Added python3 to Monte Carlo Integration

CONTRIBUTORS.md

@@ -5,3 +5,4 @@ Gathros
 Jeremie Gillet (- Jie -)
 Salim Khatib
 Hitesh C
+Jonas Vander Vennet

chapters/monte_carlo/code/python3/monte_carlo.py

@@ -0,0 +1,20 @@
+#example submitted by Jonas Vander Vennet, Python 3.5.2
+
+import random, math
+
+# function to determine whether an x, y point is inside a circle with given radius
+def in_circle(x, y, radius):
+ return x**2 + y**2 < radius**2
+
+# function to integrate a circle with given radius via monte carlo integration
+def monte_carlo(n, radius):
+ count = 0
+ for i in range(n):
+ count += in_circle(random.uniform(0,radius),random.uniform(0,radius),radius)
+ return 4*(count/n)*radius**2
+
+#estimate pi by integrating the unit circle
+estimated_pi = monte_carlo(10**6, 1)
+
+print("percent error: {:%}".format(abs(estimated_pi-math.pi)/math.pi))
+print("Estimation for pi: {:f}".format(estimated_pi))

chapters/monte_carlo/monte_carlo.md

@@ -41,6 +41,8 @@ each point is tested to see whether it's in the circle or not:
 [import:2-8, lang:"julia"](code/julia/monte_carlo.jl)
 {% sample lang="hs" %}
 [import:7-7, lang:"haskell"](code/haskell/monteCarlo.hs)
+{% sample lang="py3" %}
+[import:5-7, lang:"python"](code/python3/monte_carlo.py)
 {% endmethod %}
 
 If it's in the circle, we increase an internal count by one, and in the end,
@@ -78,6 +80,9 @@ Feel free to submit your version via pull request, and thanks for reading!
 {% sample lang="hs" %}
 ### Haskell
 [import, lang:"haskell"](code/haskell/monteCarlo.hs)
+{% sample lang="py3" %}
+### Python 3
+[import, lang:"python"](code/python3/monte_carlo.py)
 {% endmethod %}