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 !
Related
I have a code for printing all prime numbers lying in the range (2, 50)
num = 2
for i in range(2, 50):
j = 2
while(j <= (i/2)):
if (i % j == 0):
break
j += 1
if(j > i/j):
print(i,'is a prime number')
My doubt is regarding the use and significance of if(j > i/j) statement.
Based on the above, for all prime numbers by the time the code execution reaches if condition, j will always be equal to i and the if condition will become true.
Let's go through step by step for a prime and a non-prime number.
Example 1 : Prime number 7(Assume value of i is at 7 in the for loop)
i = 7
j = 2
# While statement condition check
2 <= 7/2 -> True
7%2 == 0 -> False
# Increment j, now j=3
7%3 == 0 -> False
.
.
# Increment continuously, and when j=7
7%7 == 0 -> True
#If Condition check(Current values are i=7 and j=7)
7 > 7/7 -> True
Example 2: Non-Prime Number. Let's say 9
i = 9
j = 2
# While statement condition check
2 <= 7/2 -> True
9%2 == 0 -> False
# Increment j, now j=3
9%3 == 0 -> True
#If Condition check(Current values are i=9 and j=3)
3 > 9/3 -> False
Do try for other examples step by step to understand better.
I haven't seen somebody use this algorithm for finding prime numbers before, you can do this for finding prime numbers:
for i in range(2, 50):
num, j = 0, 2
while j <= i:
if i % j == 0:
num += 1
j += 1
if num == 1:
print(i, 'is a prime number')
Honestly the best way to understand things like this, is to break it down step-by-step yourself to understand it. Take a smaller example, say range(2,6), and write the progress/steps down by hand.
Hint, do it for i = 39.
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 am trying to print prime numbers in python in the range 3-50. Here is the code:
for i in range(3,51):
flag=0
for j in range(3,i):
if(i%j==0):
flag=1
if(flag==0):
print(i)
The code is running correctly except THAT IT IS ALSO PRINTING 4. Please tell if there is anything wrong with the logic of the code (ignore the efficiency issues).
print(2)
for i in range(3,51):
flag=0
for j in range(2,int(i/2)):
if(i%j==0):
flag=1
if(flag==0):
print(i)
When i = 4, the range will be range(2,2) which is 0. It will not enter the if(i%j == 0): block. Thus your flag would always be 0 when i = 4.
One way to solve it would be to just add a check if i is 4.
if(flag==0 and i != 4):
Edit : As commented by others, by incrementing the range by 1 would be a better fix than checking if the value is 4.
for j in range(2,int(i/2)+1)
change range(3,i) to range(2,i) because if we use range(3,i), 4 is not checked with division by 2. As a result 4 is returned as prime.
for i in range(3,51):
flag=0
for j in range(2,i):
if(i%j==0):
flag=1
if(flag==0):
print(i)
However, a more structural and efficient way is to use the following:-
def isPrime(num):
for i in range(2,num/2+1):
if (num%i==0):
return False
return True
for i in range(3,51):
if isPrime(i):
print i
We don't need to check the division by all numbers till the number itself for prime. Because if we can check till the half of the given number only for increased efficiency.
start = 3
end = 50
for val in range(start, end + 1):
# If num is divisible by any number
# between 2 and val, it is not prime
if val > 1:
for n in range(2, val):
if (val % n) == 0:
break
else:
print(val)
a = int(input("enter smaller no."))
b = int(input("enter larger no."))
for i in range (a,b+1):
if i>1:
for c in range(2,i):
if i%c == 0:
break
else:
print(i)
In the code above we ran a for loop from minimum to the maximum number and storing the value in i
for i in range (a,b+1):
then we check that the number stored in i is greater than 1 or not as all numbers are divisible by 1 and we don not want to consider it.
if i>1:
after this we run another loop, this time from 2 to 1 less then the number stored in i
for c in range(2,i):
Now we check that the number stored in i is divisible by any of the number from 2 to 1 number less then itself.
if i%c == 0:
If yes then clearly it is not a prime number and we break the loop.
break
And if no we print the number.
else:
print(i)
For 4 it probably does not go inside the if loop. So I increased the range for i.
print(2)
for i in range(3,51):
flag=0
for j in range(2,int(i/2)+1):
if(i%j==0):
flag=1
if(flag==0):
print(i)
for i in range(3,51):
flag=0
for j in range(2,i):
if(i%j==0):
flag=1
if(flag==0):
print(i)
This prints
3
5
7
11
13
17
19
23
29
31
37
41
43
47
4%2 = 0
4%3 = 1
So if you start from 3 you will miss out that 4%2 is = 0
If your list would have bigger numbers, you should change
range(2,int(i/2)+1)
for
range(2,math.floor(math.sqrt(n)+1))
For instance, 400 will try numbers until 20 instead of 200.
tryfor j in range(2,i):the numbers 4 is divisible is 1,2,4 if you start from 3 and continue until 4 but not four you can't find any divisors that is why it is considering 4 as a prime
import math
for num in range(1,50):
if num>1:
if num==2:
print(num)
elif num%2==0:
continue
else:
for i in range(3,math.ceil(math.sqrt(num)),2):
if num%i==0:
break
else:
print(num)
It is more efficient as it avoids even numbers and also check for the odd divisors up-to square root of the number which is more efficient than checking up-to only half of the given numbers
I am pretty new to python, so I don't fully understand how to use loops. I am currently working on a piece of code that I have to find the first N prime numbers.
The result that is desired is if you input 5, it outputs 2, 3, 5, 7, and 11, but no matter what I input for 'max', the output always ends up being 2 and 3. Is there a way to improve this?
max=int(input("How many prime numbers do you want: "))
min=2
while(min<=(max)):
for c in range(2, min):
if min%c==0:
break
else:
print min
min=min+1
You only increment min in the else block, i.e., if min % c is nonzero for all c, i.e., if min is prime. This means that the code won't be able to move past any composite numbers. You can fix this by unindenting min=min+1 one level so that it lines up with the for and else.
number = int(input("Prime numbers between 2 and "))
for num in range(2,number + 1):
if num > 1:
for i in range(2,num):
if (num % i) == 0:
break
else:
print(num)
Solution: Get the nth prime number entry. Iterate through each natural numbers for prime number and append the prime number to a list. Terminate the program when length of a list satisfies the user nth prime number entry.
# Get the number of prime numbers entry.
try:
enterNumber = int(input("List of nth prime numbers: "))
except:
print("The entry MUST be an integer.")
exit()
startNumber = 1
primeList = []
while True:
# Check for the entry to greater than zero.
if enterNumber <= 0:
print("The entry MUST be greater than zero.")
break
# Check each number from 1 for prime unless prime number entry is satisfied.
if startNumber > 1:
for i in range(2,startNumber):
if (startNumber % i) == 0:
break
else:
primeList.append(startNumber)
if (len(primeList) == enterNumber):
print(primeList)
break
else:
startNumber = startNumber + 1
continue
Try that :
n = int(input("First N prime number, N ? "))
p = [2]
c = 2
while len(p) < n:
j = 0
c += 1
while j < len(p):
if c % p[j] == 0:
break
elif j == len(p) - 1:
p.append(c)
j += 1
print(p)
Its simple. Check the below code, am sure it works!
N = int(input('Enter the number: ')
i=1
count=0
while(count<N):
for x in range(i,i+1):
c=0
for y in range(1,x+1):
if(x%y==0):
c=c+1
if(c==2):
print(x)
count=count+1
i=i+1
The following code will give you prime numbers between 3 to N, where N is the input from user:
number = int(input("Prime numbers between 2, 3 and "))
for i in range(2,number):
for j in range(2,int(i/2)+1):
if i%j==0:
break
elif j==int(i/2):
print(i)
You can see to check a number i to be prime you only have to check its divisibility with numbers till n/2.
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).