# This program computes an approximation of pi using the proof
# of the basel problem. It requires many terms to calculate a
# good approximation.
# see: https://en.wikipedia.org/wiki/Basel_problem

# The amount of terms
terms := 1000000

# The running sum of terms
tot := 0.0

n := 1
while n <= terms {
	tot = tot + 1.0/float(n*n)
	n = n + 1
}

tot = tot * 6.0

# get the absolute value of a number
fn abs(x: float) -> float {
	if x < 0.0 {
		-x
	} else {
	    x
	}
}

# calculate an approximation of the square root of tot using
# newton's method.
# see: https://en.wikipedia.org/wiki/Newton's_method
# The required accuracy
SQRT_ACC := 0.00000001
fn sqrt(x: float) -> float {
    pg := 0.0 # previous guess
    g := 1.0 # current guess

    while abs(pg - g) >= SQRT_ACC {
        pg = g
        g = (pg + x/pg)/2.0
    }

    return g
}

pi := sqrt(tot)

# output the result
println(pi)