human
Ormiston prime Pairs | Function to check if the number is a prime or not.
Corner cases.
This is checked so that we can skip middle five numbers in below loop.
Function to update the frequency array such that freq [ i ] stores the frequency of digit i in n.
While there are digits left to process.
Update the frequency of the current digit.
Remove the last digit.
Function that returns true if a and b are anagrams of each other.
To store the frequencies of the digits in a and b.
Update the frequency of the digits in a.
Update the frequency of the digits in b.
Match the frequencies of the common digits.
If frequency differs for any digit then the numbers are not anagrams of each other.
Returns true if n1 and n2 are Ormiston primes.
Driver code"
Solution in Python:

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Write me a code to ormiston prime Pairs and the other steps required?
assistant
#Here is code to ormiston prime Pairs and the other steps required.
 def isPrime ( n ) :
	if ( n <= 1 ) :
		return False
	if ( n <= 3 ) :
		return True
	if ( n % 2 == 0 or n % 3 == 0 ) :
		return False
	i = 5
	while ( i * i <= n ) :
		if ( n % i == 0 or n % ( i + 2 ) == 0 ) :
			return False
		i = i + 6
	return True
TEN = 10
def updateFreq ( n , freq ) :
	while ( n ) :
		digit = n % TEN
		freq [ digit ] += 1
		n = n // TEN
def areAnagrams ( a , b ) :
	freqA = [ 0 ] * TEN
	freqB = [ 0 ] * TEN
	updateFreq ( a , freqA )
	updateFreq ( b , freqB )
	for i in range ( TEN ) :
		if ( freqA [ i ] != freqB [ i ] ) :
			return False
	return True
def OrmistonPrime ( n1 , n2 ) :
	return ( isPrime ( n1 ) and isPrime ( n2 ) and areAnagrams ( n1 , n2 ) )
n1 , n2 = 1913 , 1931
if ( OrmistonPrime ( n1 , n2 ) ) :
	print ( " YES " )
else :
	print ( " NO " )