How to determine if a number is power 2?

Question

I need to develop an algorithm in Visualg where I enter with a number and it tells me if it’s power of 2 or not. What strategies can I use?

Answer 1

Successively divide your x number by 2.

If the rest is always 0 and you reach the quotient 1 is because x is a power of 2.

Example:

8192/2 = 4096

4096/2 = 2048

2048/2 = 1024

1024/2 = 512

512/2 = 256

256/2 = 128

128/2 = 64

64/2 = 32

32/2 = 16

16/2 = 8

8/2 = 4

4/2 = 2

2/2 = 1

8192 is a power of 2.

Answer 2

The most efficient way is to verify that the number-1 ends in 1111... binary. In Javascript this is:

function isPowOf2(v){
    return v && !(v & (v - 1));
}

var test = [0, 1, 2, 4, 6, 8];

for(var i = 0; i < test.length; ++i){
    console.log(test[i] + " - " + isPowOf2(test[i]));
}

Applying for 8 you can see what is happening mathematically.

v = 8 = b1000 = true

v - 1 = 8 - 1 = 7 = b0111

8 & 7 = b1000 & b0111 = 0

!(8 & 7) = !(0) = true

v && !(8 & 7) = true Q.E.D

Answer 3

For this case just apply a single formula:

Log Value / Log 2.

If the result is a integer then Reported value is based on power of 2.

MATHEMATICAL EXPLANATION

We have to 2 to the power of x is equal to y, where y is the number you want to know if the base is equal to 2, and x is the power of 2 that will result in y.

By the properties of mathematics, and in particular that of logarithms, the following solution is presented.

Heed: Do not set the values as shown in this example, otherwise the result will not be an integer value!!!

Thus, without rounding up, check that the result does not display decimal places (integer), and when not displayed, you will know that this value is formed by base power 2. Good luck!

Answer 4

Divide the number by two until the rest of the division is less than or equal to 1 (use a loop for this)

• If the rest of the division is equal to 1 the number is power of 2
• If the rest of the division is less than 1 the number is not a power of 2

Answer 5

i created a c++ function that checks if a number is power 2 using base 2 logarithms. EXPLANATION: the logarithm is calculated in base 2 of the given number and all we have to do is check if the result of that calculation is an integer, to check if a number is an interim one just put that number in an integer variable X and put itlo also in a real variable Y. If the difference of X and Y is 0 then the result is interim which means that the number is a power of 2, otherwise the number is not a power of 2.

#include<bits/stdc++.h>
using namespace std;


class Solution{
    public:
    // Function to check if given number n is a power of two.
    bool isPowerofTwo(long long n){
        //declaring variables that will help to check if the result of the log is an integer 
        int i=0;
        long double x=0;
        //then a put the result of the log in an integer and the same result in a long double variable
        i=log2l(n);
        x=log2l(n);
        //now we subtract the two results
        //if the subtraction is 0 then the number is a power of two 
        if(x-i==0)return true;
        //if no, the number is not a power of two
        else return false;
        // real simple.  
        
    }
};

Answer 6

"""
 Ejercicio 5.7.6. Potencias de dos.
a) Escribir una función es_potencia_de_dos que reciba como parámetro un número 
 natural,y devuelva True si el número es una potencia de 2, y False en caso contrario.
b) Escribir una función que, dados dos números naturales pasados como parámetros, devuelva la
 suma de todas las potencias de 2 que hay en el rango formado por esos números
 (0sinohayningunapotenciade2entrelosdos).Utilizarlafunción es_potencia_de_dos ,
 descripta en el punto anterior.
"""
import os


def clear():
    """ cls de windows """
    os.system('cls' if os.name == 'nt' else 'clear')


def pausar():
    """ pausa de windows"""
    os.system('pause')


def es_potencia_de_dos(numero):
    """ Verifica si es potencia de dos, en primera instancia descartamos el 1 , que 
    seria 2 a la 0  luego vemos si podemos componer el numero que tenemos 
    como potencia de dos y de paso devolvemos que potencia de dos es , en caso de 
    serlo """

    if numero == 1:
        return True, 0

    r = 1
    for i in range(1, numero):
        r *= 2
        if r == numero:
            return True, i
        else:
            continue

    return False


def leer_numero():
    return input('\nIngrese el numero que desea verificar (* para salir) ')


x = leer_numero()
while x != '*':
    potencia, cual = es_potencia_de_dos(int(x))
    if potencia:
        print('Es potencia, 2 a la ', cual)
    else:
        print('No es potencia de dos')
    x = leer_numero()

clear()
print('Hasta luego..')
os.system('pause')