Python Program for Number of solutions to Modular Equations
Given A and B, the task is to find the number of possible values that X can take such that the given modular equation (A mod X) = B holds good. Here, X is also called a solution of the modular equation.
Examples:
Input : A = 26, B = 2
Output : 6
Explanation
X can be equal to any of {3, 4, 6, 8,
12, 24} as A modulus any of these values
equals 2 i. e., (26 mod 3) = (26 mod 4)
= (26 mod 6) = (26 mod 8) =Output:2
Input : 21 5
Output : 2
Explanation
X can be equal to any of {8, 16} as A modulus
any of these values equals 5 i.e. (21 mod
8) = (21 mod 16) = 5
If we carefully analyze the equation A mod X = B its easy to note that if (A = B) then there are infinitely many values greater than A that X can take. In the Case when (A < B), there cannot be any possible value of X for which the modular equation holds. So the only case we are left to investigate is when (A > B).So now we focus on this case in depth.
Now, in this case we can use a well known relation i.e.
Dividend = Divisor * Quotient + Remainder
We are looking for all possible X i.e. Divisors given A i.e Dividend and B i.e., remainder. So,
We can say, A = X * Quotient + B Let Quotient be represented as Y ∴ A = X * Y + B A - B = X * Y ∴ To get integral values of Y, we need to take all X such that X divides (A - B) ∴ X is a divisor of (A - B)
So, the problem reduces to finding the divisors of (A – B) and the number of such divisors is the possible values X can take.
But as we know A mod X would result in values from (0 to X – 1) we must take all such X such that X > B.
Thus, we can conclude by saying that the number of divisors of (A – B) greater than B, are the all possible values X can take to satisfy A mod X = B
# Python Program to find number of possible # values of X to satisfy A mod X = B import math # Returns the number of divisors of (A - B) # greater than B def calculateDivisors (A, B): N = A - B noOfDivisors = 0 a = math.sqrt(N) for i in range(1, int(a + 1)): # if N is divisible by i if ((N % i == 0)): # count only the divisors greater than B if (i > B): noOfDivisors +=1 # checking if a divisor isnt counted twice if ((N / i) != i and (N / i) > B): noOfDivisors += 1; return noOfDivisors # Utility function to calculate number of all # possible values of X for which the modular # equation holds true def numberOfPossibleWaysUtil (A, B): # if A = B there are infinitely many solutions # to equation or we say X can take infinitely # many values > A. We return -1 in this case if (A == B): return -1 # if A < B, there are no possible values of # X satisfying the equation if (A < B): return 0 # the last case is when A > B, here we calculate # the number of divisors of (A - B), which are # greater than B noOfDivisors = 0 noOfDivisors = calculateDivisors; return noOfDivisors # Wrapper function for numberOfPossibleWaysUtil() def numberOfPossibleWays(A, B): noOfSolutions = numberOfPossibleWaysUtil(A, B) #if infinitely many solutions available if (noOfSolutions == -1): print ("For A = " , A , " and B = " , B , ", X can take Infinitely many values" , " greater than " , A) else: print ("For A = " , A , " and B = " , B , ", X can take " , noOfSolutions , " values") # main() A = 26B = 2numberOfPossibleWays(A, B) A = 21B = 5numberOfPossibleWays(A, B) # |
Output:
For A = 26 and B = 2, X can take 6 values For A = 21 and B = 5, X can take 2 values
Time Complexity of the above approach is nothing but the time complexity of finding the number of divisors of (A – B) ie O(√(A – B))