# Program for Mobius Function

• Difficulty Level : Medium
• Last Updated : 23 Jun, 2022

Mobius Function is a multiplicative function that is used in combinatorics. It has one of three possible values -1, 0 and 1.

Examples:

Input : 6
Output : 1
Solution: Prime Factors: 2 3.
Therefore p = 2, (-1)^p = 1

Input: 49
Output: 0
Solution: Prime Factors: 7 ( occurs twice).
Since the prime factor occurs twice answer
is 0.

Input: 3
Output: -1
Solution: Prime Factors: 3. Therefore p = 1,
(-1) ^ p =-1

Input : 78
Output : 1
Solution: Prime Factors: 3, 13. Therefore p = 2,
(-1)^p = 1

Method 1 (Simple)
We iterate through all numbers i smaller than or equal to N. For every number we check if it divides N. If yes, we check if it’s also prime. If both conditions are satisfied, we check if its square also divides N. If yes, we return 0. If the square doesn’t divide, we increment count of prime factors. Finally, we return 1 if there are an even number of prime factors and return -1 if there are odd number of prime factors.

## C++

 // CPP Program to evaluate Mobius Function// M(N) = 1 if N = 1// M(N) = 0 if any prime factor of N is contained twice// M(N) = (-1)^(no of distinct prime factors)#includeusing namespace std; // Function to check if n is prime or notbool isPrime(int n){    if (n < 2)        return false;    for (int i = 2; i * i <= n; i++)        if (n % i == 0)            return false;       return true;} int mobius(int N){    // Base Case    if (N == 1)        return 1;     // For a prime factor i check if i^2 is also    // a factor.    int p = 0;    for (int i = 1; i <= N; i++) {        if (N % i == 0 && isPrime(i)) {             // Check if N is divisible by i^2            if (N % (i * i) == 0)                return 0;            else                 // i occurs only once, increase f                p++;        }    }     // All prime factors are contained only once    // Return 1 if p is even else -1    return (p % 2 != 0)? -1 : 1;} // Driver codeint main(){    int N = 17;    cout << "Mobius Functions M(N) at N = " << N << " is: "         << mobius(N) << endl;    cout << "Mobius Functions M(N) at N = " << 25 << " is: "         << mobius(25) << endl;    cout << "Mobius Functions M(N) at N = " << 6 << " is: "         << mobius(6) << endl;}

## Java

 // Java program for mobious functionimport java.io.*;public class GFG {         // C# Program to evaluate Mobius    // Function: M(N) = 1 if N = 1    // M(N) = 0 if any prime factor    // of N is contained twice    // M(N) = (-1)^(no of distinct    // prime factors)     // Function to check if n is    // prime or not    static boolean isPrime(int n)    {        if (n < 2)            return false;        for (int i = 2; i * i <= n; i++)            if (n % i == 0)                return false;        return true;    }     static int mobius(int N)    {        // Base Case        if (N == 1)            return 1;         // For a prime factor i check if        // i^2 is also a factor.        int p = 0;        for (int i = 1; i <= N; i++) {            if (N % i == 0 && isPrime(i)) {                 // Check if N is divisible by i^2                if (N % (i * i) == 0)                    return 0;                else                     // i occurs only once, increase f                    p++;            }        }         // All prime factors are contained only        // once Return 1 if p is even else -1        return (p % 2 != 0) ? -1 : 1;    }     // Driver code    static public void main(String[] args)    {        int N = 17;        System.out.println("Mobius Functions M(N) at " +                      " N = " + N + " is: "    + mobius(N));        System.out.println("Mobius Functions M(N) at " +                        " N = " + 25 + " is: " + mobius(25));        System.out.println("Mobius Functions M(N) at " +                          " N = " + 6 + " is: " + mobius(6));    }} // This code is contributed by vt_m

## Python3

 # Python Program to# evaluate Mobius def# M(N) = 1 if N = 1# M(N) = 0 if any# prime factor of# N is contained twice# M(N) = (-1)^(no of# distinct prime factors) # def to check if# n is prime or notdef isPrime(n) :     if (n < 2) :        return False    for i in range(2, n + 1) :        if (i * i <= n and n % i == 0) :            return False    return True def mobius(N) :         # Base Case    if (N == 1) :        return 1     # For a prime factor i    # check if i^2 is also    # a factor.    p = 0    for i in range(1, N + 1) :        if (N % i == 0 and                isPrime(i)) :             # Check if N is            # divisible by i^2            if (N % (i * i) == 0) :                return 0            else :                 # i occurs only once,                # increase f                p = p + 1     # All prime factors are    # contained only once    # Return 1 if p is even    # else -1    if(p % 2 != 0) :        return -1    else :        return 1 # Driver CodeN = 17print ("Mobius defs M(N) at N = {} is: {}" .         format(N, mobius(N)),end = "\n")print ("Mobius defs M(N) at N = {} is: {}" .        format(25, mobius(25)),end = "\n")print ("Mobius defs M(N) at N = {} is: {}" .          format(6, mobius(6)),end = "\n")                                     # This code is contributed by# Manish Shaw(manishshaw1)

## C#

 // C# Program to evaluate Mobius Functionusing System; public class GFG{         // M(N) = 1 if N = 1    // M(N) = 0 if any prime factor    // of N is contained twice    // M(N) = (-1)^(no of distinct    // prime factors)     // Function to check if n is    // prime or not    static bool isPrime(int n)    {        if (n == 2)        return true;         if (n % 2 == 0)        return false;        for (int i = 3; i * i <= n / 2; i += 2)            if (n % i == 0)            return false;        return true;    }     static int mobius(int N)    {                 // Base Case        if (N == 1)        return 1;         // For a prime factor i check        // if i^2 is also a factor.        int p = 0;        for (int i = 2; i <= N; i++)        {            if (N % i == 0 && isPrime(i)) {                 // Check if N is divisible by i^2                if (N % (i * i) == 0)                return 0;                else                 // i occurs only once, increase f                p++;            }        }         // All prime factors are contained only        // once Return 1 if p is even else -1        return (p % 2 != 0) ? -1 : 1;    }     // Driver code    static public void Main()    {                          Console.WriteLine("Mobius Functions M(N) at " +                         "N = " + 17 + " is: " + mobius(17));        Console.WriteLine("Mobius Functions M(N) at " +                         "N = " + 25 + " is: " + mobius(25));        Console.WriteLine("Mobius Functions M(N) at " +                          "N = " + 6 + " is: " + mobius(6));             }} // This code is contributed by vt_m



## Javascript



Output:

Mobius Functions M(N) at N = 17 is: -1
Mobius Functions M(N) at N = 25 is: 0
Mobius Functions M(N) at N = 6 is: 1

Time Complexity: O(n√n )
Auxiliary Space: O(1)

Method 2 (Efficient)
The idea is based on efficient program to print all prime factors of a given number. The interesting thing is, we do not need inner while loop here because if a number divides more than once, we can immediately return 0.

## C++

 // Program to print all prime factors# include using namespace std; // Returns value of mobius()int mobius(int n){    int p = 0;     // Handling 2 separately    if (n%2 == 0)    {        n = n/2;        p++;         // If 2^2 also divides N        if (n % 2 == 0)           return 0;    }     // Check for all other prime factors    for (int i = 3; i <= sqrt(n); i = i+2)    {        // If i divides n        if (n%i == 0)        {            n = n/i;            p++;             // If i^2 also divides N            if (n % i == 0)               return 0;        }    }     return (p % 2 == 0)? -1 : 1;} // Driver codeint main() {    int N = 17;    cout << "Mobius Functions M(N) at N = " << N << " is: "         << mobius(N) << endl;    cout << "Mobius Functions M(N) at N = " << 25 << " is: "         << mobius(25) << endl;    cout << "Mobius Functions M(N) at N = " << 6 << " is: "         << mobius(6) << endl;}

## Java

 // Java program to print all prime factorsimport java.io.*; class GFG {         // Returns value of mobius()    static int mobius(int n)    {        int p = 0;             // Handling 2 separately        if (n % 2 == 0)        {            n = n / 2;            p++;                 // If 2^2 also divides N            if (n % 2 == 0)                return 0;        }             // Check for all other prime factors        for (int i = 3; i <= Math.sqrt(n);                                    i = i+2)        {            // If i divides n            if (n % i == 0)            {                n = n / i;                p++;                     // If i^2 also divides N                if (n % i == 0)                    return 0;            }        }             return (p % 2 == 0)? -1 : 1;    }         // Driver code    public static void main (String[] args)    {        int N = 17;        System.out.println( "Mobius Functions"               + " M(N) at N = " + N + " is: "                                 + mobius(N));        System.out.println ("Mobius Functions"               + "M(N) at N = " + 25 + " is: "                                + mobius(25));        System.out.println( "Mobius Functions"                + "M(N) at N = " + 6 + " is: "                                 + mobius(6));    }} // This code is contributed by anuj_67.

## Python3

 # Python Program to evaluate# Mobius def M(N) = 1 if N = 1# M(N) = 0 if any prime factor# of N is contained twice# M(N) = (-1)^(no of distinct# prime factors)import math # def to check if n# is prime or notdef isPrime(n) :     if (n < 2) :        return False    for i in range(2, n + 1) :        if (n % i == 0) :            return False        i = i * i    return True def mobius(n) :     p = 0     # Handling 2 separately    if (n % 2 == 0) :             n = int(n / 2)        p = p + 1         # If 2^2 also        # divides N        if (n % 2 == 0) :            return 0          # Check for all    # other prime factors    for i in range(3, int(math.sqrt(n)) + 1) :             # If i divides n        if (n % i == 0) :                     n = int(n / i)            p = p + 1             # If i^2 also            # divides N            if (n % i == 0) :                return 0        i = i + 2            if(p % 2 == 0) :        return -1    else :        return 1 # Driver CodeN = 17print ("Mobius defs M(N) at N = {} is: {}\n" .                        format(N, mobius(N)));print ("Mobius defs M(N) at N = 25 is: {}\n" .                          format(mobius(25)));print ("Mobius defs M(N) at N = 6 is: {}\n" .                          format(mobius(6)));                                         # This code is contributed by# Manish Shaw(manishshaw1)

## C#

 // C# program to print all prime factorsusing System;class GFG {         // Returns value of mobius()    static int mobius(int n)    {        int p = 0;             // Handling 2 separately        if (n % 2 == 0)        {            n = n / 2;            p++;                 // If 2^2 also divides N            if (n % 2 == 0)                return 0;        }             // Check for all other prime factors        for (int i = 3; i <= Math.Sqrt(n);                                    i = i+2)        {            // If i divides n            if (n % i == 0)            {                n = n / i;                p++;                     // If i^2 also divides N                if (n % i == 0)                    return 0;            }        }             return (p % 2 == 0)? -1 : 1;    }         // Driver Code    public static void Main ()    {        int N = 17;        Console.WriteLine( "Mobius Functions"              + " M(N) at N = " + N + " is: "                                + mobius(N));        Console.WriteLine("Mobius Functions"             + "M(N) at N = " + 25 + " is: "                                + mobius(25));        Console.WriteLine( "Mobius Functions"               + "M(N) at N = " + 6 + " is: "                                + mobius(6));    }} // This code is contributed by anuj_67.



## Javascript



Output:

Mobius Functions M(N) at N = 17 is: -1
Mobius Functions M(N) at N = 25 is: 0
Mobius Functions M(N) at N = 6 is: 1

Time Complexity: O(√n)
Auxiliary Space: O(1)

Please suggest if someone has a better solution which is more efficient in terms of space and time.