Description:- In this algorithm, we are given 9 regions on the screen. Out of which one region is of the window and the rest 8 regions are around it given by 4 digit binary. The division of the regions are based on (x_max, y_max) and (x_min, y_min).
The central part is the viewing region or window, all the lines which lie within this region are completely visible. A region code is always assigned to the endpoints of the given line.
To check whether the line is visible or not.
Region Codes
Formula to check binary digits:- TBRL which can be defined as top, bottom, right, and left accordingly.
TBRL Rule
Algorithm
Steps
1) Assign the region codes to both endpoints.
2) Perform OR operation on both of these endpoints.
3) if OR = 0000,
then it is completely visible (inside the window).
else
Perform AND operation on both these endpoints.
i) if AND ≠ 0000,
then the line is invisible and not inside the window. Also, it can’t be considered for clipping.
ii) else
AND = 0000, the line is partially inside the window and considered for clipping.
4) After confirming that the line is partially inside the window, then we find the intersection with the boundary of the window. By using the following formula:-
Slope:- m= (y2-y1)/(x2-x1)
a) If the line passes through top or the line intersects with the top boundary of the window.
x = x + (y_wmax – y)/m
y = y_wmax
b) If the line passes through the bottom or the line intersects with the bottom boundary of the window.
x = x + (y_wmin – y)/m
y = y_wmin
c) If the line passes through the left region or the line intersects with the left boundary of the window.
y = y+ (x_wmin – x)*m
x = x_wmin
d) If the line passes through the right region or the line intersects with the right boundary of the window.
y = y + (x_wmax -x)*m
x = x_wmax
5) Now, overwrite the endpoints with a new one and update it.
6) Repeat the 4th step till your line doesn’t get completely clipped
Given a set of lines and a rectangular area of interest, the task is to remove lines that are outside the area of interest and clip the lines which are partially inside the area.
Input : Rectangular area of interest (Defined by
below four values which are coordinates of
bottom left and top right)
x_min = 4, y_min = 4, x_max = 10, y_max = 8
A set of lines (Defined by two corner coordinates)
line 1 : x1 = 5, y1 = 5, x2 = 7, y2 = 7
Line 2 : x1 = 7, y1 = 9, x2 = 11, y2 = 4
Line 2 : x1 = 1, y1 = 5, x2 = 4, y2 = 1
Output : Line 1 : Accepted from (5, 5) to (7, 7)
Line 2 : Accepted from (7.8, 8) to (10, 5.25)
Line 3 : Rejected
Cohen-Sutherland algorithm divides a two-dimensional space into 9 regions and then efficiently determines the lines and portions of lines that are inside the given rectangular area.
The algorithm can be outlines as follows:-
Nine regions are created, eight "outside" regions and one "inside" region. For a given line extreme point (x, y), we can quickly find its region's four bit code. Four bit code can be computed by comparing x and y with four values (x_min, x_max, y_min and y_max). If x is less than x_min then bit number 1 is set. If x is greater than x_max then bit number 2 is set. If y is less than y_min then bit number 3 is set. If y is greater than y_max then bit number 4 is set
There are three possible cases for any given line.
- Completely inside the given rectangle : Bitwise OR of region of two end points of line is 0 (Both points are inside the rectangle)
- Completely outside the given rectangle : Both endpoints share at least one outside region which implies that the line does not cross the visible region. (bitwise AND of endpoints != 0).
- Partially inside the window : Both endpoints are in different regions. In this case, the algorithm finds one of the two points that is outside the rectangular region. The intersection of the line from outside point and rectangular window becomes new corner point and the algorithm repeats
Pseudo Code:
Step 1 : Assign a region code for two endpoints of given line.
Step 2 : If both endpoints have a region code 0000
then given line is completely inside.
Step 3 : Else, perform the logical AND operation for both region codes.
Step 3.1 : If the result is not 0000, then given line is completely
outside.
Step 3.2 : Else line is partially inside.
Step 3.2.1 : Choose an endpoint of the line
that is outside the given rectangle.
Step 3.2.2 : Find the intersection point of the
rectangular boundary (based on region code).
Step 3.2.3 : Replace endpoint with the intersection point
and update the region code.
Step 3.2.4 : Repeat step 2 until we find a clipped line either
trivially accepted or trivially rejected.
Step 4 : Repeat step 1 for other lines
Below is implementation of above steps.
C++
// C++ program to implement Cohen Sutherland algorithm// for line clipping.#include <iostream>using namespace std;// Defining region codesconst int INSIDE = 0; // 0000const int LEFT = 1; // 0001const int RIGHT = 2; // 0010const int BOTTOM = 4; // 0100const int TOP = 8; // 1000// Defining x_max, y_max and x_min, y_min for// clipping rectangle. Since diagonal points are// enough to define a rectangleconst int x_max = 10;const int y_max = 8;const int x_min = 4;const int y_min = 4;// Function to compute region code for a point(x, y)int computeCode(double x, double y){ // initialized as being inside int code = INSIDE; if (x < x_min) // to the left of rectangle code |= LEFT; else if (x > x_max) // to the right of rectangle code |= RIGHT; if (y < y_min) // below the rectangle code |= BOTTOM; else if (y > y_max) // above the rectangle code |= TOP; return code;}// Implementing Cohen-Sutherland algorithm// Clipping a line from P1 = (x2, y2) to P2 = (x2, y2)void cohenSutherlandClip(double x1, double y1, double x2, double y2){ // Compute region codes for P1, P2 int code1 = computeCode(x1, y1); int code2 = computeCode(x2, y2); // Initialize line as outside the rectangular window bool accept = false; while (true) { if ((code1 == 0) && (code2 == 0)) { // If both endpoints lie within rectangle accept = true; break; } else if (code1 & code2) { // If both endpoints are outside rectangle, // in same region break; } else { // Some segment of line lies within the // rectangle int code_out; double x, y; // At least one endpoint is outside the // rectangle, pick it. if (code1 != 0) code_out = code1; else code_out = code2; // Find intersection point; // using formulas y = y1 + slope * (x - x1), // x = x1 + (1 / slope) * (y - y1) if (code_out & TOP) { // point is above the clip rectangle x = x1 + (x2 - x1) * (y_max - y1) / (y2 - y1); y = y_max; } else if (code_out & BOTTOM) { // point is below the rectangle x = x1 + (x2 - x1) * (y_min - y1) / (y2 - y1); y = y_min; } else if (code_out & RIGHT) { // point is to the right of rectangle y = y1 + (y2 - y1) * (x_max - x1) / (x2 - x1); x = x_max; } else if (code_out & LEFT) { // point is to the left of rectangle y = y1 + (y2 - y1) * (x_min - x1) / (x2 - x1); x = x_min; } // Now intersection point x, y is found // We replace point outside rectangle // by intersection point if (code_out == code1) { x1 = x; y1 = y; code1 = computeCode(x1, y1); } else { x2 = x; y2 = y; code2 = computeCode(x2, y2); } } } if (accept) { cout << "Line accepted from " << x1 << ", " << y1 << " to " << x2 << ", " << y2 << endl; // Here the user can add code to display the rectangle // along with the accepted (portion of) lines } else cout << "Line rejected" << endl;}// Driver codeint main(){ // First Line segment // P11 = (5, 5), P12 = (7, 7) cohenSutherlandClip(5, 5, 7, 7); // Second Line segment // P21 = (7, 9), P22 = (11, 4) cohenSutherlandClip(7, 9, 11, 4); // Third Line segment // P31 = (1, 5), P32 = (4, 1) cohenSutherlandClip(1, 5, 4, 1); return 0;} |
Java
// Java program to implement Cohen Sutherland algorithm// for line clipping.import java.io.*;class GFG { // Defining region codes static final int INSIDE = 0; // 0000 static final int LEFT = 1; // 0001 static final int RIGHT = 2; // 0010 static final int BOTTOM = 4; // 0100 static final int TOP = 8; // 1000 // Defining x_max, y_max and x_min, y_min for // clipping rectangle. Since diagonal points are // enough to define a rectangle static final int x_max = 10; static final int y_max = 8; static final int x_min = 4; static final int y_min = 4; // Function to compute region code for a point(x, y) static int computeCode(double x, double y) { // initialized as being inside int code = INSIDE; if (x < x_min) // to the left of rectangle code |= LEFT; else if (x > x_max) // to the right of rectangle code |= RIGHT; if (y < y_min) // below the rectangle code |= BOTTOM; else if (y > y_max) // above the rectangle code |= TOP; return code; } // Implementing Cohen-Sutherland algorithm // Clipping a line from P1 = (x2, y2) to P2 = (x2, y2) static void cohenSutherlandClip(double x1, double y1, double x2, double y2) { // Compute region codes for P1, P2 int code1 = computeCode(x1, y1); int code2 = computeCode(x2, y2); // Initialize line as outside the rectangular window boolean accept = false; while (true) { if ((code1 == 0) && (code2 == 0)) { // If both endpoints lie within rectangle accept = true; break; } else if ((code1 & code2) != 0) { // If both endpoints are outside rectangle, // in same region break; } else { // Some segment of line lies within the // rectangle int code_out; double x = 0, y = 0; // At least one endpoint is outside the // rectangle, pick it. if (code1 != 0) code_out = code1; else code_out = code2; // Find intersection point; // using formulas y = y1 + slope * (x - x1), // x = x1 + (1 / slope) * (y - y1) if ((code_out & TOP) != 0) { // point is above the clip rectangle x = x1 + (x2 - x1) * (y_max - y1) / (y2 - y1); y = y_max; } else if ((code_out & BOTTOM) != 0) { // point is below the rectangle x = x1 + (x2 - x1) * (y_min - y1) / (y2 - y1); y = y_min; } else if ((code_out & RIGHT) != 0) { // point is to the right of rectangle y = y1 + (y2 - y1) * (x_max - x1) / (x2 - x1); x = x_max; } else if ((code_out & LEFT) != 0) { // point is to the left of rectangle y = y1 + (y2 - y1) * (x_min - x1) / (x2 - x1); x = x_min; } // Now intersection point x, y is found // We replace point outside rectangle // by intersection point if (code_out == code1) { x1 = x; y1 = y; code1 = computeCode(x1, y1); } else { x2 = x; y2 = y; code2 = computeCode(x2, y2); } } } if (accept) { System.out.println("Line accepted from " + x1 + ", " + y1 + " to " + x2 + ", " + y2); // Here the user can add code to display the // rectangle along with the accepted (portion // of) lines } else System.out.println("Line rejected"); } public static void main(String[] args) { // First Line segment // P11 = (5, 5), P12 = (7, 7) cohenSutherlandClip(5, 5, 7, 7); // Second Line segment // P21 = (7, 9), P22 = (11, 4) cohenSutherlandClip(7, 9, 11, 4); // Third Line segment // P31 = (1, 5), P32 = (4, 1) cohenSutherlandClip(1, 5, 4, 1); }}// This code is contributed by jain_mudit. |
Python3
# Python program to implement Cohen Sutherland algorithm# for line clipping.# Defining region codesINSIDE = 0 # 0000LEFT = 1 # 0001RIGHT = 2 # 0010BOTTOM = 4 # 0100TOP = 8 # 1000# Defining x_max, y_max and x_min, y_min for rectangle# Since diagonal points are enough to define a rectanglex_max = 10.0y_max = 8.0x_min = 4.0y_min = 4.0# Function to compute region code for a point(x, y)def computeCode(x, y): code = INSIDE if x < x_min: # to the left of rectangle code |= LEFT elif x > x_max: # to the right of rectangle code |= RIGHT if y < y_min: # below the rectangle code |= BOTTOM elif y > y_max: # above the rectangle code |= TOP return code# Implementing Cohen-Sutherland algorithm# Clipping a line from P1 = (x1, y1) to P2 = (x2, y2)# Implementing Cohen-Sutherland algorithm# Clipping a line from P1 = (x1, y1) to P2 = (x2, y2)def cohenSutherlandClip(x1, y1, x2, y2): # Compute region codes for P1, P2 code1 = computeCode(x1, y1) code2 = computeCode(x2, y2) accept = False while True: # If both endpoints lie within rectangle if code1 == 0 and code2 == 0: accept = True break # If both endpoints are outside rectangle elif (code1 & code2) != 0: break # Some segment lies within the rectangle else: # Line needs clipping # At least one of the points is outside, # select it x = 1.0 y = 1.0 if code1 != 0: code_out = code1 else: code_out = code2 # Find intersection point # using formulas y = y1 + slope * (x - x1), # x = x1 + (1 / slope) * (y - y1) if code_out & TOP: # Point is above the clip rectangle x = x1 + (x2 - x1) * (y_max - y1) / (y2 - y1) y = y_max elif code_out & BOTTOM: # Point is below the clip rectangle x = x1 + (x2 - x1) * (y_min - y1) / (y2 - y1) y = y_min elif code_out & RIGHT: # Point is to the right of the clip rectangle y = y1 + (y2 - y1) * (x_max - x1) / (x2 - x1) x = x_max elif code_out & LEFT: # Point is to the left of the clip rectangle y = y1 + (y2 - y1) * (x_min - x1) / (x2 - x1) x = x_min # Now intersection point (x, y) is found # We replace point outside clipping rectangle # by intersection point if code_out == code1: x1 = x y1 = y code1 = computeCode(x1, y1) else: x2 = x y2 = y code2 = computeCode(x2, y2) if accept: print("Line accepted from %.2f, %.2f to %.2f, %.2f" % (x1, y1, x2, y2)) # Here the user can add code to display the rectangle # along with the accepted (portion of) lines else: print("Line rejected")# Driver script# First line segment# P11 = (5, 5), P12 = (7, 7)cohenSutherlandClip(5, 5, 7, 7)# Second line segment# P21 = (7, 9), P22 = (11, 4)cohenSutherlandClip(7, 9, 11, 4)# Third line segment# P31 = (1, 5), P32 = (4, 1)cohenSutherlandClip(1, 5, 4, 1) |
C#
using System;public class GFG{ // C# program to implement Cohen Sutherland algorithm// for line clipping.// Defining region codespublic const int INSIDE = 0; // 0000public const int LEFT = 1; // 0001public const int RIGHT = 2; // 0010public const int BOTTOM = 4; // 0100public const int TOP = 8; // 1000// Defining x_max, y_max and x_min, y_min for// clipping rectangle. Since diagonal points are// enough to define a rectanglepublic const int x_max = 10;public const int y_max = 8;public const int x_min = 4;public const int y_min = 4;// Function to compute region code for a point(x, y)public static int computeCode(double x, double y){ // initialized as being inside int code = INSIDE; if (x < x_min) // to the left of rectangle code |= LEFT; else if (x > x_max) // to the right of rectangle code |= RIGHT; if (y < y_min) // below the rectangle code |= BOTTOM; else if (y > y_max) // above the rectangle code |= TOP; return code;}// Implementing Cohen-Sutherland algorithm// Clipping a line from P1 = (x2, y2) to P2 = (x2, y2)public static void cohenSutherlandClip(double x1, double y1, double x2, double y2){ // Compute region codes for P1, P2 int code1 = computeCode(x1, y1); int code2 = computeCode(x2, y2); // Initialize line as outside the rectangular window bool accept = false; while (true) { if ((code1 == 0) && (code2 == 0)) { // If both endpoints lie within rectangle accept = true; break; } else if ((code1 & code2)!=0) { // If both endpoints are outside rectangle, // in same region break; } else { // Some segment of line lies within the // rectangle int code_out; double x=0.0; double y=0.0; // At least one endpoint is outside the // rectangle, pick it. if (code1 != 0) code_out = code1; else code_out = code2; // Find intersection point; // using formulas y = y1 + slope * (x - x1), // x = x1 + (1 / slope) * (y - y1) if ((code_out & TOP)!=0) { // point is above the clip rectangle x = x1 + (x2 - x1) * (y_max - y1) / (y2 - y1); y = y_max; } else if ((code_out & BOTTOM)!=0) { // point is below the rectangle x = x1 + (x2 - x1) * (y_min - y1) / (y2 - y1); y = y_min; } else if ((code_out & RIGHT)!=0) { // point is to the right of rectangle y = y1 + (y2 - y1) * (x_max - x1) / (x2 - x1); x = x_max; } else if ((code_out & LEFT)!=0) { // point is to the left of rectangle y = y1 + (y2 - y1) * (x_min - x1) / (x2 - x1); x = x_min; } // Now intersection point x, y is found // We replace point outside rectangle // by intersection point if (code_out == code1) { x1 = x; y1 = y; code1 = computeCode(x1, y1); } else { x2 = x; y2 = y; code2 = computeCode(x2, y2); } } } if (accept) { Console.WriteLine("Line accepted from "+x1+", " +y1+" to "+x2+", "+y2); // Here the user can add code to display the rectangle // along with the accepted (portion of) lines } else Console.WriteLine("Line rejected");}// Driver code static public void Main (){ // First Line segment // P11 = (5, 5), P12 = (7, 7) cohenSutherlandClip(5, 5, 7, 7); // Second Line segment // P21 = (7, 9), P22 = (11, 4) cohenSutherlandClip(7, 9, 11, 4); // Third Line segment // P31 = (1, 5), P32 = (4, 1) cohenSutherlandClip(1, 5, 4, 1); }}// This code is contributed by akashish__ |
Javascript
// JavaScript program to implement Cohen Sutherland algorithm// for line clipping.// Defining region codesconst INSIDE = 0; // 0000const LEFT = 1; // 0001const RIGHT = 2; // 0010const BOTTOM = 4; // 0100const TOP = 8; // 1000// Defining x_max, y_max and x_min, y_min for// clipping rectangle. Since diagonal points are// enough to define a rectangleconst x_max = 10;const y_max = 8;const x_min = 4;const y_min = 4;// Function to compute region code for a point(x, y)function computeCode(x, y){ // initialized as being inside let code = INSIDE; if (x < x_min) // to the left of rectangle code |= LEFT; else if (x > x_max) // to the right of rectangle code |= RIGHT; if (y < y_min) // below the rectangle code |= BOTTOM; else if (y > y_max) // above the rectangle code |= TOP; return code;}// Implementing Cohen-Sutherland algorithm// Clipping a line from P1 = (x2, y2) to P2 = (x2, y2)function cohenSutherlandClip(x1, y1, x2, y2){ // Compute region codes for P1, P2 let code1 = computeCode(x1, y1); let code2 = computeCode(x2, y2); // Initialize line as outside the rectangular window let accept = false; while (true) { if ((code1 == 0) && (code2 == 0)) { // If both endpoints lie within rectangle accept = true; break; } else if (code1 & code2) { // If both endpoints are outside rectangle, // in same region break; } else { // Some segment of line lies within the // rectangle let code_out; let x, y; // At least one endpoint is outside the // rectangle, pick it. if (code1 != 0) code_out = code1; else code_out = code2; // Find intersection point; // using formulas y = y1 + slope * (x - x1), // x = x1 + (1 / slope) * (y - y1) if (code_out & TOP) { // point is above the clip rectangle x = x1 + (x2 - x1) * (y_max - y1) / (y2 - y1); y = y_max; } else if (code_out & BOTTOM) { // point is below the rectangle x = x1 + (x2 - x1) * (y_min - y1) / (y2 - y1); y = y_min; } else if (code_out & RIGHT) { // point is to the right of rectangle y = y1 + (y2 - y1) * (x_max - x1) / (x2 - x1); x = x_max; } else if (code_out & LEFT) { // point is to the left of rectangle y = y1 + (y2 - y1) * (x_min - x1) / (x2 - x1); x = x_min; } // Now intersection point x, y is found // We replace point outside rectangle // by intersection point if (code_out == code1) { x1 = x; y1 = y; code1 = computeCode(x1, y1); } else { x2 = x; y2 = y; code2 = computeCode(x2, y2); } } } if (accept) { console.log("Line accepted from", x1, ",", y1, "to", x2, ",", y2); // Here the user can add code to display the rectangle // along with the accepted (portion of) lines } else console.log("Line rejected");}// Driver code// First Line segment// P11 = (5, 5), P12 = (7, 7)cohenSutherlandClip(5, 5, 7, 7);// Second Line segment// P21 = (7, 9), P22 = (11, 4)cohenSutherlandClip(7, 9, 11, 4);// Third Line segment// P31 = (1, 5), P32 = (4, 1)cohenSutherlandClip(1, 5, 4, 1);// The code is contributed by Nidhi goel |
Output:
Line accepted from 5.00, 5.00 to 7.00, 7.00 Line accepted from 7.80, 8.00 to 10.00, 5.25 Line rejected
Below is C++ code with Graphics using graphics.h
C++
// C++ program to implement Cohen Sutherland algorithm// for line clipping.// including libraries#include <bits/stdc++.h>#include <graphics.h>using namespace std;// Global Variablesint xmin, xmax, ymin, ymax;// Lines where co-ordinates are (x1, y1) and (x2, y2)struct lines { int x1, y1, x2, y2;};// This will return the sign required.int sign(int x){ if (x > 0) return 1; else return 0;}// CohenSutherLand LineClipping Algorithm As Described in theory.// This will clip the lines as per window boundaries.void clip(struct lines mylines){ // arrays will store bits // Here bits implies initial Point whereas byte implies end points int bits[4], byte[4], i, var; // setting color of graphics to be RED setcolor(RED); // Finding Bits bits[0] = sign(xmin - mylines.x1); byte[0] = sign(xmin - mylines.x2); bits[1] = sign(mylines.x1 - xmax); byte[1] = sign(mylines.x2 - xmax); bits[2] = sign(ymin - mylines.y1); byte[2] = sign(ymin - mylines.y2); bits[3] = sign(mylines.y1 - ymax); byte[3] = sign(mylines.y2 - ymax); // initial will used for initial coordinates and end for final string initial = "", end = "", temp = ""; // convert bits to string for (i = 0; i < 4; i++) { if (bits[i] == 0) initial += '0'; else initial += '1'; } for (i = 0; i < 4; i++) { if (byte[i] == 0) end += '0'; else end += '1'; } // finding slope of line y=mx+c as (y-y1)=m(x-x1)+c // where m is slope m=dy/dx; float m = (mylines.y2 - mylines.y1) / (float)(mylines.x2 - mylines.x1); float c = mylines.y1 - m * mylines.x1; // if both points are inside the Accept the line and draw if (initial == end && end == "0000") { // inbuilt function to draw the line from(x1, y1) to (x2, y2) line(mylines.x1, mylines.y1, mylines.x2, mylines.y2); return; } // this will contain cases where line maybe totally outside for partially inside else { // taking bitwise end of every value for (i = 0; i < 4; i++) { int val = (bits[i] & byte[i]); if (val == 0) temp += '0'; else temp += '1'; } // as per algo if AND is not 0000 means line is completely outside hence draw nothing and return if (temp != "0000") return; // Here contain cases of partial inside or outside // So check for every boundary one by one for (i = 0; i < 4; i++) { // if both bit are same hence we cannot find any intersection with boundary so continue if (bits[i] == byte[i]) continue; // Otherwise there exist a intersection // Case when initial point is in left xmin if (i == 0 && bits[i] == 1) { var = round(m * xmin + c); mylines.y1 = var; mylines.x1 = xmin; } // Case when final point is in left xmin if (i == 0 && byte[i] == 1) { var = round(m * xmin + c); mylines.y2 = var; mylines.x2 = xmin; } // Case when initial point is in right of xmax if (i == 1 && bits[i] == 1) { var = round(m * xmax + c); mylines.y1 = var; mylines.x1 = xmax; } // Case when final point is in right of xmax if (i == 1 && byte[i] == 1) { var = round(m * xmax + c); mylines.y2 = var; mylines.x2 = xmax; } // Case when initial point is in top of ymin if (i == 2 && bits[i] == 1) { var = round((float)(ymin - c) / m); mylines.y1 = ymin; mylines.x1 = var; } // Case when final point is in top of ymin if (i == 2 && byte[i] == 1) { var = round((float)(ymin - c) / m); mylines.y2 = ymin; mylines.x2 = var; } // Case when initial point is in bottom of ymax if (i == 3 && bits[i] == 1) { var = round((float)(ymax - c) / m); mylines.y1 = ymax; mylines.x1 = var; } // Case when final point is in bottom of ymax if (i == 3 && byte[i] == 1) { var = round((float)(ymax - c) / m); mylines.y2 = ymax; mylines.x2 = var; } // Updating Bits at every point bits[0] = sign(xmin - mylines.x1); byte[0] = sign(xmin - mylines.x2); bits[1] = sign(mylines.x1 - xmax); byte[1] = sign(mylines.x2 - xmax); bits[2] = sign(ymin - mylines.y1); byte[2] = sign(ymin - mylines.y2); bits[3] = sign(mylines.y1 - ymax); byte[3] = sign(mylines.y2 - ymax); } // end of for loop // Initialize initial and end to NULL initial = "", end = ""; // Updating strings again by bit for (i = 0; i < 4; i++) { if (bits[i] == 0) initial += '0'; else initial += '1'; } for (i = 0; i < 4; i++) { if (byte[i] == 0) end += '0'; else end += '1'; } // If now both points lie inside or on boundary then simply draw the updated line if (initial == end && end == "0000") { line(mylines.x1, mylines.y1, mylines.x2, mylines.y2); return; } // else line was completely outside hence rejected else return; }}// Driver Functionint main(){ int gd = DETECT, gm; // Setting values of Clipping window xmin = 40; xmax = 100; ymin = 40; ymax = 80; // initialize the graph initgraph(&gd, &gm, NULL); // Drawing Window using Lines line(xmin, ymin, xmax, ymin); line(xmax, ymin, xmax, ymax); line(xmax, ymax, xmin, ymax); line(xmin, ymax, xmin, ymin); // Assume 4 lines to be clipped struct lines mylines[4]; // Setting the coordinated of 4 lines mylines[0].x1 = 30; mylines[0].y1 = 65; mylines[0].x2 = 55; mylines[0].y2 = 30; mylines[1].x1 = 60; mylines[1].y1 = 20; mylines[1].x2 = 100; mylines[1].y2 = 90; mylines[2].x1 = 60; mylines[2].y1 = 100; mylines[2].x2 = 80; mylines[2].y2 = 70; mylines[3].x1 = 85; mylines[3].y1 = 50; mylines[3].x2 = 120; mylines[3].y2 = 75; // Drawing Initial Lines without clipping for (int i = 0; i < 4; i++) { line(mylines[i].x1, mylines[i].y1, mylines[i].x2, mylines[i].y2); delay(1000); } // Drawing clipped Line for (int i = 0; i < 4; i++) { // Calling clip() which in term clip the line as per window and draw it clip(mylines[i]); delay(1000); } delay(4000); getch(); // For Closing the graph. closegraph(); return 0;} |
The Cohen–Sutherland algorithm can be used only on a rectangular clip window. For other convex polygon clipping windows, Cyrus–Beck algorithm is used. We will be discussing Cyrus–Beck Algorithm in next set.
Related Post :
Polygon Clipping | Sutherland–Hodgman Algorithm
Point Clipping Algorithm in Computer Graphics
Reference:
1) https://en.wikipedia.org/wiki/Cohen–Sutherland_algorithm
2)Book:- Computer Graphics: By Donald Hearn, M. Pauline Baker
3) College notes
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