custom sqrt

[bluefrog]

I am attempting to create a sqrt function using mainly recursion with the use of a lambda.
I picked the code up from a book "Functional Python Programming 2ed".
The example is shown in the 1st chapter.

#!/usr/bin/python3

def next_(n, x):
    return (x+n/x)/2

n = 2 # converge value
f = lambda x: next_(n,x)

y=1.0 # seed value

def repeat(f, a):
    yield a
    for v in repeat(f, f(a)):
        yield v

def within (tolerance, iterable):
    def head_tail(tolerance, a, iterable):
        b = next(iterable)
        if abs(a-b) < tolerance: return b
    return head_tail(tolerance , next(iterable), iterable)

def sqrt(y, tolerance, n):
    return within(tolerance, repeat(f, y))

# show the square root of
print(sqrt(1.0, .0001, 3))

the final print however shows up as None.

[Gribouillis]

The problem is that head_tail() can return None if abs(a-b) is not smaller than tolerance. There must be something missing in that function.

[jefsummers]

Just askin' as I am not quite following the logic, in the within function, did you mean to use next() or next_()? I'd recommend avoiding defining a function with a name so close to a standard library function.

[bluefrog]

I modified

if abs(a-b) < tolerance: return b

to

if abs(a-b) <= tolerance: return b

When I run the code on my work pc, the sqrt appears as expected, for example:

But when I run the same code on my laptop, None appears.
Both are running Python 3.7.3

[ichabod801]

That's a simple fix, as shown here:
Image