Member Avatar for shubham05

I want a very short code (maximum 6 lines code) for identifying the given input is prime number or not ???

  • 3 Contributors
  • 3 Replies
  • 261 Views
  • 4 Hours Discussion Span
  • Latest Post Latest Post by shubham05

Dani AI

Generated

Building on the suggestions from (limit checks to sqrt(n)) and (loop with modulo), the following is a compact VB.NET implementation that handles common edge cases and stays within six lines:

Function IsPrime(n As Long) As Boolean
    If n < 2 Then Return False : If n = 2 Then Return True : If n Mod 2 = 0 Then Return False
    For i As Long = 3 To CLng(Math.Floor(Math.Sqrt(n))) Step 2 : If n Mod i = 0 Then Return False : Next
    Return True
End Function

Notes and cautions:

  • Early exits cover n < 2, n = 2 and even numbers, reducing work before the loop. The loop checks only odd divisors up to floor(sqrt(n)), so complexity is O(sqrt(n)).
  • Declaring i As Long and using CLng keeps the bounds consistent with n As Long. If using classic VB (VB6), replace Math.Floor(Math.Sqrt(n)) with Int(Sqr(n)).
  • For very large values (many digits) or cryptographic needs, switch from trial division to proven primality tests (e.g., Miller–Rabin or deterministic algorithms) and use BigInteger-like libraries (System.Numerics.BigInteger) to avoid overflow and performance issues.

Recommended Answers

All 3 Replies

Member Avatar for Hiroshe

We're not going to write it for you.

You should already know that a prime number is a number that is only divisible by 1 and itself.

So, the most trivial way to check if p is prime is to check to make sure it's not divisible by any number from 1 to p. This will work fine.

We can make it faster by considering that if a number is divisible by 2, then it would be disivible by 4, 6, 8, etc, so we only need to check 2 and odd numbers.

We can futher make it faster by considering for each number we check under the square-root, we are also checking the dividand above the square root. Thus, we only need to check numbers from 2 to sqrt(n).

This is probably the fastest you'll get out of a 6 line implementation without getting too fancy.

Member Avatar for hericles

One of the quickest ways is to check if n divided by 2....sqrt(n) results in an even number. If any of them do the number is not prime.
so:
detemine square root of n (rounded up)

loop through 2 to square root of n, dividing n by number. If even division occurs (modulo) n is not prime.

Member Avatar for shubham05

Thanks for the reply !!!

Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.