I have a basic understanding of how to calculate Time Complexities, but I'm not sure how to calculate it in this case due to the random nature of primes.
A quick explanation --> Essentially, I'm keeping a running count of the remainders so that I know when the next prime is.
My code:
import math
n = int(input("Enter the number:\t"))
primeList = []
checkList = []
number = 3
isPrime = True
while number <= math.sqrt(n) and isPrime:
isChanged = False
for i, checkNum in enumerate(checkList):
if checkNum == 1:
isChanged = True
checkList[i] = primeList[i]
else:
checkList[i] = checkNum - 1
if not isChanged:
primeList.append(number)
checkList.append(number)
if n % number == 0:
isPrime = False
number += 2
if isPrime:
print("Prime")
else:
print("Not Prime")
Your algorithm seems to be O(n/log(n))
There are sqrt(n) passes through the outer loop. The inner loop is bounded by the number of primes which are less than sqrt(n). By the Prime Number Theorem this is asymptotically given by sqrt(n)/log(sqrt(n)). By the laws of logarithms this is equivalent to sqrt(n)/(0.5*log(n)) = 2*sqrt(n)/log(n). The overall complexity is thus
O(sqrt(n)*2*sqrt(n)/log(n)) = O(2*n/log(n)) = O(n/log(n))
Needless to say, this isn't a very efficient way to check if n is prime. It is asymptotically little better than the O(n) naive check for divisibility by all numbers less than n.
Related
The problem asks to calculate the 1000th prime number. I am trying to solve this problem but I am stuck.
There are some guidelines on how to solve the problem.
To help you get started, here is a rough outline of the stages you should probably follow in
writing your code:
Initialize some state variables
Generate all (odd) integers > 1 as candidates to be prime
For each candidate integer, test whether it is prime
One easy way to do this is to test whether any other integer > 1 evenly
divides the candidate with 0 remainder. To do this, you can use modular
arithmetic, for example, the expression a%b returns the remainder after
dividing the integer a by the integer b.
You might think about which integers you need to check as divisors –
certainly you don’t need to go beyond the candidate you are checking, buthow much sooner can you stop checking?
If the candidate is prime, print out some information so you know where you are
in the computation, and update the state variables
Stop when you reach some appropriate end condition. In formulating this
condition, don’t forget that your program did not generate the first prime (2).
Use these ideas to guide the creation of your code.
My attempt so far is this
def calculate_thousandth_prime():
j = 0
for i in range(3,int(10e6)):
if i%2 != 0:
counter = 0
for k in range(1, i):
if i%k != 0:
counter += 1
if counter == 0:
print("This candidate is prime")
j += 1
if j == 1001:
print("The number "+str(i)+" is the thousandth prime")
break
return 0
calculate_thousandth_prime()
My code gets stuck on i%k != 0. I must be doing something wrong... any help?
You have two problems:
First, you're searching for k in range(1, i):. Because every number, including primes, is divisible by 1, you won't find any primes. Try searching range(2, i) instead.
Second, you're checking if i%k != 0:. You should be checking that i%k == 0 instead. If i is divisible by any number k, then the number is not prime.
Actually, I found a third issue: You have an off-by-one error. By initializing j=0, your code considers the first prime it finds to be the "zeroth" prime. The code will output the thousand-and-first prime, instead of the thousandth prime.
Changes I made:
Change your range to add a 2 step to skip even numbers more naturally.
Check your inner loop, you need to divide by the values range(2, i//2). Dividing by anything greater than i//2 will be less than 2.
Change your prime check to see if any number in the above range divides. If so, we know the number is false. At that point we can move onto the next number.
We want to return when the prime counter is 1000, you were returning the 1001th prime.
def calculate_thousandth_prime():
prime_counter = 0
for i in range(3,int(10e6),2):
prime = True
for k in range(2, i//2):
if i % k == 0:
prime = False
break
if prime:
print(str(i) + " is prime")
prime_counter += 1
if prime_counter == 1000:
print("The number "+str(i)+" is the thousandth prime")
break
return i
calculate_thousandth_prime()
The sieve of Eratosthenes is generally the fastest way for the early primes. You can adapt it to reach the nth prime.
For example:
def nthPrime(N):
sieve = [1]*(N**2)
p = 2
for _ in range(N):
while not sieve[p]: p += 1
sieve[p::p] = [0]*len(sieve[p::p])
return p
nthPrime(100) # 541
The prime list divisor check method is probably easier to write and understand but it is much slower (although for only 1000 primes, this wouldn't make much of a difference):
def nthPrime(N):
primes = [2]
p = 1
while len(primes)<N:
p += 2
primes += [p]*all(p%d for d in primes)
return p
This program makes a list of all prime numbers less than or equal to a given input.
Then it prints the list.
I can't understand why it includes the number 2. When I first designed the program, I initialized the list with primes = [2] because I thought, since 2 % 2 == 0,
if n % x == 0:
is_prime = False
will set is_prime to False. However, that doesn't seem to be the case.
I'm sure there is something going on with the logic of my range() in the for loops that I just don't understand.
I guess my question is: Why is 2 included in the list of primes every time?
import math
limit = int(input("Enter a positive integer greater than 1: "))
while limit < 2:
limit = int(input("Error. Please enter a positive integer greater than 1: "))
primes = []
#Check all numbers n <= limit for primeness
for n in range (2, limit + 1):
square_root = int(math.sqrt(n))
is_prime = True
for x in range(2, (square_root + 1)):
if n % x == 0:
is_prime = False
if is_prime:
primes.append(n)
#print all the primes
print("The primes less than or equal to", limit, "are:")
for num in primes:
print(num)
Because you don't enter the second for-loop when you test for n=2 and therefore you don't set is_prime = False.
# Simplified test case:
x = 2
for idx in range(2, int(math.sqrt(x))+1):
print(idx)
This doesn't print anything because range is in this case: range(2, 2) and therefore has zero-length.
Note that your approach is not really efficient because:
you test each number by all possible divisors even if you already found out it's not a prime.
you don't exclude multiples of primes in your tests: if 2 is a prime, every multiple of 2 can't be prime, etc.
There are really great functions for finding prime numbers mentioned in Fastest way to list all primes below N - so I won't go into that. But if you're interested in improvements you might want to take a look.
I'm new to programming, so please bear with me.
I wrote this code to find the largest prime factor of a number, but when I input a number of a billion or greater it doesn't solve. I've tried xrange, but I'm using Python 3.4. Can anyone point me in the right direction?
done = False
num = int(input("Enter a number: "))
for j in range(num,2,-1):
if (num % j) != 0:
continue
for i in range(2,j):
if (j % i) != 0:
continue
break
else: break
print(j)
It's an O(n^2) algorithm. It doesn't appear to solve because it's just taking a long time. Solution? Use numpy. Or find a non-naive algorithm, or write one yourself. The sieve of Eratosthenes is a good starting point.
Since you are new to programming, here is a simple solution (but know that there are substantially more complex solutions that are more efficient). First note that you only need to check divisors up to the square root of num, since if num = a * b then one is less than the square root and one is larger. Secondly you only need to check for prime divisors.
You can generate a list of primes as follows:
import math
import itertools
def primes_generator():
"""Generator for the infinite list of primes."""
primes = [2, 3]
for prime in primes:
yield prime
for c in itertools.count(5, 2):
bound = math.sqrt(c)
for p in primes:
if not (c % p):
break
if p > bound:
primes.append(c)
yield c
break
Now to find the all the prime divisors:
def prime_factorization(number, primes=None):
if not primes:
primes = primes_generator()
factorization = dict()
for p in primes:
count = 0
while not (number % p):
number = number / p
count = count + 1
if count:
factorization[p] = count
if number == 1:
return factorization
The largest prime divisor is just the largest key in the dictionary. These functions should work fine for fairly large inputs. On my machine the following takes 0.06 seconds.
print(max(prime_factorization(1000000001).keys()))
Is there any suggestion on solving next higher prime and palindrome number from a given int.
Here is the snippet I am trying but its a kind of slow, please suggest if you ve any good algorithm that i can test.
#!/usr/bin/python
def next_higher(n):
while True:
s = str(n)
if not any([n % i == 0 \
for i in range(2, int(n**0.5))]) and s == s[::-1]:
return n
n = n + 1
print next_higher(2004)
print next_higher(20)
Output:
10201
101
Updated code testing for palindrome before prime. much faster than my previous code.
I am implementing the suggestion from user2357112.
#!/usr/bin/python
def next_higher(n):
while True:
s = str(n)
if s == s[::-1]:
if not any([n % i == 0 \
for i in range(2, int(n**0.5))]):
return n
n = n + 1
print next_higher(2004111)
print next_higher(2004)
print next_higher(2004)
print next_higher(20)
There are quite a few optimizations you could do:
Like user2357.. suggested in the comments, test palindromeness first, and then check if the number is prime, since prime check is more expensive.
You don't need to check even number divisibility once you check is the number is divisible by 2. So you can change it to [2] + range(3, int(n**0.5) + 1, 2) to check only odd numbers after 2. (Also you need to do sqrt + 1 like I mentioned in the comments)
You should use () instead of []. [] generates the entire list of factors first and only then checks for any. If you use (), it creates a generator, so it stops as soon as a True value is found without calculating the entire list.
You should also use xrange instead of range for the same reason (xrange gives a generator, range gives a list)
You can use the Sieve of Eratosthenes algorithm to significantly reduce the time taken for prime number check.
You can also see if the palindrome check can be made faster. You can actually skip a whole lot of numbers instead of doing just + 1 each time.
Here is a version with most of these optimizations except the last two:
def next_higher(n):
if n % 2 == 0:
n = n - 1
while True:
n = n + 2
s = str(n)
if s == s[::-1]:
if not any((n % i == 0 for i in xrange(3, int(n**0.5) + 1, 2))):
return n
This should be pretty fast for your needs I believe. But you can do the last 2 optimizations to make it much more faster if you want.
Other than what has already been suggested,
What I suggest is that you first get the first palindrome number that is just higher than the given integer.
You can do this by trying to match the centre digits outwards.
Also, you should only check for numbers with odd number of digits, since if a number has even number of digits and it is a palindrome, then it will always be divisible by 11 and cannot be prime.
Once you get the first palindrome number that has odd number of digits and that is just higher than the current number, test it for primality and find the next palindrome number higher than this one.
You can do this by incrementing the centre digit.
Keep doing this till it rolls over to zero. In that case start incrementing the two neighbouring digits.
Continue, till you reach a prime number.
I tried optimizing the palindrome check i.e to find odd palindrome's.
Since the first digit should be odd number, i focused on that part.
Here's the code below with the assumptions its greater than 1 digit.
def next_odd_palindrome(n):
"""to check the next odd palindrome number"""
if n%2==0:
n=n-1
while True:
n=n+2
s = str(n)
if int(s[0])%2==0:
n = int(str(int(s[0])+1)+ s[1:])
s = str(n)
if s==s[::-1]:
return n
let me know if anything wrong.
Just for the fun of it, I implemented all optimizations by Hari Shankar and Abhishek Bansal.
It first finds the higher odd length palindrome, then increment the palindrome in a way that keeps its palindromity. Then checks each number using prime numbers calculated by Sieve method in the beginning.
This can process up to n=10^14 (can be higher if you increase the CACHE size) under 1 second in my computer =D
primes = []
CACHE = int(10**7) # Cache size for Sieve
# Custom class for immediate printing of output
import sys
class Unbuf:
def __init__(self,stream):
self.stream = stream
def write(self,data):
self.stream.write(data)
self.stream.flush()
sys.stdout = Unbuf(sys.stdout)
def sieve():
global primes
is_prime = [False,False]+([True]*(CACHE-1))
for i in xrange(2,int(CACHE**0.5)):
if is_prime[i]:
is_prime[i*i::i] = [False]*((CACHE-i*i+i)/i)
primes = [num for num, bool_prime in enumerate(is_prime) if bool_prime]
def is_prime(n):
"""Checks whether n is prime"""
global primes
if n<2:
return False
if n==2:
return True
for prime in primes:
if prime>n**0.5+1:
return True
if n%prime==0:
return False
# For the case that the number is bigger than the square of our largest prime
for num in xrange(primes[-1]+2,n**0.5+1,2):
if n%num==0:
return False
return True
def next_higher_odd_length_palindrome(n):
n = str(n)
if len(n)%2==0: # Even length, take the smallest odd length (10(00)*1)
n = '1'+('0'*(len(n)-1))+'1'
else:
middle_idx = len(n)/2
left = int(n[:middle_idx+1])
left_cmp = n[middle_idx::-1]
right_cmp = n[middle_idx:]
# If mirroring left part to right part
# makes the number smaller or equal, then
if right_cmp>=left_cmp:
# Increase the left half number
left = left+1
# Mirror left part to the right part
n = str(left)+str(left)[-2::-1]
return n
def next_higher(n):
if n<=1:
return 2
# Ensure the number is a palindrome of odd length
n = next_higher_odd_length_palindrome(n)
while True:
if is_prime(int(n)):
return int(n)
n = next_higher_odd_length_palindrome(n)
if int(n[0])%2==0:
new_lead = str(int(n[0])+1)
n = new_lead+n[1:-1]+new_lead
import time
print 'Sieving...',
start_time = time.time()
sieve()
print 'Done in %.3fs' % (time.time() - start_time)
print next_higher(2004111)
print next_higher(2004)
print next_higher(20)
while True:
n = int(raw_input('Enter n: '))
start_time = time.time()
result = next_higher(n)
print 'Next higher prime palindrome: %d (calculated in %.3fs)' % (result, time.time() - start_time)
Which in my computer gives this output:
Sieving... Done in 1.444s
3007003
10301
101
Enter n: 1999999999
Next higher prime palindrome: 10000500001 (calculated in 0.004s)
Enter n: 1999999999999
Next higher prime palindrome: 3000002000003 (calculated in 0.051s)
Enter n: 1000000000000
Next higher prime palindrome: 1000008000001 (calculated in 0.030s)
Enter n:
I'm trying to find the largest prime factor for a number. The code runs correctly on IDLE when used with smaller numbers, but doesn't seem to print anything to the screen at all when I assign a larger number like 600851475143 to n. Why?
def isPrime(n):
isPrime = True
for i in range(2,n-1):
if n % i == 0:
isPrime = False
return isPrime
largest = 0
n = 600851475143
for i in range(2,n-1):
if isPrime(i) and n % i == 0:
largest = i
n = n / i
continue
print("The largest prime factor is", largest)
I'm running Python 3.3, by the way.
==============================================================================
Thanks everyone!
I fixed my original code as follows:
def isPrime(n):
for i in range(2,n-1):
if n % i == 0:
return False
return True
largest = 0
n = 600851475143
for i in range(2,n-1):
if isPrime(i) and n % i == 0:
largest = i
if i == n:
break
n = n / i
print("The largest prime factor is", largest)
Like nakedfanatic said, their code runs faster, and I edited it slightly:
largest = 0
n = 600851475143
i = 2
while True:
if n % i == 0:
largest = i
if n == i:
# finished
break
n = n / i
else:
i += 1
print("The largest prime factor is", largest)
There are several areas of optimization:
all factorization only needs to got up to sqrt(n) (inclusive)
convert isPrime() to a table lookup
Initialize a lookup table using n, then you compute all primes < sqrt(n) only once and loop through them.
As comments pointed out, this takes up large memory space. We can use bit flag to cut the memory requirement to 1/8, and we can cut it by a further half if we skip all the even numbers (then have to test if n is even separately). But that may still be daunting for LARGE n.
(if using current code) return early in isPrime() (by #justhalf)
loop backwards (from sqrt(n) to 2) when looking up the largest factor
return early if the quotient is 1 after dividing by a factor (by #justhalf)
This post (suggested by #prashant) contains more complicated algorithm (making my suggestion very naive ><):
Fastest way to list all primes below N
... (edits are welcome)
It's because you keep trying even if n is already 1.
This code will help you to see the problem:
def isPrime(n):
for i in xrange(2,n-1):
if n % i == 0:
return False
return True
largest = 0
n = 600851475143
for i in xrange(2,n-1):
print 'Checking whether %d divides %d' % (i,n)
if isPrime(i) and n % i == 0:
largest = i
n = n / i
continue
print("The largest prime factor is", largest)
which will produce:
...
Checking whether 6857 divides 6857
Checking whether 6858 divides 1
Checking whether 6859 divides 1
Checking whether 6860 divides 1
Checking whether 6861 divides 1
Checking whether 6862 divides 1
Checking whether 6863 divides 1
Checking whether 6864 divides 1
Checking whether 6865 divides 1
Checking whether 6866 divides 1
Checking whether 6867 divides 1
Checking whether 6868 divides 1
Checking whether 6869 divides 1
Checking whether 6870 divides 1
Checking whether 6871 divides 1
Checking whether 6872 divides 1
Checking whether 6873 divides 1
...
You should break the loop when n becomes 1, so that it won't do unnecessary checking
n = n / i
if n==1:
break
continue
And anyway, your code might be improved by a lot, haha, see others' suggestions.
Most likely, your code isn't terminating with large n, simply because it takes so long to run through the loop.
Your code is running in O(n²) time, which means it will quickly become unreasonably slow as the size of n increases. That is why your algorithm works for small values, but hangs for large values.
This code does the same thing in O(n) time without doing any prime checking at all, and returns a result instantly:
prime_factors = []
n = 600851475143
i = 2
while True:
if n % i == 0:
prime_factors.append(i)
if n == i:
# finished
break
n = n / i
else:
i += 1
print("The largest prime factor is", prime_factors[-1])
More difficult problems may require a different algorithm.
Check this question out: Fastest way to list all primes below N
Your code looks okay, but could take a long time for a large n. Leveraging math can enable you to do this problem orders of magnitude faster.
On that link, I recommend rwh_primes1 for a pure python solution, and primesfrom3to as one that uses numpy. Both of those implementations are fairly short, relatively clear, and do basically the same thing. Those code snippets are written in Python 2, so a translation might look like this:
def rwh_primes1(n):
sieve = [True] * (n//2)
for i in range(3, int(n**0.5)+1,2):
if sieve[i//2]:
sieve[i*i//2::i] = [False] * ((n-i*i-1)//(2*i)+1)
return [2] + [2*i+1 for i in range(1,n//2) if sieve[i]]
isPrime = True
for i in range(2,n-1):
if n % i == 0:
isPrime = False
return isPrime
This loop always exits the first iteration due to the unconditional return. Try:
for i in range(2,n-1):
if n % i == 0:
return False
return True
Also the upper bound n-1 can be reduced to sqrt(n)+1.
Another aspect of your code which may be slowing it down is the second half of your code
largest = 0
n = 600851475143
for i in range(2,n-1):
if isPrime(i) and n % i == 0:
largest = i
n = n / i
continue
Specifically the statement
if isPrime(i) and n % i == 0:
Per the documentation, the second condition is only evaluated if the first one is True. In your case it would make more sense to reverse the conditions so that computationally les expensive division is performed always and the more expensive isPrime() is only called for the actual factors
largest = 0
n = 600851475143
for i in range(2,n-1):
if n % i == 0 and isPrime(i):
largest = i
n = n / i
if n == 1:
break