closest to the point N then N is considered as the next point; otherwise E. https://www.tutorialspoint.com/cgi-bin/printpage.cgi 4/5

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