I'm generating prime numbers from Fibonacci as follows (using Python, with mpmath and sympy for arbitrary precision):
from mpmath import *
def GCD(a,b):
while a:
a, b = fmod(b, a), a
return b
def generate(x):
mp.dps = round(x, int(log10(x))*-1)
if x == GCD(x, fibonacci(x-1)):
return True
if x == GCD(x, fibonacci(x+1)):
return True
return False
for x in range(1000, 2000)
if generate(x)
print(x)
It's a rather small algorithm but seemingly generates all primes (except for 5 somehow, but that's another question). I say seemingly because a very little percentage (0.5% under 1000 and 0.16% under 10K, getting less and less) isn't prime. For instance under 1000: 323, 377 and 442 are also generated. These numbers are not prime.
Is there something off in my script? I try to account for precision by relating the .dps setting to the number being calculated. Can it really be that Fibonacci and prime numbers are seemingly so related, but then when it's get detailed they aren't? :)
For this type of problem, you may want to look at the gmpy2 library. gmpy2 provides access to the GMP multiple-precision library which includes gcd() and fib() functions which calculate the greatest common divisor and the n-th fibonacci numbers quickly, and only using integer arithmetic.
Here is your program re-written to use gmpy2.
import gmpy2
def generate(x):
if x == gmpy2.gcd(x, gmpy2.fib(x-1)):
return True
if x == gmpy2.gcd(x, gmpy2.fib(x+1)):
return True
return False
for x in range(7, 2000):
if generate(x):
print(x)
You shouldn't be using any floating-point operations. You can calculate the GCD just using the builtin % (modulo) operator.
Update
As others have commented, you are checking for Fibonacci pseudoprimes. The actual test is slightly different than your code. Let's call the number being tested n. If n is divisible by 5, then the test passes if n evenly divides fib(n). If n divided by 5 leaves a remainder of either 1 or 4, then the test passes if n evenly divides fib(n-1). If n divided by 5 leaves a remainder of either 2 or 3, then the test passes if n evenly divides fib(n+1). Your code doesn't properly distinguish between the three cases.
If n evenly divides another number, say x, it leaves a remainder of 0. This is equivalent to x % n being 0. Calculating all the digits of the n-th Fibonacci number is not required. The test just cares about the remainder. Instead of calculating the Fibonacci number to full precision, you can calculate the remainder at each step. The following code calculates just the remainder of the Fibonacci numbers. It is based on the code given by #pts in Python mpmath not arbitrary precision?
def gcd(a,b):
while b:
a, b = b, a % b
return a
def fib_mod(n, m):
if n < 0:
raise ValueError
def fib_rec(n):
if n == 0:
return 0, 1
else:
a, b = fib_rec(n >> 1)
c = a * ((b << 1) - a)
d = b * b + a * a
if n & 1:
return d % m, (c + d) % m
else:
return c % m, d % m
return fib_rec(n)[0]
def is_fib_prp(n):
if n % 5 == 0:
return not fib_mod(n, n)
elif n % 5 == 1 or n % 5 == 4:
return not fib_mod(n-1, n)
else:
return not fib_mod(n+1, n)
It's written in pure Python and is very quick.
The sequence of numbers commonly known as the Fibonacci numbers is just a special case of a general Lucas sequence L(n) = p*L(n-1) - q*L(n-2). The usual Fibonacci numbers are generated by (p,q) = (1,-1). gmpy2.is_fibonacci_prp() accepts arbitrary values for p,q. gmpy2.is_fibonacci(1,-1,n) should match the results of the is_fib_pr(n) given above.
Disclaimer: I maintain gmpy2.
This isn't really a Python problem; it's a math/algorithm problem. You may want to ask it on the Math StackExchange instead.
Also, there is no need for any non-integer arithmetic whatsoever: you're computing floor(log10(x)) which can be done easily with purely integer math. Using arbitrary-precision math will greatly slow this algorithm down and may introduce some odd numerical errors too.
Here's a simple floor_log10(x) implementation:
from __future__ import division # if using Python 2.x
def floor_log10(x):
res = 0
if x < 1:
raise ValueError
while x >= 1:
x //= 10
res += 1
return res
Related
I need to determine the lowest number base system in which the input n (base 10), expressed in this number base system, is all 1s in its digits.
Examples:
7 in base 2 is 111 - fits! answer is 2
21 in base 2 is 10101 - contains 0, does not fit
21 in base 3 is 210 - contains 0 and 2, does not fit
21 in base 4 is 111 - contains only 1 it fits! answer is 4
n is always less than Number.MAX_SAFE_INTEGER or equivalent.
I have the following code, which works well with a certain range of numbers, but for huge numbers the algorithm is still time consuming:
def check_digits(number, base):
res = 1
while res == 1 and number:
res *= number % base
number //= base
return res
def get_min_base(number):
for i in range(2, int(number ** 0.5) + 2):
if check_digits(number, i) == 1:
return i
return number - 1
How can I optimize the current code to make it run faster?
The number represented by a string of x 1s in base b is b^(x-1) + b^(x-2) + ... + b^2 + b + 1.
Note that for x >= 3, this number is greater than b^(x-1) (trivially) and less than (b+1)^(x-1) (apply the binomial theorem). Thus, if a number n is represented by x 1s in base b, we have b^(x-1) < n < (b+1)^(x-1). Applying x-1'th roots, we have b < n^(1/(x-1)) < b+1. Thus, for b to exist, b must be floor(n^(1/(x-1)).
I've written things with ^ notation instead of Python-style ** syntax so far because those equations and inequalities only hold for exact real number arithmetic, not for floating point. If you try to compute b with floating point math, rounding error may throw off your calculations, especially for extremely large inputs where the ULP is greater than 1. (I think floating point is fine for the input range you're working with, but I'm not sure.)
Still, regardless of whether floating point is good enough or if you need something fancier, the idea of an algorithm is there: you can directly check if a value of x is viable by directly computing what the corresponding b would have to be, and checking if x 1s in base b really represent n.
Just some small twist, slightly faster but don't improve the time complexity.
def check_digits2(number, base):
while number % base == 1:
if number == 1:
return True
number //= base
return False
def get_min_base2(number):
for i in range(2, int(number**0.5) + 2):
if check_digits2(number, i):
return i
return number - 1
def test():
number = 100000010000001
start = time.time()
print(get_min_base(number)) # 10000000
print(f"{time.time() - start:.3f}s\n") # 3.292s
start = time.time()
print(get_min_base2(number)) # 10000000
print(f"{time.time() - start:.3f}s\n") # 1.731s
Also try to approach with some math trick, but I actually make it worse lol
def calculate_n(number, base):
return math.log(number * (base - 1) + 1, base).is_integer()
def get_min_base3(number):
for i in range(2, int(number**0.5) + 2):
if calculate_n(number, i):
return i
return number - 1
def test():
number = 100000010000001
start = time.time()
print(get_min_base3(number)) # 10000000
print(f"{time.time() - start:.3f}s\n") # 4.597s
I am an amateur Python coder trying to find an efficient solution for Project Euler Digit Sum problem. My code returns the correct result but is is inefficient for large integers such as 1234567890123456789. I know that the inefficiency lies in my sigma_sum function where there is a 'for' loop.
I have tried various alternate solutions such as loading the values into an numpy array but ran out of memory with large integers with this approach. I am eager to learn more efficient solutions.
import math
def sumOfDigits(n: int) :
digitSum = 0
if n < 10: return n
else:
for i in str(n): digitSum += int(i)
return digitSum
def sigma_sum(start, end, expression):
return math.fsum(expression(i) for i in range(start, end))
def theArguement(n: int):
return n / sumOfDigits(n)
def F(N: int) -> float:
"""
>>> F(10)
19
>>> F(123)
1.187764610390e+03
>>> F(12345)
4.855801996238e+06
"""
s = sigma_sum(1, N + 1, theArguement)
if s.is_integer():
print("{:0.0f}".format(s))
else:
print("{:.12e}".format(s))
print(F(123))
if __name__ == '__main__':
import doctest
doctest.testmod()
Try solving a different problem.
Define G(n) to be a dictionary. Its keys are integers representing digit sums and its values are the sum of all positive integers < n whose digit sum is the key. So
F(n) = sum(v / k for k, v in G(n + 1).items())
[Using < instead of ≤ simplifies the calculations below]
Given the value of G(a) for any value, how would you calculate G(10 * a)?
This gives you a nice easy way to calculate G(x) for any value of x. Calculate G(x // 10) recursively, use that to calculate the value G((x // 10) * 10), and then manually add the few remaining elements in the range (x // 10) * 10 ≤ i < x.
Getting from G(a) to G(10 * a) is mildly tricky, but not overly so. If your code is correct, you can use calculating G(12346) as a test case to see if you get the right answer for F(12345).
I'm trying to decipher the following homework question. My code is supposed to evaluate to 190 but instead evaluates to 114. So, I don't think I'm understanding the coding requirement.
The Collatz conjecture is an example of a simple computational process
whose behavior is so unpredictable that the world's best
mathematicians still don't understand it.
Consider the simple function f(n) (as defined in the Wikipedia page
above) that takes an integer n and divides it by two if n is even and
multiplies n by 3 and then adds one to the result if n is odd. The
conjecture involves studying the value of expressions of the form
f(f(f(...f(f(n))))) as the number of calls to the function f
increases. The conjecture is that, for any non-negative integer n,
repeated application of f to n yields a sequence of integers that
always includes 1.
Your task for this question is to implement the Collatz function f in
Python. The key to your implementation is to build a test that
determines whether n is even or odd by checking whether the remainder
when n is divided by 2 is either zero or one. Hint: You can compute
this remainder in Python using the remainder opertor % via the
expression n % 2. Note you will also need to use integer division //
when computing f.
Once you have implemented f, test the your implementation on the
expression f(f(f(f(f(f(f(674))))))). This expression should evaluate
to 190.
from __future__ import division
def collatz(n):
l = []
l.append(n)
while n != 1:
if n % 2 == 0:
n = n // 2
l.append(n)
else:
n = (3*n) + 1
l.append(n)
return l
print len(collatz(674))
You just misread the intermediary question. Your programs tries to answer the bigger question... This is what should return 190:
def f(n):
return n // 2 if n % 2 == 0 else 3*n + 1
print f(f(f(f(f(f(f(674)))))))
I'm using the following code for finding primitive roots modulo n in Python:
Code:
def gcd(a,b):
while b != 0:
a, b = b, a % b
return a
def primRoots(modulo):
roots = []
required_set = set(num for num in range (1, modulo) if gcd(num, modulo) == 1)
for g in range(1, modulo):
actual_set = set(pow(g, powers) % modulo for powers in range (1, modulo))
if required_set == actual_set:
roots.append(g)
return roots
if __name__ == "__main__":
p = 17
primitive_roots = primRoots(p)
print(primitive_roots)
Output:
[3, 5, 6, 7, 10, 11, 12, 14]
Code fragment extracted from: Diffie-Hellman (Github)
Can the primRoots method be simplified or optimized in terms of memory usage and performance/efficiency?
One quick change that you can make here (not efficiently optimum yet) is using list and set comprehensions:
def primRoots(modulo):
coprime_set = {num for num in range(1, modulo) if gcd(num, modulo) == 1}
return [g for g in range(1, modulo) if coprime_set == {pow(g, powers, modulo)
for powers in range(1, modulo)}]
Now, one powerful and interesting algorithmic change that you can make here is to optimize your gcd function using memoization. Or even better you can simply use built-in gcd function form math module in Python-3.5+ or fractions module in former versions:
from functools import wraps
def cache_gcd(f):
cache = {}
#wraps(f)
def wrapped(a, b):
key = (a, b)
try:
result = cache[key]
except KeyError:
result = cache[key] = f(a, b)
return result
return wrapped
#cache_gcd
def gcd(a,b):
while b != 0:
a, b = b, a % b
return a
# or just do the following (recommended)
# from math import gcd
Then:
def primRoots(modulo):
coprime_set = {num for num in range(1, modulo) if gcd(num, modulo) == 1}
return [g for g in range(1, modulo) if coprime_set == {pow(g, powers, modulo)
for powers in range(1, modulo)}]
As mentioned in comments, as a more pythoinc optimizer way you can use fractions.gcd (or for Python-3.5+ math.gcd).
Based on the comment of Pete and answer of Kasramvd, I can suggest this:
from math import gcd as bltin_gcd
def primRoots(modulo):
required_set = {num for num in range(1, modulo) if bltin_gcd(num, modulo) }
return [g for g in range(1, modulo) if required_set == {pow(g, powers, modulo)
for powers in range(1, modulo)}]
print(primRoots(17))
Output:
[3, 5, 6, 7, 10, 11, 12, 14]
Changes:
It now uses pow method's 3-rd argument for the modulo.
Switched to gcd built-in function that's defined in math (for Python 3.5) for a speed boost.
Additional info about built-in gcd is here: Co-primes checking
In the special case that p is prime, the following is a good bit faster:
import sys
# translated to Python from http://www.bluetulip.org/2014/programs/primitive.js
# (some rights may remain with the author of the above javascript code)
def isNotPrime(possible):
# We only test this here to protect people who copy and paste
# the code without reading the first sentence of the answer.
# In an application where you know the numbers are prime you
# will remove this function (and the call). If you need to
# test for primality, look for a more efficient algorithm, see
# for example Joseph F's answer on this page.
i = 2
while i*i <= possible:
if (possible % i) == 0:
return True
i = i + 1
return False
def primRoots(theNum):
if isNotPrime(theNum):
raise ValueError("Sorry, the number must be prime.")
o = 1
roots = []
r = 2
while r < theNum:
k = pow(r, o, theNum)
while (k > 1):
o = o + 1
k = (k * r) % theNum
if o == (theNum - 1):
roots.append(r)
o = 1
r = r + 1
return roots
print(primRoots(int(sys.argv[1])))
You can greatly improve your isNotPrime function by using a more efficient algorithm. You could double the speed by doing a special test for even numbers and then only testing odd numbers up to the square root, but this is still very inefficient compared to an algorithm such as the Miller Rabin test. This version in the Rosetta Code site will always give the correct answer for any number with fewer than 25 digits or so. For large primes, this will run in a tiny fraction of the time it takes to use trial division.
Also, you should avoid using the floating point exponentiation operator ** when you are dealing with integers as in this case (even though the Rosetta code that I just linked to does the same thing!). Things might work fine in a particular case, but it can be a subtle source of error when Python has to convert from floating point to integers, or when an integer is too large to represent exactly in floating point. There are efficient integer square root algorithms that you can use instead. Here's a simple one:
def int_sqrt(n):
if n == 0:
return 0
x = n
y = (x + n//x)//2
while (y<x):
x=y
y = (x + n//x)//2
return x
Those codes are all in-efficient, in many ways, first of all you do not need to iterate for all co-prime reminders of n, you need to check only for powers that are dividers of Euler's function from n. In the case n is prime Euler's function is n-1. If n i prime, you need to factorize n-1 and make check with only those dividers, not all. There is a simple mathematics behind this.
Second. You need better function for powering a number imagine the power is too big, I think in python you have the function pow(g, powers, modulo) which at each steps makes division and getting the remainder only ( _ % modulo ).
If you are going to implement the Diffie-Hellman algorithm it is better to use safe primes. They are such primes that p is a prime and 2p+1 is also prime, so that 2p+1 is called safe prime. If you get n = 2*p+1, then the dividers for that n-1 (n is prime, Euler's function from n is n-1) are 1, 2, p and 2p, you need to check only if the number g at power 2 and g at power p if one of them gives 1, then that g is not primitive root, and you can throw that g away and select another g, the next one g+1, If g^2 and g^p are non equal to 1 by modulo n, then that g is a primitive root, that check guarantees, that all powers except 2p would give numbers different from 1 by modulo n.
The example code uses Sophie Germain prime p and the corresponding safe prime 2p+1, and calculates primitive roots of that safe prime 2p+1.
You can easily re-work the code for any prime number or any other number, by adding a function to calculate Euler's function and to find all divisors of that value. But this is only a demo not a complete code. And there might be better ways.
class SGPrime :
'''
This object expects a Sophie Germain prime p, it does not check that it accept that as input.
Euler function from any prime is n-1, and the order (see method get_order) of any co-prime
remainder of n could be only a divider of Euler function value.
'''
def __init__(self, pSophieGermain ):
self.n = 2*pSophieGermain+1
#TODO! check if pSophieGermain is prime
#TODO! check if n is also prime.
#They both have to be primes, elsewhere the code does not work!
# Euler's function is n-1, #TODO for any n, calculate Euler's function from n
self.elrfunc = self.n-1
# All divisors of Euler's function value, #TODO for any n, get all divisors of the Euler's function value.
self.elrfunc_divisors = [1, 2, pSophieGermain, self.elrfunc]
def get_order(self, r):
'''
Calculate the order of a number, the minimal power at which r would be congruent with 1 by modulo p.
'''
r = r % self.n
for d in self.elrfunc_divisors:
if ( pow( r, d, self.n) == 1 ):
return d
return 0 # no such order, not possible if n is prime, - see small Fermat's theorem
def is_primitive_root(self, r):
'''
Check if r is a primitive root by modulo p. Such always exists if p is prime.
'''
return ( self.get_order(r) == self.elrfunc )
def find_all_primitive_roots(self, max_num_of_roots = None):
'''
Find all primitive roots, only for demo if n is large the list is large for DH or any other such algorithm
better to stop at first primitive roots.
'''
primitive_roots = []
for g in range(1, self.n):
if ( self.is_primitive_root(g) ):
primitive_roots.append(g)
if (( max_num_of_roots != None ) and (len(primitive_roots) >= max_num_of_roots)):
break
return primitive_roots
#demo, Sophie Germain's prime
p = 20963
sggen = SGPrime(p)
print (f"Safe prime : {sggen.n}, and primitive roots of {sggen.n} are : " )
print(sggen.find_all_primitive_roots())
Regards
How would I generate a non-prime random number in a range in Python?
I am confused as to how I can create an algorithm that would produce a non-prime number in a certain range. Do I define a function or create a conditional statement? I would like each number in the range to have the same probability. For example, in 1 - 100, each non-prime would not have a 1% chance but instead has a ~1.35% chance.
Now, you didn't say anything about efficiency, and this could surely be optimized, but this should solve the problem. This should be an efficient algorithm for testing primality:
import random
def isPrime(n):
if n % 2 == 0 and n > 2:
return False
return all(n % i for i in range(3, int(math.sqrt(n)) + 1, 2))
def randomNonPrime(rangeMin, rangeMax):
nonPrimes = filter(lambda n: not isPrime(n), xrange(rangeMin, rangeMax+1))
if not nonPrimes:
return None
return random.choice(nonPrimes)
minMax = (1000, 10000)
print randomNonPrime(*minMax)
After returning a list of all non-primes in range, a random value is selected from the list of non-primes, making the selection of any non-prime in range just as likely as any other non-prime in the range.
Edit
Although you didn't ask about efficiency, I was bored, so I figured out a method of doing this that makes a range of (1000, 10000000) take a little over 6 seconds on my machine instead of over a minute and a half:
import numpy
import sympy
def randomNonPrime(rangeMin, rangeMax):
primesInRange = numpy.fromiter(
sympy.sieve.primerange(rangeMin, rangeMax),
dtype=numpy.uint32,
count=-1
)
numbersInRange = numpy.arange(rangeMin, rangeMax+1, dtype=numpy.uint32)
nonPrimes = numbersInRange[numpy.invert(numpy.in1d(numbersInRange, primesInRange))]
if not nonPrimes.size:
return None
return numpy.random.choice(nonPrimes)
minMax = (1000, 10000000)
print randomNonPrime(*minMax)
This uses the SymPy symbolic mathematics library to optimize the generation of prime numbers in a range, and then uses NumPy to filter our output and select a random non-prime.
The algorithm and ideas to choose is very dependent on your exact use-case, as mentioned by #smarx.
Assumptions:
Each non-prime within the range has the same probability of beeing chosen / uniformity
It is sufficient that the sampled number is not a prime with a very high probability (algorithmic false positives are less likely than CPU-bugs & co.)
The sampling-range could be big (sieve-like approaches are slow)
High performance of a single sample is desired (no caching; no sampling without replacement)
Method:
Sample random-number in range
Check if this number is prime with a very fast probabilistic primality test
Stop when observing first non-prime number
If no number is found, stop algorithm after max_trials
max_trials-value is set by an approximation to the Coupon-Collectors-Problem (wiki): expected number of samples to observe each candidate once
Characteristics of method
Fast for single samples (10000 samples per second on single CPU; given range as in example)
Easy to prove uniformity
Good asymptotic behaviour regarding range-size and range-position (number sizes)
Code
import random
import math
""" Miller-Rabin primality test
source: https://jeremykun.com/2013/06/16/miller-rabin-primality-test/
"""
def decompose(n):
exponentOfTwo = 0
while n % 2 == 0:
n = n//2 # modified for python 3!
exponentOfTwo += 1
return exponentOfTwo, n
def isWitness(possibleWitness, p, exponent, remainder):
possibleWitness = pow(possibleWitness, remainder, p)
if possibleWitness == 1 or possibleWitness == p - 1:
return False
for _ in range(exponent):
possibleWitness = pow(possibleWitness, 2, p)
if possibleWitness == p - 1:
return False
return True
def probablyPrime(p, accuracy=100):
if p == 2 or p == 3: return True
if p < 2: return False
exponent, remainder = decompose(p - 1)
for _ in range(accuracy):
possibleWitness = random.randint(2, p - 2)
if isWitness(possibleWitness, p, exponent, remainder):
return False
return True
""" Coupon-Collector Problem (approximation)
How many random-samplings with replacement are expected to observe each element at least once
"""
def couponcollector(n):
return int(n*math.log(n))
""" Non-prime random-sampling
"""
def get_random_nonprime(min, max):
max_trials = couponcollector(max-min)
for i in range(max_trials):
candidate = random.randint(min, max)
if not probablyPrime(candidate):
return candidate
return -1
# TEST
print(get_random_nonprime(1000, 10000000))