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
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; } |
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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 = 10.0y_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) 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) |
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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; } |
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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:
https://en.wikipedia.org/wiki/Cohen–Sutherland_algorithm
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Improved By : imkiller
