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
Related
n = int(input())
counter = 0
while n > 0:
if (n // 2) > 1:
counter = counter +1
print (counter)
Hi,
I am a python learner and I am having problems with this homework I was given.
Read a natural number from the input.
Find out how many times in a row this number can be divided by two
(e.g. 80 -> 40 -> 20 -> 10 -> 5, the answer is 4 times)
And I should use while loop to do it.
Any Ideas, because I really don't have any idea how to do it.
This is my best try
You are not changing n. I would write it like this:
while (n % 2) == 0:
n //= 2
counter += 1
Try this, we take the value from "n" and check whether it is divisible by two or not. If it is divisible by two, we increment the counter and then divide that number by 2. If not, it will print the output.
n = int(input("Input your number: "))
counter = 0
while n % 2 != 1:
counter = counter + 1
n = n/2
print(counter)
Your while loop condition is wrong.
While the number is evenly divisible by 2, divide it by 2 and increment counter
num = int(input('Enter a number: '))
counter = 0
while num % 2 == 0 and num != 0:
num = num / 2
counter = counter + 1
print(counter)
The code above will not work as intended. The intended functionality is to take an input natural number and find out how many times in a row the number can be divided by 2. However, the code will only divide the number by 2 once.
To fix this, you can change the code to the following:
n = int(input())
counter = 0
while n > 0:
if (n % 2) == 0:
counter = counter +1
n = n // 2
print (counter)
You need to test whether a number is divisible by 2. You can do this in one of two ways...
x % 2 == 0 # will be True if it's divisible by 2
x & 1 == 0 # will be True if it's divisible by 2
So, you need a loop where you test for divisibility by 2, if True divide your original value by 2 (changing its value) and increment a counter
N = int(input())
counter = 0
if N != 0:
while N % 2 == 0:
counter += 1
N //= 2
print(counter)
Or, if you prefer more esoteric programming, then how about this:
N = int(input())
b = bin(N)
print(0 if (o := b.rfind('1')) < 0 else b[o:].count('0'))
My code for Project Euler problem 7 seems correct because it works for small numbers, but it takes forever for large numbers. Where did I go wrong
prime_numbers = []
for number in (range(2,2000000000)):
if number > 1:
for i in range(2, number):
if number % i == 0:
break
else:
prime_numbers.append(number)
print(prime_numbers[10001])
Look at comments and other answers for inspiration on using a faster algorithm. This answer suggests some small speed optimizations to your algorithm, which may be enough to get the result in time.
Stop as soon as you have your result:
prime_numbers = []
number = 1
while True:
number += 1
for i in range(2, number):
if number % i == 0:
break
else:
prime_numbers.append(number)
if len(prime_numbers) > 10001:
break
print(prime_numbers[10001])
No need to look at even numbers or divisors:
prime_numbers = [2]
number = 1
while True:
number += 2
for i in range(2, number, 2):
if number % i == 0:
break
else:
prime_numbers.append(number)
if len(prime_numbers) > 10001:
break
print(prime_numbers[10001])
The upper limit of i can be lower: the square root of number:
import math
prime_numbers = [2]
number = 1
while True:
number += 2
for i in range(2, math.isqrt(number) + 1, 2):
if number % i == 0:
break
else:
prime_numbers.append(number)
if len(prime_numbers) > 10001:
break
print(prime_numbers[10001])
Only look at prime divisors:
import math
prime_numbers = [2]
number = 1
while True:
number += 2
divisor_limit = math.isqrt(number)
no_divisor = True
for d in prime_numbers:
if d > divisor_limit:
break
if number % d == 0:
no_divisor = False
break
if no_divisor:
prime_numbers.append(number)
if len(prime_numbers) > 10001:
break
print(prime_numbers[10001])
Your code is valid, but not very fast - and therefore for large numbers, your program is just going to take far too long. You could try to make some changes, skip even numbers, don't range from 2 to the number but rather 2 to the square root and so on, but you could also look at a different, faster approach called the Sieve of Eratosthenes, you can read about it here.
The issue is Prime Numbers -that the solution is not implemented effectively.
I hear about eratosthenes sieve.
What are the other methods of implementing prime numbers - in a more efficient way?
n = int(input())
suma = 0
m = 0
while m < n:
if n > 100000:
break
x = int(input())
if 1 < x < 10000:
for i in range(x):
if x % (i + 1) == 0:
suma += 1
if suma == 2 and x != 2:
m += 1
print('o')
suma = 0
else:
m += 1
print('x')
suma = 0
The one of solution: https://medium.com/#dhruvpatel1057/generate-prime-numbers-in-python-using-segmented-sieve-of-eratosthenes-245b79da6687
You are using a very naive approach for primality checking.
As a general naive but not so much method, I'd recommend using Wilson's theorem as prime checker. Using math.factorial instead of a python loop should provide you with some reasonable speed increase while keeping the code fairly simple.
I hear about eratosthenes sieve - but not idea, how to implement it.
That's not the sieve of eratosthenes proper but what people usually talk about when they mention it is that when you find a prime you go through all your candidates and remove its factors ("sieving" them out hence the name) and the new first candidate is the next prime in the sequence.
There are more efficient primality tests than sieving everything though, check the "primality test" wikipedia page for examples.
This code will generate the amount of prime numbers the user asked for. It is very efficient, and can work out the first thousand prime numbers in milliseconds.
amount = int(input("Enter the amount of prime numbers you would like to see: "))
primes = []
num = 1
while len(primes) < amount:
num += 1
# If num is bigger than 1, and is 2 or is odd, and when divided by all the number from 3 to num's square root plus 1 (excluding even numbers), there is always a remainder.
if num > 1 and (num == 2 or num % 2 != 0) and all(num % divisor != 0 for divisor in range(3, int(num ** 0.5) + 1, 2)):
primes.append(num)
print(f"The first {amount} prime numbers are:\n{primes}")
this is sample code , make a list of prime, and from this list check the number to process is getting divided or not , if it is not getting divided then it is prime else it is not
# your code goes here
x = int(input())
prime =[]
for i in range(2,x):
if i not in prime:
if prime == []:
prime.append(i)
else:
check = 1
for j in prime:
if i%j==0:
check = 0
break
if check:
prime.append(i)
print(prime)
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.
The Following code keeps telling me a wrong number and I can't see why, know it's brute force but it should still work... also the number it returns has indeed over 500 divisors, 512 to be exact, help would be much appreciated
Number = 1
Count = 2
Found = False
while Found == False:
Divisors = 0
if (Number % 2) != 0:
for i in range(1, int(Number**(1/2)), 2):
if Number % i == 0:
Divisors += 1
else:
for i in range(1, int(Number**(1/2))):
if Number % i == 0:
Divisors += 1
if Divisors >= 500:
print (Number)
Found = True
else:
Number += Count
Count += 1
For reference: Problem 12 from the Euler Challange
The number of divisors of an integer is just the product of (1 + exponent) for each pure power in the factor decomposition of an integer.
As an example: 28 = 2^2 * 7
The powers are 2 and 1, so the number of divisors is (2+1)*(1+1) = 3*2 = 6. Easy one
Bigger one: 2047 * 2048 / 2 = 2^10 * 23 * 89
The powers are 10, 1 and 1, so the number of divisors is 11*2*2 = 44
Easier: 100 = 2^2 * 5^2
The powers are 2, 2 so there are 3*3=9 divisors. The same applies to 36=2^2*3^2. The only interesting part is the exponents.
So, use any prime factor decomposition (use a sieve, you don't need a primality test) it would be much faster and more reliable than trying each of the possible numbers.
def factorize(i):
# returns an array of prime factors
whatever
def number_of_divisors(i):
n = 1
for v in Counter(factorize(i)).values():
n *= v + 1
return n
I'm not sure what Euler Challenge 12 is, but one obvious issue is the (1/2). If you try typing that in a Python prompt, you'll get 0. The reason why is that it will try to do integer math. I suggest just putting (0.5), or alternatively you could do (1/2.0).
Your divisor counting method is wrong. 12 has 6 divisors, but your code only counts 2.
Problems:
a number often has divisors larger than its square root
range doesn't include its upper bound, so you're stopping too early
the code you have been write is searching till number**0.5 and it is wrong you must search until number/2
so the corrected answer is like below:
Note : I add some extra code to show the progress. and they are not affect the solution.
Another Note: since the Nubmer itself is not counted like in the problem example, I add once to perform that.
Number = 1
Count = 2
Found = False
big_Devisor = 0
print "Number Count Divisors"
while Found == False:
Divisors = 1 # because the Number is itself Devisor
if (Number % 2) != 0:
for i in range(1, int(Number/2), 2):
if Number % i == 0:
Divisors += 1
else:
for i in range(1, int(Number/2)):
if Number % i == 0:
Divisors += 1
if Divisors >= 500:
print (Number)
Found = True
else:
if Divisors > big_Devisor:
big_Devisor = Divisors
print Number,'\t', Count, '\t', Divisors
Number += Count
Count += 1