I have to define a function called is_prime that takes a number x as input, then for each number n from 2 to x - 1, test if x is evenly divisible by n.
If it is, return False. If none of them are, then return True.
The system I'm using (codecademy) has given my code the error message "Oops, try again. Does your is_prime function take exactly one argument (an integer)? Your code threw a "unsupported operand type(s) for %: 'int' and 'list'" error."
Please could someone to fix my code with an explanation of how my code is wrong?
def is_prime(x):
n = range(2, x-1)
if x % n == 0:
return False
else:
return True
You need to loop over the range:
def is_prime(x):
if x < 2:
return False
for i in range(2,x):
if x % i == 0:
return False
return True
if x % i == 0 never evaluates to True, you have a prime so you return True outside the loop after you have checked each i. You could simply return a gen exp:
def is_prime(x):
return x > 1 and not any(x % i == 0 for i in range(2, x))
That's not how you use range.
You want
def is_prime(x):
for n in range(2, x-1):
if x % n == 0:
return False
return True
Range returns a list, so your original code was checking to see if the list object was evenly divisible by n, instead of checking each element of the list.
Related
Hey so i have this function to check if a number is a prime number
def is_prime(n):
flag = True
for i in range(2, n ):
if (n % i) == 0:
flag = False
return flag
print(is_prime(1))
However when i test the number 1, it skips the for loop and returns True which isn't correct because 1 is not a prime number.
How could i fix this?
You can first start by checking if n is greater than 1 the code should proceed, else it should return False. If n passes the first condition, only then the code can proceed to verify if n is indeed prime or not.
def is_prime(n):
flag = True
if n > 1:
for i in range(2, n ):
if (n % i) == 0:
flag = False
return flag # Returns this flag after check whether n is prime or not
# Returns False if n <= 1
return False
print(is_prime(1))
output:
False
2 is also skipped by the loop, but the function returns true thanks to the flag, and we know that's right.
Also you can exit early the loop if the condition is met:
def is_prime(n: int) -> bool:
if n > 1:
for i in range(2, n): # if n == 2, there is no loop, is never checked
if (n % i) == 0:
return False # can return early once we meet the condition, don't need to finish the loop
return True
print(is_prime(7534322224))
print(is_prime(5))
An alternative approach (though much slower on bigger numbers):
def is_prime(n: int) -> bool:
if n < 2: return False
return n == 2 or True not in [True for i in range(2, n) if (n % i) == 0]
print(is_prime(75343224))
print(is_prime(5))
I am trying to create a function that will test whether a number is a prime number and then return True or False. I am an extreme beginner at Python, so please make your code as simple as possible. Here's what I tried so far (only returns True):
def isPrime(x):
x = int(x)
for i in range(2, x):
if(x % i == 0):
x == False
else:
x == True
return x
print(isPrime(input("Enter a prime number.")))
You almost made it, just need to change it like this:
def isPrime(x):
x = int(x)
for i in range(2, x):
if x % i == 0:
return False
return x >= 2
print(isPrime(input("Enter a prime number.")))
EDIT: As pointed by #MarkRansom, when the i variable reaches the square root of x, it is safe to assume that there are no more possible divisors for it:
from math import sqrt
def isPrime(x):
x = int(x)
for i in range(2, int(sqrt(x)) + 1):
...
I believe the simplest form of your program would be along the lines:
def isPrime(x):
if x > 1:
for i in range(2, x):
if x % i == 0:
return False
return True
return False
print(isPrime(int(input("Enter a number: "))))
But this is not as efficient as it could be. To do better, we would treat 2 (and all even numbers) as a special case and only allow the divisor to cover the odd numbers from 3 to the square root of the number we're testing. The more divisions we can avoid, the faster it goes.
Ultimately using a Sieve of Eratosthenes will beat this approach, even with optimizations.
When you typed this, I believe you meant as follows below:
def isPrime(x):
x = int(x)
for i in range(2, x):
if (x % i) == 0:
x = False
else:
x = True
return x
print(isPrime(input("Enter a prime number.")))
## Which runs as follows in python IDLE:
>>> def isPrime(x):
x = int(x)
for i in range(2, x):
if (x % i) == 0:
x = False
else:
x = True
return x
>>> print(isPrime(input("Enter a prime number.")))
Enter a prime number.11
True
>>> print(isPrime(input("Enter a prime number.")))
Enter a prime number.100
False
I know when I started out, I made many typing errors.
By rereading my code and lots of practice, I slowly got better.
Here's a how I would write out your task.
def isPrime(h):
h = int(h)
test = None
for x in range(2, h):
if (h%x) == 0:
test = False
print(f"{test}, the number {h} is not a prime!")
break
else:
test = True
print(f"{test}ly, the number {h} is a prime!")
## Then I'll run the call to the function in python IDLE as:
>>>
>>> isPrime(input("Enter a number.\n"))
Enter a number.
11
Truely, the number 11 is a prime!
>>>
>>> isPrime(input("Enter a number.\n"))
Enter a number.
110
False, the number 110 is not a prime!
>>>
One of most important things, that really helped me was:
reading others code (and the docs!!)
then play with their code in IDLE
repeat the above <--------(I still do this all the time!!)
I have a little problem when i need to determine if a number , x is a prime number or not.
x will be a randomly generated positive integer and i get the following message when i execute code:
Your function fails on is_prime(2). It returns None when it should return True.
My code:
def is_prime(x):
if x < 2:
return False
else:
for n in range(2, x):
while x % n == 0:
return False
break
else:
return True
i want the while loop to iterate through n == 2 up to n == (x-1) but it doesnt seem to do it!
what am i doing wrong?
You could write this function much more simplier:
import math
def is_prime(n):
if n % 2 == 0 and n > 2:
return False
for i in range(3, int(math.sqrt(n)) + 1, 2):
if n % i == 0:
return False
return True
So your call for is_prime(2) wouldn't pass the if test and wouldn't be used in the for run and would just return True.
All other prime numbers should be tested if they are greater than 3 and not even.
I have a bit of a problem. I am writing an is_prime function on, but whenever I run it, it fails on is_prime(9), and I cannot see why:
def is_prime(x):
if x < 2: ##because negative numbers, 0 and 1 are not prime##
return False
elif x == 2:
return True
else:
for n in range(2, x):
if x % n == 0:
return False
else:
return True
it returns True for some reason on is_prime(9)?
That is because the function does not check all eligible divisors until it returns.
Instead, it exits early with True if x is not divisible by 2, which is not what you want for odd numbers (e.g. 9 is not divisible by 2, yet it's not prime).
Instead, you want to try all possible divisors from 2 to x-1, and then return if x is divisible by none of them.
To do so, rewrite as such:
def is_prime(x):
if x < 2: ##because negative numbers, 0 and 1 are not prime##
return False
elif x == 2:
return True
else:
for n in range(2, x):
if x % n == 0:
return False
return True
Below is my code to find the prime number using Python which is not working. Here the function prime will take an integer as input and return whether its a prime number or not. Could you please sort out the problem and explain it.
def prime(x):
if x == 0 or 1:
return False
elif x == 2:
return True
else:
for n in range(2, x):
if x % n == 0:
return False
else:
return True
I think i have sorted out the first issue, the first "if" statement should be if x == 0 or x == 1. Now what about the rest.
What does your for loop?
if x % n == 0:
return False
else:
return True
which by the way eqals return bool(x % n)
So, you return in first iteration, when n == 2.
The whole for loop equals return bool(x % 2), which simply checks if x is diviseable by 2.
That's not what you want.
So, what do you want?
You want to check if x is not diviseable by any numer from range(2, x).
You know that x is not prime, if you find one n from range(2, x), for which x % n == 0 is True.
You know that x is prime, when there is no n in range(2, x), for which x % n == 0 is True.
When can you say that none of n from range is a divisor of x?
After checking all ns from range!
After is the key here.
After the loop, in which you try to find divisor, you can only tell that x is prime.
I hope you now understand the code others posted without explanation.
Note: alternative syntax
Code others posted is correct. However, in Python there is second way writing the for, using for .. else:
for x in range(2, x):
if x % n == 0:
return False
else:
return True
The problem is that the return true should not happen until the for loop has completed.
what we have in the original code is a cuple of tests for trivial cases (x less than 3)
and then a loop for testing all the larger numbers.
In the loop an attempt is made to divide x by a smaller number (starting with 2) and then if it divides evenly False is returned, if it does not True is returned, that is the mistake, instead of returning true, the loop should be allowed to repeat, and division should be tried again with the next number, and only after the supply of divisors (from the for loop) has been exhausted should true be returned.
here's a fixed version:
def prime(x):
if x <= 1:
return False
elif x == 2:
return True
else:
for n in range(2, x):
if x % n == 0:
return False
return True
Others have commented that the loop need not continue all the way up to x and that stopping at sqrt(x) is sufficient, they are correct. doing that will make it faster in almost all cases.
Another speed up can be had if you have a list of small primes (upto sqrt(x)) - you need only test divisibility by the primes below sqrt(x),and not every integer in that range.
The below code is for finding the prime number from 2 to nth number.
For an example, the below code will print the prime number from 2 to 50 and also it will print the number in between 2 to 5o which is not prime.
import time
i=2
j=2
count=0
while(i<50):
while (i>j):
if (i%j)==0:
count=count+1
j=j+1
else:
j=j+1
if count==0:
print i," is a prime"
else:
print i," is not a prime"
i=i+1
j=2
count=0
time.sleep(2)