How to find integer nth roots?

Question 1

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

Question 2

How about:

def nth_root(val, n):
    ret = int(val**(1./n))
    return ret + 1 if (ret + 1) ** n == val else ret

print nth_root(124, 3)
print nth_root(125, 3)
print nth_root(126, 3)
print nth_root(1, 100)

Here, both val and n are expected to be integer and positive. This makes the return expression rely exclusively on integer arithmetic, eliminating any possibility of rounding errors.

Note that accuracy is only guaranteed when val**(1./n) is fairly small. Once the result of that expression deviates from the true answer by more than 1, the method will no longer give the correct answer (it'll give the same approximate answer as your original version).

Still I am wondering why int(125**(1/3)) is 4

In [1]: '%.20f' % 125**(1./3)
Out[1]: '4.99999999999999911182'

int() truncates that to 4.

Question 3

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

Question 4

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.

Question 5

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
>>>

Question 6

I wonder if starting off with a method based on logarithms can help pin down the sources of rounding error. For example:

import math
def power_floor(n, k):
    return int(math.exp(1.0 / k * math.log(n)))

def nth_root(val, n):
    ret = int(val**(1./n))
    return ret + 1 if (ret + 1) ** n == val else ret

cases = [
    (124, 3),
    (125, 3),
    (126, 3),
    (1, 100),
    ]


for n, k in cases:
    print "{0:d} vs {1:d}".format(nth_root(n, k), power_floor(n, k))

prints out

4 vs 4
5 vs 5
5 vs 5
1 vs 1

Question 7

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

Question 8

You can round to nearest integer instead of rounding down / to zero (I don't know what Python specifies) :

def rtn (x):
    return int (x + 0.5)

>>> rtn (125 ** (1/3))
5

Question 9

int(125**(1/3)) should clearly be 5, i.e. the right answer, so this must be standard computer rounding error, i.e internally the result is 4.9999999999 which gets rounded down to 4. This problem will exist with whatever algorithm you use. One simple ad-hoc solution is to add a tiny number e.g. int((125**(1/3)) + 0.00000001)

Question 10

Do this before everything:

from __future__ import division

and then run any of the above specified techniques to have your results.

Question 11

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.