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Jul 21, 2018 · @Servy What do you mean with "If it's sufficiently small it's not even going to be inefficient"? If you sieve up to sqrt(n) to get the primes you need for trial division, the sieving is more work than the unnecessary divisions by composites, if you avoid multiples of 2, 3, and maybe 5, if you're enterprisy.
Jun 20, 2018 · It works on the Euclidean algorithm to calculate H.C.F. Basically, I check if the HCF of the number AND the consecutively second number is 1; and if the number itself is divisible by either 2 or 3. Don't ask how I mathematically reached the solution, it just struck me :D.
If a number is prime it will have 2 factors (1 and number itself). If it's not a prime they will have 1, number itself and more, you need not run the loop till the number, may be you can consider running it till the square root of the number. You can either do it by euler's prime logic. Check following snippet:
Oct 20, 2017 · Both return either True or False. Function isPrime1 is very fast to return False is a number is not a prime. For example with a big number. But it is slow in testing True for big prime numbers. Function isPrime2 is faster in returning True for prime numbers. But if a number is big and it is not prime, it takes too long to return a value.
Aug 16, 2016 · The simplest test is to start with trial division by small primes. Your statement that it is 6n + 1 6 n + 1 represents trial division by 2 2 and 3 3. You can keep going until you get tired. Then try Fermat's little theorem, which says that for p p prime and a a coprime to p p, ap−1 ≡ 1 (mod p) a p − 1 ≡ 1 (mod p), so see if ...
using your code, and focusing on code structure: def is_prime(x): # function contents must be indented if x == 0: return False elif x == 1: return False # your base cases need to check X, not n, and move them out of the loop elif x == 2: return True for n in range(3, x-1): if x % n == 0: return False # only return true once you've checked ALL the numbers(for loop done) return True
Dec 2, 2018 · I wouldn't burden your user with that mysterious second argument but rather present a different method of just one argument, that first deals with numbers less than 2 and even numbers, and then calls into your recursive method with the proper arguments:
Apr 25, 2015 · I just swapped these values. The rest of the code is for checking number 2 (which is prime) and number less than 2 (that are not primes) I forgot to mention that the comparison Y < X is buggy, because you want to test for all numbers between 2 and X-1, that comparison includes X.
Oct 23, 2016 · 8. Your method for finding if your number is prime is the correct method. To make it so that it does not consistently print out whether or not the number is prime, you could have an external variable, which represents whether or not the number is prime. Such as. boolean prime = true; for (int p = 2; p < sum; p++) {.
Dec 13, 2010 · 2. Use mathematics first find square root of number then start loop till the number ends which you get after square rooting. check for each value whether the given number is divisible by the iterating value .if any value divides the given number then it is not a prime number otherwise prime. Here is the code.
