simplifying julia code for monte carlo

chapters/monte_carlo/code/julia/monte_carlo.jl

@@ -1,5 +1,8 @@
 # function to determine whether an x, y point is in the unit circle
-function in_circle(x_pos::Float64, y_pos::Float64, radius::Float64)
+function in_circle(x_pos::Float64, y_pos::Float64)
+
+    # Setting radius to 1 for unit circle
+    radius = 1
     if (x_pos^2 + y_pos^2 < radius^2)
         return true
     else
@@ -8,20 +11,24 @@ function in_circle(x_pos::Float64, y_pos::Float64, radius::Float64)
 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

@@ -27,7 +27,7 @@ $$
 This means,
 
 $$
-\text{Area}_{\text{circle}} = \text{Area}_{\text{square}}\times\text{Ratio} = 4r^2 \times \text{ratio}
+\text{Area}_{\text{circle}} = \text{Area}_{\text{square}}\times\text{Ratio} = 4r^2 \times \text{Ratio}
 $$
 
 So, if we can find the $$\text{Ratio}$$ and we know $$r$$, we should be able to easily find the $$\text{Area}_{\text{circle}}$$.