I'm a newbie in algorithms. I have recently started studying binary search and tryed to implement it on my own. The task is simple: we have an array of integers a and an integer x. If a contains x the result should be its index, otherwise the function should return -1.
Here is the code I have written:
def binary_search(a, x):
l = 0
r = len(a)
while r - l > 0:
m = (l + r) // 2
if a[m] < x:
l = m
else:
r = m
if a[l] == x:
return l
return -1
But this code stucks in infinite cycle on a = [1, 2] and x = 2. I suppose, that I have incorrect cycle condition (probably, should be r - l >= 0), but this solution does not help. Where am I wrong?
Let me do some desk checking. I'll assume a = [1, 2] and we are searching for a 2
So we start with
l = 0
r = 2
Since r - l = 2 > 0, we enter the while-loop.
m = (l + r) / 2 = (0 + 2) / 2 = 1
a[m] = a[1] = 2 == x (hence not less than x)
r = m = 1 (and l remains the same)
Now r - l = 1 - 0 = 1 > 0, so we continue
m = (l + r) / 2 = (0 + 1) / 2 = 0
a[m] = a[0] = 1 < x
l = m = 0 (and r remains the same)
After this iteration both r and l have the same value as before, which then produces an endless loop.
Ashok's answer is a great fix. But I think it'll be educational to do some desk checking on the fixed code and look what improves it.
Basically the problematic situation arises, when l + 1 = r.
Then m will always evaluate to l, a[l] < x and l is set to m again, which doesn't change the situation.
In a larger piece of code it'll make sense to make a table that contains a column for each variable to watch and a column to write down the code line that was evaluated. A column for remarks won't harm either.
As Mani mentioned you are not considering when A[m]==x. Include that case (at that point you've found a so just return m), and once you have that case we can let l=m+1 when we are still below x. Like this:
def binary_search(a, x):
l = 0
r = len(a)
while r - l > 0:
m = (l + r) // 2
if a[m] < x:
l = m + 1
elif a[m]==x:
return m
else:
r = m
if l<len(a) and a[l] == x:
return l
return -1
Given an arbitrary range of 1 to F and a starting point S with an ending point G such that the only directions we could go is L Left steps and R Right Steps (also arbitrary), create a general solution that will return the number of steps it would take to go from S to R if it is possible otherwise return not possible.
You are bound to the range [1, F] which means that you cannot move L or R steps if the next move will be more than F or less than 1
Example:
F = 100
S = 2
G = 1
L = 0
R = 1
Output: not possible
F = 10
S = 1
G = 10
L = 1
R = 2
Output: 6
Explanation: [1 -> 3(R) -> 5(R) -> 7(R) -> 9(R) -> 8(L) -> 10(R)]
I've been given this problem in our class and our current topic is binary search and divide and conquer. Here's my approach but this does not solve one hidden case.
F = int(input())
S = int(input())
G = int(input())
L = int(input())
R = int(input())
count = 0
while S != G:
dist = abs(S - G) # Takes the current distance from S to G
if S > G:
if S-L > 0:
S -= L
count += 1
else:
S += R
count += 1
else:
if S+R <= F:
S += R
count += 1
else:
S -= L
count += 1
if dist == abs(S - G): # If distance doesn't change after trying
print("not possible") # a move, conclude that it is not possible.
break
if S == G: print(count)
Mathematically this problem means that we are looking for
integer solutions (for x and y) of the following equation:
x * R - y * L = G - S
we can start by creating a function to check if there is a solution quickly:
def path(S, G, F, L, R):
x=0
while True:
y = (x * R - G + S) / L
if y>=0 and int(y)==y:
return(x,y)
else:
x+=1
This will work if there are solutions, but not if they are not.
It can be proved mathematically that there is no solution when L devides R but not G-S. Here is the proof:
If
R mod L =0 (L devides R)
(G - S)/L != 0 (L doesn't devide G-S)
then by deviding the whole equation (x * R - y * L = G - S) by L we take:
x * R/L - y = (G - S)/L <=>
y= (x * R/L) - (G - S)/L
Now, we want y mod 1 = 0 (means that y is integer) for x mod 1 =0 (x integers). Using common modulo operations we take:
y mod 1 = [(x * R/L) - (G - S)/L] mod 1 =
[(x * R/L) mod 1 - ((G - S)/L) mod 1] mod 1 =
[(x mod 1 * (R/L) mod 1) mod 1 - ((G - S)/L) mod 1] mod 1 =
[(x mod 1 * 0) mod 1 - ((G - S)/L) mod 1] mod 1 =
[((G - S)/L) mod 1] mod 1
This cannot be 0 if L doesn't devide G-S which eventally means that there are no pair of integers x,y that can satisfy the original condition.
Programatically this means for our code, the following additions:
def path(S, G, F, L, R):
if R%L==0 and (G-S)%L != 0 :
return 'No solutions'
x=0
while True:
y = (x * R - G + S) / L
if y>=0 and int(y)==y:
return(x,y)
else:
x+=1
I don;t know if mathematically we can prove that the above if is the only exception, it can probably be proved with some more modulo operations. Programatically we can add some clock in our code so that if there are no solutions, which means that it will go to infitive loop, we can return False after some time. Here is how we can do this:
How would I stop a while loop after n amount of time?
If the distance didn't change that doesn't mean it's impossible, you could have just jumped over the point and made it to the other side with the same distance, imagine L=2, R=1, S=3, G=2, you start distance 1 from goal, jump left (still distance 1) then jump right and win. What you need to check is whether you have gone in a loop and ended up at a location you have already tried before. You can either keep track of these locations (say in a set) or do some math ahead of time and figure out how many Ls and Rs it takes before you have definitely looped (probably not intended to figure this out).
F=int(input())
S=int(input())
G=int(input())
L=int(input())
R=int(input())
L*=-1
Fl=1-L
Fr=F-R
h0=set()
n=0
while True:
if S<G:
S+= R if S<=Fr else L
elif G<S:
S+= L if Fl<=S else R
else:
print(n)
break
if S in h0:
print('not possible')
break
h0.add(S)
n+=1
I have the following code:
def isPP(n):
pos = [int(i) for i in range(n+1)]
pos = pos[2:] ##to ignore the trivial n** 1 == n case
y = []
for i in pos:
for it in pos:
if i** it == n:
y.append((i,it))
#return list((i,it))
#break
if len(y) <1:
return None
else:
return list(y[0])
Which works perfectly up until ~2000, since I'm storing far too much in memory. What can I do to make it work efficiently for large numbers (say, 50000 or 100000). I tried to make it end after finding one case, but my algorithm is still far too inefficient if the number is large.
Any tips?
A number n is a perfect power if there exists a b and e for which b^e = n. For instance 216 = 6^3 = 2^3 * 3^3 is a perfect power, but 72 = 2^3 * 3^2 is not.
The trick to determining if a number is a perfect power is to know that, if the number is a perfect power, then the exponent e must be less than log2 n, because if e is greater then 2^e will be greater than n. Further, it is only necessary to test prime es, because if a number is a perfect power to a composite exponent it will also be a perfect power to the prime factors of the composite component; for instance, 2^15 = 32768 = 32^3 = 8^5 is a perfect cube root and also a perfect fifth root.
The function isPerfectPower shown below tests each prime less than log2 n by first computing the integer root using Newton's method, then powering the result to check if it is equal to n. Auxiliary function primes compute a list of prime numbers by the Sieve of Eratosthenes, iroot computes the integer kth-root by Newton's method, and ilog computes the integer logarithm to base b by binary search.
def primes(n): # sieve of eratosthenes
i, p, ps, m = 0, 3, [2], n // 2
sieve = [True] * m
while p <= n:
if sieve[i]:
ps.append(p)
for j in range((p*p-3)/2, m, p):
sieve[j] = False
i, p = i+1, p+2
return ps
def iroot(k, n): # assume n > 0
u, s, k1 = n, n+1, k-1
while u < s:
s = u
u = (k1 * u + n // u ** k1) // k
return s
def ilog(b, n): # max e where b**e <= n
lo, blo, hi, bhi = 0, 1, 1, b
while bhi < n:
lo, blo, hi, bhi = hi, bhi, hi+hi, bhi*bhi
while 1 < (hi - lo):
mid = (lo + hi) // 2
bmid = blo * pow(b, (mid - lo))
if n < bmid: hi, bhi = mid, bmid
elif bmid < n: lo, blo = mid, bmid
else: return mid
if bhi == n: return hi
return lo
def isPerfectPower(n): # x if n == x ** y, or False
for p in primes(ilog(2,n)):
x = iroot(p, n)
if pow(x, p) == n: return x
return False
There is further discussion of the perfect power predicate at my blog.
IIRC, it's far easier to iteratively check "Does it have a square root? Does it have a cube root? Does it have a fourth root? ..." You will very quickly get to the point where putative roots have to be between 1 and 2, at which point you can stop.
I think a better way would be implementing this "hack":
import math
def isPP(n):
range = math.log(n)/math.log(2)
range = (int)(range)
result = []
for i in xrange(n):
if(i<=1):
continue
exponent = (int)(math.log(n)/math.log(i))
for j in [exponent-1, exponent, exponent+1]:
if i ** j == n:
result.append([i,j])
return result
print isPP(10000)
Result:
[[10,4],[100,2]]
The hack uses the fact that:
if log(a)/log(b) = c,
then power(b,c) = a
Since this calculation can be a bit off in floating points giving really approximate results, exponent is checked to the accuracy of +/- 1.
You can make necessary adjustments for handling corner cases like n=1, etc.
a relevant improvement would be:
import math
def isPP(n):
# first have a look at the length of n in binary representation
ln = int(math.log(n)/math.log(2)) + 1
y = []
for i in range(n+1):
if (i <= 1):
continue
# calculate max power
li = int(math.log(i)/math.log(2))
mxi = ln / li + 1
for it in range(mxi):
if (it <= 1):
continue
if i ** it == n:
y.append((i,it))
# break if you only need 1
if len(y) <1:
return None
else:
return list(y[0])
I'm new to python and i'm having a hard time trying to find the root of a polynomial via using the bisection method. So far I have 2 methods. One for evaluating the polynomial at value x
def eval(x, poly):
"""
Evaluate the polynomial at the value x.
poly is a list of coefficients from lowest to highest.
:param x: Argument at which to evaluate
:param poly: The polynomial coefficients, lowest order to highest
:return: The result of evaluating the polynomial at x
"""
result = poly[0]
for i in range(1, len(poly)):
result = result + poly[i] * x**i
return result
The next method is supposed to use bisection to find the root of the polynomials given
def bisection(a, b, poly, tolerance):
poly(a) <= 0
poly(b) >= 0
try:
if
"""
Assume that poly(a) <= 0 and poly(b) >= 0.
:param a: poly(a) <= 0 Raises an exception if not true
:param b: poly(b) >= 0 Raises an exception if not true
:param poly: polynomial coefficients, low order first
:param tolerance: greater than 0
:return: a value between a and b that is within tolerance of a root of the polynomial
"""
How would I find the root using bisection? I have been provided a test script to test these out.
EDIT: I followed the pseudocode and ended up with this:
def bisection(a, b, poly, tolerance):
#poly(a) <= 0
#poly(b) >= 0
difference = abs(a-b)
xmid = (a-b)/2
n = 1
nmax = 60
while n <= nmax:
mid = (a-b) / 2
if poly(mid) == 0 or (b - a)/2 < tolerance:
print(mid)
n = n + 1
if sign(poly(mid)) == sign(poly(a)):
a = mid
else:
b = mid
return xmid
is this correct? I havent been able to test it because of indentation errors with the return xmid statement.
Your code seems fine, besides the mess with xmid and mid. mid = (a + b) / 2 instead of mid = (a - b) / 2 and you don't need the difference variable.
Cleaned it up a bit:
def sign(x):
return -1 if x < 0 else (1 if x > 0 else 0)
def bisection(a, b, poly, tolerance):
mid = a # will be overwritten
for i in range(60):
mid = (a+b) / 2
if poly(mid) == 0 or (b - a)/2 < tolerance:
return mid
if sign(poly(mid)) == sign(poly(a)):
a = mid
else:
b = mid
return mid
print(bisection(-10**10, 10**10, lambda x: x**5 - x**4 - x**3 - x**2 - x + 9271, 0.00001))
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.