Excellent tutorials. You only need to go up to n/2 to determine whether a number is prime. Moreover, if you store all the prime numbers identified up to n in an array, these can be tested up the largest before going on to consecutive integers larger than the largest prime as in the range 2 to the largest prime number identified up to n/2 at least one prime factor, if any factor will exist.
Yes because if it is not divisible by any number smaller than the square root of n there could be no one higher. Look if divide 21 by 3 you get the result 7. 7 is higher than sqrt(21), but still a divider. On the other hand side you alswase need to have a num smaller than sqrt(n) to allow this. Let's do 16 instead: 2 and 8 try to get closer: 4 and 4 there could not be a higher num because if there would the other pair had to be smaller than 4. 👍👌🖖
As soon as you said Modulus and remainder is 0 for prime I knew how to code it.Thanks again for the tutorials they actually taught C better than other methods I tried.
Finally, a video that has helped me complete the prime number loop for my lab. So many videos left me lost and confused. This is so much simpler when talked about at the simplest level.
This is why braces are the dumbest way to organize code. Here's the same code in Python. def isPrime(n): for x in range(2,int(sqrt(n)+1)): if not(n%x): return False return True
I'm clinging, but braces are not used to organize code. It is simply a grouping instruction, it collects several instructions into one, which can be then used at the same time, for example, in a conditional statement. For example in java: public static boolean isPrime(int n){ if( n
What if the input here is 2, which should be a primenumber. The program would mark it as not prime since 2%2=0. Is there a way of implementation without an additional if statement, that explicitly checks if the input is 2?
@@izanyibukayvette381 There should be a way to do it mathematically, as we're not mathematicians use an if statement. I said that there's a way because computer scientists work a lot when the subject is prime numbers, research for an algorithm that is 100% solid mathematically
Hey everyone, I tried to code this little project myself but only the first number's check is correct can you help me with the problem? the ./a.out file is below: 5 is a prime number 4 is not a prime number 3 is not a prime number 2 is not a prime number 1 is not a prime number #include int main() { int d=0;
for(int i=4; i>0;i--) { for(int k=i;k>0; k--) { if( 0 == i % k) { d=d+1; } else if( 0 != i%k) { d=d; } } if(d==2) { printf("%i is a prime number ",i);
} else if(d != 2) { printf("%i is not a prime number",i);
I think you need to reset d = 0 at the beginning of each iteration of the outer loop, like this: ... for(int i = 4; i > 0; i--) { d = 0; for(int k = i; k > 0; k--) { ... If you don't reset d, then it will just keep increasing for each number. Therefore it would never be equal to 2 again, and every number (except for possibly the first number) would seem not prime (i.e. "composite").
I googled and found this: "Prime numbers are a mathematical concept that describes POSITIVE whole numbers that can only be divided evenly by two other whole numbers (or factors)."
input should always be above 1 by default considering its already mathematically proven to not be prime, I understand concern tho, but result is due to 1 being only other number requiring diff algo to determine whether prime or not--confused me too
i like how you not so tense on your videos appreciated maa`n
Excellent tutorials. You only need to go up to n/2 to determine whether a number is prime. Moreover, if you store all the prime numbers identified up to n in an array, these can be tested up the largest before going on to consecutive integers larger than the largest prime as in the range 2 to the largest prime number identified up to n/2 at least one prime factor, if any factor will exist.
You actually only need to check up to sqrt(n)!
@@cheerios623 yeah 💛
Yes because if it is not divisible by any number smaller than the square root of n there could be no one higher. Look if divide 21 by 3 you get the result 7. 7 is higher than sqrt(21), but still a divider. On the other hand side you alswase need to have a num smaller than sqrt(n) to allow this. Let's do 16 instead: 2 and 8 try to get closer: 4 and 4 there could not be a higher num because if there would the other pair had to be smaller than 4. 👍👌🖖
As soon as you said Modulus and remainder is 0 for prime I knew how to code it.Thanks again for the tutorials they actually taught C better than other methods I tried.
It's good if it's small numbers but for greater numbers they're more efficient algorithm to find prime numbers
Check if a number is prime? More like watch these videos for a good time! Thanks again for making and uploading all this amazing content.
That's very helpful
Finally, a video that has helped me complete the prime number loop for my lab. So many videos left me lost and confused. This is so much simpler when talked about at the simplest level.
Kardesim selaminaleykum, cs50'den bahsediyorsun sanirim. Nasil gidiyor bitirdin mi?
currently doing one for my lab.
Thank you very much!
Thanks, this video really made it simple to understand!
You're such an amazing teacher
Omg thank you so much!!
you´re the best
the best teacher
Thanks man
bool isPrime(n) {
for(let x=2; x
This is why braces are the dumbest way to organize code.
Here's the same code in Python.
def isPrime(n):
for x in range(2,int(sqrt(n)+1)):
if not(n%x):
return False
return True
I'm clinging, but braces are not used to organize code. It is simply a grouping instruction, it collects several instructions into one, which can be then used at the same time, for example, in a conditional statement. For example in java:
public static boolean isPrime(int n){
if( n
bro you love pizza more than anythingggg lolololo
ua-cam.com/video/0ABnO_WoexE/v-deo.html
를 이용하여
1부터 1,000까지 사이에 있는
소수의 개수(counting prime numbers)를 구하는 영상 입니다.
What if the input here is 2, which should be a primenumber. The program would mark it as not prime since 2%2=0.
Is there a way of implementation without an additional if statement, that explicitly checks if the input is 2?
yes
There are lot ways to do it, man
Check here
ua-cam.com/video/Eyl17WgH218/v-deo.html
Subscribe if you like the code
@@tiagodmota5840 but how me too if input is 2 then the output will look like 2 is not prime number
@@izanyibukayvette381 There should be a way to do it mathematically, as we're not mathematicians use an if statement. I said that there's a way because computer scientists work a lot when the subject is prime numbers, research for an algorithm that is 100% solid mathematically
The Best
Hey everyone, I tried to code this little project myself but only the first number's check is correct can you help me with the problem?
the ./a.out file is below:
5 is a prime number
4 is not a prime number
3 is not a prime number
2 is not a prime number
1 is not a prime number
#include
int main()
{
int d=0;
for(int i=4; i>0;i--)
{
for(int k=i;k>0; k--)
{
if( 0 == i % k)
{
d=d+1;
}
else if( 0 != i%k)
{
d=d;
}
}
if(d==2)
{
printf("%i is a prime number
",i);
}
else if(d != 2)
{
printf("%i is not a prime number",i);
}
printf("
");
}
I think you need to reset d = 0 at the beginning of each iteration of the outer loop, like this:
...
for(int i = 4; i > 0; i--)
{
d = 0;
for(int k = i; k > 0; k--)
{
...
If you don't reset d, then it will just keep increasing for each number. Therefore it would never be equal to 2 again, and every number (except for possibly the first number) would seem not prime (i.e. "composite").
I DID IT!!!!
-7 is a prime number?
I googled and found this: "Prime numbers are a mathematical concept that describes POSITIVE whole numbers that can only be divided evenly by two other whole numbers (or factors)."
LIKE :)
Great video thanks You!
please can you code it
hi guys, what is wrong with my code?
#include
#include
int main()
{
int input, i;
printf("enter a number:");
scanf("%d", &input);
for(int i=2; i
you need a ; after i
2?
input=1
Result 1 is a prime number :(
input should always be above 1 by default considering its already mathematically proven to not be prime, I understand concern tho, but result is due to 1 being only other number requiring diff algo to determine whether prime or not--confused me too