Given any random integer, create a function to find the next number that is a prime number and also a palindrome.
My attempt
def golf(number):
x = number + 1
for i in range(2, x):
if x % i == 0 or str(x) != str(x)[::-1]:
golf(number + 1)
return x
E.g golf(13) = 101
I'm actually looking for an alternative option than the recursion method i used. How can this best be accomplished without using recursion?
Thanks
Here's a variation on byron he's answer which adds several optimizations:
We can eliminate all even x values (other than 2) before doing any elaborate tests, since we can trivially tell they are not prime.
A small improvement is to only call str(x) once, and reuse the value later.
We can take advantage of the fact that all even-length palindromes are multiples of 11, which means that (except for 11 itself) they're not prime. We can jump ahead to the next odd-length x value.
Since we've already eliminated even numbers, our prime test only needs to test odd divisors. Further we can stop our loop when we reach sqrt(x), rather than going all the way to x itself.
Finally, there's no need to use a Boolean flag variable to carry the primeness out of the loop. If we don't break, the else block attached to the loop will be run.
The code:
import math
def next_prime_palindrome(x):
while True:
x += 1
if x > 2 and x % 2 == 0: # even numbers greater than 2 are non-prime
continue
s = str(x) # compute str(x) just once
if x > 11 and len(s) % 2 == 0: # all even-length palindromes are multiples of 11
x = 10 ** len(s) # so jump to the next odd-length integer
continue
if s != s[::-1]: # palindrome test
continue
for i in xrange(3, round(math.sqrt(x))+1, 2): # loop over odd potential divisors
if x % i == 0: # prime test
break
else: # this else block runs only if no break happened in the loop, so x is prime
return x
Here are some tests runs, showing a few cases where the optimizations save significant time:
>>> next_prime_palindrome(1)
2
>>> next_prime_palindrome(3)
5
>>> next_prime_palindrome(9)
11
>>> next_prime_palindrome(11)
101
>>> next_prime_palindrome(99999)
1003001
>>> next_prime_palindrome(999999999)
10000500001
A further improvement might be to directly generate palindromes, rather than working with integers to start with, and doing a palindrome test to filter them. That would get quite a bit further from your original design, so I'll leave that for someone else.
Palindrome are a sparser set of numbers than primes, and you can generate palindromes directly.
Consider the sequence 98.102
These are palidrome numbers you can base on these
989, 9889, 999, 9999, 10001, 100001, 10101, 101101, 10201, 102201
ADDED
Not also that all of the palidromes with an odd number of digits will come before the palidromes with an even number of digits.
If you write this as a generator (ie using yield) get get a straightforward algorithm for generating palindromic numbers in order.
For 1..9 you generate either 9 or 18 palindromes depending upon whether you consider 1 digit numbers palindromic.
For 10..99 you generate 90 even digit and 90 odd digit palindromes.
For 100..999 you generate 900 even digit and 900 odd digit palindromes.
You have just generated all 1989 (or 1997 if including single digit numbers) of the palindromic numbers less than 1 million. There are 78,498 primes less than 1 million
Any algorithm that is based on generating primes then testing for a palindrome will be much slower that generating palindromes and then testing for primes
def golf(number):
primes = []
i = 2
while i <= number:
if isPrime(i, primes):
primes.append(i)
i += 1
answer = primes[-1] + 1
while True:
if isPrime(answer, primes):
primes.append(answer)
if str(answer) == str(answer)[::-1]:
return answer
answer += 1
def isPrime(n, primes):
for (p for p in primes if p<=n**0.5):
if n%p == 0:
return False
return True
Your solution can be slightly modified in order to create an iterative solution:
def golf(number):
x = number + 1
while True:
is_golf = True
for i in range(2, x):
if x % i == 0 or str(x) != str(x)[::-1]:
is_golf = False
break
if is_golf:
return x
x += 1
improved according to Blckknght's advice, thanks.
def golf(number):
x = number
while True:
x += 1
if str(x) != str(x)[::-1]:
continue
for i in xrange(2, x):
if x % i == 0 :
break
else:
return x
I'm doing this problem on a site that I found (project Euler), and there is a question that involves finding the largest prime factor of a number. My solution fails at really large numbers so I was wondering how this code could be streamlined?
""" Find the largest prime of a number """
def get_factors(number):
factors = []
for integer in range(1, number + 1):
if number%integer == 0:
factors.append(integer)
return factors
def test_prime(number):
prime = True
for i in range(1, number + 1):
if i!=1 and i!=2 and i!=number:
if number%i == 0:
prime = False
return prime
def test_for_primes(lst):
primes = []
for i in lst:
if test_prime(i):
primes.append(i)
return primes
################################################### program starts here
def find_largest_prime_factor(i):
factors = get_factors(i)
prime_factors = test_for_primes(factors)
print prime_factors
print find_largest_prime_factor(22)
#this jams my computer
print find_largest_prime_factor(600851475143)
it fails when using large numbers, which is the point of the question I guess. (computer jams, tells me I have run out of memory and asks me which programs I would like to stop).
************************************ thanks for the answer. there was actually a couple bugs in the code in any case. so the fixed version of this (inefficient code) is below.
""" Find the largest prime of a number """
def get_factors(number):
factors = []
for integer in xrange(1, number + 1):
if number%integer == 0:
factors.append(integer)
return factors
def test_prime(number):
prime = True
if number == 1 or number == 2:
return prime
else:
for i in xrange(2, number):
if number%i == 0:
prime = False
return prime
def test_for_primes(lst):
primes = []
for i in lst:
if test_prime(i):
primes.append(i)
return primes
################################################### program starts here
def find_largest_prime_factor(i):
factors = get_factors(i)
print factors
prime_factors = test_for_primes(factors)
return prime_factors
print find_largest_prime_factor(x)
From your approach you are first generating all divisors of a number n in O(n) then you test which of these divisors is prime in another O(n) number of calls of test_prime (which is exponential anyway).
A better approach is to observe that once you found out a divisor of a number you can repeatedly divide by it to get rid of all of it's factors. Thus, to get the prime factors of, say 830297 you test all small primes (cached) and for each one which divides your number you keep dividing:
830297 is divisible by 13 so now you'll test with 830297 / 13 = 63869
63869 is still divisible by 13, you are at 4913
4913 doesn't divide by 13, next prime is 17 which divides 4913 to get 289
289 is still a multiple of 17, you have 17 which is the divisor and stop.
For further speed increase, after testing the cached prime numbers below say 100, you'll have to test for prime divisors using your test_prime function (updated according to #Ben's answer) but go on reverse, starting from sqrt. Your number is divisible by 71, the next number will give an sqrt of 91992 which is somewhat close to 6857 which is the largest prime factor.
Here is my favorite simple factoring program for Python:
def factors(n):
wheel = [1,2,2,4,2,4,2,4,6,2,6]
w, f, fs = 0, 2, []
while f*f <= n:
while n % f == 0:
fs.append(f)
n /= f
f, w = f + wheel[w], w+1
if w == 11: w = 3
if n > 1: fs.append(n)
return fs
The basic algorithm is trial division, using a prime wheel to generate the trial factors. It's not quite as fast as trial division by primes, but there's no need to calculate or store the prime numbers, so it's very convenient.
If you're interested in programming with prime numbers, you might enjoy this essay at my blog.
My solution is in C#. I bet you can translate it into python. I've been test it with random long integer ranging from 1 to 1.000.000.000 and it's doing good. You can try to test the result with online prime calculator Happy coding :)
public static long biggestPrimeFactor(long num) {
for (int div = 2; div < num; div++) {
if (num % div == 0) {
num \= div
div--;
}
}
return num;
}
The naive primality test can be improved upon in several ways:
Test for divisibility by 2 separately, then start your loop at 3 and go by 2's
End your loop at ceil(sqrt(num)). You're guaranteed to not find a prime factor above this number
Generate primes using a sieve beforehand, and only move onto the naive way if you've exhausted the numbers in your sieve.
Beyond these easy fixes, you're going to have to look up more efficient factorization algorithms.
Use a Sieve of Eratosthenes to calculate your primes.
from math import sqrt
def sieveOfEratosthenes(n):
primes = range(3, n + 1, 2) # primes above 2 must be odd so start at three and increase by 2
for base in xrange(len(primes)):
if primes[base] is None:
continue
if primes[base] >= sqrt(n): # stop at sqrt of n
break
for i in xrange(base + (base + 1) * primes[base], len(primes), primes[base]):
primes[i] = None
primes.insert(0,2)
return filter(None, primes)
The point to prime factorization by trial division is, the most efficient solution for factorizing just one number doesn't need any prime testing.
You just enumerate your possible factors in ascending order, and keep dividing them out of the number in question - all thus found factors are guaranteed to be prime. Stop when the square of current factor exceeds the current number being factorized. See the code in user448810's answer.
Normally, prime factorization by trial division is faster on primes than on all numbers (or odds etc.), but when factorizing just one number, to find the primes first to test divide by them later, will might cost more than just going ahead with the increasing stream of possible factors. This enumeration is O(n), prime generation is O(n log log n), with the Sieve of Eratosthenes (SoE), where n = sqrt(N) for the top limit N. With trial division (TD) the complexity is O(n1.5/(log n)2).
Of course the asymptotics are to be taken just as a guide, actual code's constant factors might change the picture. Example, execution times for a Haskell code derived from here and here, factorizing 600851475149 (a prime):
2.. 0.57 sec
2,3,5,... 0.28 sec
2,3,5,7,11,13,17,19,... 0.21 sec
primes, segmented TD 0.65 sec first try
0.05 sec subsequent runs (primes are memoized)
primes, list-based SoE 0.44 sec first try
0.05 sec subsequent runs (primes are memoized)
primes, array-based SoE 0.15 sec first try
0.06 sec subsequent runs (primes are memoized)
so it depends. Of course factorizing the composite number in question, 600851475143, is near instantaneous, so it doesn't matter there.
Here is an example in JavaScript
function largestPrimeFactor(val, divisor = 2) {
let square = (val) => Math.pow(val, 2);
while ((val % divisor) != 0 && square(divisor) <= val) {
divisor++;
}
return square(divisor) <= val
? largestPrimeFactor(val / divisor, divisor)
: val;
}
I converted the solution from #under5hell to Python (2.7x). what an efficient way!
def largest_prime_factor(num, div=2):
while div < num:
if num % div == 0 and num/div > 1:
num = num /div
div = 2
else:
div = div + 1
return num
>> print largest_prime_factor(600851475143)
6857
>> print largest_prime_factor(13195)
29
Try this piece of code:
from math import *
def largestprime(n):
i=2
while (n>1):
if (n % i == 0):
n = n/i
else:
i=i+1
print i
strinput = raw_input('Enter the number to be factorized : ')
a = int(strinput)
largestprime(a)
Old one but might help
def isprime(num):
if num > 1:
# check for factors
for i in range(2,num):
if (num % i) == 0:
return False
return True
def largest_prime_factor(bignumber):
prime = 2
while bignumber != 1:
if bignumber % prime == 0:
bignumber = bignumber / prime
else:
prime = prime + 1
while isprime(prime) == False:
prime = prime+1
return prime
number = 600851475143
print largest_prime_factor(number)
I Hope this would help and easy to understand.
A = int(input("Enter the number to find the largest prime factor:"))
B = 2
while (B <(A/2)):
if A%B != 0:
B = B+1
else:
A = A/B
C = B
B = 2
print (A)
This code for getting the largest prime factor, with nums value of prime_factor(13195) when I run it, will return the result in less than a second.
but when nums value gets up to 6digits it will return the result in 8seconds.
Any one has an idea of what is the best algorithm for the solution...
def prime_factor(nums):
if nums < 2:
return 0
primes = [2]
x = 3
while x <= nums:
for i in primes:
if x%i==0:
x += 2
break
else:
primes.append(x)
x += 2
largest_prime = primes[::-1]
# ^^^ code above to gets all prime numbers
intermediate_tag = []
factor = []
# this code divide nums by the largest prime no. and return if the
# result is an integer then append to primefactor.
for i in largest_prime:
x = nums/i
if x.is_integer():
intermediate_tag.append(x)
# this code gets the prime factors [29.0, 13.0, 7.0, 5.0]
for i in intermediate_tag:
y = nums/i
factor.append(y)
print(intermediate_tag)
print(f"prime factor of {nums}:==>",factor)
prime_factor(13195)
[455.0, 1015.0, 1885.0, 2639.0]
prime factor of 13195:==> [29.0, 13.0, 7.0, 5.0]
Overall Problem: Project Euler 12 - What is the value of the first triangle number to have over five hundred divisors?
Focus of problem: The divisor function
Language: Python
Description: The function I used is brute and the time it take for the program to find a number with more divisors than x increases almost exponentially with each 10 or 20 numbers highers. I need to get to 500 or more divisors. I've identified that the divisor function is what is hogging down the program. The research I did lead me to divisor functions and specifically the divisor function which is supposed to be a function that will count all the divisors of any integer. Every page I've looked at seems to be directed toward mathematics majors and I only have high-school maths. Although I did come across some page that mentioned allot about primes and the Sieve of Atkins but I could not make the connection between primes and finding all the divisors of any integer nor find anything on the net about it.
Main Question: Could someone explain how to code the divisor function or even provide a sample? Maths concepts make more sense to me when I look at them with code. So much appreciated.
brute force divisor function:
def countdiv(a):
count = 0
for i in range(1,(a/2)+1):
if a % i == 0:
count += 1
return count + 1 # +1 to account for number itself as a divisor
If you need a bruteforce function to calculate Number of Divisors (also known as tau(n))
Here's what it looks like
def tau(n):
sqroot,t = int(n**0.5),0
for factor in range(1,sqroot+1):
if n % factor == 0:
t += 2 # both factor and N/factor
if sqroot*sqroot == n: t = t - 1 # if sqroot is a factor then we counted it twice, so subtract 1
return t
The second method involves a decomposing n into its prime factors (and its exponents).
tau(n) = (e1+1)(e2+1)....(em+1) where n = p1^e1 * p2^e2 .... pm^em and p1,p2..pm are primes
More info here
The third method and much more simpler to understand is simply using a Sieve to calculate tau.
def sieve(N):
t = [0]*(N+1)
for factor in range(1,N+1):
for multiple in range(factor,N+1,factor):
t[multiple]+=1
return t[1:]
Here's it in action at ideone
I agree with the two other answers submitted here in that you will only need to search up to the square root of the number. I have one thing to add to this however. The solutions offered will get you the correct answer in a reasonable amount of time. But when the problems start getting tougher, you will need an even more powerful function.
Take a look at Euler's Totient function. Though it only indirectly applies here, it is incredibly useful in later problems. Another related concept is that of Prime Factorization.
A quick way to improve your algorithm is to find the prime factorization of the number. In the Wikipedia article, they use 36 as an example, whose prime factorization is 2^2 * 3^2. Therefore, knowing this, you can use combinatorics to find the number of factors of 36. With this, you will not actually be computing each factor, plus you'd only have to check divisors 2 and 3 before you're complete.
When searching for divisors of n you never have to search beyond the square root of the number n. Whenever you find a divisor that's less than sqrt(n) there is exactly one matching divisor which is greater than the root, so you can increment your count by 2 (if you find divisor d of n then n/d will be the counterpart).
Watch out for square numbers, though. :) The root will be a divisor that doesn't count twice, of course.
If you're going to solve the Project Euler problems you need some functions that deal with prime numbers and integer factorization. Here is my modest library, which provides primes(n), is_prime(n) and factors(n); the focus is on simplicity, clarity and brevity at the expense of speed, though these functions should be sufficient for Project Euler:
def primes(n):
"""
list of primes not exceeding n in ascending
order; assumes n is an integer greater than
1; uses Sieve of Eratosthenes
"""
m = (n-1) // 2
b = [True] * m
i, p, ps = 0, 3, [2]
while p*p < n:
if b[i]:
ps.append(p)
j = 2*i*i + 6*i + 3
while j < m:
b[j] = False
j = j + 2*i + 3
i += 1; p += 2
while i < m:
if b[i]:
ps.append(p)
i += 1; p += 2
return ps
def is_prime(n):
"""
False if n is provably composite, else
True if n is probably prime; assumes n
is an integer greater than 1; uses
Miller-Rabin test on prime bases < 100
"""
ps = [2,3,5,7,11,13,17,19,23,29,31,37,41,
43,47,53,59,61,67,71,73,79,83,89,97]
def is_spsp(n, a):
d, s = n-1, 0
while d%2 == 0:
d /= 2; s += 1
if pow(a,d,n) == 1:
return True
for r in xrange(s):
if pow(a, d*pow(2,r), n) == n-1:
return True
return False
if n in ps: return True
for p in ps:
if not is_spsp(n,p):
return False
return True
def factors(n):
"""
list of prime factors of n in ascending
order; assumes n is an integer, may be
positive, zero or negative; uses Pollard's
rho algorithm with Floyd's cycle finder
"""
def gcd(a,b):
while b: a, b = b, a%b
return abs(a)
def facts(n,c,fs):
f = lambda(x): (x*x+c) % n
if is_prime(n): return fs+[n]
t, h, d = 2, 2, 1
while d == 1:
t = f(t); h = f(f(h))
d = gcd(t-h, n)
if d == n:
return facts(n, c+1, fs)
if is_prime(d):
return facts(n//d, c+1, fs+[d])
return facts(n, c+1, fs)
if -1 <= n <= 1: return [n]
if n < -1: return [-1] + factors(-n)
fs = []
while n%2 == 0:
n = n//2; fs = fs+[2]
if n == 1: return fs
return sorted(facts(n,1,fs))
Once you know how to factor a number, it is easy to count the number of divisors. Consider 76576500 = 2^2 * 3^2 * 5^3 * 7^1 * 11^1 * 13^1 * 17^1. Ignore the bases and look at the exponents, which are 2, 2, 3, 1, 1, 1, and 1. Add 1 to each exponent, giving 3, 3, 4, 2, 2, 2, and 2. Now multiply that list to get the number of divisors of the original number 76576500: 3 * 3 * 4 * 2 * 2 * 2 * 2 = 576. Here's the function:
def numdiv(n):
fs = factors(n)
f = fs.pop(0); d = 1; x = 2
while fs:
if f == fs[0]:
x += 1
else:
d *= x; x = 2
f = fs.pop(0)
return d * x
You can see these functions at work at http://codepad.org/4j8qp60u, and learn more about how they work at my blog. I'll leave it to you to work out the solution to Problem 12.
I've recently been working on Project Euler problems in Python. I am fairly new to Python, and still somewhat new as a programmer.
In any case, I've ran into a speed-related issue coding a solution for problem #5. The problem is,
"2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?"
I've checked around some, and I haven't been able to find anything on this problem pertaining to Python specifically. There were some completed scripts, but I want to avoid looking at other's code in full, if possible, instead wanting to improve my own.
The code I have written runs successfully for the example of 2520 and the range 1 to 10, and should be directly modifiable to work with the question. However, upon running it, I do not get an answer. Presumably, it is a very high number, and the code is not fast enough. Printing the current number being checked seems to support this, reaching several million without getting an answer.
The code, in it's current implementation is as follows:
rangemax = 20
def div_check(n):
for i in xrange(11,rangemax+1):
if n % i == 0:
continue
else:
return False
return True
if __name__ == '__main__':
num = 2
while not div_check(num):
print num
num += 2
print num
I have already made a couple changes which I think should help the speed. For one, for a number to be divisible by all numbers 1 to 20, it must be even, as only even numbers are divisible by 2. Hence, I can increment by 2 instead of 1. Also, although I didn't think of it myself, I found someone point out that a number divisible by 11 to 20 is divisible by 1 to 10. (Haven't checked that one, but it seems reasonable)
The code still, however is not fast enough. What optimisations, either programmatic, or mathematics, can I make to make this code run faster?
Thanks in advance to any who can help.
Taking the advice of Michael Mior and poke, I wrote a solution. I tried to use a few tricks to make it fast.
Since we need a relatively short list of numbers tested, then we can pre-build the list of numbers rather than repeatedly calling xrange() or range().
Also, while it would work to just put the numbers [1, 2, 3, ..., 20] in the list, we can think a little bit, and pull numbers out:
Just take the 1 out. Every integer is evenly divisible by 1.
If we leave the 20 in, there is no need to leave the 2 in. Any integer evenly divisible by 20 is evenly divisible by 2 (but the reverse might not be true). So we leave the 20 and take out the 2, the 4, and the 5. Leave the 19, as it's prime. Leave the 18, but now we can take out the 3 and the 6. If you repeat this process, you wind up with a much shorter list of numbers to try.
We start at 20 and step numbers by 20, as Michael Mior suggested. We use a generator expression inside of all(), as poke suggested.
Instead of a while loop, I used a for loop with xrange(); I think this is slightly faster.
The result:
check_list = [11, 13, 14, 16, 17, 18, 19, 20]
def find_solution(step):
for num in xrange(step, 999999999, step):
if all(num % n == 0 for n in check_list):
return num
return None
if __name__ == '__main__':
solution = find_solution(20)
if solution is None:
print "No answer found"
else:
print "found an answer:", solution
On my computer, this finds an answer in under nine seconds.
EDIT:
And, if we take advice from David Zaslavsky, we realize we can start the loop at 2520, and step by 2520. If I do that, then on my computer I get the correct answer in about a tenth of a second.
I made find_solution() take an argument. Try calling find_solution(2520).
My first answer sped up the original calculation from the question.
Here's another answer that solves it a different way: just find all the prime factors of each number, then multiply them together to go straight to the answer. In other words, this automates the process recommended by poke in a comment.
It finishes in a fraction of a second. I don't think there is a faster way to do this.
I did a Google search on "find prime factors Python" and found this:
http://www.stealthcopter.com/blog/2009/11/python-factors-of-a-number/
From that I found a link to factor.py (written by Mike Hansen) with some useful functions:
https://gist.github.com/weakish/986782#file-factor-py
His functions didn't do quite what I wanted, so I wrote a new one but used his pull_prime_factors() to do the hard work. The result was find_prime_factors() which returns a list of tuples: a prime number, and a count. For example, find_prime_factors(400) returns [(2,4), (5,2)] because the prime factors of 400 are: (2*2*2*2)*(5*5)
Then I use a simple defaultdict() to keep track of how many we have seen so far of each prime factor.
Finally, a loop multiplies everything together.
from collections import defaultdict
from factor import pull_off_factors
pf = defaultdict(int)
_primes = [2,3,5,7,11,13,17,19,23,29]
def find_prime_factors(n):
lst = []
for p in _primes:
n = pull_off_factors(n, p, lst)
return lst
def find_solution(low, high):
for num in xrange(low, high+1):
lst = find_prime_factors(num)
for n, count in lst:
pf[n] = max(pf[n], count)
print "prime factors:", pf
solution = 1
for n, count in pf.items():
solution *= n**count
return solution
if __name__ == '__main__':
solution = find_solution(1, 20)
print "answer:", solution
EDIT: Oh wow, I just took a look at #J.F. Sebastian's answer to a related question. His answer does essentially the same thing as the above code, only far more simply and elegantly. And it is in fact faster than the above code.
Least common multiple for 3 or more numbers
I'll leave the above up, because I think the functions might have other uses in Project Euler. But here's the J.F. Sebastian solution:
def gcd(a, b):
"""Return greatest common divisor using Euclid's Algorithm."""
while b:
a, b = b, a % b
return a
def lcm(a, b):
"""Return lowest common multiple."""
return a * b // gcd(a, b)
def lcmm(*args):
"""Return lcm of args."""
return reduce(lcm, args)
def lcm_seq(seq):
"""Return lcm of sequence."""
return reduce(lcm, seq)
solution = lcm_seq(xrange(1,21))
print "lcm_seq():", solution
I added lcm_seq() but you could also call:
lcmm(*range(1, 21))
Since your answer must be divisible by 20, you can start at 20 and increment by 20 instead of by two. In general, you can start at rangemax and increment by rangemax. This reduces the number of times div_check is called by an order of magnitude.
Break down the number as a prime factorization.
All primes less than 20 are:
2,3,5,7,11,13,17,19
So the bare minimum number that can be divided by these numbers is:
2*3*5*7*11*13*17*19
Composites:
4,6,8,9,10,12,14,15,16,18,20 = 2^2, 2*3, 2^3, 3^2, 2*5, 2^2*3, 2*7, 3*5, 2*3^2, 2^2*5
Starting from the left to see which factors needed:
2^3 to build 4, 8, and 16
3 to build 9
Prime factorization: 2^4 * 3^2 * 5 * 7 * 11 * 13 * 17 * 19 = 232,792,560
I got the solution in 0.066 milliseconds (only 74 spins through a loop) using the following procedure:
Start with smallest multiple for 1, which = 1. Then find the smallest multiple for the next_number_up. Do this by adding the previous smallest multiple to itself (smallest_multiple = smallest_multiple + prev_prod) until next_number_up % smallest_multiple == 0. At this point smallest_multiple is the correct smallest multiple for next_number_up. Then increment next_number_up and repeat until you reach the desired smallest_multiple (in this case 20 times). I believe this finds the solution in roughly n*log(n) time (though, given the way numbers seem to work, it seems to complete much faster than that usually).
For example:
1 is the smallest multiple for 1
Find smallest multiple for 2
Check if previous smallest multiple works 1/2 = .5, so no
previous smallest multiple + previous smallest multiple == 2.
Check if 2 is divisible by 2 - yes, so 2 is the smallest multiple for 2
Find smallest multiple for 3
Check if previous smallest multiple works 2/3 = .667, so no
previous smallest multiple + previous smallest multiple == 4
Check if 4 is divisible by 3 - no
4 + previous smallest multiple == 6
Check if 6 is divisible by 3 - yes, so 6 is the smallest multiple for 3
Find smallest multiple for 4
Check if previous smallest multiple works 6/4 = 1.5, so no
previous smallest multiple + previous smallest multiple == 12
Check if 12 is divisble by 4 - yes, so 12 is the smallest multiple for 4
repeat until 20..
Below is code in ruby implementing this approach:
def smallestMultiple(top)
prod = 1
counter = 0
top.times do
counter += 1
prevprod = prod
while prod % counter != 0
prod = prod + prevprod
end
end
return prod
end
List comprehensions are faster than for loops.
Do something like this to check a number:
def get_divs(n):
divs = [x for x in range(1,20) if n % x == 0]
return divs
You can then check the length of the divs array to see if all the numbers are present.
Two different types of solutions have been posted here. One type uses gcd calculations; the other uses prime factorization. I'll propose a third type, which is based on the prime factorization approach, but is likely to be much faster than prime factorization itself. It relies on a few simple observations about prime powers -- prime numbers raised to some integral exponent. In short, it turns out that the least common multiple of all numbers below some number n is equal to the product of all maximal prime powers below n.
To prove this, we begin by thinking about the properties that x, the least common multiple of all numbers below n, must have, and expressing them in terms of prime powers.
x must be a multiple of all prime powers below n. This is obvious; say n = 20. 2, 2 * 2, 2 * 2 * 2, and 2 * 2 * 2 * 2 are all below 20, so they all must divide x. Likewise, 3 and 3 * 3 are both below n and so both must divide x.
If some number a is a multiple of the prime power p ** e, and p ** e is the maximal power of p below n, then a is also a multiple of all smaller prime powers of p. This is also quite obvious; if a == p * p * p, then a == (p * p) * p.
By the unique factorization theorem, any number m can be expressed as a multiple of prime powers less than m. If m is less than n, then m can be expressed as a multiple of prime powers less than n.
Taken together, the second two observations show that any number x that is a multiple of all maximal prime powers below n must be a common multiple of all numbers below n. By (2), if x is a multiple of all maximal prime powers below n, it is also a multiple of all prime powers below n. So by (3), it is also a multiple of all other numbers below n, since they can all be expressed as multiples of prime powers below n.
Finally, given (1), we can prove that x is also the least common multiple of all numbers below n, because any number less than x could not be a multiple of all maximal prime powers below n, and so could not satisfy (1).
The upshot of all this is that we don't need to factorize anything. We can just generate primes less than n!
Given a nicely optimized sieve of eratosthenes, one can do that very quickly for n below one million. Then all you have to do is find the maximal prime power below n for each prime, and multiply them together.
prime_powers = [get_max_prime_power(p, n) for p in sieve(n)]
result = reduce(operator.mul, prime_powers)
I'll leave writing get_max_prime_power as an exercise. A fast version, combined with the above, can generate the lcm of all numbers below 200000 in 3 seconds on my machine.
The result is a 86871-digit number!
This solution ran pretty quickly for me (imports numpy).
t0 = time.time()
import numpy
ints = numpy.array(range(1,21))
primes = [2,3,5,7,11,13,17,19] # under 20
facts = []
for p in primes:
counter = 0
nums = ints
while any(nums % p == 0):
nums = nums / float(p)
counter += 1
facts.append(counter)
facts = numpy.array(facts)
mults = primes**facts
ans = 1
for m in mults:
ans = m * ans
t1 =time.time()
perf = t1 - t0
print "Problem 5\nAnswer:",ans, "runtime:", perf, "seconds"
"""Problem 5
Answer: 232792560 runtime: 0.00505399703979 seconds"""
Here i have also done using prime factorization way.
#!/usr/bin/env python
import math
def is_prime(num):
if num > 1:
if num == 2:
return True
if num%2 == 0:
return False
for i in range(3, int(math.sqrt(num))+1, 2):
if num%i == 0:
return False
return True
return False
def lcm(number):
prime = []
lcm_value = 1
for i in range(2,number+1):
if is_prime(i):
prime.append(i)
final_value = []
for i in prime:
x = 1
while i**x < number:
x = x + 1
final_value.append(i**(x-1))
for j in final_value:
lcm_value = j * lcm_value
return lcm_value
if __name__ == '__main__':
print lcm(20)
After checking how much time it has taken, it was not bad at all.
root#l-g6z6152:~/learn/project_euler# time python lcm.py
232792560
real 0m0.019s
user 0m0.008s
sys 0m0.004s
I wrote a solution to euler5 that:
Is orders of magnitude faster than most of the solutions here when n=20 (though not all respondents report their time) because it uses no imports (other than to measure time for this answer) and only basic data structures in python.
Scales much better than most other solutions. It will give the answer for n=20 in 6e-05 seconds, or for n=100 in 1 millisec, faster than most of the responses for n=20 listed here.
import time
a=time.clock() # set timer
j=1
factorlist=[]
mydict={}
# change second number to desired number +1 if the question were changed.
for i in range(2,21,1):
numberfactors=[]
num=i
j=2
# build a list of the prime factors
for j in range(j,num+1,1):
counter=0
if i%j==0:
while i%j==0:
counter+=1
numberfactors.append(j)
i=i/j
# add a list of factors to a dictionary, with each prime factor as a key
if j not in mydict:
mydict[j] = counter
# now, if a factor is already present n times, including n times the factor
# won't increase the LCM. So replace the dictionary key with the max number of
# unique factors if and only if the number of times it appears is greater than
# the number of times it has already appeared.
# for example, the prime factors of 8 are 2,2, and 2. This would be replaced
# in the dictionary once 16 were found (prime factors 2,2,2, and 2).
elif mydict[j] < counter:
mydict[j]=counter
total=1
for key, value in mydict.iteritems():
key=int(key)
value=int(value)
total=total*(key**value)
b=time.clock()
elapsed_time=b-a
print total, "calculated in", elapsed_time, "seconds"
returns:
232792560 calculated in 6e-05 seconds
# does not rely on heuristics unknown to all users, for instance the idea that
# we only need to include numbers above 10, etc.
# For all numbers evenly divisible by 1 through 100:
69720375229712477164533808935312303556800 calculated in 0.001335 seconds
Here is program in C language. Cheers
#include <stdio.h>
#include <stdlib.h>
//2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
//What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
bez_ost(int q)
{
register br=0;
for( register i=1;i<=20;i++)
if(q%i==0)
br++;
if(br==20)
return 1;
return 0;
}
int main()
{
register j=20;
register ind=0;
while(ind!=1)
{
j++;
if(bez_ost(j))
break;
}
fprintf(stdout,"\nSmallest positive number that is evenlu divisible by all of the numbers from 1 to 20 is: %d\n\a",j);
system("Pause");
}
I've had the same problem. The algorithm seems to be quite slow, but it does work nonetheless.
result = list()
xyz = [x for x in range(11, 21)]
number = [2520]
count = 0
while len(result) == 0:
for n in number:
print n
for x in xyz:
if n % x == 0:
count += 1
elif n % x != 0:
count = 0
break
if count == 10:
result.append(number[0])
elif count != 10:
number[0] += 1
print result
This was the algorithm I made.
How about this? The required number is, after all, the LCM of the given numbers.
def lcm(a,b):
lcm1 = 0
if a == b:
lcm1 = a
else:
if a > b:
greater = a
else:
greater = b
while True:
if greater % a == 0 and greater % b == 0:
lcm1 = greater
break
greater += 1
return lcm1
import time
start_time = time.time()
list_numbers = list(range(2,21))
lcm1 = lcm(list_numbers[0],list_numbers[1])
for i in range(2,len(list_numbers)):
lcm1 = lcm(lcm1,list_numbers[i])
print(lcm1)
print('%0.5f'%(time.time()-start_time))
This code took a full 45 s to get the answer to the actual question! Hope it helps.
import time
primes = [11,13,17,19]
composites = [12,14,15,16,18,20]
def evenlyDivisible(target):
evenly = True
for n in composites:
if target % n > 0:
evenly = False
break
return evenly
step = 1
for p in primes:
step *= p
end = False
number = 0
t1 = time.time()
while not end:
number += step
if evenlyDivisible(number):
end = True
print("Smallest positive evenly divisible number is",number)
t2 = time.time()
print("Time taken =",t2-t1)
Executed in 0.06 seconds
Here is my Python solution, it has 12 iteration so compiled quite fast:
smallest_num = 1
for i in range (1,21):
if smallest_num % i > 0: # If the number is not divisible by i
for k in range (1,21):
if (smallest_num * k) % i == 0: # Find the smallest number divisible by i
smallest_num = smallest_num * k
break
print (smallest_num)
Here's an observation on this problem. Ultimately, it takes 48 iterations to find the solution.
Any number that is divisible by all of the numbers from 1..20 must be divisible by the product of the primes in that range, namely 2, 3, 5, 7, 11, 13, 17, and 19. It cannot be smaller than the product of these primes, so let's use that number, 232,792,560, as the increment, rather than 20, or 2,520, or some other number.
As it turns out, 48 * 232,792,560 is divisible by all numbers 1..20. By the way, the product of all of the non-primes between 1..20 is 66. I haven't quite figured out the relationship between 48 and 66 in this context.
up = int(input('Upper limit: '))
number = list(range(1, up + 1))
n = 1
for i in range(1, up):
n = n * number[i]
for j in range(i):
if number[i] % number[j] == 0:
n = n / number[j]
number[i] = number[i] / number[j]
print(n)
How I can reduce the complexity of this
num = 1
found = False
while not found:
count =0
for i in range(1, 21):
if num %i == 0:
count+=1
if count ==10:
print(num)
found = True
num+=1
Here is the code in C++ to find the solution for this question.
What we have to do is to run a loop from 1 until we got that number so just iterate through the loop and once the number get evenly divisble(remainder 0) flag value dont get change and flag remains 1 and we got that number and break through outer loop and print the answer
#include <bits/stdc++.h>
using namespace std;
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int i,j,flag=1;
for(i=1;;i++) //iterate until we got the number
{
flag=1;
for(j=2;j<=20;j++) //check form 1 to 20 for that number
{
if(i%j!=0) //if the number is not evenly divisible we break loop and
{
flag=0;break; // initilize flag as 0 i.e. that number is not what we want
}
}
if(flag==1) //if any number we got that is evenly divisible i.e. flag value doesnt change we got that number we break through the loop and print the answer
break;
} // after ending of the loop as we jump to next number make flag also 1 again so that again inner loop conditions apply on it
cout<<i;
return 0;
}
A typescript variant that seems to be relatively quick, leveraging recursion and known facts.
describe(`2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?`,
() => {
it("prooves the example: 10", () => smallestWithoutRemainder(10).should.be.equal(2520));
it("prooves 1", () => smallestWithoutRemainder(1).should.be.equal(1));
it("prooves 2", () => smallestWithoutRemainder(2).should.be.equal(2));
it("prooves 3", () => smallestWithoutRemainder(3).should.be.equal(6));
it("prooves 4", () => smallestWithoutRemainder(4).should.be.equal(12));
it("prooves 5", () => smallestWithoutRemainder(5).should.be.equal(60));
it("prooves 6", () => smallestWithoutRemainder(6).should.be.equal(60));
it("prooves 7", () => smallestWithoutRemainder(7).should.be.equal(420));
it("prooves 8", () => smallestWithoutRemainder(8).should.be.equal(840));
it("prooves 9", () => smallestWithoutRemainder(9).should.be.equal(2520));
it("prooves 12", () => smallestWithoutRemainder(12).should.be.equal(27720));
it("prooves 20", () => smallestWithoutRemainder(20).should.be.equal(232792560));
it("prooves 30", () => smallestWithoutRemainder(30).should.be.equal(2329089562800));
it("prooves 40", () => smallestWithoutRemainder(40).should.be.equal(5342931457063200));
});
let smallestWithoutRemainder = (end: number, interval?: number) => {
// What do we know?
// - at 10, the answer is 2520
// - can't be smaller than the lower multiple of 10
// - must be an interval of the lower multiple of 10
// so:
// - the interval and the start should at least be divisable by 'end'
// - we can recurse and build on the results before it.
if (!interval) interval = end;
let count = Math.floor(end / 10);
if (count == 1) interval = 2520;
else if (count > 1) interval = smallestWithoutRemainder((count - 1) * 10, interval);
for (let i = interval; true; i += interval) {
let failed = false;
for (let j = end; j > 1; j--) {
if (i % j != 0) {
failed = true;
break;
}
}
if (!failed) return i;
}
}
I think this the answer:
primes = [11, 13, 17, 19]
result = 2520
for i in primes:
result *= i
print (result * 2)