Monday, January 25, 2016

Check if a number can be expressed as x raised to power y


Check if a number can be expressed as x^y (x raised to power y) - GeeksforGeeks
Given a positive integer n, find if it can be expressed as xy where y > 1 and x > 0. x and y both are integers.

Check if the number is even or odd, initialize X with 2 or 3 based on that, and then increment X by 2. It will directly reduce the calculations to half.
Also, better approach is to find the prime factors of N, and then use them to find all the X such that X^Y=N.
An alternative approach to the problem could be finding the prime factorization of number n,
i.e. say the prime factorization of n= p1^x1*p2^x2*.......*pk^xk ,
then the number could only be expressed as x^y if and only if gcd(x1,x2,....,xk)!=1;
i.e. if the gcd is 1 or number is prime then it can't be expressed as x^y .
Moreover the time complexity of finding prime factorization is O(sqrt(n)) and gcd is log(n).Finding prime factorization could be improved using some sieve like erasthothenes.
int gcd(int a,int b){
if(a<b){
int t=a;a=b;b=t;
}
if(b==0)
return a;
return gcd(b,a%b);
}
int test_if_power(int n){
bool flag=true;
int t=n;
int curr_gcd=INT_MAX;
// if n==1 then it can't be expresses as x^y for y>1
if(n==1)
    flag=false;
if(n%2==0){
int p=0;
while(n%2==0)
{
n=n/2;
p++;
}
// find the power of 2 in the prime factorization of n
if(p==1)
flag=false;
if(curr_gcd==INT_MAX)
curr_gcd=p;
}

for(int i=3;i<=sqrt(n);i+=2){
if(n%i==0){
int p=0;
while(n%i==0){
n/=i;
p++;
}
// if for any prime factor p the power is 1 then n can not be expressed as x^y
if(p==1)
flag=false;
// if i is the first prime factor of p then set the gcd to power of i;
if(curr_gcd==INT_MAX)
curr_gcd=p;
else
curr_gcd=gcd(p,curr_gcd);
//find the gcd of curr_gcd and the power of prime factor i;

}
}
if(n>2){
flag=false;
}
if(!flag){
// return in case flag is false or curr_gc is 1;
return 0;
}
if(curr_gcd!=1)
cout<<t<<" ";
}



One optimization in above solution is to avoid call to pow() by multiplying p with x one by one.
bool isPower(unsigned int n)
{
    // Base case
    if (n <= 1) return true;
    // Try all numbers from 2 to sqrt(n) as base
    for (int x=2; x<=sqrt(n); x++)
    {
        unsigned  p = x;
        // Keep multiplying p with x while is smaller
        // than or equal to x
        while (p <= n)
        {
            p *= x;
            if (p == n)
                return true;
        }
    }
    return false;
}
The idea is simple try all numbers x starting from 2 to square root of n (given number). For every x, try x^y where y starts from 2 and increases one by one until either x^y becomes n or greater than n.
// Returns true if n can be written as x^y
bool isPower(unsigned n)
{
    if (n==1)  return true;
    // Try all numbers from 2 to sqrt(n) as base
    for (int x=2; x<=sqrt(n); x++)
    {
        unsigned y = 2;
        unsigned p = pow(x, y);
        // Keep increasing y while power 'p' is smaller
        // than n.
        while (p<=n && p>0)
        {
            if (p==n)
                return true;
            y++;
            p = pow(x, y);
         }
    }
    return false;
}
Check if a number can be expressed as x raised to power y
Given a positive integer, find if it can be expressed by integers x and y as x^y where
y > 1 and x > 0.

Example:
Consider the number: 125
Since 125 can be written as x^y where x = 5, y = 3, so return true.
Consider the number: 120
As 120 cannot be expressed as x^y for any integers x > 0 and y > 1, so return false.

    public static boolean isPower(int a) {
        int factor = 2;
        while (factor <= Math.sqrt(a)) {
            int number = a;
            while (number % factor == 0) {
                number /= factor;
            }
            if (number == 1) {
                return true;
            } else {
                factor++;
            }
        }
        return false;
    }
http://www.ideserve.co.in/learn/check-if-a-number-can-be-expressed-as-x-raised-to-power-y-set-2
Instead of repeated divisions as given in previous method, we can apply mathematical formula as described:
Given a number a, we have to find integers x and y, such that:
     a = x^y
If we take log on both sides, we get:
     log a = log (x^y)
     log a = y log x
     y = log a / log x
Hence, we have to find an integer x for which RHS gives an integer y.

Algorithm:
1: Starting with i = 2, if (log a / log i) is an integer then return true.
2: Else increment i by 1 till i < √a.
3: If no such i is found, return false.
    public static boolean isPowerMath(int a) {
        if (a == 1)
            return true;
        for (int i = 2; i <= Math.sqrt(a); i++) {
            double value = Math.log(a) / Math.log(i);
            if ((value - (int) value) < 0.000000001) {
                return true;
            }
        }
        return false;
    }
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