Good suggestion. But one thing to notice is that the solution given in the video runs in O(sqrt(n)) whereas your solution runs in O(n). While O(n) is efficient enough, it still leads to a huge performance difference when the input is big enough.

Super bro

that code is not at all efficient... suppose I enter the value 10000 then you program loop will run 10000 times which is unnecessary and useless bcz we know that the factor of any number is present within the square root range of that number..

yes,3 is no needed, thank you for the information

I was thinking the same

But can you tell me why we are calculating square root here????

#include int main() { int a,b,i; printf("enter a no. :"); scanf("%d",&a); for(i=2;i

👍

Yes not needed

yes i just asked the same question! did you find the answer?? i really want to know.

Neso is great bro.

bhai ye class 6 mein hota h

@@uraharakisuke5305 haan, class 6 ka questions UPSC exams me b atta he.. kisi b knowledge ko chota ya bara maat samjo..

@@andistheinforitbutso7513 pata hai bhai maine woh nahi bola...maine just bola ki ye matlab school mein bhi padhaya jaata hai koi out of the world cheez nahi hai...

I had the same question and a friend pointed out to the solution: If n is not prime then there are two positive integers a and b such that: n = ab If a = b then n = a² and a is the root If a != b then either a

@@Buddharta thanks buddy understood

@@Buddharta Bro can you explain by giving integer examples? Why we are square root the number first?

@@mdf1259 sure, bro. For every non prime there is an interger product that expresses it, for example: 6=2x3 20=2x10 9=3x3 27=3x9 Every divisor comes in pairs since you need two numbers to make a product, if you see in the examples above, (2,3) (2,10) (3,3)(3,9) at least one is samller than the square root and if you have one you have the other by dividing

@@Buddharta Bro I didn't get it. Still had a dought For non prime two product needs that is ok. Then what is your logic ? Thanks for your effort and time!

sir firstly tell us that only enter a positive number but zero is a non positive number

The printf part won't be evaluated...suppose val2 = 2; and i = 2;

Wrong

maybe it just for precaution? what i know is tht we need to do tht if the operation going to make change to the ori value.

Code execution time pdh Jayega kyuki loop phir boht Baar chlega

Have you got java tutorials

Sure, just name it "Program to print prime numbers" 🤓🤓

Try to change last conditional statement to: if(count>0) printf(“Composite”); else printf(“Prime”); You can also check for i/p 0 or 1 at first and print neither prime nor composite

it worked fine for me, re-check your code. :)

The algorithm will look something along the lines of : prime = 1 for i=2 to (n/2) { if(n % i == 0){ prime = 0 break } } if(prime){ print n is prime } else{ print n is not prime } Less computation, no special cases or anything. EDIT: I've completely missed the sqrt() part in the video. It is already computationally efficient. You just need a flag to decrease typographical complexity then.

No it doesn't say 4 is a prime number. At the first iteration, 4 is divisible by 2 and the reminder is 0. So the flag is reset. And yes 1 is the only special case. Also, I have written about sqrt (n) in the previous post, see the 'EDIT' part .

Yup that was my mistake. Was in a hurry. And I've never said you're wrong. And was never in fear to admit my mistakes too. I said there is a better way. And yes, what you're doing is very helpful indeed.

The neso academy has taken time complexity in consideration while writing the program on the other hand there are different methods for finding whether a number is prime or not but this is the most efficient one.

@@subhranil_mukherjee your way is much more inefficient. If you pick large 3 digit numbers like 659 (tested to be a prime number), then you would be testing up to the 300s (n/2) to see if it is prime. You only have to test up to the square root, which is literally up to 26 (much fewer cases). The reason you find the square root is because if there is a pair of factors other than 1 and the number you are testing, the square root would be the center point of two factors being multiplied. Take 26, you don't need to test all the way to 13 because anything below 6 will be multiplied with a number higher than 6 since 6 is the approximate central integer factor. You start at 2, and you've already found out it works because 2 and 13 work. This happens the same way as you get to larger and larger integers.