Python Prime Number for-else range
lower = int(input("from:"))
upper = int(input("to:"))
for num in range(lower,upper + 1):
if num > 1:
for i in range(2,num):
if (num % i) == 0:
break
else:
print(num)
Why does this code print "2" as a prime number? (it is but it should not print it)
2%2==0 so it should skip it...
python python-3.x algorithm primes
asked Nov 20 '18 at 19:22
Sotiris LiagasSotiris Liagas
367
add a comment |
lower = int(input("from:"))
upper = int(input("to:"))
for num in range(lower,upper + 1):
if num > 1:
for i in range(2,num):
if (num % i) == 0:
break
else:
print(num)
Why does this code print "2" as a prime number? (it is but it should not print it)
2%2==0 so it should skip it...
python python-3.x algorithm primes
asked Nov 20 '18 at 19:22
Sotiris LiagasSotiris Liagas
367
What happens wheniis 3?
– Woody1193
Nov 20 '18 at 19:25
1
for i in range(2,num)if num is 2 that's empy list so you wont get 2%2==0
– Filip Młynarski
Nov 20 '18 at 19:31
1
Well.... two is a prime number. It can only be divided by itself and one. However, I do see the error in your code causing the unexpected response:range(start,end)needs theendto be greater than thestart. Python's duck-typing means a check won't be performed and your if statment doesn't get checked. I wish python didn't do that, but unfortunately it does.
– Mark_Anderson
Nov 20 '18 at 19:38
add a comment |
lower = int(input("from:"))
upper = int(input("to:"))
for num in range(lower,upper + 1):
if num > 1:
for i in range(2,num):
if (num % i) == 0:
break
else:
print(num)
Why does this code print "2" as a prime number? (it is but it should not print it)
2%2==0 so it should skip it...
python python-3.x algorithm primes
asked Nov 20 '18 at 19:22
Sotiris LiagasSotiris Liagas
367
lower = int(input("from:"))
upper = int(input("to:"))
for num in range(lower,upper + 1):
if num > 1:
for i in range(2,num):
if (num % i) == 0:
break
else:
print(num)
Why does this code print "2" as a prime number? (it is but it should not print it)
2%2==0 so it should skip it...
python python-3.x algorithm primes
python python-3.x algorithm primes
asked Nov 20 '18 at 19:22
Sotiris LiagasSotiris Liagas
367
asked Nov 20 '18 at 19:22
Sotiris LiagasSotiris Liagas
367
edited Nov 20 '18 at 19:45
Sotiris Liagas
asked Nov 20 '18 at 19:22
Sotiris LiagasSotiris Liagas
367
asked Nov 20 '18 at 19:22
Sotiris LiagasSotiris Liagas
367
asked Nov 20 '18 at 19:22
Sotiris LiagasSotiris Liagas
367
367
What happens wheniis 3?
– Woody1193
Nov 20 '18 at 19:25
1
for i in range(2,num)if num is 2 that's empy list so you wont get 2%2==0
– Filip Młynarski
Nov 20 '18 at 19:31
1
Well.... two is a prime number. It can only be divided by itself and one. However, I do see the error in your code causing the unexpected response:range(start,end)needs theendto be greater than thestart. Python's duck-typing means a check won't be performed and your if statment doesn't get checked. I wish python didn't do that, but unfortunately it does.
– Mark_Anderson
Nov 20 '18 at 19:38
add a comment |
What happens wheniis 3?
– Woody1193
Nov 20 '18 at 19:25
1
for i in range(2,num)if num is 2 that's empy list so you wont get 2%2==0
– Filip Młynarski
Nov 20 '18 at 19:31
1
Well.... two is a prime number. It can only be divided by itself and one. However, I do see the error in your code causing the unexpected response:range(start,end)needs theendto be greater than thestart. Python's duck-typing means a check won't be performed and your if statment doesn't get checked. I wish python didn't do that, but unfortunately it does.
– Mark_Anderson
Nov 20 '18 at 19:38
What happens when
i is 3?– Woody1193
Nov 20 '18 at 19:25
What happens when
i is 3?– Woody1193
Nov 20 '18 at 19:25
1
1
for i in range(2,num) if num is 2 that's empy list so you wont get 2%2==0– Filip Młynarski
Nov 20 '18 at 19:31
for i in range(2,num) if num is 2 that's empy list so you wont get 2%2==0– Filip Młynarski
Nov 20 '18 at 19:31
1
1
Well.... two is a prime number. It can only be divided by itself and one. However, I do see the error in your code causing the unexpected response:
range(start,end) needs the end to be greater than the start. Python's duck-typing means a check won't be performed and your if statment doesn't get checked. I wish python didn't do that, but unfortunately it does.– Mark_Anderson
Nov 20 '18 at 19:38
Well.... two is a prime number. It can only be divided by itself and one. However, I do see the error in your code causing the unexpected response:
range(start,end) needs the end to be greater than the start. Python's duck-typing means a check won't be performed and your if statment doesn't get checked. I wish python didn't do that, but unfortunately it does.– Mark_Anderson
Nov 20 '18 at 19:38
add a comment |
2 Answers
2
active
oldest
votes
When num is 2, range(2, num) is empty, so the if (num % i) == 0: check is not performed, and the else block executes.
answered Nov 20 '18 at 19:30
Patrick HaughPatrick Haugh
29.2k92747
add a comment |
Others have noted the error in the range(start,end) code. Correcting that, your code for primes could be rewritten as:
lower = int(input("from:"))
upper = int(input("to:"))
for num in range(lower,upper + 1):
if num > 1:
for i in range(2,max(num,3)):
if (num % i) == 0:
break
else:
print(num)
This isn't the fastest way to get primes mind you, each potential prime must be tested against EVERY smaller number as a possible divisor. It's much faster to instead count upwards and work out the multiples of smaller numbers. That way we only need to do the maths once for each possible divisor.
For completeness, here is a program that can produce primes efficiently (uses the sieve of Eratosthenes method).
#### INPUTS
lower = int(input("from:"))
upper = int(input("to:"))
### Code
n = upper
prime_booleans = [True for i in range(n+1)]
p = 2
while (p * p <= n):
# Is current number a prime, eliminate the numbers that are multiples of it
if (prime_booleans[p] == True):
for i in range(p * 2, n+1, p):
prime_booleans[i] = False
p += 1
# Print all prime numbers
for p in range(lower, n):
if prime_booleans[p]:
print p,
answered Nov 20 '18 at 20:03
Mark_AndersonMark_Anderson
426216
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
When num is 2, range(2, num) is empty, so the if (num % i) == 0: check is not performed, and the else block executes.
answered Nov 20 '18 at 19:30
Patrick HaughPatrick Haugh
29.2k92747
add a comment |
When num is 2, range(2, num) is empty, so the if (num % i) == 0: check is not performed, and the else block executes.
answered Nov 20 '18 at 19:30
Patrick HaughPatrick Haugh
29.2k92747
add a comment |
When num is 2, range(2, num) is empty, so the if (num % i) == 0: check is not performed, and the else block executes.
answered Nov 20 '18 at 19:30
Patrick HaughPatrick Haugh
29.2k92747
When num is 2, range(2, num) is empty, so the if (num % i) == 0: check is not performed, and the else block executes.
answered Nov 20 '18 at 19:30
Patrick HaughPatrick Haugh
29.2k92747
answered Nov 20 '18 at 19:30
Patrick HaughPatrick Haugh
29.2k92747
answered Nov 20 '18 at 19:30
Patrick HaughPatrick Haugh
29.2k92747
answered Nov 20 '18 at 19:30
Patrick HaughPatrick Haugh
29.2k92747
29.2k92747
add a comment |
add a comment |
Others have noted the error in the range(start,end) code. Correcting that, your code for primes could be rewritten as:
lower = int(input("from:"))
upper = int(input("to:"))
for num in range(lower,upper + 1):
if num > 1:
for i in range(2,max(num,3)):
if (num % i) == 0:
break
else:
print(num)
This isn't the fastest way to get primes mind you, each potential prime must be tested against EVERY smaller number as a possible divisor. It's much faster to instead count upwards and work out the multiples of smaller numbers. That way we only need to do the maths once for each possible divisor.
For completeness, here is a program that can produce primes efficiently (uses the sieve of Eratosthenes method).
#### INPUTS
lower = int(input("from:"))
upper = int(input("to:"))
### Code
n = upper
prime_booleans = [True for i in range(n+1)]
p = 2
while (p * p <= n):
# Is current number a prime, eliminate the numbers that are multiples of it
if (prime_booleans[p] == True):
for i in range(p * 2, n+1, p):
prime_booleans[i] = False
p += 1
# Print all prime numbers
for p in range(lower, n):
if prime_booleans[p]:
print p,
answered Nov 20 '18 at 20:03
Mark_AndersonMark_Anderson
426216
add a comment |
Others have noted the error in the range(start,end) code. Correcting that, your code for primes could be rewritten as:
lower = int(input("from:"))
upper = int(input("to:"))
for num in range(lower,upper + 1):
if num > 1:
for i in range(2,max(num,3)):
if (num % i) == 0:
break
else:
print(num)
This isn't the fastest way to get primes mind you, each potential prime must be tested against EVERY smaller number as a possible divisor. It's much faster to instead count upwards and work out the multiples of smaller numbers. That way we only need to do the maths once for each possible divisor.
For completeness, here is a program that can produce primes efficiently (uses the sieve of Eratosthenes method).
#### INPUTS
lower = int(input("from:"))
upper = int(input("to:"))
### Code
n = upper
prime_booleans = [True for i in range(n+1)]
p = 2
while (p * p <= n):
# Is current number a prime, eliminate the numbers that are multiples of it
if (prime_booleans[p] == True):
for i in range(p * 2, n+1, p):
prime_booleans[i] = False
p += 1
# Print all prime numbers
for p in range(lower, n):
if prime_booleans[p]:
print p,
answered Nov 20 '18 at 20:03
Mark_AndersonMark_Anderson
426216
add a comment |
Others have noted the error in the range(start,end) code. Correcting that, your code for primes could be rewritten as:
lower = int(input("from:"))
upper = int(input("to:"))
for num in range(lower,upper + 1):
if num > 1:
for i in range(2,max(num,3)):
if (num % i) == 0:
break
else:
print(num)
This isn't the fastest way to get primes mind you, each potential prime must be tested against EVERY smaller number as a possible divisor. It's much faster to instead count upwards and work out the multiples of smaller numbers. That way we only need to do the maths once for each possible divisor.
For completeness, here is a program that can produce primes efficiently (uses the sieve of Eratosthenes method).
#### INPUTS
lower = int(input("from:"))
upper = int(input("to:"))
### Code
n = upper
prime_booleans = [True for i in range(n+1)]
p = 2
while (p * p <= n):
# Is current number a prime, eliminate the numbers that are multiples of it
if (prime_booleans[p] == True):
for i in range(p * 2, n+1, p):
prime_booleans[i] = False
p += 1
# Print all prime numbers
for p in range(lower, n):
if prime_booleans[p]:
print p,
answered Nov 20 '18 at 20:03
Mark_AndersonMark_Anderson
426216
Others have noted the error in the range(start,end) code. Correcting that, your code for primes could be rewritten as:
lower = int(input("from:"))
upper = int(input("to:"))
for num in range(lower,upper + 1):
if num > 1:
for i in range(2,max(num,3)):
if (num % i) == 0:
break
else:
print(num)
This isn't the fastest way to get primes mind you, each potential prime must be tested against EVERY smaller number as a possible divisor. It's much faster to instead count upwards and work out the multiples of smaller numbers. That way we only need to do the maths once for each possible divisor.
For completeness, here is a program that can produce primes efficiently (uses the sieve of Eratosthenes method).
#### INPUTS
lower = int(input("from:"))
upper = int(input("to:"))
### Code
n = upper
prime_booleans = [True for i in range(n+1)]
p = 2
while (p * p <= n):
# Is current number a prime, eliminate the numbers that are multiples of it
if (prime_booleans[p] == True):
for i in range(p * 2, n+1, p):
prime_booleans[i] = False
p += 1
# Print all prime numbers
for p in range(lower, n):
if prime_booleans[p]:
print p,
answered Nov 20 '18 at 20:03
Mark_AndersonMark_Anderson
426216
answered Nov 20 '18 at 20:03
Mark_AndersonMark_Anderson
426216
answered Nov 20 '18 at 20:03
Mark_AndersonMark_Anderson
426216
answered Nov 20 '18 at 20:03
Mark_AndersonMark_Anderson
426216
426216
add a comment |
add a comment |
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What happens when
iis 3?– Woody1193
Nov 20 '18 at 19:25
1
for i in range(2,num)if num is 2 that's empy list so you wont get 2%2==0– Filip Młynarski
Nov 20 '18 at 19:31
1
Well.... two is a prime number. It can only be divided by itself and one. However, I do see the error in your code causing the unexpected response:
range(start,end)needs theendto be greater than thestart. Python's duck-typing means a check won't be performed and your if statment doesn't get checked. I wish python didn't do that, but unfortunately it does.– Mark_Anderson
Nov 20 '18 at 19:38