Develop Reference

Why is my Python prime number checker not working?

Why is my code below giving me wrong results? When I give it numbers like 2 and 3, it says they're not prime numbers. Only some numbers work but for the most part it gives wrong answers.
def prompt_input(input_msg, error_msg):
while True:
userinput = input(input_msg)
try:
integer = int(userinput)
if integer > 1:
return integer
print(error_msg)
except ValueError:
print(error_msg)
def check_prime(number):
for i in range(2, number):
if number % i == 0:
return False
return True
primenum = prompt_input(
"Give an integer that's bigger than 1: ",
"You had one job"
)
if check_prime(primenum):
print("This is a prime.")
else:
print("This is not a prime.")

First issue: Your for loop returns after one iteration, so the correct return logic would be:
def check_prime(number):
for i in range(2, number):
if number % i == 0:
return False
return True
However you are implementing this incorrectly as well, the loop needs to iterate up to the sqrt of the number.
def check_prime(number):
for i in range(2, int(sqrt(number))+1):
if number % i == 0:
return False
return True
Hope this helps!

Need to unindent the return True.
def check_prime(number):
for i in range(2, number):
if number % i == 0:
return False
return True

Python Perfect Number

In this Python Question, I should get False if the the number is not perfect. instead, I'm getting "None". What should I change?
def perfect(number):
sum = 0
is_perfect = False
if number < 0:
return is_perfect
for i in range(1, number):
if(number % i == 0):
sum = sum + i
if (sum == number):
is_perfect = True
return is_perfect
print(perfect(8))

You should return False at the End otherwise None is getting returned implicitly:
def perfect(number):
...
for i in range(1, number):
...
return False

Hi your problem is that you have no return when you don't match if (sum == number):..
so, the simplest solution is to add an return False at the end. But also your sum condition is inside the for that's not fine because you can encounter a fake perfect
like 24... so be carefull about that condition
edit 0 is not a perfect number so add the condition to first if
def perfect(number):
sum = 0
is_perfect = False
if number <= 0:
return is_perfect
for i in range(1, number):
if(number % i == 0):
sum = sum + i
if (sum == number):
is_perfect = True
return is_perfect
return False
I would totally reshape your code to make it a litle more clear:
def is_perfect(number):
acum= 0
if number <= 0:
return False
for i in range(1, number):
if(number % i == 0):
acum += i
return (acum == number) #return true if the condition match, False if not
#just to check function
print([(i,is_perfect(i)) for i in range(1,100) if is_perfect(i)])
#output
[(6, True), (28, True)]

why doesn't memoization work in this Python code?

i am following a dynamic programming video. However the memoization for my code is not working. It is not storing any True/False when i print(memo) it is blank. Please advise
def cansum(targetsum,numbers):
memo = dict()
print(memo)
# check in memory
if targetsum in memo:
return memo[targetsum]
if targetsum < 0: return False
if targetsum == 0: return True
for number in numbers:
remainder = targetsum - number
if cansum(remainder,numbers) == True:
memo[targetsum] = True
return True
memo[targetsum] = False
return False
print(cansum(7,[2,3])) #True
print(cansum(7,[5,3,4,7])) #True

I think this is what you might want to do:
def cansum(targetsum, numbers):
memo = dict()
def cansum_helper(targetsum, numbers):
# check in memory
if targetsum in memo:
return memo[targetsum]
if targetsum < 0:
return False
if targetsum == 0:
return True
for number in numbers:
remainder = targetsum - number
if cansum_helper(remainder, numbers) == True:
memo[targetsum] = True
return True
memo[targetsum] = False
return False
result = cansum_helper(targetsum, numbers)
print(memo)
return result
print(cansum(7, [2, 3])) # True
print(cansum(7, [5, 3, 4, 7])) # True
If you put
memo = dict()
into recursive function, you are creating one dict for every recursive function, and once the memo is set, the return statement follows, so you won't be able to see changes. But what is intended is that you only need one dict for your whole function.

Python - function returns true when a given number is a prime number or else false

Hi I'm a beginner and I'm stuck on this question that wants me to use only while loop to solve. The question wants me to write a function that returns True when the given number is a prime number and it returns False if the given number is not a prime number.
My code so far:
def is_prime(n):
i = 2
while i <= n//2:
if n%i != 0:
return True
else:
return False
i+=1
The problem I have is I think my code displays the correct output for numbers 4 and above and it returns 'None' for 1, 2, and 3. I've debugged it and I think the problem is the while loop condition. But I don't know how to fix it. I would appreciate it if any of you pros can help me out!
edit:
I changed the while condition but 1 still returns None.. and 2 returns False when it's supposed to return True
def is_prime(n):
i = 2
while i <= n:
if n%i != 0:
return True
else:
return False
i+=1

import math;
def is_prime(n):
i = 2
while i < max(math.sqrt(n),2):
if n%i != 0:
return True
else:
return False
if i == 2:
i+=1
else
i+=2

You could hard-code these 3 cases, in case you dont want to use sqrt:
def is_prime(n):
i = 2
if n in (1,3):
return True
elif n == 2:
return False
while i <= n//2:
if n%i != 0:
return True
else:
return False
i+=1
for x in range(1, 5):
print(x, '=', is_prime(x))
Output:
(1, '=', True)
(2, '=', False)
(3, '=', True)
(4, '=', False)

Want to get really fancy? Make a Sieve of Eratosthenes:
def is_prime(n):
a = list()
# Assume all are prime
a[0:n+1] = (n+1)*[1]
# Start with removing even numbers
i = 2
while i*i <= n:
print ("I: ", i)
# Set all divisible by i to 0
a[0:n+1:i] = len(a[0:n+1:i])*[0]
# If a[n] is zero, return False
if a[n] == 0:
return False
# Increment i until we have a prime number
while a[i] == 0:
i+=1
if a[n] == 0:
return False
else:
return True

If you want to impress your lesson teacher you can show him a fast probabilistic prime number isprime for numbers larger than 2**50. I haven't found any errors in it after weeks of cpu time stress testing it on a 6 core AMD:
import random
import math
def lars_last_modulus_powers_of_two(hm):
return math.gcd(hm, 1<<hm.bit_length())
def fast_probabilistic_isprime(hm):
if hm < 2**50:
return "This is to only be used on numbers greater than 2**50"
if lars_last_modulus_powers_of_two(hm+hm) != 2:
return False
if pow(2, hm-1, hm) == 1:
return True
else:
return False
def fast_probabilistic_next_prime(hm):
if hm < 2**50:
return "This is to only be used on numbers greater than 2**50"
if hm % 2 == 0:
hm = hm + 1
hm += 2
while fast_probabilistic_isprime(hm) == False:
hm += 2
return hm
""" hm here is bitlength, which must be larger than 50.
usage is create_probabilistic_prime(1000)
"""
def create_probabilistic_prime(hm):
if 2**hm < 2**50:
return "This is to only be used on numbers greater than 2**50"
num = random.randint(2**hm,2**(hm+1))
return fast_probabilistic_next_prime(num)

'None' is printed when I want 'True' for input 5

In this primality test program, 'None' is printed when I input a prime, instead of 'True'. How can I get it to print 'True'?.
def main():
import math
def check_n(n):
n_s = int(math.sqrt(n))
for i in range(2, n_s):
if (n_s % i) == 0:
return False
break
else:
return True
def msg():
n = int(input('Enter a number, I will return True if it is a prime'))
return n
print(check_n(msg()))
main()

You need to change int(math.sqrt(n)) to int(math.sqrt(n)+1), because range runs until n_s-1. So if the input is 5, range(2,int(math.sqrt(5))) is just range(2,2), which is empty.
In addition, you need to take the return True outside of the for loop, otherwise your code may stop in a too early stage. You also don't need the break statement after return False (the function will never arrive to that line, as it will return False if it enters to that if statement).
Finally, change if (n_s % i) == 0: to if (n % i) == 0:, as you need to check if n is divisible by i (and not its square root).
Here is a more clean version:
import math
def check_n(n):
n_s = int(math.sqrt(n)+1)
for i in range(2, n_s):
if (n % i) == 0:
return False
return True
def msg():
n = int(input('Enter a number, I will return True if it is a prime'))
return n
print(check_n(msg()))

First: Your break statement is redundant.
Second: For values such as 3 the for loop is never executing because value
n_s is less than 2 and since the for loop isn't executing the
python is returning the default value None(which is returned when
no value is specified).
Hence your check_n(n) function has to be
def check_n(n):
n_s = int(math.sqrt(n))
for i in range(2, n_s + 1):
if (n_s % i) == 0:
return False
return True

one liner :
check_n = lambda n : sum([i for i in range(2, int(math.sqrt(n)+1)) if n % i == 0]) == 0
don't overcomplicate things ..

Your range is (2,2) or None when you choose anything less than 9.
So to solve your first problem: add 2 to n_s (for input 3)
You also have a problem with your logic.
Your for loop should be checking that n mod i is 0, not n_s.
This should work:
def main():
import math
def check_n(n):
n_s = int(math.sqrt(n)+1)
for i in range(2, n_s):
if (n % i) == 0:
return False
return True
def msg():
n = int(input('Enter a number, I will return True if it is a prime'))
return n
print(check_n(msg()))
main()