I'm reading a book about python programming for begginers. One of it's tasks is to write a prime number calcutator that calculates 'n' prime numbers. So far I've studied strings, logic gates, while and conditions.
The idea is to make it using only those operators.
I need help because I'm stuck with this code.
Here's what I've done:
odd = 3
number = 2
limit = int(input('How many primes do you need: '))
remnant = number % odd
even_remnant = number % 2
counter = 0
while counter <= limit:
if number == 2:
print('2')
number = 2
elif (number % 2) != 0:
remnant = number % odd
while odd < number:
print('while2')
remnant = number % odd
if (number % odd) != 0 and odd == (number - 1):
print(f'{number}.')
odd = 3
number = 1
counter = 1
break
elif (number % odd) == 0:
break
odd = 2
elif (number % 2) == 0:
number = 1
odd = 3
What do you think?
Thanks everyone.
CodePudding user response：
Put your debugging pants on, we're going in.
First, the code doesn't run as it's written. The variables counter and impar are undefined. First step is to remove syntax errors like that. Looks like we want to start counter at 0 and the line that uses impar isn't necessary so we can delete it.
odd = 3
number = 2
limit = int(input('How many primes do you need: '))
counter = 0
while counter < limit:
if number == 2:
print('2')
number = 2
elif (number % 2) != 0:
while odd < number:
print('while2')
remnant = number % odd
if (number % odd) != 0 and odd == (number - 1):
print(f'{number}.')
odd = 3
number = 1
break
elif (number % odd) == 0:
break
odd = 2
elif (number % 2) == 0:
number = 1
odd = 3
Now the code runs without error, but all it does is print
2
while2
And then fails to terminate.
So we know we enter the while odd < number loop only once and we don't print anything during that loop. If we also print the value of odd and number while we are in there we see odd = 3 and number = 5. Neither of the if conditions are met and the odd = 2 line is hit. Now odd = 5 and the while loop exits without printing 5 even though 5 is prime. If we want to hit our print statement by meeting the condition odd == (number - 1) we better go in steps of 1 when incrementing odd. Let's change to odd = 1 and re-run the code.
Now when I say I need 2 primes it prints
2
5
7
And then prints while2 forever. At least it prints prime numbers! But it skipped 3 and printed too many, and I had to use Ctrl-C to quit the program. Too many primes were printed because the outer loop while counter <= limit: condition was never reached. Inside the loop, we never increase the value of counter. Whenever we print a prime, we need to increase counter.
Also, to make sure we print 3, take a look at the first if condition in the loop.
if number == 2:
print('2')
number = 2 # Oops, we skipped over 3
Let's update this:
if number == 2:
print('2')
print('3')
counter = 2 # Let's count both of these!
number = 2
Also adding counter = 1 after the other print, re-running the code we get
How many primes do you need: 2
2
3
5.
How many primes do you need: 3
2
3
5.
7.
Oops, we are getting one more than we need. This is because when counter == limit we run the while loop one more time. Let's change our while loop condition to while counter < limit:. That change gets us just the right number of primes.
How many primes do you need: 4
2
3
5.
7.
But if we ask for 5
How many primes do you need: 5
2
3
5.
7.
And the program never exits. If we check the values of odd and number, we see that the loops is running with odd=3 and number=9 over and over again.
Reason through the code when odd=3 and number=9. We break out of the while odd < number while loop when we hit this code
elif (number % odd) == 0
break
But we never increase the value of number, so it is still equal to 9 the next time through the loop. Let's update this to
elif (number % odd) == 0
number = 1
break
Now when we re-run the code we get
How many primes do you need: 5
2
3
5.
7.
11.
Huzzah! And it works when asking for more primes as well. Here is the code as it is currently:
odd = 3
number = 2
limit = int(input('How many primes do you need: '))
counter = 0
while counter < limit:
if number == 2:
print('2')
print('3')
counter = 2
number = 2
elif (number % 2) != 0:
while odd < number:
if (number % odd) != 0 and odd == (number - 1):
print(f'{number}.')
odd = 3
number = 1
break
elif (number % odd) == 0:
number = 1
break
odd = 1
elif (number % 2) == 0:
number = 1
odd = 3
Now that we have working code, let's improve it! One of our bugs was that we forgot to increase number by 1 in one case. Notice that no matter how we exit the outer while loop while counter <= limit: we want to increment number. So, instead of doing it in many places, let's move all of those to the end of the while block.
We also set odd=3 whenever exiting the while block. What we want to ensure is that odd=3 at the start of the while loop, so let's move that there. Now there is no more code in the elif (number % 2) == 0: block, so we can remove that line.
number = 2
limit = int(input('How many primes do you need: '))
counter = 0
while counter < limit:
odd = 3
if number == 2:
print('2')
print('3')
counter = 2
elif (number % 2) != 0:
while odd < number:
if (number % odd) != 0 and odd == (number - 1):
print(f'{number}.')
counter = 1
break
elif (number % odd) == 0:
break
odd = 1
number = 1
I think the code is more clear if the while loop ends when the condition is met, rather than on break statements. We want the while loop to end if we find the number is divisible by something, or we run out of numbers to check.
`while number % odd != 0 and odd < number:`
And the only thing we need to do in the while loop is increment odd. Then after the loop, we can check the value of odd to see which condition was met.
number = 2
limit = int(input('How many primes do you need: '))
counter = 0
while counter < limit:
odd = 3
if number == 2:
print('2')
print('3')
counter = 2
elif (number % 2) != 0:
while number % odd != 0 and odd < number:
odd = 1
if odd == number: # No divisor was found!
print(f'{number}.')
counter = 1
number = 1
Notice that we are "hard coding" the divisibility by 2 (number % 2) != 0 and then using the variable odd to check divisibility by everything else. If we start odd at 2 instead of 3, we don't have to do the hard coding.
number = 2
limit = int(input('How many primes do you need: '))
counter = 0
while counter < limit:
odd = 2
if number == 2:
print('2')
print('3')
counter = 2
while number % odd != 0 and odd < number:
odd = 1
if odd == number: # No divisor was found!
print(f'{number}.')
counter = 1
number = 1
When we make this change, we also notice that we find the primes 2 and 3 twice, so we can remove the hard coded version of those:
number = 2
limit = int(input('How many primes do you need: '))
counter = 0
while counter < limit:
odd = 2
while number % odd != 0 and odd < number:
odd = 1
if odd == number: # No divisor was found!
print(f'{number}.')
counter = 1
number = 1
CodePudding user response：
When I try to run this code it tells me that Counter is not set, so right before entering the loop set Counter to 0.
Another problem is that you start by finding 2 in the first case of your loop, this is nice. Here after the loop runs again, now with number set to 4. Because of your = 2 instruction.
It then runs the last elif case. where (number % 2) == 0. here it set number = 5, and odd = 3. But it doesn't print 3. I think you mean to do this.
Now it runs the loop again, and enter the second elif case (number % 2) != 0. The first line in the elif clause the variable impar is not defined so it will fail.
CodePudding user response：
To calculate 'n' number for prime numbers you needn't use so many statements, if you make use of the the arithmetic and logical or bit-wise operators, which you will be learning in the future chapters of the python book you're referring.
I shall help you by editing the code for you.
number = int(input("Enter range: "))
print("Prime numbers:", end=' ')
for n in range(2, number):
for i in range(2, n):
if n % i == 0:
break
else:
print(n, end=' ')