☃️Gap in Primes

The prime numbers are not regularly spaced. For example from 2 to 3 the gap is 1. From 3 to 5 the gap is 2. From 7 to 11 it is 4. Between 2 and 50 we have the following pairs of 2-gaps primes: 3-5, 5-7, 11-13, 17-19, 29-31, 41-43

A prime gap of length n is a run of n-1 consecutive composite numbers between two successive primes (see: http://mathworld.wolfram.com/PrimeGaps.html).

We will write a function gap with parameters:

g (integer >= 2) which indicates the gap we are looking for

m (integer > 2) which gives the start of the search (m inclusive)

n (integer >= m) which gives the end of the search (n inclusive)

In the example above gap(2, 3, 50) will return [3, 5] or (3, 5) or {3, 5} which is the first pair between 3 and 50 with a 2-gap.

So this function should return the first pair of two prime numbers spaced with a gap of g between the limits m, n if these numbers exist otherwise nil or null or None or Nothing (depending on the language).

In C++ return in such a case {0, 0}. In F# return [||]. In Kotlin return []

Examples:

gap(2, 5, 7) --> [5, 7] or (5, 7) or {5, 7}

gap(2, 5, 5) --> nil. In C++ {0, 0}. In F# [||]. In Kotlin return[]`

gap(4, 130, 200) --> [163, 167] or (163, 167) or {163, 167}

([193, 197] is also such a 4-gap primes between 130 and 200 but it's not the first pair)

gap(6,100,110) --> nil or {0, 0} : between 100 and 110 we have 101, 103, 107, 109 but 101-107is not a 6-gap because there is 103in between and 103-109is not a 6-gap because there is 107in between.

Note for Go

For Go: nil slice is expected when there are no gap between m and n. Example: gap(11,30000,100000) --> nil

Ref https://en.wikipedia.org/wiki/Prime_gap

Best Practices

Py First:

def gap(g, m, n):
    previous_prime = n
    for i in range(m, n + 1):
        if is_prime(i):
            if i - previous_prime == g: 
                return [previous_prime, i]
            previous_prime = i
    return None


def is_prime(n):
    for i in range(2, int(n**.5 + 1)):
        if n % i == 0:
            return False
    return True

Py Second:

def gap(g, m, n):
    prev = 2
    for x in range(m|1, n + 1, 2):
        if all(x%d for d in range(3, int(x**.5) + 1, 2)):
            if x - prev == g: return [prev, x]
            prev = x

Py Third:

def is_prime(num):
    square_num =int(num**0.5)+1
    for divisor in range(2,square_num):
        if num % divisor == 0:
            return False
    return True

def gap(g, m, n):
    for num in range(m,n+1):
        if (is_prime(num) and is_prime(num+g) 
            and not any(is_prime(num1) for num1 in range(num+1,num+g))):
            return [num,num+g]
            break
    return None