Is_prime(X):- X > 2,%0 and 1 aren't primes, 2 is dealt with above 1 is mod(X,2),%number is odd N is floor(X/2),%I want to only divide X from 1 to X/2 forall( between(1,N,Z), mod(X,Z) > 0 ).

• C Program To Check Prime Or Not
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Code for Prolog program to check whether a given list is palindrome or not without using a reverse operation. Prolog program to reverse and append the list-1. Code, Example for Program to check whether the number inputted is prime or not in Assembly Language.

I'm a beginner in Prolog but managed to fix your problem. Divisible(X,Y):- 0 is X mod Y,!

Divisible(X,Y):- X Y+1, divisible(X, Y+1). IsPrime(2):- true,!

IsPrime(X):- X. This answer is a follow-up to. IsPrime2/1 is as close as possible to isPrime1/1 with a few crucial changes (highlighted below): isPrime2(2):-!

IsPrime2(X):- X 3, X mod 2 = = 0, Nmax is ceiling(sqrt(X)), isPrime2(X,3,Nmax). IsPrime2(X,N,Nmax):- ( N Nmax - true; 0 = = X mod N, M is N + 2, isPrime2(X,M,Nmax) ). Let's query!?- time(isPrime1(99999989)).% 99,999,990 inferences, 5.082 CPU in 5.078 seconds (100% CPU, 19678881 Lips) true.?- time(isPrime2(99999989)).% 20,002 inferences, 0.001 CPU in 0.001 seconds (100% CPU, 13615185 Lips) true. Does not perform well on large numbers. This is because it is not tail recursive (I think).

Also, you can speed everything up with a few facts about prime numbers. 1) 2 is the only even prime number 2) Any number greater than half the original does not divide evenly isPrime1(2):-!

IsPrime1(X):- X 3, ( 0 is X mod 2 - false; Half is X/2, isPrime1(X,3,Half) ). IsPrime1(X,N,Half):- ( N Half - true; 0 is X mod N - false; M is N + 2, isPrime1(X,M,Half) ).

1?- time(isPrime1(999983)).% 1,249,983 inferences, 0.031 CPU in 0.039 seconds (80% CPU, 39999456 Lips) true. EDIT1 Is it possible to take it a step further? IsPrime/3 is more efficient than because it compares only to previously known primes. However, the problem is generating this list. AllPrimes(Max,Y):- allPrimes(3,Max,2,Y). AllPrimes(X,Max,L,Y):- Z is X+2, Nmax is ceiling(sqrt(X)), ( X = Max - Y = L; ( isPrime(X,L,Nmax) - append(L,X,K),%major bottleneck allPrimes(Z,Max,K,Y); allPrimes(Z,Max,L,Y) )). IsPrime(X,P Ps,Nmax):- ( P Nmax - true%could append here but still slow; 0 = = X mod P, isPrime(X,Ps,Nmax) ).

C Program To Check Prime Or Not

@lefunction i never heard and used trace before. Thank you.1?- trace. True.trace 1?- isPrime(6). Call: (6) isPrime(6)? Creep Call: (7) help(6, 6)? Creep Call: (8) 62? Creep Exit: (8) 62?

Creep Call: (8) G1570 is 6+ -1? Creep Exit: (8) 5 is 6+ -1? Creep Call: (8) G1573 is 6 mod 5? Creep Exit: (8) 1 is 6 mod 5? Creep Call: (8) 10? Creep Exit: (8) 10?

Creep Call: (8) help(6, 5)? Creep Call: (9) 52? Creep Exit: (9) 52?

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Creep Call: (9) G1576 is 5+ -1? Creep Exit: (9) 4 is 5+ -1? – Jul 8 '15 at 20:52. This code is attempting to determine if X is a prime number, by doing the following: let Y = X initially 1. Check to see if the number (Y) is greater than 2.

Assign a new variable (LOW) one-less than the starting number (Y-1) 3. If X mod LOW is greater than zero, then recurse with LOW as the new Y Repeat this until X mod LOW is greater than zero and your mod is 1 (Y=2), then if I'm reading this (and remembering the formula) correctly, you should have X as a prime. If at some point X mod LOW equals zero, then X is a non-prime.

Example: X=6 (non-prime) Y=6, LOW=5, Z = 6 mod 5 = 1 - help(6,5) Y=5, LOW=4, Z = 6 mod 4 = 2 - help(6,4) Y=4, LOW=3 Z = 6 mod 3 = 0 - non prime because it's divisible by 3 in this case Example: X=5 (prime) Y=5, LOW=4, Z= 5 mod 4 = 1 - help(5,4) Y=4, LOW=3, Z= 5 mod 3 = 2 - help(5,3) Y=3, LOW=2 Z= 5 mod 2 = 3 - help(5,2) Y=2, - once you get to this point, X is prime, because LOW=1, and any number mod 1 is greater than zero, and you can't 'X mod 0'. It's effectively iterating over numbers less than X to see if it divides equally (mod = 0). The definition of a prime number: An integer n is prime if the following holds true:. N is greater than 1, and. N is divisible only by itself and 1. Since every integer is (A) divisible by itself, and (B) divisible by 1, there is no real need to check that these conditions hold true.

The real constraint that must be checked is this: For a number n to be prime, the following constraints must be satisfied:. n must be greater than 1, and. No m must exist in the domain 1 m 2,% - N is greater than 2, and 0 = = N mod 2% - and N is odd (no real need to check even numbers, eh? Notdivisible(3,N)% - and N is not divisible any other odd number in the range 3 - (N-1). Notdivisible(N,N).% if we've counted up to N, we're good.

Notdivisible(M,N):-% otherwise.