How to check if a given number is a power of two?

Question

The code below isn't working right for some inputs.

a, i = set(), 1
while i <= 10000:
    a.add(i)
    i <<= 1

N = int(input())
if N in a:
    print("True")
else:
    print("False")

My initial idea was to check for each input if it's a power of 2 by starting from 1 and multiplying by 2 until exceeding the input number, comparing at each step. Instead, I store all the powers of 2 in a set beforehand, in order to check a given input in O(1) . How can this be improved?

Answer 1

Bit Manipulations

One approach would be to use bit manipulations :

(n & (n-1) == 0) and n != 0

Explanation: every power of 2 has exactly 1 bit set to 1 (the bit in that number's log base-2 index). So when subtracting 1 from it, that bit flips to 0 and all preceding bits flip to 1. That makes these 2 numbers the inverse of each other so when AND-ing them, we will get 0 as the result.

For example:

                    n = 8

decimal |   8 = 2**3   |  8 - 1 = 7   |   8 & 7 = 0
        |          ^   |              |
binary  |   1 0 0 0    |   0 1 1 1    |    1 0 0 0
        |   ^          |              |  & 0 1 1 1
index   |   3 2 1 0    |              |    -------
                                           0 0 0 0
-----------------------------------------------------
                    n = 5

decimal | 5 = 2**2 + 1 |  5 - 1 = 4   |   5 & 4 = 4
        |              |              |
binary  |    1 0 1     |    1 0 0     |    1 0 1
        |              |              |  & 1 0 0
index   |    2 1 0     |              |    ------
                                           1 0 0

So, in conclusion, whenever we subtract one from a number, AND the result with the number itself, and that becomes 0 - that number is a power of 2!

Of course, AND-ing anything with 0 will give 0, so we add the check for n != 0 .

---

math functions

You could always use math functions, but notice that using them without care could cause incorrect results :

• math.log(x[, base]) with base=2 :

   import math math.log(n, 2).is_integer()

• math.log2(x) :

   math.log2(n).is_integer()

Worth noting that for any n <= 0 , both functions will throw a ValueError as it is mathematically undefined (and therefore shouldn't present a logical problem).

• math.frexp(x) :

   abs(math.frexp(n)[0]) == 0.5

As noted above, for some numbers these functions are not accurate and actually give FALSE RESULTS :

• math.log(2**29, 2).is_integer() will give False
• math.log2(2**49-1).is_integer() will give True
• math.frexp(2**53+1)[0] == 0.5 will give True !!

This is because math functions use floats, and those have an inherent accuracy problem .

---

(Expanded) Timing

Some time has passed since this question was asked and some new answers came up with the years. I decided to expand the timing to include all of them.

According to the math docs , the log with a given base, actually calculates log(x)/log(base) which is obviously slow. log2 is said to be more accurate, and probably more efficient. Bit manipulations are simple operations, not calling any functions.

So the results are:

Ev : 0.28 sec

log with base=2 : 0.26 sec

count_1 : 0.21 sec

check_1 : 0.2 sec

frexp : 0.19 sec

log2 : 0.1 sec

bit ops: 0.08 sec

The code I used for these measures can be recreated in this REPL (forked from this one ).

Answer 2

Refer to the excellent and detailed answer to "How to check if a number is a power of 2" — for C#. The equivalent Python implementation, also using the "bitwise and" operator & , is this:

def is_power_of_two(n):
    return (n != 0) and (n & (n-1) == 0)

As Python has arbitrary-precision integers , this works for any integer n as long as it fits into memory.

To summarize briefly the answer cited above: The first term, before the logical and operator, simply checks if n isn't 0 — and hence not a power of 2. The second term checks if it's a power of 2 by making sure that all bits after that bitwise & operation are 0. The bitwise operation is designed to be only True for powers of 2 — with one exception: if n (and thus all of its bits) were 0 to begin with.

To add to this: As the logical and "short-circuits" the evaluation of the two terms, it would be more efficient to reverse their order if, in a particular use case, it is less likely that a given n be 0 than it being a power of 2.

Answer 3

In binary representation, a power of 2 is a 1 (one) followed by zeros. So if the binary representation of the number has a single 1, then it's a power of 2. No need here to check num != 0 :

print(1 == bin(num).count("1")) 

Answer 4

Note: this should be a comment on Tomerikoo's answer (currently the most upvoted) but unfortunately Stack Overflow won't let me comment due to reputation points.

Tomerikoo's answer is very well explained and thought-out. While it covers most applications, but I believe needs a slight modification to make it more robust against a trivial case. Their answer is:

(n & (n-1) == 0) and n != 0

The second half checks if the input is an actual 0 which would invalidate the bitwise-and logic. There is another one trivial case when this could happen: input is 1 and the bitwise-and takes place with 0 again, just on the second term. Strictly speaking, 2^0=1 of course but I doubt that it's useful for most applications. A trivial modification to account for that would be:

(n & (n-1) == 0) and (n != 0 and n-1 != 0)

Answer 5

The bin builtin returns a string "0b1[01]?" (regex notation) for every strictly positive integer (if system memory suffices, that is), so that we can write the Boolean expression

'1' not in bin(abs(n))[3:]

that yields True for n that equals 0 , 1 and 2**k .

1 is 2**0 so it is unquestionably a power of two, but 0 is not, unless you take into account the limit of x=2**k for k → -∞ . Under the second assumption we can write simply

check0 = lambda n: '1' not in bin(abs(n))[3:]

and under the first one (excluding 0 )

check1 = lambda n: '1' not in bin(abs(n))[3:] and n != 0

^{_{Of course the solution here proposed is just one of the many possible ones that
you can use to check if a number is a power of two... and for sure not the most
efficient one but I'm posting it in the sake of completeness :-)}}

Answer 6

The following code checks whether n is a power of 2 or not:

def power_of_two(n):
    count = 0
    st = str(bin(n))
    st = st[2:]

    for i in range(0,len(st)):
        if(st[i] == '1'):
            count += 1
        
    if(count == 1):
        print("True")
    else:
        print("False")

Many beginners won't know how code like (n != 0) and (n & (n-1) == 0) works. But if we want to check whether a number is a power of 2 or not, we can convert the number to binary format and see it pretty clearly.

For Example:

      ^ (to the power of)

2^0 = 1    (Bin Value : 0000 0001)  
2^1 = 2    (Bin Value : 0000 0010)  
2^2 = 4    (Bin Value : 0000 0100)
2^3 = 8    (Bin Value : 0000 1000)
2^4 = 16   (Bin Value : 0001 0000)
2^5 = 32   (Bin Value : 0010 0000)
2^6 = 64   (Bin Value : 0100 0000)
2^7 = 128  (Bin Value : 1000 0000)

If you look at the binary values of all powers of 2, you can see that there is only one bit True . That's the logic in this program.

So If we count the number of 1 bit's in a binary number and if it is equal to 1, then the given number is power of 2, otherwise it is not.

Answer 7

n = int(input())
if '1' in list(bin(n))[3:]: #also can use if '1' in bin(n)[3:]  OR can also use format(n, 'b')[1:]
    print("False")
else:
    print("True")

For every number which is power of 2 say(N = 2^n), where n = +integer bin(N)=bin(2^(+int)) will have string of form: 0b10000000 ei 0b1.....zero only if not 0, N is not power of 2.

Also, format(n, 'b') returns bin(n)[2:] so can be used

• Source

  >>> format(14, '#b'), format(14, 'b') ('0b1110', '1110') >>> f'{14:#b}', f'{14:b}' ('0b1110', '1110')

Answer 8

Use *2 instead of bit shifts. Multiplication or addition are much more readable.

Answer 9

Most of the above answers use bin() of format(int(input()), "b")

The below code also works: Ev(x) returns True if x is power of 2

# Ev(x) ~ ispoweroftwo(x)
def Ev(x):
    if x==2:        return True
    elif x%2:       return False
    return Ev(x//2)

The above code is based on generating bin()

#This function returns binary of integers
def binary(x):
        a = ""
        while x!= 0:
                a += str(x%2)
                x = x//2
        return a[::-1]

I = int(input())
print(format(I, "b")) # To cross-check if equal of not
print(binary(I))

I have tried to add my answer because I found what we are doing using bin(x)[3:] or format(x, "b") is almost like asking the boolean answer of whether or not a given number x is divisible by two.....and we keep asking the same

Answer 10

I have written a python function that will check the power of any number:

import math
def checkPowTwo(num):
  x = int(input("The power of the number to be calculated is: "))
  output = math.log(num, x)
  residue = output - int(output)
  if residue == 0:
    print (num, " is a power of the number desired.")
  else:
    print (num, " is not a power of the number desired.")
y = checkPowTwo(int(input()))