def is_prime(x):
x = abs(x)
if x == 0 or x == 1:
return False
elif x == 2:
return True
for i in range(2, x):
if x%i == 0:
return False
return True
The tester says the above code shouldn't return true for x = -7.
Please help.
Thank you :)
**Edit: **
im so sorry, just read somewhere that any number less than 2 (i.e 0, 1 and all negative numbers are not prime
It does return True for me. Check the value of x after you call abs(x), it will maybe help.
It does return True for negative numbers for me.
Also you can run the loop from 2 to x/2 and reduce the complexity.
My Python is not good, but something like this should work better:
def is_prime(x):
# Handle 1, 0 and negatives.
if x < 2:
return False
# Handle even numbers.
elif x % 2 == 0:
return x == 2
# Test with odd divisors.
for i in range(3, sqrt(x), 2):
if x % i == 0:
return False
return True
The sqrt(x) limit is a big time saver, and dealing with even numbers separately can halve the time with well mixed inputs.
Related
For context, I am trying to solve Project Euler problem 3 using Python:
What is the largest prime factor of the number 600851475143?
As a first step to this, I am trying to write a function that returns whether or not a number is prime as a Boolean. I made a first attempt, and checked out how this has been written previously. I have ended up with the following code:
def isprime(x):
limit = x**0.5
i = 2
if x < 2:
return False
elif x == 2:
return True
else:
while i <= limit:
if x%i == 0:
return False
i = i + 1
else:
return True
For some reason, the code above does not work perfectly. For example, isprime(99) would return True.
Please, can someone help me understand why this isn't working? I am trying to avoid just copying and pasting someone else's code, as I want to understand exactly what is going on here.
To me, it looks like the issue is with the final else statement. I say this because the logic reads "in the event that x%i == 0, this number is not prime" but it doesn't explicitly say what to do in the event that no x%i iterations == 0.
Any help on this would be appreciated! I'm not necessarily looking for the cleanest, neatest way of doing this, but more just trying to first make this code work.
Just to show an alternative, what you could do is checking from number 2 to your number if the operation (x % i) is equal to zero. If it never happend, it will be a prime.
def isprime(x):
# check for factors
for i in range(2,x):
if (x % i) == 0:
return False
else:
return True
print(isprime(99))
Try this :
def isprime(x):
limit = x**0.5
i = 2
if x <= 2:
return False
while i <= limit:
if x%i == 0:
return False
i = i + 1
return True
I've changed many things. have this point in your mind that there is no need to else clauses when you return at the end of if block.
you need to tell what happens when x%i==0 condition not met and the value of i remain constant and also need to see when all conditions not met, then it is a prime
# your code goes here
def isprime(x):
limit = x**0.5
i = 2
if x < 2:
return False
elif x == 2:
return True
else:
while i <= limit:
if x%i == 0:
return False
i+=1
return True
print(isprime(144)) # false
print(isprime(99)) # false
print(isprime(131)) # true
I am new to programming and I face an issue while trying to write a program in finding out prime number. Here is my code:
def is_prime(x):
if x < 2:
return False
elif x == 2:
return True
else:
for n in range (2,x-1):
if x % n == 0:
return False
else:
return True
I received an error stating "Your function fails on is_prime(3). It returns None when it should return True."
Can someone please explain the flaw in this code?
Thank you!
range() has an exclusive upper bound, so it's trying to get the range between 2 and 2 (3 - 1), which is no elements. Since you can't iterate over nothing, the for loop never runs, so None is returned (this is the default return type of a function if none is specified).
The solution to your immediate problem would be to use range(2, x) rather than range(2, x - 1). You'll find that you'll have problems at x > 3 though because as #khelwood said, you're returning True or False immediately after checking the first value. Instead, only return True after checking all values in the range.
def is_prime(x):
if x < 2:
return False
elif x == 2:
return True
else:
for n in range (2,x): # range function will iterate till x-1
if x % n == 0:
return False
# return true only at the end after making sure it is not divisible by any number in the middle
return True
I started getting back to python coding and realized I couldn't quite figure this out. I'm trying to code a prime number function. Could someone help with this?
Here is my code:
def is_prime(x):
a = True
for n in range(2, x-1):
while n < x:
n+=1
if x % n == 0:
a = False
elif n < 2:
a = False
else:
a = True
break
break
return a
If anyone has an idea on what I'm doing wrong, please let me know. A month ago I tried this and couldn't get the logic down. I think I was stumped and didn't ever ask for help... Also, how long do you think I should try to do this for before I ask for help on average?
As, it has been said, you can optimize the code by just checking the odd numbers and iterating upto the sqrt of the num
import math
def isPrime(num):
if(num==1):
return False
if(num==2):
return True
if(num%2==0):
return False
i = 3
while(i<math.sqrt(num)+1):
if num%i==0:
return False
i += 2
return True
#do the inputs and check if isPrime
#print(isPrime(2))
using your code, and focusing on code structure:
def is_prime(x):
# function contents must be indented
if x == 0:
return False
elif x == 1:
return False
# your base cases need to check X, not n, and move them out of the loop
elif x == 2:
return True
for n in range(3, x-1):
if x % n == 0:
return False
# only return true once you've checked ALL the numbers(for loop done)
return True
adding some optimizations:
def is_prime(x):
if x <= 1:
return False
if x == 2:
return True
for n in range(3, x**(0.5)+1, 2): # this skips even numbers and only checks up to sqrt(x)
if x % n == 0:
return False
return True
def prime(number):
for i in range(2,number,1):
if number % i == 0:
return False
return True
entry = int(input("Please enter the number: "))
while True:
if prime(entry):
print ("It's a prime number. ")
continue
else:
print ("It's not a prime number.. ")
continue
You will entry a number then this function will give you if its a prime or not. Check the function it will solve your problem.
ALSO You can upgrade the speed of your program. You know prime numbers can not be even, so you you dont have to check even numbers
like 4-6-8-26. So in the range function, which is (2,number) add "2"
end of it. like (3,number,2) then program will not check even numbers.
ALSO a factor can not be bigger than that numbers square root. So you dont have to check all of numbers till your main number, its
enough to check your numbers square root. like: (2,number**0.5) that
means from 2 to your number square root. So double up for program
speed.
So the function will be:
def prime(number):
for i in range(3,(number**0.5)+1),2):
if number % i == 0:
return False
return True
The first function enough for you actually. I am working with huge numbers, so I have to upgrade the speed of my program.
def is_prime(x):
for n in range(2, x-1):
if n == 0:
return False
elif n == 1:
return False
elif n == 2:
return True
elif x % n == 0:
return False
else:
return True
What you are doing wrong is , the first three if elif blocks are executed every time in the loop. So, they should be outside the for loop. Also.
if x%n==0:
return True
It is fine. But the later
else:
return True
is wrong because whenever the first n does not divide the x, it will return Composite. So,
return True
block must be outside the for loop.
Other optimizations can be done by yourself.
This is how I solved it
def is_prime(n):
if n==1:
print("It's not a Prime number")
for z in range(2,int(n/2)):
if n%z==0:
print("It's not a Prime number")
break
else:
print("It's a prime number")
or if you want to return a boolean value
def is_prime(n):
if n==1:
return False
for z in range(2,int(n/2)):
if n%z==0:
return False
break
else:
return True
This was the question our teacher gave us:
"One way to determine whether or not a number is a prime number is as follows:
if the number < 2, then return False
if the number is 2, then return True
for each value of i, where i >=2 and i < number:
if i divides the number evenly, then return False
return True"
I've managed to work through most of it, but was stuck when it said 'for each value of i, where i >=2 and i < number:'
how do I write this in code?
if number < 2:
return False
if number == 2:
return True
?????????
if i%2 == 0:
return False
return True
Yo need a loop to check all numbers from 2 to one less than the number being checked. There are better ways to do it (such as only checking up to the square root of the number) but a simplistic algorithm would be:
def isPrime (n):
if n < 2:
return False
for x in range (2, n):
if n % x == 0:
return False
return True
So, in terms of what you need to add to your code:
a loop iterating some variable from two up to one less than the number.
checking modulo with that variable rather than the hard-coded 2.
You will need to start a loop from 2 to the number
for i in range(2,number)
if number%i == 0:
return false
def isprime(num):
#this is the part you are missing
for divider in range(2,num):
if num%divider == 0:
return False
#end of missing part
return not num%2 == 0 or num==2 or num==0
for i in range(0,200):
if isprime(i): print i, isprime(i)
So I'm relatively new to Python and trying to define a function which will check to see if a number is prime or not. The code is as follows:
def prime(x):
if x < 2:
return False
elif x == 2:
return True
else:
for i in range(3, int((x**0.5)+1)):
if x % i == 0:
return False
else:
return True
This seems to work for most values however it fails on certain values such as 25, can anyone help to explain to me why? Thanks!
return leaves your function once it is reached. Let's look at the case of 25.
Is x<2 no, so continue.
Is it x==2 no, so continue.
Is x divisible by i(=3)? No, so go to the else clause, return True leave the function.
See the issue?
To put it another way, for sufficiently large x, your function is equivalent to:
def prime_prime(x):
if x < 2:
return False
elif x == 2:
return True
else:
if x % 3 == 0:
return False
else:
return True
It's the
else:
clause, the first i it checks will make it return true.
Instead, write
else:
for i in range(3, int((x**0.5)+1)):
if x % i == 0:
return False
return True
EDIT:
And as #mata points out, you actually need to start at 2, so it should be for i in range(2, int((x**0.5)+1)):