monte carlo in crystal

contents/monte_carlo_integration/code/crystal/monte_carlo.cr

@@ -0,0 +1,29 @@
+# Method to determine whether an x, y point is in the unit circle.
+def in_circle(x_pos : Float64, y_pos : Float64)
+  # Setting radius to 1 for unit circle
+  radius = 1
+  x_pos ** 2 + y_pos ** 2 < radius ** 2
+end
+
+# Method to integrate a unit circle to find pi via monte_carlo
+def monte_carlo(n : Int)
+  pi_count = 0
+  n.times do
+    point_x = rand
+    point_y = rand
+
+    if in_circle(point_x, point_y)
+      pi_count += 1
+    end
+  end
+
+  # This is using a quarter of the unit sphere in a 1x1 box.
+  # The formula is pi = (box_length^2 / radius^2) * (pi_count / n), but we
+  #     are only using the upper quadrant and the unit circle, so we can use
+  #     4*pi_count/n instead
+  4 * pi_count / n
+end
+
+pi_estimate = monte_carlo(10000000)
+puts "The pi estimate is: #{pi_estimate}"
+puts "Percent error is: #{100 * (pi_estimate - Math::PI).abs / Math::PI} %"

contents/monte_carlo_integration/monte_carlo_integration.md

@@ -101,6 +101,8 @@ each point is tested to see whether it's in the circle or not:
 [import:4-9, lang:"coconut"](code/coconut/monte_carlo.coco)
 {% sample lang="ps1" %}
 [import:1-3, lang:"powershell"](code/powershell/MonteCarlo.ps1)
+{% sample lang="crystal" %}
+[import:2-6, lang:"crystal"](code/crystal/monte_carlo.cr)
 {% endmethod %}
 
 If it's in the circle, we increase an internal count by one, and in the end,
@@ -210,6 +212,8 @@ The code snippets were taken from this [scratch project](https://scratch.mit.edu
 [import, lang:"coconut"](code/coconut/monte_carlo.coco)
 {% sample lang="ps1" %}
 [import, lang:"powershell"](code/powershell/MonteCarlo.ps1)
+{% sample lang="crystal" %}
+[import, lang:"crystal"](code/crystal/monte_carlo.cr)
 {% endmethod %}
 
 <script>