Mid-Point Line Generation Algorithm

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.

  1. DDA algorithm for line drawing
  2. Introduction to Bresenhams’s algorithm for line drawing.

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.

  1. 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).
  2. If M is above the line, then choose E as next point.
  3. If M is below the line, then choose NE as next point.

midpoint

How to find if a point is above a line or below a line?
Below are some assumptions to keep algorithm simple.

  1. We draw line from left to right.
  2. x1 < x2 and y1< y2
  3. 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. 
midpoint

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

Difference (Or delta) of two distances:
DELd = dnew – dold
= a(Xp+2)- a(Xp+1)+ b(Yp+1/2)- b(Yp+1/2)+ c-c
= a(Xp) +2a – a(Xp) – a
= a.
Therefore, dnew = dold + dy. (as a = dy)
mid point line


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)

Algorithm:

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.

C++

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// C++ program for Mid-point line generation
#include<bits/stdc++.h>
using namespace std;
  
// Header file for including graphics functions
// #include<graphics.h>
  
// midPoint function for line generation
void midPoint(int X1, int Y1, int X2, int Y2)
{
    // calculate dx & dy
    int dx = X2 - X1;
    int dy = Y2 - Y1;
  
    // initial value of decision parameter d
    int d = dy - (dx/2);
    int x = X1, y = Y1;
  
    // Plot initial given point
    // putpixel(x,y) can be used to print pixel
    // of line in graphics
    cout << x << "," << y << "\n";
  
    // iterate through value of X
    while (x < X2)
    {
        x++;
  
        // E or East is chosen
        if (d < 0)
            d = d + dy;
  
        // NE or North East is chosen
        else
        {
            d += (dy - dx);
            y++;
        }
  
        // Plot intermediate points
        // putpixel(x,y) is used to print pixel
        // of line in graphics
        cout << x << "," << y << "\n";
    }
}
  
// Driver program
int main()
{
    // graphics driver and mode
    // used in graphics.h
    // int gd = DETECT, gm;
  
    // Initialize graphics function
    // initgraph (&gd, &gm, "");
  
    int X1 = 2, Y1 = 2, X2 = 8, Y2 = 5;
    midPoint(X1, Y1, X2, Y2);
    return 0;
}

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Java

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// Java program for Mid-point 
// line generation
class GFG
{
// midPoint function for line generation
static void midPoint(int X1, int Y1,
                     int X2, int Y2)
{
    // calculate dx & dy
    int dx = X2 - X1;
    int dy = Y2 - Y1;
  
    // initial value of decision
    // parameter d
    int d = dy - (dx/2);
    int x = X1, y = Y1;
  
    // Plot initial given point
    // putpixel(x,y) can be used to
    // print pixel of line in graphics
    System.out.print(x +"," + y + "\n");
  
    // iterate through value of X
    while (x < X2)
    {
        x++;
  
        // E or East is chosen
        if (d < 0)
            d = d + dy;
  
        // NE or North East is chosen
        else
        {
            d += (dy - dx);
            y++;
        }
  
        // Plot intermediate points
        // putpixel(x,y) is used to print
        // pixel of line in graphics
        System.out.print(x +"," + y + "\n");
    }
}
  
// Driver code 
public static void main (String[] args)
{
    int X1 = 2, Y1 = 2, X2 = 8, Y2 = 5;
    midPoint(X1, Y1, X2, Y2);
}
}
  
// This code is contributed by Anant Agarwal.

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Python 3

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# Python3 program for Mid-point
# line generation 
  
  
# midPoint function for line generation 
def midPoint(X1,Y1,X2,Y2): 
    # calculate dx & dy 
    dx = X2 - X1 
    dy = Y2 - Y1 
  
    # initial value of decision parameter d 
    d = dy - (dx/2
    x = X1
    y = Y1 
  
    # Plot initial given point 
    # putpixel(x,y) can be used to print pixel 
    # of line in graphics 
    print(x,",",y,"\n")
    # iterate through value of X 
    while (x < X2):
        x=x+1
        # E or East is chosen
        if(d < 0):
            d = d + dy 
  
        # NE or North East is chosen 
        else:
            d = d + (dy - dx) 
            y=y+1
      
  
        # Plot intermediate points 
        # putpixel(x,y) is used to print pixel 
        # of line in graphics 
        print(x,",",y,"\n"
      
  
# Driver program 
  
if __name__=='__main__':
    X1 = 2
    Y1 = 2
    X2 = 8
    Y2 = 5
    midPoint(X1, Y1, X2, Y2) 
  
# This code is contributed by ash264

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C#

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// C# program for Mid-point 
// line generation
using System;
  
class GFG {
      
    // midPoint function for line
    // generation
    static void midPoint(int X1, int Y1,
                         int X2, int Y2)
    {
          
        // calculate dx & dy
        int dx = X2 - X1;
        int dy = Y2 - Y1;
      
        // initial value of decision
        // parameter d
        int d = dy - (dx/2);
        int x = X1, y = Y1;
      
        // Plot initial given point
        // putpixel(x,y) can be used
        // to print pixel of line in
        // graphics
        Console.Write(x + "," + y + "\n");
      
        // iterate through value of X
        while (x < X2)
        {
            x++;
      
            // E or East is chosen
            if (d < 0)
                d = d + dy;
      
            // NE or North East is chosen
            else
            {
                d += (dy - dx);
                y++;
            }
      
            // Plot intermediate points
            // putpixel(x,y) is used to print
            // pixel of line in graphics
            Console.Write(x + "," + y + "\n");
        }
    }
      
    // Driver code 
    public static void Main ()
    {
        int X1 = 2, Y1 = 2, X2 = 8, Y2 = 5;
        midPoint(X1, Y1, X2, Y2);
    }
}
  
// This code is contributed by nitin mittal.

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PHP

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<?php
// PHP program for Mid-point
// line generation
  
// midPoint function for
// line generation
function midPoint($X1, $Y1
                  $X2, $Y2)
{
      
    // calculate dx & dy
    $dx = $X2 - $X1;
    $dy = $Y2 - $Y1;
  
    // initial value of 
    // decision parameter d
    $d = $dy - ($dx/2);
    $x = $X1;
    $y = $Y1;
  
    // Plot initial given point
    // putpixel(x,y) can be used
    // to print pixel of line 
    // in graphics
    echo $x , "," , $y , "\n";
  
    // iterate through 
    // value of X
    while ($x < $X2)
    {
        $x++;
  
        // E or East is chosen
        if ($d < 0)
            $d = $d + $dy;
  
        // NE or North East 
        // is chosen
        else
        {
            $d += ($dy - $dx);
            $y++;
        }
  
        // Plot intermediate points
        // putpixel(x,y) is used 
        // to print pixel of 
        // line in graphics
        echo $x , "," ,$y , "\n";
    }
}
  
    // Driver Code
    $X1 = 2; 
    $Y1 = 2;
    $X2 = 8; 
    $Y2 = 5;
    midPoint($X1, $Y1, $X2, $Y2);
  
// This code is contributed by nitin mittal.
?>

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Output:

2,2
3,3
4,3
5,4
6,4
7,5
8,5

References:
http://www.eng.utah.edu/~cs5600/slides/Wk%202%20Lec02_Bresenham.pdf

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 contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : nitin mittal, ash264



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