How to find integer nth roots?

Question

I want to find the greatest integer less than or equal to the kth root of n. I tried

int(n**(1/k))

But for n=125, k=3 this gives the wrong

Answer 1

Here it is in Lua using Newton-Raphson method

> function nthroot (x, n) local r = 1; for i = 1, 16 do r = (((n - 1) * r) + x / (r ^ (n -   1))) / n end return r end
> return nthroot(125,3)
5
> 

Python version

>>> def nthroot (x, n):
...     r = 1
...     for i in range(16):
...             r = (((n - 1) * r) + x / (r ** (n - 1))) / n
...     return r
... 
>>> nthroot(125,3)
5
>>> 

Answer 2

def nthrootofm(a,n):
    a= pow(a,(1/n))
    return 'rounded:{},'.format(round(a))
a=125
n=3
q=nthrootofm(a,n)
print(q)

just used a format string , maybe this helps.

Answer 3

Why not to try this :

125 ** (1 / float(3)) 

or

pow(125, 1 / float(3))

It returns 5.0, so you can use int(), to convert to int.

Answer 4

One solution first brackets the answer between lo and hi by repeatedly multiplying hi by 2 until n is between lo and hi, then uses binary search to compute the exact answer:

def iroot(k, n):
    hi = 1
    while pow(hi, k) < n:
        hi *= 2
    lo = hi // 2
    while hi - lo > 1:
        mid = (lo + hi) // 2
        midToK = pow(mid, k)
        if midToK < n:
            lo = mid
        elif n < midToK:
            hi = mid
        else:
            return mid
    if pow(hi, k) == n:
        return hi
    else:
        return lo

A different solution uses Newton's method, which works perfectly well on integers:

def iroot(k, n):
    u, s = n, n+1
    while u < s:
        s = u
        t = (k-1) * s + n // pow(s, k-1)
        u = t // k
    return s

Answer 5

My cautious solution after being so badly burned:

def nth_root(N,k):
    """Return greatest integer x such that x**k <= N"""
    x = int(N**(1/k))      
    while (x+1)**k <= N:
        x += 1
    while x**k > N:
        x -= 1
    return x

Answer 6

def nth_root(n, k):
    x = n**(1./k)
    y = int(x)
    return y + 1 if y != x else y