how to use for loop to calculate a prime number
0
for i in range(2, 101):
for j in range(2, i):
if (i % j) == 0:
print(i,"is a composite number")
break
I tried making the if (i%j) != 0: but then it wouldn't work (4 is not a prime number)
python loops sieve
edited Nov 11 '18 at 6:33
smci
14.8k672104
asked Nov 11 '18 at 5:56
Haoyang SongHaoyang Song
10210
add a comment |
0
for i in range(2, 101):
for j in range(2, i):
if (i % j) == 0:
print(i,"is a composite number")
break
I tried making the if (i%j) != 0: but then it wouldn't work (4 is not a prime number)
python loops sieve
edited Nov 11 '18 at 6:33
smci
14.8k672104
asked Nov 11 '18 at 5:56
Haoyang SongHaoyang Song
10210
1
I'm not sure what you're asking here. That program appears to correctly (albeit not efficiently) differentiate between prime and composite numbers.
– Matt
Nov 11 '18 at 5:58
well when I addif (i%j) != 0:it dosen't work
– Haoyang Song
Nov 11 '18 at 6:06
1
if (i%j) != 0:tests for divisibility of i by j, specifically that i is not divisible by j. If i were divisible by j (for some 2 <= j < i), it couldn't be prime. There are 520 existing questions on [python] divisibility; this is a duplicate and should be closed; please read through the others.
– smci
Nov 11 '18 at 13:30
I did try the others, the others wasn't what I was asking for
– Haoyang Song
Nov 11 '18 at 17:41
add a comment |
0
0
0
2
for i in range(2, 101):
for j in range(2, i):
if (i % j) == 0:
print(i,"is a composite number")
break
I tried making the if (i%j) != 0: but then it wouldn't work (4 is not a prime number)
python loops sieve
edited Nov 11 '18 at 6:33
smci
14.8k672104
asked Nov 11 '18 at 5:56
Haoyang SongHaoyang Song
10210
for i in range(2, 101):
for j in range(2, i):
if (i % j) == 0:
print(i,"is a composite number")
break
I tried making the if (i%j) != 0: but then it wouldn't work (4 is not a prime number)
python loops sieve
python loops sieve
edited Nov 11 '18 at 6:33
smci
14.8k672104
asked Nov 11 '18 at 5:56
Haoyang SongHaoyang Song
10210
edited Nov 11 '18 at 6:33
smci
14.8k672104
asked Nov 11 '18 at 5:56
Haoyang SongHaoyang Song
10210
edited Nov 11 '18 at 6:33
smci
14.8k672104
edited Nov 11 '18 at 6:33
smci
14.8k672104
edited Nov 11 '18 at 6:33
smci
14.8k672104
14.8k672104
asked Nov 11 '18 at 5:56
Haoyang SongHaoyang Song
10210
asked Nov 11 '18 at 5:56
Haoyang SongHaoyang Song
10210
asked Nov 11 '18 at 5:56
Haoyang SongHaoyang Song
10210
10210
1
I'm not sure what you're asking here. That program appears to correctly (albeit not efficiently) differentiate between prime and composite numbers.
– Matt
Nov 11 '18 at 5:58
well when I addif (i%j) != 0:it dosen't work
– Haoyang Song
Nov 11 '18 at 6:06
1
if (i%j) != 0:tests for divisibility of i by j, specifically that i is not divisible by j. If i were divisible by j (for some 2 <= j < i), it couldn't be prime. There are 520 existing questions on [python] divisibility; this is a duplicate and should be closed; please read through the others.
– smci
Nov 11 '18 at 13:30
I did try the others, the others wasn't what I was asking for
– Haoyang Song
Nov 11 '18 at 17:41
add a comment |
1
I'm not sure what you're asking here. That program appears to correctly (albeit not efficiently) differentiate between prime and composite numbers.
– Matt
Nov 11 '18 at 5:58
well when I addif (i%j) != 0:it dosen't work
– Haoyang Song
Nov 11 '18 at 6:06
1
if (i%j) != 0:tests for divisibility of i by j, specifically that i is not divisible by j. If i were divisible by j (for some 2 <= j < i), it couldn't be prime. There are 520 existing questions on [python] divisibility; this is a duplicate and should be closed; please read through the others.
– smci
Nov 11 '18 at 13:30
I did try the others, the others wasn't what I was asking for
– Haoyang Song
Nov 11 '18 at 17:41
1
1
I'm not sure what you're asking here. That program appears to correctly (albeit not efficiently) differentiate between prime and composite numbers.
– Matt
Nov 11 '18 at 5:58
I'm not sure what you're asking here. That program appears to correctly (albeit not efficiently) differentiate between prime and composite numbers.
– Matt
Nov 11 '18 at 5:58
well when I add
if (i%j) != 0: it dosen't work– Haoyang Song
Nov 11 '18 at 6:06
well when I add
if (i%j) != 0: it dosen't work– Haoyang Song
Nov 11 '18 at 6:06
1
1
if (i%j) != 0: tests for divisibility of i by j, specifically that i is not divisible by j. If i were divisible by j (for some 2 <= j < i), it couldn't be prime. There are 520 existing questions on [python] divisibility; this is a duplicate and should be closed; please read through the others.– smci
Nov 11 '18 at 13:30
if (i%j) != 0: tests for divisibility of i by j, specifically that i is not divisible by j. If i were divisible by j (for some 2 <= j < i), it couldn't be prime. There are 520 existing questions on [python] divisibility; this is a duplicate and should be closed; please read through the others.– smci
Nov 11 '18 at 13:30
I did try the others, the others wasn't what I was asking for
– Haoyang Song
Nov 11 '18 at 17:41
I did try the others, the others wasn't what I was asking for
– Haoyang Song
Nov 11 '18 at 17:41
add a comment |
2 Answers
2
active
oldest
votes
2
The for loop you've used is correct for finding prime numbers. I would just another condition to it: if i > 1:. Also, you would want to print the prime number
for i in range(2, 101):
if i > 1: # Prime numbers are greater than 1
for j in range(2, i):
if (i % j) == 0:
print(i,"is a composite number")
break
else:
print(i,"is a prime number")
answered Nov 11 '18 at 6:04
Mayank PorwalMayank Porwal
4,8822724
You are welcome. Thanks for accepting.
– Mayank Porwal
Nov 11 '18 at 6:16
1
if the range start at 2 thei > 1check becomes unnecessary though
– Anton vBR
Nov 11 '18 at 6:27
Yes, in this case it does.
– Mayank Porwal
Nov 11 '18 at 6:35
The extrai > 1test is unnecessary and just obfuscates. You always run a sieve starting at 2 or higher, never 1. This answer adds nothing. Also, the question is a duplicate.
– smci
Nov 11 '18 at 7:22
add a comment |
-2
You can fix up your original algorithm like this:
for i in range(2, 101):
if all([(i % j) for j in range(2, i)]):
print(i,"is a prime number")
In general, you're probably better off using/learning from the established algorithms in cases like these. Here's a Python implementation of a well-known algorithm (the Sieve of Eratosthenes) for generating the first n primes (credit to tech.io for the code):
def sieve(n):
primes = 2*[False] + (n-1)*[True]
for i in range(2, int(n**0.5+1.5)):
for j in range(i*i, n+1, i):
primes[j] = False
return [prime for prime, checked in enumerate(primes) if checked]
Some test output:
print(sieve(100))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
answered Nov 11 '18 at 6:01
teltel
7,34121431
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
2
The for loop you've used is correct for finding prime numbers. I would just another condition to it: if i > 1:. Also, you would want to print the prime number
for i in range(2, 101):
if i > 1: # Prime numbers are greater than 1
for j in range(2, i):
if (i % j) == 0:
print(i,"is a composite number")
break
else:
print(i,"is a prime number")
answered Nov 11 '18 at 6:04
Mayank PorwalMayank Porwal
4,8822724
You are welcome. Thanks for accepting.
– Mayank Porwal
Nov 11 '18 at 6:16
1
if the range start at 2 thei > 1check becomes unnecessary though
– Anton vBR
Nov 11 '18 at 6:27
Yes, in this case it does.
– Mayank Porwal
Nov 11 '18 at 6:35
The extrai > 1test is unnecessary and just obfuscates. You always run a sieve starting at 2 or higher, never 1. This answer adds nothing. Also, the question is a duplicate.
– smci
Nov 11 '18 at 7:22
add a comment |
2
The for loop you've used is correct for finding prime numbers. I would just another condition to it: if i > 1:. Also, you would want to print the prime number
for i in range(2, 101):
if i > 1: # Prime numbers are greater than 1
for j in range(2, i):
if (i % j) == 0:
print(i,"is a composite number")
break
else:
print(i,"is a prime number")
answered Nov 11 '18 at 6:04
Mayank PorwalMayank Porwal
4,8822724
You are welcome. Thanks for accepting.
– Mayank Porwal
Nov 11 '18 at 6:16
1
if the range start at 2 thei > 1check becomes unnecessary though
– Anton vBR
Nov 11 '18 at 6:27
Yes, in this case it does.
– Mayank Porwal
Nov 11 '18 at 6:35
The extrai > 1test is unnecessary and just obfuscates. You always run a sieve starting at 2 or higher, never 1. This answer adds nothing. Also, the question is a duplicate.
– smci
Nov 11 '18 at 7:22
add a comment |
2
2
2
The for loop you've used is correct for finding prime numbers. I would just another condition to it: if i > 1:. Also, you would want to print the prime number
for i in range(2, 101):
if i > 1: # Prime numbers are greater than 1
for j in range(2, i):
if (i % j) == 0:
print(i,"is a composite number")
break
else:
print(i,"is a prime number")
answered Nov 11 '18 at 6:04
Mayank PorwalMayank Porwal
4,8822724
The for loop you've used is correct for finding prime numbers. I would just another condition to it: if i > 1:. Also, you would want to print the prime number
for i in range(2, 101):
if i > 1: # Prime numbers are greater than 1
for j in range(2, i):
if (i % j) == 0:
print(i,"is a composite number")
break
else:
print(i,"is a prime number")
answered Nov 11 '18 at 6:04
Mayank PorwalMayank Porwal
4,8822724
answered Nov 11 '18 at 6:04
Mayank PorwalMayank Porwal
4,8822724
answered Nov 11 '18 at 6:04
Mayank PorwalMayank Porwal
4,8822724
answered Nov 11 '18 at 6:04
Mayank PorwalMayank Porwal
4,8822724
4,8822724
You are welcome. Thanks for accepting.
– Mayank Porwal
Nov 11 '18 at 6:16
1
if the range start at 2 thei > 1check becomes unnecessary though
– Anton vBR
Nov 11 '18 at 6:27
Yes, in this case it does.
– Mayank Porwal
Nov 11 '18 at 6:35
The extrai > 1test is unnecessary and just obfuscates. You always run a sieve starting at 2 or higher, never 1. This answer adds nothing. Also, the question is a duplicate.
– smci
Nov 11 '18 at 7:22
add a comment |
You are welcome. Thanks for accepting.
– Mayank Porwal
Nov 11 '18 at 6:16
1
if the range start at 2 thei > 1check becomes unnecessary though
– Anton vBR
Nov 11 '18 at 6:27
Yes, in this case it does.
– Mayank Porwal
Nov 11 '18 at 6:35
The extrai > 1test is unnecessary and just obfuscates. You always run a sieve starting at 2 or higher, never 1. This answer adds nothing. Also, the question is a duplicate.
– smci
Nov 11 '18 at 7:22
You are welcome. Thanks for accepting.
– Mayank Porwal
Nov 11 '18 at 6:16
You are welcome. Thanks for accepting.
– Mayank Porwal
Nov 11 '18 at 6:16
1
1
if the range start at 2 the
i > 1 check becomes unnecessary though– Anton vBR
Nov 11 '18 at 6:27
if the range start at 2 the
i > 1 check becomes unnecessary though– Anton vBR
Nov 11 '18 at 6:27
Yes, in this case it does.
– Mayank Porwal
Nov 11 '18 at 6:35
Yes, in this case it does.
– Mayank Porwal
Nov 11 '18 at 6:35
The extra
i > 1 test is unnecessary and just obfuscates. You always run a sieve starting at 2 or higher, never 1. This answer adds nothing. Also, the question is a duplicate.– smci
Nov 11 '18 at 7:22
The extra
i > 1 test is unnecessary and just obfuscates. You always run a sieve starting at 2 or higher, never 1. This answer adds nothing. Also, the question is a duplicate.– smci
Nov 11 '18 at 7:22
add a comment |
-2
You can fix up your original algorithm like this:
for i in range(2, 101):
if all([(i % j) for j in range(2, i)]):
print(i,"is a prime number")
In general, you're probably better off using/learning from the established algorithms in cases like these. Here's a Python implementation of a well-known algorithm (the Sieve of Eratosthenes) for generating the first n primes (credit to tech.io for the code):
def sieve(n):
primes = 2*[False] + (n-1)*[True]
for i in range(2, int(n**0.5+1.5)):
for j in range(i*i, n+1, i):
primes[j] = False
return [prime for prime, checked in enumerate(primes) if checked]
Some test output:
print(sieve(100))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
answered Nov 11 '18 at 6:01
teltel
7,34121431
add a comment |
-2
You can fix up your original algorithm like this:
for i in range(2, 101):
if all([(i % j) for j in range(2, i)]):
print(i,"is a prime number")
In general, you're probably better off using/learning from the established algorithms in cases like these. Here's a Python implementation of a well-known algorithm (the Sieve of Eratosthenes) for generating the first n primes (credit to tech.io for the code):
def sieve(n):
primes = 2*[False] + (n-1)*[True]
for i in range(2, int(n**0.5+1.5)):
for j in range(i*i, n+1, i):
primes[j] = False
return [prime for prime, checked in enumerate(primes) if checked]
Some test output:
print(sieve(100))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
answered Nov 11 '18 at 6:01
teltel
7,34121431
add a comment |
-2
-2
-2
You can fix up your original algorithm like this:
for i in range(2, 101):
if all([(i % j) for j in range(2, i)]):
print(i,"is a prime number")
In general, you're probably better off using/learning from the established algorithms in cases like these. Here's a Python implementation of a well-known algorithm (the Sieve of Eratosthenes) for generating the first n primes (credit to tech.io for the code):
def sieve(n):
primes = 2*[False] + (n-1)*[True]
for i in range(2, int(n**0.5+1.5)):
for j in range(i*i, n+1, i):
primes[j] = False
return [prime for prime, checked in enumerate(primes) if checked]
Some test output:
print(sieve(100))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
answered Nov 11 '18 at 6:01
teltel
7,34121431
You can fix up your original algorithm like this:
for i in range(2, 101):
if all([(i % j) for j in range(2, i)]):
print(i,"is a prime number")
In general, you're probably better off using/learning from the established algorithms in cases like these. Here's a Python implementation of a well-known algorithm (the Sieve of Eratosthenes) for generating the first n primes (credit to tech.io for the code):
def sieve(n):
primes = 2*[False] + (n-1)*[True]
for i in range(2, int(n**0.5+1.5)):
for j in range(i*i, n+1, i):
primes[j] = False
return [prime for prime, checked in enumerate(primes) if checked]
Some test output:
print(sieve(100))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
answered Nov 11 '18 at 6:01
teltel
7,34121431
edited Nov 11 '18 at 6:16
answered Nov 11 '18 at 6:01
teltel
7,34121431
answered Nov 11 '18 at 6:01
teltel
7,34121431
answered Nov 11 '18 at 6:01
teltel
7,34121431
7,34121431
add a comment |
add a comment |
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1
I'm not sure what you're asking here. That program appears to correctly (albeit not efficiently) differentiate between prime and composite numbers.
– Matt
Nov 11 '18 at 5:58
well when I add
if (i%j) != 0:it dosen't work– Haoyang Song
Nov 11 '18 at 6:06
1
if (i%j) != 0:tests for divisibility of i by j, specifically that i is not divisible by j. If i were divisible by j (for some 2 <= j < i), it couldn't be prime. There are 520 existing questions on [python] divisibility; this is a duplicate and should be closed; please read through the others.– smci
Nov 11 '18 at 13:30
I did try the others, the others wasn't what I was asking for
– Haoyang Song
Nov 11 '18 at 17:41