Question

What estimate of pi would our randomized algorithm compute if our source of random numbers gives the following 10 pairs of points (x,y):
-0.8322,-0.4899
-0.5224,0.5117
0.7583,0.8756
0.7782,-0.8260
0.2597,0.6354
-0.0072,-0.6112
-0.5218,-0.3444
-0.9582,0.9481
-0.0192,-0.2660
0.9353,-0.2829

Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer.

a

2.2

b

2.4

c

2.6

d

2.8

e

3.0

f

3.2

g

3.4

h

3.6

i

3.8

j

4.0

Answer 1

Answer:

d) 2.8

We can estimating the value of Pi using Monte Carlo ,methods. The idea is to simulate random (x, y) points in a 2-D plane with domain as a square of side 1 unit. Imagine a circle inside the same domain with same diameter and inscribed into the square. We then calculate the ratio of number points that lied inside the circle and total number of generated points.

We know that area of the square is 1 unit sq while that of circle is $\pi \ast (\frac{1}{2})^{2} = \frac{\pi}{4}$ .

Now for a very large number of generated points,

area of circle / area of square = points generated inside circle / total points inside square

therefore, = 4 * (points generated inside circle / total points inside square)

7

Here I have written simple python program to calculate estimate of pi from the given set of random points.

estimate of pi.py

points = [ [-0.8322,-0.4899], [-0.5224,0.5117], [0.7583,0.8756], [0.7782,-0.8260], [0.2597,0.6354], [-0.0072,-0.6112], [-0.5218,-0.3444], [-0.9582,0.9481], [-0.0192,-0.2660], [0.9353,-0.2829]]

circle_points = 0
square_points = 0

for p in points:
    # Distance between (x, y) from the origin 
    origin_dist= p[0]**2 + p[1]**2
    # Checking if (x, y) lies inside the circle
    if (origin_dist<= 1): 
        circle_points+= 1
  
    square_points+= 1

# Estimating value of pi, 
# pi= 4*(no. of points inside the circle)/ (no. of points inside the square) 
print(4* circle_points/square_points)

Output:

二.8 >>> >|

Image of code:

1 points = [ [-0.8322,-0.4899], [-0.5224,0.5117], [0.7583,0.8756], [0.7782,-0.8260], [0.2597,0.6354], [-0.0072,-0.6112], [-6.5218,-0.3444], [-0.9582,0.9481], [-0.0192,-0.2660], [0.9353,-0.2829]] circle_points square_points = 0 = @ 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 for p in points: # Distance between (x, y) from the origin origin_dist= p[@]**2 + P[1]**2 # Checking if (x, y) lies inside the circle if (origin_dist<= 1): circle_points += 1 square_points = 1 # Estimating value of pi, # pi= 4*(no. of points inside the circle)/ (no. of points inside the square) print(4* circle_points/square_points)

Hope you like it.

If you have any doubts ask me in comment section.
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All the best.