Line Clipping | Set 1 (Cohen–Sutherland Algorithm)

Difficulty Level : Medium
Last Updated : 31 Mar, 2020

Given a set of lines and a rectangular area of interest, the task is to remove lines which 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

cohen-sutherland-algo1

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 codes
const int INSIDE = 0; // 0000
const int LEFT = 1; // 0001
const int RIGHT = 2; // 0010
const int BOTTOM = 4; // 0100
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 rectangle
const 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 code
int 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;
}

Python

# Python program to implement Cohen Sutherland algorithm
# for line clipping.
  
# Defining region codes
INSIDE = 0  # 0000
LEFT = 1    # 0001
RIGHT = 2   # 0010
BOTTOM = 4  # 0100
TOP = 8     # 1000
  
# Defining x_max, y_max and x_min, y_min for rectangle
# Since diagonal points are enough to define a rectangle
x_max = 10.0
y_max = 8.0
x_min = 4.0
y_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)
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)

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++ program to implement Cohen Sutherland algorithm
// for line clipping.
// including libraries
#include <bits/stdc++.h>
#include <graphics.h>
using namespace std;
  
// Global Variables
int 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 Algorith As Described in theory.
// This will clip the lines as per window boundries.
void clip(struct lines mylines)
{
    // arrays will store bits
    // Here bits impiles initial Point whereas bite implies end points
    int bits[4], bite[4], i, var;
    // setting color of graphics to be RED
    setcolor(RED);
  
    // Finding Bits
    bits[0] = sign(xmin - mylines.x1);
    bite[0] = sign(xmin - mylines.x2);
    bits[1] = sign(mylines.x1 - xmax);
    bite[1] = sign(mylines.x2 - xmax);
    bits[2] = sign(ymin - mylines.y1);
    bite[2] = sign(ymin - mylines.y2);
    bits[3] = sign(mylines.y1 - ymax);
    bite[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 (bite[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") {
        // inbuild 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] & bite[i]);
            if (val == 0)
                temp += '0';
            else
                temp += '1';
        }
        // as per algo if AND is not 0000 means line is completely outside hene draw nothing and retrurn
        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 boths bit are same hence we cannot find any intersection with boundary so continue
            if (bits[i] == bite[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 && bite[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 && bite[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 && bite[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 && bite[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);
            bite[0] = sign(xmin - mylines.x2);
            bits[1] = sign(mylines.x1 - xmax);
            bite[1] = sign(mylines.x2 - xmax);
            bits[2] = sign(ymin - mylines.y1);
            bite[2] = sign(ymin - mylines.y2);
            bits[3] = sign(mylines.y1 - ymax);
            bite[3] = sign(mylines.y2 - ymax);
        } // end of for loop
        // Inialize 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 (bite[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 Function
int main()
{
    int gd = DETECT, gm;
  
    // Setting values of Clipping window
    xmin = 40;
    xmax = 100;
    ymin = 40;
    ymax = 80;
  
    // intialize 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;
  
    // Drwaing 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);
    }
  
    // Drwaing 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.

Reference:
https://en.wikipedia.org/wiki/Cohen–Sutherland_algorithm

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