How to compute the nth root of a very big integer

Question

I need a way to compute the nth root of a long integer in Python.

I tried pow(m, 1.0/n), but it doesn't work:

OverflowError: long int too large to convert to float

Any ideas?

By long integer I mean REALLY long integers like:

11968003966030964356885611480383408833172346450467339251 196093144141045683463085291115677488411620264826942334897996389 485046262847265769280883237649461122479734279424416861834396522 819159219215308460065265520143082728303864638821979329804885526 557893649662037092457130509980883789368448042961108430809620626 059287437887495827369474189818588006905358793385574832590121472 680866521970802708379837148646191567765584039175249171110593159 305029014037881475265618958103073425958633163441030267478942720 703134493880117805010891574606323700178176718412858948243785754 898788359757528163558061136758276299059029113119763557411729353 915848889261125855717014320045292143759177464380434854573300054 940683350937992500211758727939459249163046465047204851616590276 724564411037216844005877918224201569391107769029955591465502737 961776799311859881060956465198859727495735498887960494256488224 613682478900505821893815926193600121890632

Answer 1

If it's a REALLY big number. You could use a binary search.

def find_invpow(x,n):
    """Finds the integer component of the n'th root of x,
    an integer such that y ** n <= x < (y + 1) ** n.
    """
    high = 1
    while high ** n <= x:
        high *= 2
    low = high/2
    while low < high:
        mid = (low + high) // 2
        if low < mid and mid**n < x:
            low = mid
        elif high > mid and mid**n > x:
            high = mid
        else:
            return mid
    return mid + 1

For example:

>>> x = 237734537465873465
>>> n = 5
>>> y = find_invpow(x,n)
>>> y
2986
>>> y**n <= x <= (y+1)**n
True
>>>
>>> x = 119680039660309643568856114803834088331723464504673392511960931441>
>>> n = 45
>>> y = find_invpow(x,n)
>>> y
227661383982863143360L
>>> y**n <= x < (y+1)**n
True
>>> find_invpow(y**n,n) == y
True
>>>

Answer 2

Gmpy is a C-coded Python extension module that wraps the GMP library to provide to Python code fast multiprecision arithmetic (integer, rational, and float), random number generation, advanced number-theoretical functions, and more.

Includes a root function:

x.root(n): returns a 2-element tuple (y,m), such that y is the (possibly truncated) n-th root of x; m, an ordinary Python int, is 1 if the root is exact (x==y**n), else 0. n must be an ordinary Python int, >=0.

For example, 20th root:

>>> import gmpy
>>> i0=11968003966030964356885611480383408833172346450467339251 
>>> m0=gmpy.mpz(i0)
>>> m0
mpz(11968003966030964356885611480383408833172346450467339251L)
>>> m0.root(20)
(mpz(567), 0)

Answer 3

If you are looking for something standard, fast to write with high precision. I would use decimal and adjust the precision (getcontext().prec) to at least the length of x.

Code (Python 3.0)

from decimal import *

x =   '11968003966030964356885611480383408833172346450467339251\
196093144141045683463085291115677488411620264826942334897996389\
485046262847265769280883237649461122479734279424416861834396522\
819159219215308460065265520143082728303864638821979329804885526\
557893649662037092457130509980883789368448042961108430809620626\
059287437887495827369474189818588006905358793385574832590121472\
680866521970802708379837148646191567765584039175249171110593159\
305029014037881475265618958103073425958633163441030267478942720\
703134493880117805010891574606323700178176718412858948243785754\
898788359757528163558061136758276299059029113119763557411729353\
915848889261125855717014320045292143759177464380434854573300054\
940683350937992500211758727939459249163046465047204851616590276\
724564411037216844005877918224201569391107769029955591465502737\
961776799311859881060956465198859727495735498887960494256488224\
613682478900505821893815926193600121890632'

minprec = 27
if len(x) > minprec: getcontext().prec = len(x)
else:                getcontext().prec = minprec

x = Decimal(x)
power = Decimal(1)/Decimal(3)

answer = x**power
ranswer = answer.quantize(Decimal('1.'), rounding=ROUND_UP)

diff = x - ranswer**Decimal(3)
if diff == Decimal(0):
    print("x is the cubic number of", ranswer)
else:
    print("x has a cubic root of ", answer)

Answer

x is the cubic number of 22873918786185635329056863961725521583023133411 451452349318109627653540670761962215971994403670045614485973722724603798 107719978813658857014190047742680490088532895666963698551709978502745901 704433723567548799463129652706705873694274209728785041817619032774248488 2965377218610139128882473918261696612098418

Answer 4

You can make it run slightly faster by avoiding the while loops in favor of setting low to 10 ** (len(str(x)) / n) and high to low * 10. Probably better is to replace the len(str(x)) with the bitwise length and using a bit shift. Based on my tests, I estimate a 5% speedup from the first and a 25% speedup from the second. If the ints are big enough, this might matter (and the speedups may vary). Don't trust my code without testing it carefully. I did some basic testing but may have missed an edge case. Also, these speedups vary with the number chosen.

If the actual data you're using is much bigger than what you posted here, this change may be worthwhile.

from timeit import Timer

def find_invpow(x,n):
    """Finds the integer component of the n'th root of x,
    an integer such that y ** n <= x < (y + 1) ** n.
    """
    high = 1
    while high ** n < x:
        high *= 2
    low = high/2
    while low < high:
        mid = (low + high) // 2
        if low < mid and mid**n < x:
            low = mid
        elif high > mid and mid**n > x:
            high = mid
        else:
            return mid
    return mid + 1

def find_invpowAlt(x,n):
    """Finds the integer component of the n'th root of x,
    an integer such that y ** n <= x < (y + 1) ** n.
    """
    low = 10 ** (len(str(x)) / n)
    high = low * 10

    while low < high:
        mid = (low + high) // 2
        if low < mid and mid**n < x:
            low = mid
        elif high > mid and mid**n > x:
            high = mid
        else:
            return mid
    return mid + 1

x = 237734537465873465
n = 5
tests = 10000

print "Norm", Timer('find_invpow(x,n)', 'from __main__ import find_invpow, x,n').timeit(number=tests)
print "Alt", Timer('find_invpowAlt(x,n)', 'from __main__ import find_invpowAlt, x,n').timeit(number=tests)

Norm 0.626754999161

Alt 0.566340923309

Answer 5

Oh, for numbers that big, you would use the decimal module.

ns: your number as a string

ns = "11968003966030964356885611480383408833172346450467339251196093144141045683463085291115677488411620264826942334897996389485046262847265769280883237649461122479734279424416861834396522819159219215308460065265520143082728303864638821979329804885526557893649662037092457130509980883789368448042961108430809620626059287437887495827369474189818588006905358793385574832590121472680866521970802708379837148646191567765584039175249171110593159305029014037881475265618958103073425958633163441030267478942720703134493880117805010891574606323700178176718412858948243785754898788359757528163558061136758276299059029113119763557411729353915848889261125855717014320045292143759177464380434854573300054940683350937992500211758727939459249163046465047204851616590276724564411037216844005877918224201569391107769029955591465502737961776799311859881060956465198859727495735498887960494256488224613682478900505821893815926193600121890632"
from decimal import Decimal
d = Decimal(ns)
one_third = Decimal("0.3333333333333333")
print d ** one_third

and the answer is: 2.287391878618402702753613056E+305

TZ pointed out that this isn't accurate... and he's right. Here's my test.

from decimal import Decimal

def nth_root(num_decimal, n_integer):
    exponent = Decimal("1.0") / Decimal(n_integer)
    return num_decimal ** exponent

def test():
    ns = "11968003966030964356885611480383408833172346450467339251196093144141045683463085291115677488411620264826942334897996389485046262847265769280883237649461122479734279424416861834396522819159219215308460065265520143082728303864638821979329804885526557893649662037092457130509980883789368448042961108430809620626059287437887495827369474189818588006905358793385574832590121472680866521970802708379837148646191567765584039175249171110593159305029014037881475265618958103073425958633163441030267478942720703134493880117805010891574606323700178176718412858948243785754898788359757528163558061136758276299059029113119763557411729353915848889261125855717014320045292143759177464380434854573300054940683350937992500211758727939459249163046465047204851616590276724564411037216844005877918224201569391107769029955591465502737961776799311859881060956465198859727495735498887960494256488224613682478900505821893815926193600121890632"
    nd = Decimal(ns)
    cube_root = nth_root(nd, 3)
    print (cube_root ** Decimal("3.0")) - nd

if __name__ == "__main__":
    test()

It's off by about 10**891

Answer 6

Possibly for your curiosity:

http://en.wikipedia.org/wiki/Hensel_Lifting

This could be the technique that Maple would use to actually find the nth root of large numbers.

Pose the fact that x^n - 11968003.... = 0 mod p, and go from there...

Answer 7

I may suggest four methods for solving your task. First is based on Binary Search. Second is based on Newton's Method. Third is based on Shifting n-th Root Algorithm. Fourth is called by me Chord-Tangent method described by me in picture here.

Binary Search was already implemented in many answers above. I just introduce here my own vision of it and its implementation.

As alternative I also implement Optimized Binary Search method (marked Opt). This method just starts from range [hi / 2, hi) where hi is equal to 2^(num_bit_length / k) if we're computing k-th root.

Newton's Method is new here, as I see it wasn't implemented in other answers. It is usually considered to be faster than Binary Search, although my own timings in code below don't show any speedup. Hence this method here is just for reference/interest.

Shifting Method is 30-50% faster than optimized binary search method, and should be even faster if implemented in C++, because C++ has fast 64 bit arithemtics which is partially used in this method.

Chord-Tangent Method:

Chord-Tangent Method is invented by me on piece of paper (see image above), it is inspired and is an improvement of Newton method. Basically I draw a Chord and a Tangent Line and find intersection with horizontal line y = n, these two intersections form lower and upper bound approximations of location of root solution (x0, n) where n = x0 ^ k. This method appeared to be fastest of all, while all other methods do more than 2000 iterations, this method does just 8 iterations, for the case of 8192-bit numbers. So this method is 200-300x times faster than previous (by speed) Shifting Method.

As an example I generate really huge random integer of 8192 bits in size. And measure timings of finding cubic root with both methods.

In test() function you can see that I passed k = 3 as root's power (cubic root), you can pass any power instead of 3.

Try it online!

def binary_search(begin, end, f, *, niter = [0]):
    while begin < end:
        niter[0] += 1
        mid = (begin + end) >> 1
        if f(mid):
            begin = mid + 1
        else:
            end = mid
    return begin

def binary_search_kth_root(n, k, *, verbose = False):
    # https://en.wikipedia.org/wiki/Binary_search_algorithm
    niter = [0]
    res = binary_search(0, n + 1, lambda root: root ** k < n, niter = niter)
    if verbose:
        print('Binary Search iterations:', niter[0])
    return res

def binary_search_opt_kth_root(n, k, *, verbose = False):
    # https://en.wikipedia.org/wiki/Binary_search_algorithm
    niter = [0]
    hi = 1 << (n.bit_length() // k - 1)
    while hi ** k <= n:
        niter[0] += 1
        hi <<= 1
    res = binary_search(hi >> 1, hi, lambda root: root ** k < n, niter = niter)
    if verbose:
        print('Binary Search Opt iterations:', niter[0])
    return res

def newton_kth_root(n, k, *, verbose = False):
    # https://en.wikipedia.org/wiki/Newton%27s_method
    f = lambda x: x ** k - n
    df = lambda x: k * x ** (k - 1)
    x, px, niter = n, 2 * n, [0]
    while abs(px - x) > 1:
        niter[0] += 1
        px = x
        x -= f(x) // df(x)
    if verbose:
        print('Newton Method iterations:', niter[0])
    mini, minv = None, None
    for i in range(-2, 3):
        v = abs(f(x + i))
        if minv is None or v < minv:
            mini, minv = i, v
    return x + mini

def shifting_kth_root(n, k, *, verbose = False):
    # https://en.wikipedia.org/wiki/Shifting_nth_root_algorithm
    B_bits = 64
    
    r, y = 0, 0
    B = 1 << B_bits
    Bk_bits = B_bits * k
    Bk_mask = (1 << Bk_bits) - 1
    niter = [0]
    
    for i in range((n.bit_length() + Bk_bits - 1) // Bk_bits - 1, -1, -1):
        alpha = (n >> (i * Bk_bits)) & Bk_mask
        B_y = y << B_bits
        Bk_yk = (y ** k) << Bk_bits
        Bk_r_alpha = (r << Bk_bits) + alpha
        Bk_yk_Bk_r_alpha = Bk_yk + Bk_r_alpha
        beta = binary_search(1, B, lambda beta: (B_y + beta) ** k <= Bk_yk_Bk_r_alpha, niter = niter) - 1
        y, r = B_y + beta, Bk_r_alpha - ((B_y + beta) ** k - Bk_yk)

    if verbose:
        print('Shifting Method iterations:', niter[0])

    return y

def chord_tangent_kth_root(n, k, *, verbose = False):
    niter = [0]
    hi = 1 << (n.bit_length() // k - 1)
    while hi ** k <= n:
        niter[0] += 1
        hi <<= 1
    f = lambda x: x ** k
    df = lambda x: k * x ** (k - 1)
    # https://i.stack.imgur.com/et9O0.jpg
    x_begin, x_end = hi >> 1, hi
    y_begin, y_end = f(x_begin), f(x_end)
    for icycle in range(1 << 30):
        if x_end - x_begin <= 1:
            break
        niter[0] += 1
        if 0: # Do Binary Search step if needed
            x_mid = (x_begin + x_end) >> 1
            y_mid = f(x_mid)
            if y_mid > n:
                x_end, y_end = x_mid, y_mid
            else:
                x_begin, y_begin = x_mid, y_mid
        # (y_end - y_begin) / (x_end - x_begin) = (n - y_begin) / (x_n - x_begin) ->
        x_n = x_begin + (n - y_begin) * (x_end - x_begin) // (y_end - y_begin)
        y_n = f(x_n)
        tangent_x = x_n + (n - y_n) // df(x_n) + 1
        
        chord_x = x_n + (n - y_n) * (x_end - x_n) // (y_end - y_n)
        
        assert chord_x <= tangent_x, (chord_x, tangent_x)
        x_begin, x_end = chord_x, tangent_x
        y_begin, y_end = f(x_begin), f(x_end)
        assert y_begin <= n, (chord_x, y_begin, n, n - y_begin)
        assert y_end > n, (icycle, tangent_x - binary_search_kth_root(n, k), y_end, n, y_end - n)
    if verbose:
        print('Chord Tangent Method iterations:', niter[0])
    return x_begin

def test():
    import random, timeit
    
    nruns = 3
    bits = 8192
    n = random.randrange(1 << (bits - 1), 1 << bits)
    
    a = binary_search_kth_root(n, 3, verbose = True)
    b = binary_search_opt_kth_root(n, 3, verbose = True)
    c = newton_kth_root(n, 3, verbose = True)
    d = shifting_kth_root(n, 3, verbose = True)
    e = chord_tangent_kth_root(n, 3, verbose = True)
    assert abs(a - b) <= 0 and abs(a - c) <= 1 and abs(a - d) <= 1 and abs(a - e) <= 1, (a - b, a - c, a - d, a - e)

    print()
    print('Binary Search timing:', round(timeit.timeit(lambda: binary_search_kth_root(n, 3), number = nruns) / nruns, 3), 'sec')
    print('Binary Search Opt timing:', round(timeit.timeit(lambda: binary_search_opt_kth_root(n, 3), number = nruns) / nruns, 3), 'sec')
    print('Newton Method timing:', round(timeit.timeit(lambda: newton_kth_root(n, 3), number = nruns) / nruns, 3), 'sec')
    print('Shifting Method timing:', round(timeit.timeit(lambda: shifting_kth_root(n, 3), number = nruns) / nruns, 3), 'sec')
    print('Chord Tangent Method timing:', round(timeit.timeit(lambda: chord_tangent_kth_root(n, 3), number = nruns) / nruns, 3), 'sec')

if __name__ == '__main__':
    test()

Output:

Binary Search iterations: 8192
Binary Search Opt iterations: 2732
Newton Method iterations: 9348
Shifting Method iterations: 2752
Chord Tangent Method iterations: 8

Binary Search timing: 0.506 sec
Binary Search Opt timing: 0.05 sec
Newton Method timing: 2.09 sec
Shifting Method timing: 0.03 sec
Chord Tangent Method timing: 0.001 sec

Answer 8

I came up with my own answer, which takes @Mahmoud Kassem's idea, simplifies the code, and makes it more reusable:

def cube_root(x):
    return decimal.Decimal(x) ** (decimal.Decimal(1) / decimal.Decimal(3))

I tested it in Python 3.5.1 and Python 2.7.8, and it seemed to work fine.

The result will have as many digits as specified by the decimal context the function is run in, which by default is 28 decimal places. According to the documentation for the power function in the decimal module, "The result is well-defined but only “almost always correctly-rounded”.". If you need a more accurate result, it can be done as follows:

with decimal.localcontext() as context:
    context.prec = 50
    print(cube_root(42))

Answer 9

In older versions of Python, 1/3 is equal to 0. In Python 3.0, 1/3 is equal to 0.33333333333 (and 1//3 is equal to 0).

So, either change your code to use 1/3.0 or switch to Python 3.0 .

Answer 10

Try converting the exponent to a floating number, as the default behaviour of / in Python is integer division

n**(1/float(3))

Answer 11

Well, if you're not particularly worried about precision, you could convert it to a sting, chop off some digits, use the exponent function, and then multiply the result by the root of how much you chopped off.

E.g. 32123 is about equal to 32 * 1000, the cubic root is about equak to cubic root of 32 * cubic root of 1000. The latter can be calculated by dividing the number of 0s by 3.

This avoids the need for the use of extension modules.