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Program for Mobius Function in C++
Given a number n; the task is to find the Mobius function of the number n.
What is Mobius Function?
A Mobius function is number theory function which is defined by
$$\mu(n)\equiv\begin{cases}0\1\(-1)^{k}\end{cases}$$
n= 0 If n has one or more than one repeated factors
n= 1 If n=1
n= (-1)k If n is product of k distinct prime numbers
Example
Input: N = 17 Output: -1 Explanation: Prime factors: 17, k = 1, (-1)^k 🠠(-1)^1 = -1 Input: N = 6 Output: 1 Explanation: prime factors: 2 and 3, k = 2 (-1)^k 🠠(-1)^2 = 1 Input: N = 25 Output: 0 Explanation: Prime factor is 5 which occur twice so the answer is 0
Approach we will be using to solve the given problem −
- Take an input N.
- Iterate i from 1 to less than N check the divisible number of N and check if it is a prime or not.
- If the both conditions satisfy we will check if the square of the number also divides N then the return 0.
- Else we increment the count of prime factors, if the count number is even then return 1 else if it’s odd return -1.
- Print the result.
Algorithm
Start Step 1→ In function bool isPrime(int n) Declare i If n < 2 then, Return false Loop For i = 2 and i * i <= n and i++ If n % i == 0 Return false End If Return true Step 2→ In function int mobius(int N) Declare i and p = 0 If N == 1 then, Return 1 End if Loop For i = 1 and i <= N and i++ If N % i == 0 && isPrime(i) If (N % (i * i) == 0) Return 0 Else Increment p by 1 End if End if Return (p % 2 != 0)? -1 : 1 Step 3→ In function int main() Declare and set N = 17 Print the results form mobius(N) Stop
Example
#include<iostream>
using namespace std;
// Function to check if n is prime or not
bool isPrime(int n) {
int i;
if (n < 2)
return false;
for ( i = 2; i * i <= n; i++)
if (n % i == 0)
return false;
return true;
}
int mobius(int N) {
int i;
int p = 0;
//if n is 1
if (N == 1)
return 1;
// For a prime factor i check if i^2 is also
// a factor.
for ( 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 p
p++;
}
}
// All prime factors are contained only once
// Return 1 if p is even else -1
return (p % 2 != 0)? -1 : 1;
}
// Driver code
int main() {
int N = 17;
cout << mobius(N) << endl;
}Output
N = -1
Updated on: 20-Dec-2019
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