I am trying to write a code in python, to calculate the maximum graphs sets by using this formula for n vertices:
2**(n(n-1)/2) (i hope i wrote it correctly)
i am trying to do it in the lowest complexity/running time buy using % of 1000000007- is recursive the right way? or iterative?
I've read the Wikipedia article regarding exponential squaring- but couldn't make the leap from there to my problem :(
after some pen and pen and paper work I've discovered that in this case- n(n-1)/2 is always even - so i am removing the block that deals with odd n values.
This is the code a wrote so far for the recursion-
x for the base (2) and n for the number of vertices:
def graphs_num(x, n):
n = (n*n-n)/2
if n == 0:
return 0
elif n == 1:
return 1
else:
y = graphs_num(x, n/2)
return y*y
no results so far -can you assist please?
2nd edit: (i forgot the x in the last line)
here's #alec's code:
def count(n, total=1):
n = (n*n-1)/2
if n < 2:
return total
total *= 2 ** (n - 1) % 1000000007
return count(x, n-1, total)
count(4)
now I am looking to have x already defined as 2- so the only input need by the function will be n
I'm not completely familiar with this problem but based on the formula you gave this will return the same result for n >= 0 (I assume you wouldn't have a negative number of vertices).
def count(n, total=1):
if n < 2:
return total
total *= 2 ** (n - 1)
return count(n-1, total)
>>> count(5)
1024
Edit (explanation):
Basically you will notice that the values increase by a factor of 2**n.
>>> [2 ** (n * (n-1) // 2) for n in range(1, 6)]
[1, 2, 8, 64, 1024]
This can be illustrated by the following:
f(1) = 1 = 2**0
f(2) = 2 = 2**1 * 2**0
f(3) = 8 = 2**2 * 2**1 * 2**0
...
And simplified to show how it might be used recursively:
f(1) = 1 = 2**0
f(2) = 2 = 2**1 * f(1)
f(3) = 8 = 2**2 * f(2)
...
By this point, it is easy to notice the relationship is f(n) = 2**(n-1) * f(n-1). In the recursive function, the lines underneath the base case do exactly this:
total *= 2 ** (n - 1)
return count(n-1, total)
I multiple the total by 2**(n-1), and recurse to the next value for n, remembering to decrement by 1. The base case ensures we bottom out below 2, and returns the total.
Now if you want to change the base of the exponent and use a mod, it will behave accordingly.
def count(x, n, total=1):
if n < 2:
return total
total *= x ** (n - 1) % 1000000007
return count(x, n-1, total)
>>> count(2, 3)
8
>>> count(3, 3)
27
I'm trying to translate a loop to a recursive algorithm. Fairly simple, I've just hadn't been able to make it ignore the n value when summing up the values, like range does.
This is the iterative function:
def function(n):
total=0
for i in range(1,n,2):
total += i
print(total)
function(5) # Output: 4
This is the recursive I've tried:
def function1(n):
if n==1:
return n
else:
return n+function1(n-2)
function(5) # Output: 9
So function1 does sum the n when it should be ignored. Cause range() does not include the stop number.
Then, I tried:
def f1(n):
def f_recursive(n):
if n==1 or n==2:
return 1
elif n==0:
return 0
else:
return n + f_recursive(n - 2)
return f_recursive(n) - n
print(f1(5)) # Output: 4 Yeiii!!
But then I realised, that only works for odd numbers. Not for even. If f1(6) then you get 4 when it should be 9, because it ends up being 11-6= 9.
So silly me I tried:
def f1(n):
def f_recursive(n):
if n==1 or n==2:
return 1
elif n==0:
return 0
elif n%2 == 0:
return n + f_recursive(n - 3)
elif n%2 == 1:
return n + f_recursive(n - 2)
return f_recursive(n) - n
print(f1(6))
Which of course also did not work. Am I not understanding recursion properly here?
The tricky part is excluding the upper bound. If the upper bound is your only parameter n, you have to know when it's the first call, and when it's an intermediate (recursive) call. Alternatively, if inner functions are okay, you could instead just count from 1 up until you hit n:
def function1(n):
def inner(i):
return 0 if i >= n else i + inner(i + 2)
return inner(1)
You want to compute the sum of all odd integers from 1 up to, but not including, n.
This leaves 2 possibilities:
If n is <= 1, there are no numbers to sum, so the sum is 0.
The highest number that might be included in the list is n-1, but only if it is odd. Either way, the rest of the sum is "the sum of all odd integers from 1 up to, but not including, n-1" (sound familiar?)
This translates to:
def f1(n):
if n <= 1:
return 0
else:
isOdd = (n-1)%2==1
return f1(n-1) + (n-1 if isOdd else 0)
The problem with your recursion is that you're returning n rather than the value in the range (list) that you're currently on, this poses a problem since n is not inclusive within the range and should not be added to the final total
Ideally you need to reverse the logic and traverse it the same way your range does
def func(start,end, step):
if(start >= end):
return 0
return start + func(start + step, end, step)
You just have to recognize the three types of ranges you might be adding up.
range(1, n, 2) where n <= 1: The empty range, so the sum is 0
range(1, n, 2) where n > 1 and n is even: the range is 1, ..., n-1. (E.g. range(1, 6, 2) == [1, 3, 5])
range(1, n, 2) where n > 1 and n is odd: the range is 1, ..., n-2 (E.g., range(1, 5, 2) == [1, 3]
Translating this to code is straightforward:
def f_recursive1(n):
if n <= 1:
return 0
elif n % 2 == 0:
return n - 1 + f_recursive1(n-2)
else: # n odd
return n - 2 + f_recursive1(n-2)
However, this does more work than is strictly necessary, since subtracting 2 from n will never change its parity; you don't need to check n is even or odd in every recursive call.
def f_recursive2(n):
def f_helper(x):
if x <= 0:
return 0
return x + f_helper(x-2)
if n % 2 == 0:
return f_helper(n-1)
else:
return f_helper(n-2)
If we are allowed multiplication and division, I hope you realise that this particular task does not require more than just a base case.
Python code:
def f(n):
total=0
for i in range(1,n,2):
total += i
return total
def g(n):
half = n // 2
return half * half
for n in xrange(100):
print f(n), g(n)
Since
*
* * *
* * * * *
* * * * * * *
can be seen as nested, folded rows. Here are the top two folded rows:
*
* * *
* * * *
* * * *
Let's rotate counterclockwise 45 degrees
* * * *
* * * *
* *
* *
and add the other two folded rows,
*
* *
* * * *
* * * *
* * * *
* * *
and
*
to get
* * * *
* * * *
* * * *
* * * *
the area of a square.
Stern's Diatomic Sequence can be read about in more details over here; however, for my purpose I will define it now.
Definition of Stern's Diatomic Sequence
Let n be a number to generate the fusc function out of. Denoted fusc(n).
If n is 0 then the returned value is 0.
If n is 1 then the returned value is 1.
If n is even then the returned value is fusc(n / 2).
If n is odd then the returned value is fusc((n - 1) / 2) + fusc((n + 1) / 2).
Currently, my Python code brute forces through most of the generation, other than the dividing by two part since it will always yield no change.
def fusc (n):
if n <= 1:
return n
while n > 2 and n % 2 == 0:
n /= 2
return fusc((n - 1) / 2) + fusc((n + 1) / 2)
However, my code must be able to handle digits in the magnitude of 1000s millions of bits, and recursively running through the function thousands millions of times does not seem very efficient or practical.
Is there any way I could algorithmically improve my code such that massive numbers can be passed through without having to recursively call the function so many times?
With memoization for a million bits, the recursion stack would be extremely large. We can first try to look at a sufficiently large number which we can work by hand, fusc(71) in this case:
fusc(71) = fusc(35) + fusc(36)
fusc(35) = fusc(17) + fusc(18)
fusc(36) = fusc(18)
fusc(71) = 1 * fusc(17) + 2 * fusc(18)
fusc(17) = fusc(8) + fusc(9)
fusc(18) = fusc(9)
fusc(71) = 1 * fusc(8) + 3 * fusc(9)
fusc(8) = fusc(4)
fusc(9) = fusc(4) + fusc(5)
fusc(71) = 4 * fusc(4) + 3 * fusc(5)
fusc(4) = fusc(2)
fusc(3) = fusc(1) + fusc(2)
fusc(71) = 7 * fusc(2) + 3 * fusc(3)
fusc(2) = fusc(1)
fusc(3) = fusc(1) + fusc(2)
fusc(71) = 11 * fusc(1) + 3 * fusc(2)
fusc(2) = fusc(1)
fusc(71) = 14 * fusc(1) = 14
We realize that we can avoid recursion completely in this case as we can always express fusc(n) in the form a * fusc(m) + b * fusc(m+1) while reducing the value of m to 0. From the example above, you may find the following pattern:
if m is odd:
a * fusc(m) + b * fusc(m+1) = a * fusc((m-1)/2) + (b+a) * fusc((m+1)/2)
if m is even:
a * fusc(m) + b * fusc(m+1) = (a+b) * fusc(m/2) + b * fusc((m/2)+1)
Therefore, you may use a simple loop function to solve the problem in O(lg(n)) time
def fusc(n):
if n == 0: return 0
a = 1
b = 0
while n > 0:
if n%2:
b = b + a
n = (n-1)/2
else:
a = a + b
n = n/2
return b
lru_cache works wonders in your case. make sure maxsize is a power of 2. may need to fiddle a bit with that size for your application. cache_info() will help with that.
also use // instead of / for integer division.
from functools import lru_cache
#lru_cache(maxsize=512, typed=False)
def fusc(n):
if n <= 1:
return n
while n > 2 and n % 2 == 0:
n //= 2
return fusc((n - 1) // 2) + fusc((n + 1) // 2)
print(fusc(1000000000078093254329870980000043298))
print(fusc.cache_info())
and yes, this is just meomization as proposed by Filip Malczak.
you might gain an additional tiny speedup using bit-operations in the while loop:
while not n & 1: # as long as the lowest bit is not 1
n >>= 1 # shift n right by one
UPDATE:
here is a simple way of doing meomzation 'by hand':
def fusc(n, _mem={}): # _mem will be the cache of the values
# that have been calculated before
if n in _mem: # if we know that one: just return the value
return _mem[n]
if n <= 1:
return n
while not n & 1:
n >>= 1
if n == 1:
return 1
ret = fusc((n - 1) // 2) + fusc((n + 1) // 2)
_mem[n] = ret # store the value for next time
return ret
UPDATE
after reading a short article by dijkstra himself a minor update.
the article states, that f(n) = f(m) if the fist and last bit of m are the same as those of n and the bits in between are inverted. the idea is to get n as small as possible.
that is what the bitmask (1<<n.bit_length()-1)-2 is for (first and last bits are 0; those in the middle 1; xoring n with that gives m as described above).
i was only able to do small benchmarks; i'm interested if this is any help at all for the magitude of your input... this will reduce the memory for the cache and hopefully bring some speedup.
def fusc_ed(n, _mem={}):
if n <= 1:
return n
while not n & 1:
n >>= 1
if n == 1:
return 1
# https://www.cs.utexas.edu/users/EWD/transcriptions/EWD05xx/EWD578.html
# bit invert the middle bits and check if this is smaller than n
m = n ^ (1<<n.bit_length()-1)-2
n = m if m < n else n
if n in _mem:
return _mem[n]
ret = fusc(n >> 1) + fusc((n >> 1) + 1)
_mem[n] = ret
return ret
i had to increase the recursion limit:
import sys
sys.setrecursionlimit(10000) # default limit was 1000
benchmarking gave strange results; using the code below and making sure that i always started a fresh interperter (having an empty _mem) i sometimes got significantly better runtimes; on other occasions the new code was slower...
benchmarking code:
print(n.bit_length())
ti = timeit('fusc(n)', setup='from __main__ import fusc, n', number=1)
print(ti)
ti = timeit('fusc_ed(n)', setup='from __main__ import fusc_ed, n', number=1)
print(ti)
and these are three random results i got:
6959
24.117448464001427
0.013900151001507766
6989
23.92404893300045
0.013844672999766772
7038
24.33894686200074
24.685758719999285
that is where i stopped...
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.