Given coordinate of two points A(x1, y1) and B(x2, y2) such that x1 < x2 and y1 < y2. The task to find all the intermediate points required for drawing line AB on the computer screen of pixels. Note that every pixel has integer coordinates.
We have discussed below algorithms for this task.
In this post, Mid-Point Line drawing algorithm is discussed which is a different way to represent Bresenham’s algorithm introduced in previous post.
As discussed in previous post, for any given/calculated previous pixel P(Xp,Yp), there are two candidates for the next pixel closest to the line, E(Xp+1, Yp) and NE(Xp+1, Yp+1) (E stands for East and NE stands for North-East).
In Mid-Point algorithm we do following.
- Find middle of two possible next points. Middle of E(Xp+1, Yp) and NE(Xp+1, Yp+1) is M(Xp+1, Yp+1/2).
- If M is above the line, then choose E as next point.
- If M is below the line, then choose NE as next point.
How to find if a point is above a line or below a line?
Below are some assumptions to keep algorithm simple.
- We draw line from left to right.
- x1 < x2 and y1< y2
- Slope of the line is between 0 and 1. We draw a line from lower left to upper right.
Cases other than above assumptions can be handled using reflection.
Let us consider a line y = mx + B. We can re-write the equation as : y = (dy/dx)x + B or (dy)x + B(dx) - y(dx) = 0 Let F(x, y) = (dy)x - y(dx) + B(dx) -----(1) Let we are given two end points of a line (under above assumptions) -> For all points (x,y) on the line, the solution to F(x, y) is 0. -> For all points (x,y) above the line, F(x, y) result in a negative number. -> And for all points (x,y) below the line, F(x, y) result in a positive number.
This relationship is used to determine the relative
position of M
M = (Xp+1, Yp+1/2)
So our decision parameter d is,
d = F(M) = F(Xp+1, Yp+1/2)
How to efficiently find new value of d from its old value?
For simplicity, let as write F(x, y) as ax + by + c.
Where a = dy
b = -dx
c = B*dx
We got these values from above equation (1)
Case 1: If E is chosen then for next point :
dnew = F(Xp+2, Yp+1/2)
= a(Xp+2) + b(Yp+1/2) + c
dold = a(Xp+1) + b(Yp+1/2) + c
Case 2: If NE is chosen then for next point :
dnew = F(Xp+2, Yp+3/2)
= a(Xp+2) + b(Yp+3/2) + c
dold = a(Xp+1) + b(Yp+1/2) + c
Difference (Or delta) of two distances:
DELd = dnew -dold
= a(Xp+2)- a(Xp+1)+ b(Yp+3/2)- b(Yp+1/2)+ c-c
= a(Xp) + 2a – a(Xp) – a + b(Yp) + 3/2b – b(Yp) -1/2b
= a + b
Therefore, dnew = dold + dy – dx. (as a = dy , b = -dx)
Calculation For initial value of decision parameter d0:
d0 = F(X1+1 , Y1+1/2)
= a(X1 + 1) + b(Y1 + 1/2) +c
= aX1+ bY1 + c + a + b/2
= F(X1,Y1) + a + b/2
= a + b/2 (as F(X1, Y1) = 0 )
d0 = dy – dx/2. (as a = dy, b = -dx)
Input (X1,Y1) and (X2,Y2) dy = Y2- Y1 dx = X2 - X1 // initial value of // decision parameter d d = dy - (dx/2) x = X1 , y = Y1 // plot initial given point Plot(x , y) // iterate through value of X while(x < X2) x = x+1 // 'E' is chosen if (d < 0) d = d + dy // 'NE' is chosen else d = d + dy - dx y = y+1 Plot(x,y)
Below is the implementation of above idea.
2,2 3,3 4,3 5,4 6,4 7,5 8,5
This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to firstname.lastname@example.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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- Comparions between DDA and Bresenham Line Drawing algorithm
- Line Clipping | Set 1 (Cohen–Sutherland Algorithm)
- Klee's Algorithm (Length Of Union Of Segments of a line)
- Chain Code for 2D Line
- Draw a line in C++ graphics
- Program to find line passing through 2 Points
- Find X and Y intercepts of a line passing through the given points
- Scan-line Polygon filling using OPENGL in C
- Find Partition Line such that sum of values on left and right is equal