I have a series of numbers I need to find the sum of. The value of the first iterative operation is 1, the second is 20. Every iteration which follows then uses the previous result in the formula n * (n + 1) / 2, so the third iteration, say i03 = 20 * (20 + 1) / 2, and the fourth, i04 = i03 * (i03 + 1) / 2. This continues until the 20th iteration of i20 = i19 * (i19 + 1) / 2. I want to do this using memoization. This is my code:
def outFun():
def sumFun(squares, total = 0, CONST = 20):
if squares > 2:
total = sumFun(squares - 1) * int((sumFun(squares - 1) + 1) / 2)
elif not squares - 2:
total = CONST
return total
return 1 + sumFun(20)
What am I doing wrong?
Here is how I understand your problem: You have a formula x_n = x_{n-1} * (x_{n-1} + 1)/2 with recursion base defined as x_1 = 20 (or x_2 = 20? Not clear from your description). The most efficient way to solve the recursion is bottom-up approach, when you start with x_1, then calculate x_2, etc. Alternative is to use dynamic programming/memorization:
mem={}
def f(x):
if x == 1: # base case
return 20
if not x in mem: # if we did not calculate it before - calculate
mem[x] = f(x-1) * (f(x-1) +1) / 2
return mem[x] # otherwise return it
print f(1)
print f(2)
print f(3)
prints
20
210
22155
f(20) is a little large to print, so I will print the number of digits in it:
print "number of digits: %s" % len(str(f(20)))
number of digits: 530115
The code took about 9 seconds to run on my desktop:
import timeit
mem={}
print "Execution time: %s" % timeit.Timer("len(str(f(20)))",
setup = "from __main__ import f").timeit(1)
you're calling
sumFun(squares - 1)
twice!
Why not introduce a variable to store the result? Something like:
if squares > 2:
nextResult = sumFun(squares - 1)
total = nextResult * ((nextResult + 1) / 2)
Since factoring a quadratic equation in my head just happens, and has done that since I learned it - how would I go about starting to write a quadratic factorer in Python?
Improving Keiths's answer:
Start with a polynomial P(x) = a*x^2 + b*x + c.
Use the quadratic formula (or another method of your choice) to find the roots r1 and r2 to P(x) = 0.
You can now factor P(x) as a*(x-r1)(x-r2).
If your factor (3x - 4)(x - 9) the solution will be 3*(x - 4/3)(x - 9).
You might want to find a way to multiply the 3 into the factors to get rid of fractions / look pretty. In this case, it might help to use fraction arithmetic instead of doubles so you can know the denominators better.
Use the quadratic formula.
I tried implementing hugomg's approach. I stole the "gcd" and "simplify fraction" function from online. Here is my sloppy approach:
from math import sqrt
def gcd(a, b):
while b:
a, b = b, a % b
return a
def simplify_fraction(numer, denom):
if denom == 0:
return "Division by 0 - result undefined"
# Remove greatest common divisor:
common_divisor = gcd(numer, denom)
(reduced_num, reduced_den) = (numer / common_divisor, denom / common_divisor)
# Note that reduced_den > 0 as documented in the gcd function.
if common_divisor == 1:
return (numer, denom)
else:
# Bunch of nonsense to make sure denominator is negative if possible
if (reduced_den > denom):
if (reduced_den * reduced_num < 0):
return(-reduced_num, -reduced_den)
else:
return (reduced_num, reduced_den)
else:
return (reduced_num, reduced_den)
def quadratic_function(a,b,c):
if (b**2-4*a*c >= 0):
x1 = (-b+sqrt(b**2-4*a*c))/(2*a)
x2 = (-b-sqrt(b**2-4*a*c))/(2*a)
# Added a "-" to these next 2 values because they would be moved to the other side of the equation
mult1 = -x1 * a
mult2 = -x2 * a
(num1,den1) = simplify_fraction(a,mult1)
(num2,den2) = simplify_fraction(a,mult2)
if ((num1 > a) or (num2 > a)):
# simplify fraction will make too large of num and denom to try to make a sqrt work
print("No factorization")
else:
# Getting ready to make the print look nice
if (den1 > 0):
sign1 = "+"
else:
sign1 = ""
if (den2 > 0):
sign2 = "+"
else:
sign2 = ""
print("({}x{}{})({}x{}{})".format(int(num1),sign1,int(den1),int(num2),sign2,int(den2)))
else:
# if the part under the sqrt is negative, you have a solution with i
print("Solutions are imaginary")
return
# This function takes in a, b, and c from the equation:
# ax^2 + bx + c
# and prints out the factorization if there is one
quadratic_function(7,27,-4)
If I run this I get the output:
(7x-1)(1x+4)
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
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
>>>
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)
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
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
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
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...
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
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))
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 .
Try converting the exponent to a floating number, as the default behaviour of / in Python is integer division
n**(1/float(3))
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.