0 Members and 2 Guests are viewing this topic.
boolean isPrime=true;for(int k = 2 ; k < number ; k++){if ( number/k == 0) isPrime=false;}
I know that But I need to write a formula so Java can know if the number is prime or not
Does it exist, if it does, can you tell me?(I need it to do mah APCS Homework )
I know that But I need to write a formula so Java can know if the number is prime or notEDIT: never mind. I got itCode: [Select]boolean isPrime=true;for(int k = 2 ; k < number ; k++){if ( number/k == 0) isPrime=false;}
boolean isPrime=true;for(int k = 2 ; k < number ; k++){if (number==2 || number/k == 0) isPrime=false;}
oops O_oCode: [Select]boolean isPrime=true;for(int k = 2 ; k < number ; k++){if (number==2 || number/k == 0) isPrime=false;}EDIT: Aww
boolean isPrime=true;for(int k = 2 ; k < number ; k++) if ( number/k == 0) isPrime=false;
It has been proven, by the way, that there is no largest prime number
I once wrote a TI Basic program to find prime numbers that used this pattern I found among the spacing between primes. I think it was able to count out about 1000 primes in rhe first hour and could also save its state at any time so you wouldn't have to start over.
Arrange your numbers in a matrix. The width determines the numbers you can eliminate.Here's the basic concept. The width is a multiple of several prime numbers. If you multiply 3,5,and 7 together you get 105. In the 105 columns, you can eliminate every 3rd starting at 3, every 5th starting at 5, and every 7th starting at 7. You don't need to put in two, since you can simply eliminate every other going down each column (-210 instead of -105)Example using 3*5 (15)0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 2930 31 32 33 34 35 36 37 38 39 40 41 42 43 4445 46 47 48 49 50 51 52 53 54 55 56 57 58 59 1 2 4 7 8 11 13 14 16 17 19 22 23 26 28 29 31 32 34 37 38 41 43 44 46 47 49 52 53 56 58 59The more primes you multiply in (nonprimes are useless, eliminate no more rows) the more you can eliminate! yay!Now for checkerboarding for multiples of 2: 1 7 11 13 17 19 23 29 31 37 41 43 47 49 53 59True, you eliminate two, but who cares? You don't lose any primes.By the way, doing this exactly this way with 105 (0-104) give you 31.428% remaining, whereas 3 gives 33.333% It's exponentially less effective, but still waaay better than just -2.
I have a nice theory working too.It'll take a bit of space, so....Spoiler For stuff you care about: Code: [Select]Arrange your numbers in a matrix. The width determines the numbers you can eliminate.Here's the basic concept. The width is a multiple of several prime numbers. If you multiply 3,5,and 7 together you get 105. In the 105 columns, you can eliminate every 3rd starting at 3, every 5th starting at 5, and every 7th starting at 7. You don't need to put in two, since you can simply eliminate every other going down each column (-210 instead of -105)Example using 3*5 (15)0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 2930 31 32 33 34 35 36 37 38 39 40 41 42 43 4445 46 47 48 49 50 51 52 53 54 55 56 57 58 59 1 2 4 7 8 11 13 14 16 17 19 22 23 26 28 29 31 32 34 37 38 41 43 44 46 47 49 52 53 56 58 59The more primes you multiply in (nonprimes are useless, eliminate no more rows) the more you can eliminate! yay!Now for checkerboarding for multiples of 2: 1 7 11 13 17 19 23 29 31 37 41 43 47 49 53 59True, you eliminate two, but who cares? You don't lose any primes.By the way, doing this exactly this way with 105 (0-104) give you 31.428% remaining, whereas 3 gives 33.333% It's exponentially less effective, but still waaay better than just -2.Edited for better readabilityEdit: Also, I find it worth noting that since the math involved in this is fairly simple, and not based on the size of the number, it is practical for use on RSA cracking
Author
Topic: Formula for getting prime number? (Read 17884 times)
