My objective of the below code is
check if entered number is a prime
if not print the next biggest prime
def primetest (num):
for c in range (2, num):
if num % c == 0:
repeattest (num) #not prime? increment number
else :
print (num,"is a prime number")
break
def repeattest (num): # check prime if not increment number by 1
for z in range (2, num):
num = num+1
primetest (num)
if num % z == 0:
num = num+1
else:
print ("Next Prime:", num+1)
break
num = int (input ("enter a number:")) # main code:
for y in range (2, num):
if num % y == 0:
repeattest (num)
else:
print (num,"is a prime number")
break
I think the logic is fine, but not sure why im not getting any output.
Time comlexity of your code is O(N) when it find a number which is prime or not.
There is no pointing on dividing from 2 to len(num)-1. It is enough to loop from 2 to sqrt of the given number. Therefore time complexity reduce to O(n) to O(log(n)).
import math
num = int (input ("enter a number:"))
def primeTest(num):
isPrime = 0
for i in range(2,int(math.sqrt(num)+1)):
if num%i == 0:
isPrime = isPrime + 1
break
if isPrime == 0:
print(num, "is a prime number")
else:
num = num + 1
repeatTest(num)
def repeatTest (num):
isPrime = 0
for i in range(2,int(math.sqrt(num))):
if num%i == 0:
isPrime = isPrime + 1
break
if isPrime == 0:
print("Next Prime: ", num)
else:
num = num + 1
repeatTest(num)
primeTest(num)
The logic which you are using to find if a number if prime seems wrong .
Taking a integer like 9 prints "9 is a prime number" .
And also you are checking for next prime numbers from 2 to Num .
Num being the input , you cant get a number greater than that .
It exits from the loop without even getting in , therefore not printing anything when you are searching for next prime .
You need to change the logic .
write a separate function for checking prime and end the loop when you find the next prime number instead of stopping at num .
Related
#Write your code below this line 👇
def prime_checker(number):
for num in range (2, number):
if num % number == 0:
print("It is not a prime number")
else:
print("It is a prime number")
#Write your code above this line 👆
#Do NOT change any of the code below👇
n = int(input("Check this number: "))
prime_checker(number=n)
How can I print a text that number is prime or not only once?
Fix
number % num == 0 and not num % number == 0
A number isn't prime the moment you find a number that doesn't divide it, but only when you have tested all and none divides it
Use for/else construction, it goes into else if no break has been used
def prime_checker(number):
for num in range(2, number):
if number % num == 0:
print("It is not a prime number")
break
else:
print("It is a prime number")
Note that this only fixes your way to do, but that isn't the optimal way to check if a numbre is a prime one, at least, ending range at square root of number, and directly verifying division by small numbers like 2,3,5,7
There are few mistakes in your code. I have modified them.
def prime_checker(number):
for num in range(2, number):
if number % num == 0:
print('Not prime')
return
print('Prime number')
# Write your code above this line 👆
# Do NOT change any of the code below👇
n = int(input("Check this number: "))
prime_checker(number=n)
For loop is to check if any of the number starting from 2 is a factor of number or not.
First, a slightly more efficient prime check by
going until the sqrt of the number only
going in steps of 2
import math
def is_prime(n: int) -> bool:
if n in (2, 3, 5):
return True
if n < 2 or n % 2 == 0:
return False
for i in range(3, math.ceil(math.sqrt(n)), 2):
if n % i == 0:
return False
return True
Now you can wrap that function in yours
def prime_checker(n: int):
msg = "%d is prime" if is_prime(n) else "%d is not prime"
print(msg % n)
prime_checker(11)
# 11 is prime
I think it's good to check a number is prime or not
def is_prime(n):
st = "prime" # being prime status
for i in range(2,n):
if n % i == 0: # if number is prime
st = "not prime"
break;
return st
n = int(input("enter n: "))
print (is_prime(n))
You only need to change the end of your code
your function is true except for these lines:
if num % number == 0:
print("It is not a prime number")
else:
print("It is a prime number")
you should change these to:
st = "Prime"
if number % num == 0:
st = "not prime"
return st
I am pretty new to python, so I don't fully understand how to use loops. I am currently working on a piece of code that I have to find the first N prime numbers.
The result that is desired is if you input 5, it outputs 2, 3, 5, 7, and 11, but no matter what I input for 'max', the output always ends up being 2 and 3. Is there a way to improve this?
max=int(input("How many prime numbers do you want: "))
min=2
while(min<=(max)):
for c in range(2, min):
if min%c==0:
break
else:
print min
min=min+1
You only increment min in the else block, i.e., if min % c is nonzero for all c, i.e., if min is prime. This means that the code won't be able to move past any composite numbers. You can fix this by unindenting min=min+1 one level so that it lines up with the for and else.
number = int(input("Prime numbers between 2 and "))
for num in range(2,number + 1):
if num > 1:
for i in range(2,num):
if (num % i) == 0:
break
else:
print(num)
Solution: Get the nth prime number entry. Iterate through each natural numbers for prime number and append the prime number to a list. Terminate the program when length of a list satisfies the user nth prime number entry.
# Get the number of prime numbers entry.
try:
enterNumber = int(input("List of nth prime numbers: "))
except:
print("The entry MUST be an integer.")
exit()
startNumber = 1
primeList = []
while True:
# Check for the entry to greater than zero.
if enterNumber <= 0:
print("The entry MUST be greater than zero.")
break
# Check each number from 1 for prime unless prime number entry is satisfied.
if startNumber > 1:
for i in range(2,startNumber):
if (startNumber % i) == 0:
break
else:
primeList.append(startNumber)
if (len(primeList) == enterNumber):
print(primeList)
break
else:
startNumber = startNumber + 1
continue
Try that :
n = int(input("First N prime number, N ? "))
p = [2]
c = 2
while len(p) < n:
j = 0
c += 1
while j < len(p):
if c % p[j] == 0:
break
elif j == len(p) - 1:
p.append(c)
j += 1
print(p)
Its simple. Check the below code, am sure it works!
N = int(input('Enter the number: ')
i=1
count=0
while(count<N):
for x in range(i,i+1):
c=0
for y in range(1,x+1):
if(x%y==0):
c=c+1
if(c==2):
print(x)
count=count+1
i=i+1
The following code will give you prime numbers between 3 to N, where N is the input from user:
number = int(input("Prime numbers between 2, 3 and "))
for i in range(2,number):
for j in range(2,int(i/2)+1):
if i%j==0:
break
elif j==int(i/2):
print(i)
You can see to check a number i to be prime you only have to check its divisibility with numbers till n/2.
I'm a beginning programmer in python and have a question about my code I'm writing:
number = int(input("Enter a random number: "))
for num in range(1, number + 1) :
for i in range(2, num) :
if (num % i) == 0 :
break
else :
print(num)
break
When I run this program I also get 9, 15 21 as output. But these are not prime numbers.
What is wrong with my code?
Thanks!
With if (num % i) == 0: you go to the else block for every num, which is not a multiply of 2, as you start i with 2, thus printing num. I got all odd numbers printed with your code.
You can use something like this:
number = int(input("Enter a random number: "))
for num in range(1, number + 1):
prime = True
for i in range(2, num):
if (num % i) == 0:
prime = False
break
if prime:
print(num)
It sets prime to False when it encounters a divisor without rest.
the trouble is in your else: statement which has to run outside of the first loop. I recommend the following:
def isprime(num):
for i in range(2, num):
if (num % i) == 0:
return False
return True
for num in range(1, number + 1) :
if isprime(num):
print num
Your problem
The else statement that you put inside the for loop means that if that number(num) is divisible by this current no represented by i, return true considering it as a prime which IS NOT CORRECT.
Suggestion
Your outer loop starts with 1 which should change to 2, as 1 is not a prime number
Solution
def fun(number):
#Not counting 1 in the sequence as its not prime
for num in range(2, number + 1) :
isPrime = True
for i in range(2, num) :
if (num % i) == 0 :
isPrime = False
break
if isPrime:
print num
fun(10)
Output
1
2
3
5
7
I am assuming the random number is the range you want the numbers to be within.
I found that the variable i is always equal to 2 in your code.This destroys the purpose of having a second for loop
Prime numbers are numbers that cannot be divisible by 2, 3 or 7, excluding 2, 3 or 7!
With this knowledge I adapted your code to show the true prime numbers.
solution
number = int(input("Enter a random number: "))
for num in range(2,number +1) :
printnum = True
if num == 2:
printnum = True
elif num == 3:
printnum = True
elif num == 7:
printnum = True
elif num % 2 == 0:
printnum = False
elif num % 3 == 0:
printnum = False
elif num % 7 == 0:
printnum = False
if printnum == True:
print(num)
This is my method to generate first ten prime numbers with no imports, just simple code with components we are learning on college. I save those numbers into list.
def main():
listaPrim = []
broj = 0
while len(listaPrim) != 10:
broj+= 1
brojac = 0
for i in range(1,100):
if broj % i == 0:
brojac += 1
if brojac == 2:
listaPrim.append(broj)
print(listaPrim)
if __name__=='__main__':
main()
Problem
Here is the problem I am trying to solve.
The prime factors of 13195 are 5, 7, 13, 29. What is the largest prime factor of the number 600851475143?
def prime_calc():
num = raw_input("What is the number you want the primes for?")
prim_num = []
x = 2
while num/x > 1:
new_num = num / x
if num % x == 0:
return prim_num.append(x)
elif num % x != 0:
new_num = num/x += 1
return prim_num.append(x)
else:
break
I keep getting an invalid syntax error starting from the bottom up to the fourth line that doesnt like my "+=" operator
This line:
new_num = num/x += 1
should be broken into two lines:
x += 1
new_num = num/x
The statement x += 1 in python doesn't return anything, so you can't use it as part of an expression.
Similarly, the two instances of:
return prim_num.append(x)
Also won't work, since the statement: prime_num.append(x) doesn't return anything.
You need to break this into:
prime_num.append(x)
return prime_num
num=int(input("Please enter number to calculate prime factor"))
k=0
item=[]
if(num%2==0):
prime=2
item.insert(k,prime)
k=k+1
j=3
flag=int(num/2)
while(j<flag and num>2):
if num%j==0:
prime=j
item.insert(k,prime)
k=k+1
num=num/j
j=j+2
else:
j=j+2
if k==0:
print("sorry no prime factor for this number")
else:
print("Please find the largest prime factor below")
print(item.pop())
I wrote a code in python to find the nth prime number.
print("Finds the nth prime number")
def prime(n):
primes = 1
num = 2
while primes <= n:
mod = 1
while mod < (num - 1):
ptrue = 'true'
if num%(num-mod) == 0:
ptrue = 'false'
break
mod += 1
if ptrue == 'true':
primes += 1
return(num)
nth = int(input("Enter the value of n: "))
print(prime(nth)
The code looked fine to me, but it returns an error when I run it:
Traceback (most recent call last):
File "C:/Users/AV/Documents/Python/nth Prime.py", line 17, in <module>
print(prime(nth))
File "C:/Users/AV/Documents/Python/nth Prime.py", line 13, in prime
if ptrue == 'true':
UnboundLocalError: local variable 'ptrue' referenced before assignment
It appears to me as if it is trying to say that I am referring to ptrue in the last line even though I am not. What is the problem here... Can anyone help?
how about using Boolean ? and initalize ptrue out of while loop
print("Finds the nth prime number")
def prime(n):
primes = 1
num = 2
while primes <= n:
mod = 1
ptrue = True
while mod < (num - 1):
if num%(num-mod) == 0:
ptrue = False
break
mod += 1
if ptrue == True:
primes += 1
return(num)
nth = int(input("Enter the value of n: "))
print prime(nth)
ptrue is local to your while loop which goes out of scope as soon as the while loop ends. so declare ptrue before the start of your inner while loop
Get rid of ptrue entirely and use else with your inner loop. For example:
while mod < (num - 1):
if num % (num - mod) == 0:
break
mod += 1
else:
primes += 1 # only executes if loop terminates normally, without `break`
You can try this:
#This program finds nth prime number
import math
def is_prime(number):
if number < 2:
return False
if number % 2 == 0:
return False
else:
for i in range(3, number):
if not number % i:
return False
return True
n = input('Enter n: ')
#This array stores all the prime numbers found till n
primes = []
for i in range(100000):
if is_prime(i):
primes.append(i)
if len(primes) == n:
break
print("nth prime number is: " + str(primes[n-1]))
The first part is define a function that calculates the next prime number given any number.
import math
def is_prime(x): # function
for i in range(2,int(math.sqrt(x))+1):
if x%i == 0:
return is_prime(x+1)
return x
For example, is_prime(10) will return 11.
The next step is to write a generator that returns a list of prime numbers.
def get_prime(k): # generator
cnt = 1
n = 2
while cnt <= k:
yield(is_prime(n))
n = is_prime(n) + 1
cnt += 1
For example, get_prime(5) will return [2,3,5,7,11].
The code below can help you test the results.
a = get_prime(50)
lists = list(a)[:]
for idx, value in enumerate(lists):
print("The {idx}th value of prime is {value}.".format(idx = idx+1, value = value))
All answers depends on user input, but here is a simple code to give nth number, no matter how big the n is ....
def isprime(n): # First the primality test
if n<2:
return False
for i in range(2,n):
if n%i==0:
return False
break
else:
return True
def nthprime(n): # then generic code for nth prime number
x=[]
j=2
while len(x)<n:
if (isprime(j)) == True:
x.append(j)
j =j+1
print(x[n-1])
n=int(input('enter n'))
a=[2,3,5,7]
i=3
j=9
while i<n:
flag=0
j=j+2
for k in range(len(a)):
if (a[k]<=int(j**0.5) and j%a[k]==0):
flag=1
break
if flag==0:
a=a+[j]
i=i+1
print(a[n-1])
Try this.
n = int(input())
count=1
u=2
prime=[]
while(count<=n):
temp=0
for i in range(2,u):
if(u%i==0):
temp=1
if(temp==0):
count+=1
prime.append(u)
u+=1
print(prime[-1])
Program to find nth Prime Number.
def nth_Prime(num):
Semi = num*num
Res_1 = [True for i in range(Semi+1)]
prime = 2
while prime*prime <= Semi:
if Res_1[prime] == True:
for i in range(prime*prime, Semi+1, prime):
Res_1[i] = False
prime += 1
Res_2 = []
for i in range(2, Semi+1):
if Res_1[i]:
Res_2.append(i)
return Res_2[num-1]
if __name__ == "__main__":
num = int(input("Enter nth Number: "))
print(nth_Prime(num))
Try this out ,I just made few changes in yours.
Here I am checking for each prime number using all(num%i!=0 for i in range(2,num)) checking its remainder not equal to zero so if it is true for that range (starting from 2 and less than itself) it is a prime and for that all() function helps me later if its a prime I increment the 'p' count and check till 'p' is less than the 'n'(Input Number) so when it equates the condition its the nth prime we are looking for.
n=raw_input("enter the nth prime ")
num=4
p=2
while p <int(n):
if all(num%i!=0 for i in range(2,num)):
p=p+1
num=num+1
print "nTH prime number: ",num-1