simplifying Julia code for Monte Carlo

chapters/monte_carlo/code/julia/monte_carlo.jl

@@ -1,27 +1,30 @@
 # function to determine whether an x, y point is in the unit circle
-function in_circle(x_pos::Float64, y_pos::Float64, radius::Float64)
-    if (x_pos^2 + y_pos^2 < radius^2)
-        return true
-    else
-        return false
-    end
+function in_circle(x_pos::Float64, y_pos::Float64)
+
+    # Setting radius to 1 for unit circle
+    radius = 1
+    return x_pos^2 + y_pos^2 < radius^2
 end
 
 # function to integrate a unit circle to find pi via monte_carlo
-function monte_carlo(n::Int64, radius::Float64)
+function monte_carlo(n::Int64)
 
     pi_count = 0
     for i = 1:n
         point_x = rand()
         point_y = rand()
 
-        if (in_circle(point_x, point_y, radius))
+        if (in_circle(point_x, point_y))
             pi_count += 1
         end
     end
 
-    pi_estimate = 4*pi_count/(n*radius^2)
+    # 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
+    pi_estimate = 4*pi_count/n
     println("Percent error is: ", signif(100*(pi - pi_estimate)/pi, 3), " %")
 end
 
-monte_carlo(10000000, 0.5)
+monte_carlo(10000000)

chapters/monte_carlo/monte_carlo.md

@@ -38,7 +38,7 @@ each point is tested to see whether it's in the circle or not:
 
 {% method %}
 {% sample lang="jl" %}
-[import:2-8, lang:"julia"](code/julia/monte_carlo.jl)
+[import:2-7, lang:"julia"](code/julia/monte_carlo.jl)
 {% sample lang="c" %}
 [import:7-9, lang:"c_cpp"](code/c/monte_carlo.c)
 {% sample lang="hs" %}