多边形区域（递归使用Python）

Question

I'm trying a solve an exercise from Exploring Python book. But, I guess I don't understand concept of the recursion. I've written some Recursively function. Therefore I know some of the aspects. But, I don't have enough experience. And I've stopped to study programming about one year.

Anyway, let me give you the full question:

A polygon can be represented by a list of (x, y) pairs where each pair is a tuple: [ (x1, y1), (x2, y2), (x3, y3) , ... (xn, yn)]. Write a recursive function to compute the area of a polygon. This can be accomplished by “cutting off” a triangle, using the fact that a triangle with corners (x1, y1), (x2, y2), (x3, y3) has area (x1y1 + x2y2 + x3y2 – y1x2 –y2x3 – y3x1) / 2.

Despite the fact that, the question already gave the formula, I used another formula. Because, I made some research about area of a polygon. And if you look at here the formula is different.

And describing my program step by step would be better, in order to explain what I want. OK, I had to declare global scopes, because of recursion:

area = 0
x = [0] * 3
y = [0] * 3 

And then, I created a recursively function. Zero is always returned by this function as a result. So my real problem is this:

def areaofpolygon(polygon, i):
    global area, x, y # My variables 
    try: # I prefered using try statement from using if-else statements. So it is the easier I guess.
        x[i], y[i] = polygon[i] # X and Y coordinates from tuple
        area += (x[i]*y[i+1] - x[i+1]*y[i]) #My formula
    except IndexError:
        return area/2

    areaofpolygon(polygon, i+1)   # Here, this is my weird recursion

And my main function:

  def main():
      mypolygon = [(1,2), (2,5), (1,4)] # I declared polygon as tuples
      # I called my function and started to count from zero, and the result will be prompted.
      print(areaofpolygon(mypolygon,0))

      return 0
  if __name__ == '__main__':
      main()

And here is my full code without comments:

'''
Created on Feb 24, 2012

@author: msarialp
'''
area = 0
x = [0] * 3
y = [0] * 3
def areaofpolygon(polygon, i):
    global area, x, y
    try:
        x[i], y[i] = polygon[i]
        area += (x[i]*y[i+1] - x[i+1]*y[i])
    except IndexError:
        return area/2

    areaofpolygon(polygon, i+1)   
def main():
    mypolygon = [(1,2), (2,5), (1,4)]
    print(areaofpolygon(mypolygon,0))

    return 0
if __name__ == '__main__':
    main()

EDIT One

After reading your answers, I've understood what was wrong with my code. So I decided to share last version of my program in order to get some other helps. Again, I had to declare global variables. How can I apply ( lop_triangle ) function from senderle

area = 0
x = [0] * 3
y = [0] * 3

My function that divides tuple and to get x and y coordinates.

def sides_of_polygon(polygon, i):
    global x, y
    try:
        x[i], y[i] = polygon[i]
        return sides_of_polygon(polygon, i+1)
    except IndexError:
        return x, y

My function calculate area of polygon( Same as before )

def area_of_polygon(x, y, i):
    global area
    try:
        area += x[i]*y[i+1] - x[i+1]*y[i]
        return area_of_polygon(x, y, i+1)
    except IndexError:
        return area/2.0

My main function...

def main():
    mypolygon = [(1,2), (2,5), (1,4)]
    dx, dy = sides_of_polygon(mypolygon, 0)
    print(area_of_polygon(dx,dy,0))

    return 0
if __name__ == '__main__':
    main()

Please help me to improve my code without giving full solution.

EDIT Two

After making discussion with senderle, I understood where is the problem and senderle's solution is better than mine so I suggest that you should use it. Anyway, He helped me to make my code correct.And I had to change my formula again.

area += x[i]*y[(i+1) % 3] - x[(i+1) % 3]*y[i]

He also added for longer polygons 3 must be len(vertices). Thanks everyone for their time.

Answer 1

The implementation of your formula is flawed. It looks ahead to values in your x, y lists that haven't been set yet with (x[i]*y[i+1] - x[i+1]*y[i])

If you put a print statement inside your try-except block you will see that you are simply multiplying by zero and getting zero area:

try:
    x[i], y[i] = polygon[i]
    area += (x[i]*y[i+1] - x[i+1]*y[i])
    print x[i], y[i+1], x[i+1], y[i]
except IndexError, e:
    return area/2

#1 0 0 2
#2 0 0 5

Also, you are not returning the results of your recursive call to areaofpolygon, so you will never get that area/2 . You want: return areaofpolygon(polygon, i+1) . And make sure you actually divide by 2.0 so that you get float precision since your points are ints.

Try just using the formula you found or that was suggested in another question.

Update

Here is a fixed version of your code:

#!/usr/bin/env python

from random import randint
from shapely.geometry import Polygon

area = 0

def areaofpolygon(polygon, i):
    global area
    if i == 0: 
        area = 0

    try:
        x1, y1 = polygon[i]
        x2, y2 = polygon[i+1]
        area += (x1*y2) - (x2*y1)

    except IndexError, e:
        x1, y1 = polygon[0]
        x2, y2 = polygon[-1]
        area += (x2*y1) - (x1*y2)
        return abs(area/2.0)

    return areaofpolygon(polygon, i+1)   

def main():
    mypolygon = [(randint(0, 100), randint(0, 100)) for _ in xrange(10)]
    print mypolygon

    area = areaofpolygon(mypolygon, 0)
    assert area == Polygon(mypolygon).area

    print "Test passed."
    return 0

if __name__ == '__main__':
    main()

### Output ###
$ ./test.py 
[(29, 76), (85, 49), (27, 80), (94, 98), (19, 1), (75, 6), (55, 38), (74, 62), (0, 25), (93, 94)]
Test passed.
$ ./test.py 
[(13, 37), (98, 74), (42, 58), (32, 64), (95, 97), (34, 62), (34, 59), (21, 76), (55, 32), (76, 31)]
Test passed.
$ ./test.py 
[(38, 67), (66, 59), (16, 71), (53, 100), (64, 52), (69, 31), (45, 23), (52, 37), (27, 21), (42, 74)]
Test passed.

Notice that you didn't need global x,y lists. And you also missed the last part of the equation where you use the last point and the first point.

Answer 2

The essence of recursion is as follows:

1. Identify a simple base case and solve for that.
2. Conceive of a process that, when repeated, reduces a more complex case to that base case.
3. Apply the process in #2 to the problem until you reach the base case.

In your case, the first step is easy. The smallest polygon is a triangle. The area of a triangle is (x1y2 + x2y3 + x3y1 – y1x2 –y2x3 – y3x1) / 2 . (It looks like they misstated it in the problem though...)

The second step is also easy, because the problem statement gives it to you: given an n-vertex polygon, lop off a triangle, determine its area, and add it to the area of the resulting (n-1)-vertex polygon.

We'll break it down into parts. First, a function to solve #1:

def area_of_triangle(points):
    (x1, y1), (x2, y2), (x3, y3) = points
    return abs(x1 * y2 + x2 * y3 + x3 * y1 - y1 * x2 - y2 * x3 - y3 * x1) / 2

Easy. Now a function to solve #2. All we need is a function that lops off a triangle and returns both it and the resulting smaller polygon:

def lop_triangle(points):
    triangle = [points[0], points[-1], points[-2]]
    polygon = points[:-1]
    return triangle, polygon

If it's not obvious, this simply creates a triangle using the first and the last two points of the polygon. Then it removes the last point of the polygon, which is now equivalent to chopping off the triangle. (Draw a n-polygon and label its vertices from 0 to n to see how it works.) So now we have a triangle and a simpler polygon.

Now, let's put it all together. This third step is in some ways the hardest, but because we solved the first two problems already, the third is easier to understand.

def area_of_polygon(points):
    if len(points) == 3:
        return area_of_triangle(points)
    else:
        triangle, polygon = lop_triangle(points)
        return area_of_triangle(triangle) + area_of_polygon(polygon)

All the magic happens in that last line. Every time area_of_polygon gets a triangle, it just returns the area of a triangle. But when it gets a larger polygon, it lops off a triangle, takes the area of that triangle, and adds it to... the area of a smaller polygon. So say the polygon has 5 vertices. The first time area_of_polygon is called (c1), it lops off a triangle, takes its area, and then calls area_of_polygon (c2) again , but this time with a 4-vertex polygon. Then area_of_polygon lops of a triangle, and calls area_of_polygon (c3) again , but this time with a 3-vertex polygon. And then it doesn't have to call area_of_polygon again. It just returns the area of a triangle to the previous call (c2). That sums the result with the triangle in (c2) and returns that value to (c1). And then you have your answer.

Also, for what it's worth, the wolfram formula can be written with great clarity in three lines:

def area_of_polygon(vertices):
    pairs = zip(vertices, vertices[1:] + vertices[0:1])
    return sum(x1 * y2 - y1 * x2 for (x1, y1), (x2, y2) in pairs) / 2

Answer 3

Use this formula.

https://upload.wikimedia.org/wikipedia/en/math/c/b/b/cbb6a25439b51061adb913c2a6706484.png

You accomplish your task in one for loop.