I found this code in a website for checking prime number
def isPrime(number):
if (number <= 1):
return False
elif (number <= 3):
return True
elif (number % 2 == 0 or number % 3 == 0):
return False
i = 5
while(i * i <= number):
if (number % i == 0 or number % (i + 2) == 0):
return False
i += 6
return True
but I can't understand the logic of if statement under while loop, that is if (number % i == 0 or number % (i + 2) == 0)
why need i+2??? when i is even, i+2 is also even and odd while i is odd. So, why need to check for i+2???
Except for 2 and 3, all prime numbers are of the form 6n±1. This code checks for 2 and 3 explicitly and then checks higher numbers in pairs: 6n-1, 6n+1, starting with 5. So it checks 5, 7 then 11, 13 then 17, 19 and so on. It steps 6 between pairs and steps 2 within each pair. Doing it this way avoids ever having to check multiples of 2 or 3.
i starts from 5, so adding 2 means taking in count only evens numbers.
It makes the check faster.
If you want to find out if a number is prime or not, here is a short sample
# Check if a positive integer number is a prime number. Check only till square root +1.
import math
def isPrime(n):
if ( n > 3 ):
if (n % 2 == 0):
return False
for i in range(3, math.isqrt(n), 2):
if (n % i == 0):
return False
return True
else:
return True
counter = 0
numList = []
beginning = 1
primeOrNot = False
number = int(input("Type a number to check if it's prime."))
while True:
if number == 2:
primeOrNot = True
elif "2" in str(number) or "4" in str(number) or "6" in str(number) or "8" in str(number) or "0" in str(number) or number == 1:
primeOrNot = False
numList = [''.join(p) for p in permutations(str(number))]
for p in range(0, len(numList)):
numList[p] = int(numList[p])
for u in range(2, number):
if all(o % u == 0 for o in numList):
counter = counter + 1
if counter == 0:
print(numList)
break
I'm trying to make it so it prints a prime if all permutations of the number are also primes.
I've tried to do that with all but what it's doing is checking if atleast 1 of the elements in the list fits the o % u = 0. I want it to check if all of the elements work, not if one. Anything I can do using all or using something else?
Example of output now:
Input: 12
Output: Nothing, Neither 12 nor 21 are prime
Input: 35
Output: 35, 53
Even though 35 isn't prime it prints because 53 is.
Your logic is faulty:
for u in range(2, number):
if all(o % u == 0 for o in numList):
counter = counter + 1
For each potential divisor in the range, you check to see whether that divisor goes into every permutation of the input. What value of u do you envision is going to divide both 35 and 53?
For starters, you need this to be any, not all.
Second, I strongly recommend that you look up how to find primes and/or factor numbers. You can learn better ways to handle this, especially when your input string is longer.
Also, note that a 0 or 5 in the input will also ensure that some of the permutations will be composite; assuming that the input is at least two digits, it must be entirely composed of 1379 digits.
if number in [2, 3, 5, 7]:
print("number is prime")
exit(0)
if any (d not in "1379" for d in str(number)):
print("This input has composite permutations")
I'm trying to print the factors of the number 20 in python so it goes:
20
10
5
4
2
1
I realize this is a pretty straightforward question, but I just had a question about some specifics in my attempt. If I say:
def factors(n):
i = n
while i < 0:
if n % i == 0:
print(i)
i-= 1
When I do this it only prints out 20. I figure there's something wrong when I assign i=n and then decremented i, is it also affecting n? How does that work?
Also I realize this could probably be done with a for loop but when I use a for loop I can only figure out how to print the factors backwards so that I get: 1, 2, 5, 10....
Also I need to do this using just iteration. Help?
Note: This isn't a homework question I'm trying to relearn python on my own since it's been a while so I feel pretty silly being stuck on this question :(
while i < 0:
This will be false right from the start, since i starts off positive, presumably. You want:
while i > 0:
In words, you want to "start i off at n, and decrement it while it is still greater than 0, testing for factors at each step".
>>> def factors(n):
... i = n
... while i > 0: # <--
... if n % i == 0:
... print(i)
... i-= 1
...
>>> factors(20)
20
10
5
4
2
1
The while condition should be i > 0 and not i < 0 because it will never satisfy it as i begins in 20 (or more in other cases)
Hope my answer helps!
#The "while True" program allows Python to reject any string or characters
while True:
try:
num = int(input("Enter a number and I'll test it for a prime value: "))
except ValueError:
print("Sorry, I didn't get that.")
continue
else:
break
#The factor of any number contains 1 so 1 is in the list by default.
fact = [1]
#since y is 0 and the next possible factor is 2, x will start from 2.
#num % x allows Python to see if the number is divisible by x
for y in range(num):
x = y + 2
if num % x is 0:
fact.append(x)
#Lastly you can choose to print the list
print("The factors of %s are %s" % (num, fact))
What I want to do is:
create a "list" of odd numbers
then test if they are prime
the test will be done with multiples up to half of the value of the odd number hence halfodd
put the prime number into a list
print that list
however my result is a list of numbers from 1 to 1003 that skip 3, 4, and 5
Is there a semantic error here?
#Prime number generator
def primenumber():
primelist = [1, 2]
num = 3
even = num%2
multi = 0
result = 0
while len(primelist) < 1000:
if even != 0:
oddnum = num
i = 2
halfodd = ((oddnum + 1)/2)
while i < halfodd:
i =+ 1
multi = oddnum%i
if multi == 0:
result += 1
if result != 0:
primelist.append(oddnum)
prime_num = oddnum
num += 1
print primelist
primenumber()
if result != 0:
Since result is (supposed to be) a count of how many factors you found, you want
if result == 0:
You're also not resetting result when you advance to a new candidate prime, you never recompute even, and if you did, you'd be considering the odd numbers twice each.
This should work:
def primenumber():
primelist = [1, 2]
candidate = 3
while len(primelist) < 1000:
isCandidatePrime = True
if (candidate % 2 == 0): # if candidate is even, not a prime
isCandidatePrime = False
else:
for i in range(3, (candidate+1)/2, 2): # else check odds up to 1/2 candidate
if (candidate % i == 0): # if i divides it, not a prime
isCandidatePrime = False
break
if (isCandidatePrime):
primelist.append(candidate)
candidate += 1
print primelist
I think there are several improvements (check only numbers less than or equal to the square root of the candidate, only check prime numbers, not all numbers less than the square root of the candidate) but I'll leave that alone for now.
count = 0
i = 11
while count <= 1000 and i <= 10000:
if i%2 != 0:
if (i%3 == 0 or i%4 == 0 or i%5 == 0 or i%6 == 0 or i%7 == 0 or i%9 == 0):
continue
else:
print i,'is prime.'
count += 1
i+=1
I'm trying to generate the 1000th prime number only through the use of loops. I generate the primes correctly but the last prime i get is not the 1000th prime. How can i modify my code to do so. Thank in advance for the help.
EDIT: I understand how to do this problem now. But can someone please explain why the following code does not work ? This is the code I wrote before I posted the second one on here.
count = 1
i = 3
while count != 1000:
if i%2 != 0:
for k in range(2,i):
if i%k == 0:
print(i)
count += 1
break
i += 1
Let's see.
count = 1
i = 3
while count != 1000:
if i%2 != 0:
for k in range(2,i):
if i%k == 0: # 'i' is _not_ a prime!
print(i) # ??
count += 1 # ??
break
i += 1 # should be one space to the left,
# for proper indentation
If i%k==0, then i is not a prime. If we detect that it's not a prime, we should (a) not print it out, (b) not increment the counter of found primes and (c) we indeed should break out from the for loop - no need to test any more numbers.
Also, instead of testing i%2, we can just increment by 2, starting from 3 - they will all be odd then, by construction.
So, we now have
count = 1
i = 3
while count != 1000:
for k in range(2,i):
if i%k == 0:
break
else:
print(i)
count += 1
i += 2
The else after for gets executed if the for loop was not broken out of prematurely.
It works, but it works too hard, so is much slower than necessary. It tests a number by all the numbers below it, but it's enough to test it just up to its square root. Why? Because if a number n == p*q, with p and q between 1 and n, then at least one of p or q will be not greater than the square root of n: if they both were greater, their product would be greater than n.
So the improved code is:
from math import sqrt
count = 1
i = 1
while count < 1000:
i += 2
for k in range(2, 1+int(sqrt(i+1))):
if i%k == 0:
break
else:
# print(i) ,
count += 1
# if count%20==0: print ""
print i
Just try running it with range(2,i) (as in the previous code), and see how slow it gets. For 1000 primes it takes 1.16 secs, and for 2000 – 4.89 secs (3000 – 12.15 ses). But with the sqrt it takes just 0.21 secs to produce 3000 primes, 0.84 secs for 10,000 and 2.44 secs for 20,000 (orders of growth of ~ n2.1...2.2 vs. ~ n1.5).
The algorithm used above is known as trial division. There's one more improvement needed to make it an optimal trial division, i.e. testing by primes only. An example can be seen here, which runs about 3x faster, and at better empirical complexity of ~ n1.3.
Then there's the sieve of Eratosthenes, which is quite faster (for 20,000 primes, 12x faster than "improved code" above, and much faster yet after that: its empirical order of growth is ~ n1.1, for producing n primes, measured up to n = 1,000,000 primes):
from math import log
count = 1 ; i = 1 ; D = {}
n = 100000 # 20k:0.20s
m = int(n*(log(n)+log(log(n)))) # 100k:1.15s 200k:2.36s-7.8M
while count < n: # 400k:5.26s-8.7M
i += 2 # 800k:11.21-7.8M
if i not in D: # 1mln:13.20-7.8M (n^1.1)
count += 1
k = i*i
if k > m: break # break, when all is already marked
while k <= m:
D[k] = 0
k += 2*i
while count < n:
i += 2
if i not in D: count += 1
if i >= m: print "invalid: top value estimate too small",i,m ; error
print i,m
The truly unbounded, incremental, "sliding" sieve of Eratosthenes is about 1.5x faster yet, in this range as tested here.
A couple of problems are obvious. First, since you're starting at 11, you've already skipped over the first 5 primes, so count should start at 5.
More importantly, your prime detection algorithm just isn't going to work. You have to keep track of all the primes smaller than i for this kind of simplistic "sieve of Eratosthanes"-like prime detection. For example, your algorithm will think 11 * 13 = 143 is prime, but obviously it isn't.
PGsimple1 here is a correct implementatioin of what the prime detection you're trying to do here, but the other algorithms there are much faster.
Are you sure you are checking for primes correctly? A typical solution is to have a separate "isPrime" function you know that works.
def isPrime(num):
i = 0
for factor in xrange(2, num):
if num%factor == 0:
return False
return True
(There are ways to make the above function more effective, such as only checking odds, and only numbers below the square root, etc.)
Then, to find the n'th prime, count all the primes until you have found it:
def nthPrime(n):
found = 0
guess = 1
while found < n:
guess = guess + 1
if isPrime(guess):
found = found + 1
return guess
your logic is not so correct.
while :
if i%2 != 0:
if (i%3 == 0 or i%4 == 0 or i%5 == 0 or i%6 == 0 or i%7 == 0 or i%9 == 0):
this cannot judge if a number is prime or not .
i think you should check if all numbers below sqrt(i) divide i .
Here's a is_prime function I ran across somewhere, probably on SO.
def is_prime(n):
return all((n%j > 0) for j in xrange(2, n))
primes = []
n = 1
while len(primes) <= 1000:
if is_prime(n):
primes.append(n)
n += 1
Or if you want it all in the loop, just use the return of the is_prime function.
primes = []
n = 1
while len(primes) <= 1000:
if all((n%j > 0) for j in xrange(2, n)):
primes.append(n)
n += 1
This is probably faster: try to devide the num from 2 to sqrt(num)+1 instead of range(2,num).
from math import sqrt
i = 2
count = 1
while True:
i += 1
prime = True
div = 2
limit = sqrt(i) + 1
while div < limit:
if not (i % div):
prime = False
break
else:
div += 1
if prime:
count += 1
if count == 1000:
print "The 1000th prime number is %s" %i
break
Try this:
def isprime(num):
count = num//2 + 1
while count > 1:
if num %count == 0:
return False
count -= 1
else:
return True
num = 0
count = 0
while count < 1000:
num += 1
if isprime(num):
count += 1
if count == 1000:
prime = num
Problems with your code:
No need to check if i <= 10000.
You are doing this
if i%2 != 0:
if (i%3 == 0 or i%4 == 0 or i%5 == 0 or i%6 == 0 or i%7 == 0 or i%9 == 0):
Here, you are not checking if the number is divisible by a prime number greater than 7.
Thus your result: most probably divisible by 11
Because of 2. your algorithm says 17 * 13 * 11 is a prime(which it is not)
How about this:
#!/usr/bin/python
from math import sqrt
def is_prime(n):
if n == 2:
return True
if (n < 2) or (n % 2 == 0):
return False
return all(n % i for i in xrange(3, int(sqrt(n)) + 1, 2))
def which_prime(N):
n = 2
p = 1
while True:
x = is_prime(n)
if x:
if p == N:
return n
else:
p += 1
n += 1
print which_prime(1000)
n=2 ## the first prime no.
prime=1 ## we already know 2 is the first prime no.
while prime!=1000: ## to get 1000th prime no.
n+=1 ## increase number by 1
pon=1 ## sets prime_or_not(pon) counter to 1
for i in range(2,n): ## i varies from 2 to n-1
if (n%i)==0: ## if n is divisible by i, n is not prime
pon+=1 ## increases prime_or_not counter if n is not prime
if pon==1: ## checks if n is prime or not at the end of for loop
prime+=1 ## if n is prime, increase prime counter by 1
print n ## prints the thousandth prime no.
Here is yet another submission:
ans = 0;
primeCounter = 0;
while primeCounter < 1000:
ans += 1;
if ans % 2 != 0:
# we have an odd number
# start testing for prime
divisor = 2;
isPrime = True;
while divisor < ans:
if ans % divisor == 0:
isPrime = False;
break;
divisor += 1;
if isPrime:
print str(ans) + ' is the ' + str(primeCounter) + ' prime';
primeCounter += 1;
print 'the 1000th prime is ' + str(ans);
Here's a method using only if & while loops. This will print out only the 1000th prime number. It skips 2. I did this as problem set 1 for MIT's OCW 6.00 course & therefore only includes commands taught up to the second lecture.
prime_counter = 0
number = 3
while(prime_counter < 999):
divisor = 2
divcounter = 0
while(divisor < number):
if(number%divisor == 0):
divcounter = 1
divisor += 1
if(divcounter == 0):
prime_counter+=1
if(prime_counter == 999):
print '1000th prime number: ', number
number+=2
I just wrote this one. It will ask you how many prime number user wants to see, in this case it will be 1000. Feel free to use it :).
# p is the sequence number of prime series
# n is the sequence of natural numbers to be tested if prime or not
# i is the sequence of natural numbers which will be used to devide n for testing
# L is the sequence limit of prime series user wants to see
p=2;n=3
L=int(input('Enter the how many prime numbers you want to see: '))
print ('# 1 prime is 2')
while(p<=L):
i=2
while i<n:
if n%i==0:break
i+=1
else:print('#',p,' prime is',n); p+=1
n+=1 #Line X
#when it breaks it doesn't execute the else and goes to the line 'X'
This will be the optimized code with less number of executions, it can calculate and display 10000 prime numbers within a second.
it will display all the prime numbers, if want only nth prime number, just set while condition and print the prime number after you come out of the loop. if you want to check a number is prime or not just assign number to n, and remove while loop..
it uses the prime number property that
* if a number is not divisible by the numbers which are less than its square root then it is prime number.
* instead of checking till the end(Means 1000 iteration to figure out 1000 is prime or not) we can end the loop within 35 iterations,
* break the loop if it is divided by any number at the beginning(if it is even loop will break on first iteration, if it is divisible by 3 then 2 iteration) so we iterate till the end only for the prime numbers
remember one thing you can still optimize the iterations by using the property *if a number is not divisible with the prime numbers less than that then it is prime number but the code will be too large, we have to keep track of the calculated prime numbers, also it is difficult to find a particular number is a prime or not, so this will be the Best logic or code
import math
number=1
count = 0
while(count<10000):
isprime=1
number+=1
for j in range(2,int(math.sqrt(number))+1):
if(number%j==0):
isprime=0
break
if(isprime==1):
print(number,end=" ")
count+=1