for num in range(2,101):
for i in range(2,num):
if (num % i) == 0:
break
else :
print(num, end = ' ')
I don't understand why the output starts from 2
I thought num starts with 2, and the i also starts with 2
so, (num % i) == 0 is right
What's wrong with me?
When num == 2, range(2, num) is empty, because ranges stop before the end number.
So for i in range(2, num): never executes the body in that case, and therefore it never tests num % i == 0. The loop ends immediately, without executing break, so it goes into the else: block.
Related
integer = int(input("Enter an integer: "))
if integer >= 2:
print(2)
for i in range(2, integer + 1): # range 2,3,...,integer (excludes 1)
for j in range(2, i): # we are going to try dividing by these
if i % j == 0: # not prime
break
else: # is prime
print(i)
input:
7
output:
2
3
5
5
5
7
7
7
7
7
output I want:
2
3
5
7
adding more detail to get past error:
It looks like your post is mostly code; please add some more details.
You're printing i for every value of j that doesn't divide it, until you get a value that does divide it and you execute break.
You should only print i when you don't break out of the loop. Put the else: statement on the for loop, not if. This else: statement is executed when the loop finishes normally instead of breaking.
for i in range(2, integer + 1): # range 2,3,...,integer (excludes 1)
for j in range(2, i): # we are going to try dividing by these
if i % j == 0: # not prime
break
else: # is prime
print(i)
Put the print(i) in the outer for loop with a flag that keeps track of the result.
for i in range(2, integer + 1): # range 2,3,...,integer (excludes 1)
isPrime = True
for j in range(2, i): # we are going to try dividing by these
if i % j == 0:
isPrime = False
break
if isPrime:
print(i)
Your logic is slightly wrong. I have added the correct code below with comments :
integer = int(input("Enter an integer: "))
if integer >= 2:
print(2)
for i in range(2, integer + 1): # range 2,3,...,integer (excludes 1)
prime=true #assume that i is prime
for j in range(2, i): # we are going to try dividing by these
if i % j == 0: # not prime
prime=false # we now know it is not prime
break
if prime==true: # if we didn't do prime=0, then it's a prime
print(i)
What you were doing is printing i for every j from 2 to i that did not divide i. But instead, what had to be done is print i only once when none of the j from 2 to i divided i.
Hope you understand the mistake and this clears your doubt !
In summary, I know 2 is a prime number because it is divisible by 1 and itself but why does it still print 2 if 2%2 = 0 and the range is between 2 and the number you're checking, which in this case is 2. Shouldn't it start at 3?
for num in range (1, 1000):
if num > 1:
for i in range (2, num):
if (num % i ) == 0:
break
else:
print (num)
I expected 3 to be the first output.
2 is printed because for i in range(2, num) performs no iterations when num == 2. This is because the second argument to range() is not included in the list that range() generates. So for the 2 iteration of your outer loop, the equivalent iteration code is:
num = 2
if num > 1:
for i in []:
if (num % i ) == 0:
break
else:
print (num)
which will print 2 since the 'break' line won't be executed.
Python for loops include the first number, but exclude the last number. Because of this, the clause for i in range(2,2) won't actually return any numbers. This is why the number 2 gets printed in your else block.
It also helps to keep your if and else blocks at the same indent level to avoid confusing the computer.
I am about to execute a function which aim is to return a Prime/Not prime statement if its argument is or isn't a prime number. I succeeded using a for loop:
def prime1(n):
z = []
for i in range (1, n+1):
if (n/i).is_integer():
z.append(i)
i=i+1
if len(z) == 2:
print ("Prime")
else:
print ("Not prime")`
Then I tried to do the same but using the while loop:
def prime2(n):
z = []
i = 1
while i < int(len(range(1, n+1))):
if (n/i).is_integer():
z.append(i)
i=i+1
if len(z) == 2:
print ("Prime")
else:
print ("Not prime")
Unfortunately, my system continues to calculating without printing me an output.
Can you explain me where I have made a mistake?
The i = i + 1 does nothing in your for loop, since the value of i is overwritten with the next value of the iterator; effectively, the for loop is performing i = i + 1 for you on every iteration, whether or not i divides n. You need to do the same thing in your while loop:
while i < n + 1:
if (n/i).is_integer():
z.append(i)
i = i + 1
The most pythonic way I could think of is below:
def isPrime(n):
return all(n % i for i in range(2, int(n ** 0.5) + 1)) and n > 1
for i in range(1, 20):
print(isPrime(i))
Explanation:
all makes sure that every item in the given expression returns True
n % i returns True if n != 0 (even negative numbers are allowed)
int(n ** 0.5) is equivalent to sqrt(n) and as range always returns numbers up to n - 1 you must add 1
n > 1 makes sure that n is not 1
The problem in your code is that your i = i + 1 in the wrong scope
Your program checks if (n/i).is_integer() which returns False as n / 2 is not a integer
Improving your code:
Instead of (n/i).is_integer() you can use n % i == 0, which returns the remainder equals 0
Next you must place i = i + 1 in the outer scope
And personally, I was never a fan of i = i + 1. Use i += 1
I think the best way is using the code I have shown above.
Hope this helps!
Edit:
You can make it print 'Prime' or 'Not Prime' as follows:
def isPrime(n):
print('Prime' if all(n % i for i in range(2, int(n ** 0.5) + 1))
and n > 1 else 'Not Prime')
for i in range(1, 20):
isPrime(i)
You are increment the iterable variable i inside the if statement, so the variable is never incremented, stucking on an infinite loop.
When you used for it worked because the iterable changes itself after every full iteration of the block.
Moving the incremenf of i one identation block to the left (inside the while instead of for) will work just fine
Code adapted from https://www.programiz.com/python-programming/examples/prime-number
You don't need to check until num and you can go 2 by 2 if you don't have a database of prime numbers
import math
#num = 437
if num > 1:
# check for factors
for i in range(2, int(math.ceil(math.sqrt(num))), 2):
if (num % i) == 0:
print(num, "is not a prime number")
print(i, "times", num // i, "is", num)
break
else:
print(num, "is a prime number")
# if input number is less than
# or equal to 1, it is not prime
else:
print(num, "is not a prime number")
Our preferred method should not be while loops if we don't need to use them, that being said, we could use list comprehensions:
def prime(n):
z = []
[z.append(i) for i in range(1, n+1) if (n/i).is_integer()]
[print("Prime") if len(z) == 2 else print("Not Prime")]
prime(101)
But lets take a loop at what you got your for loop:
for i in range (1, n+1):
if (n/i).is_integer():
z.append(i)
i=i+1
The line i = i + doesn't serve the purpose you intend, the loop is going to iterate from 1 to n+1 no matter what
Now the while loop:
while i < int(len(range(1, n+1))):
if (n/i).is_integer():
z.append(i)
# i=i+1 does not go inside if statement
i += 1
Now you do need to increment i but if that incrementing only occurs when the if conditions are met, then if the if condition is not met you are going to be stuck at the same i looping over it.
Also try using i += 1 means the same thing
I have written a python code to find prime numbers between 2 and 30. But my code is not evaluating for 2 and 3. Can anyone tell me what is wrong in this code?
for i in range(2, 30):
for j in range(2, i-1):
if ((i % j) == 0):
print(i, "is not a prime number")
break
else:
print(i, "is a prime number")
break
The logic of your code is wrong. The else clause should be attached to the inner for loop, so it's only executed if the loop is exhausted without finding a divisor.
for i in range(2, 30):
for j in range(2, i-1):
if ((i % j) == 0):
print(i, "is not a prime number")
break
else:
print(i, "is a prime number")
Also note that the outer loop only runs up to 29, since the upper boundary is not included in a range. The inner loop does not include i - 1, but that's totatlly fine, since any non-trivial divisor is less than i - 1.
The inner loop won't be entered at all for 2 and 3, since the range will be empty in these cases. This is fine as well, since the else clause will be immediately entered.
for i in range(2, 30):
prime = True
for j in range(2, i-1):
if ((i % j) == 0):
prime = False
# print(i, "is not a prime number")
break
# else:
# print(i, "is a prime number")
# break
if prime:
print(i, "is a prime number")
else :
print(i, "is not a prime number")
There are lot of online link to solve the prime number problem. To improve yourself search yourself and understand the. Hope this and this link help you a lot. Happy codding
It isn't working because the nested for declaration:
for i in range(2, 30):
for j in range(2, i-1):
Uses range(2, i-1) and until i is 4, range returns no value:
i = 2 --> range(2, 1) # No value
i = 3 --> range(2, 2) # No value
i = 4 --> range(2, 3) # 2
This is because the range function returns values from the first parameter (included) to the second parameter (not included).
I am supposed to write a program that displays numbers from 100 to 200, ten per line, that are divisible by 5 or 6 but NOT both. This is my code so far. I know it's a basic problem so can you tell me the basic code that I'm missing instead of the "shortcut" steps. Any help is appreciated!
def main():
while (num >= 100) and (num <= 200):
for (num % 5 == 0) or (num % 6 == 0)
print (num)
main()
This is how I would go about it. I would recommend using a for loop over a while loop if you know the range you need. You are less likely to get into an endless loop. The reason for the n variable is since you said you needed 10 numbers per line. The n variable will track how many correct numbers you find so that you know when you have ten results and can use a normal print statement which automatically includes a newline. The second print statement will not add a newline.
n = 0
for i in range(100,201):
if (i%5 == 0 or i%6 == 0) and not (i%5 == 0 and i%6 == 0):
n += 1
if n%10 == 0:
print(i)
else:
print(str(i) + ", ", end="")
You should init every variable using in the code
While (condition) will break when the condition false. Since your condition depends on num, but num is never changed in your code, infinity loop will happen. You need to add num = num + 1 at the end of your loop block.
It's supposed to use if not for for each iterator here. And the condition you used for your problem is wrong to.
Should be like this:
def main():
num = 100
while (num >= 100) and (num <= 200):
if ((num % 5 == 0) or (num % 6 == 0)) and (num % 30 != 0):
print (num)
num = num + 1
main()