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Nodata
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Hi!
There is an algorithm called the sieve of erastothenes to find a list of prime numbers on a list from 2 to n. I found an implementation of this algorithm saying that you need to iterate prove all the posible divisor from 2 to the square root of n.
Why is that?
 2 years ago
 2 years ago
Nodata Group Title
Hi! There is an algorithm called the sieve of erastothenes to find a list of prime numbers on a list from 2 to n. I found an implementation of this algorithm saying that you need to iterate prove all the posible divisor from 2 to the square root of n. Why is that?
 2 years ago
 2 years ago

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Nodata Group TitleBest ResponseYou've already chosen the best response.0
The reason taht I need to iterate from 2 to the square root of n, is a mathematical reason.
 2 years ago

FoolAroundMath Group TitleBest ResponseYou've already chosen the best response.2
Consider any number 'x'. If you start iterating divisors from 2 onwards, say you get the following: I'll take an example case here, say 96 So, 96%2 = 0 and 96/2 = 48 So from one iteration of divisors which perfectly divides the given number, we get two divisors. Here in this case when we iterate for 2, we get two divisors 2, and 48. Next 96%3 = 0 and 96/3 = 32 So, 3,32 is another pair of divisors Note that with each pair, the 'distance' between the two divisors gets closer and closer. What is the limit of this difference? Clearly, if x is a perfect square, say 625 then the limiting case will be 625%25 = 0 and 625/25 = 25. Where the two divisors are the same. In the case of 96, [sqrt(96)]+1 = 10, so all we need to do is evaluate from 2 to 10. The factors of 96 are (2,48),(3,32),(4,24),(6,16),(8,12). Note that once we reach 10, any higher divisor that exists say for example 24 has already been discovered when we checked for x%4. Thus, it is sufficient to check only upto [sqrt(x)]+1 where [.] = smallest integer less than or equal to x Hence instead of iterating over all values from 2 to n1 or 2 to n/2, we can simply iterate from 2 to sqrt(x).
 2 years ago

Nodata Group TitleBest ResponseYou've already chosen the best response.0
I just read that I did not understand, but I will work on that thank you @FoolAroundMath =)
 2 years ago

Nodata Group TitleBest ResponseYou've already chosen the best response.0
Is this a mathematical equality[sqrt(96)] + 1 = 10?
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
yup.. [] is for integer value
 2 years ago

FoolAroundMath Group TitleBest ResponseYou've already chosen the best response.2
[x] = the greatest integer that is equal to or less than x. http://en.wikibooks.org/wiki/Discrete_Mathematics/Sieve_of_Eratosthenes This link might help
 2 years ago

Nodata Group TitleBest ResponseYou've already chosen the best response.0
ohh I didn't know that, that
 2 years ago
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