Given an integer, how do I find the next largest power of two using bit-twiddling?

Question

If I have a integer number n, how can I find the next number k > n such that k = 2^i, with some i element of N by bitwise shifting or logic.

Example: If I have n = 123, how can I find k = 128, which is a power of two, and not 124 which is only divisible by two. This should be simple, but it eludes me.

Solution 1

For 32-bit integers, this is a simple and straightforward route:

unsigned int n;

n--;
n |= n >> 1;   // Divide by 2^k for consecutive doublings of k up to 32,
n |= n >> 2;   // and then or the results.
n |= n >> 4;
n |= n >> 8;
n |= n >> 16;
n++;           // The result is a number of 1 bits equal to the number
               // of bits in the original number, plus 1. That's the
               // next highest power of 2.

Here's a more concrete example. Let's take the number 221, which is 11011101 in binary:

n--;           // 1101 1101 --> 1101 1100
n |= n >> 1;   // 1101 1100 | 0110 1110 = 1111 1110
n |= n >> 2;   // 1111 1110 | 0011 1111 = 1111 1111
n |= n >> 4;   // ...
n |= n >> 8;
n |= n >> 16;  // 1111 1111 | 1111 1111 = 1111 1111
n++;           // 1111 1111 --> 1 0000 0000

There's one bit in the ninth position, which represents 2^8, or 256, which is indeed the next largest power of 2. Each of the shifts overlaps all of the existing 1 bits in the number with some of the previously untouched zeroes, eventually producing a number of 1 bits equal to the number of bits in the original number. Adding one to that value produces a new power of 2.

Another example; we'll use 131, which is 10000011 in binary:

n--;           // 1000 0011 --> 1000 0010
n |= n >> 1;   // 1000 0010 | 0100 0001 = 1100 0011
n |= n >> 2;   // 1100 0011 | 0011 0000 = 1111 0011
n |= n >> 4;   // 1111 0011 | 0000 1111 = 1111 1111
n |= n >> 8;   // ... (At this point all bits are 1, so further bitwise-or
n |= n >> 16;  //      operations produce no effect.)
n++;           // 1111 1111 --> 1 0000 0000

And indeed, 256 is the next highest power of 2 from 131.

If the number of bits used to represent the integer is itself a power of 2, you can continue to extend this technique efficiently and indefinitely (for example, add a n >> 32 line for 64-bit integers).

Solution 2

There is actually a assembly solution for this (since the 80386 instruction set).

You can use the BSR (Bit Scan Reverse) instruction to scan for the most significant bit in your integer.

bsr scans the bits, starting at the most significant bit, in the doubleword operand or the second word. If the bits are all zero, ZF is cleared. Otherwise, ZF is set and the bit index of the first set bit found, while scanning in the reverse direction, is loaded into the destination register

(Extracted from: http://dlc.sun.com/pdf/802-1948/802-1948.pdf)

And than inc the result with 1.

so:

bsr ecx, eax  //eax = number
jz  @zero
mov eax, 2    // result set the second bit (instead of a inc ecx)
shl eax, ecx  // and move it ecx times to the left
ret           // result is in eax

@zero:
xor eax, eax
ret

In newer CPU's you can use the much faster lzcnt instruction (aka rep bsr). lzcnt does its job in a single cycle.

Solution 3

A more mathematical way, without loops:

public static int ByLogs(int n)
{
    double y = Math.Floor(Math.Log(n, 2));

    return (int)Math.Pow(2, y + 1);
}

Solution 4

Here's a logic answer:

function getK(int n)
{
  int k = 1;
  while (k < n)
    k *= 2;
  return k;
}

Solution 5

Here's John Feminella's answer implemented as a loop so it can handle Python's long integers:

def next_power_of_2(n):
    """
    Return next power of 2 greater than or equal to n
    """
    n -= 1 # greater than OR EQUAL TO n
    shift = 1
    while (n+1) & n: # n+1 is not a power of 2 yet
        n |= n >> shift
        shift <<= 1
    return n + 1

It also returns faster if n is already a power of 2.

For Python >2.7, this is simpler and faster for most N:

def next_power_of_2(n):
    """
    Return next power of 2 greater than or equal to n
    """
    return 2**(n-1).bit_length()

View more solutions

Comments

[AndreasT]

Thanks. But I was searching for a "bit-twiddle" way. ;-)

[AndreasT]

Ah great thanks! I had a hard time to understand why you could double k every step, but since you "double" every '1' it became obvious. Thanks for the example.

[John Feminella]

@AndreasT: No problem! (BTW, to see why you need to go all the way up to n >> 16, consider what happens if n is already a power of 2. You'll only have a single 1 bit that needs to cover all of the previous bits, which is why all the shifting is necessary.)

[AndreasT]

wtf! 8-) , will check it out. Thx

[zxcat]

+1 yes, good trick, I like it :) Some more bit-twiddling tricks can be found here: graphics.stanford.edu/~seander/bithacks.html

[zildjohn01]

+1, not what the OP wanted, but strangely enough I needed an answer that would fit in a single inline expression: 2 ^ (floor(log(x) / log(2)) + 1)

[That Realty Programmer Guy]

Is this idea: fill out the all the 1s up to the largest, currently 'on' bit? If so, would it be n+min+1?

[Ian Durkan]

Great example, this algorithm appears in at least one place in the Java APIs and was confusing me.

[endolith]

What if the input is 0? Also, OP asked for the next largest power of 2, floor finds the next lower power of 2, no?

[DanDan]

It gets you the existing value. The next part, y+1 gets you the next largest power.

[Divick]

Though this works but could anyone please give an explanation on why it works?

[Olathe]

This fails. (5 & -5) = 1; 5 + 1 = 6.

[Griffork]

I think there is a typo in your second example, the number after the pipe should be 0100 0001.

[endolith]

If your input is already a power of 2, this will not return it though

[endolith]

Why not just bit shift until it equals zero? Seems like that would be faster.

[John Feminella]

@endolith That takes log N shifts. This takes log(log N) shifts, so it uses fewer shifts.

[endolith]

@JohnFeminella: Yeah, I probably should have read the code more carefully before commenting...

[endolith]

@DanDan: but if n is already a power of 2, it gives the next power of 2 rather than returning n

[jwalker]

It does k >= n, not k > n

[subrat71]

You can safely replace your while with do/while and save one test, as you've already confirmed the initial test with the preceding if statement.

[Derek Illchuk]

Thanks @seh; improved with do while.

[Chris Middleton]

If you're going to use bit shifts, you might as well do shift <<= 1 instead of shift *= 2. Not sure if that's actually faster in Python, but it should be.

[nemesit]

if you just add a n >> 32 line you'd get illegal instruction for values n > 4611686018427387904 , so for 64-bit you need a check whether n is beyond the largest power of two that actually fits in the 64-bit

[yyny]

Talking about assembly in a Python answer adds no value since Python uses byte code...

[Wauzl]

@endolith: If n is a power of two and you want to obtain n instead of "the next power of two", you can use 2 ^ ( ceil(log(x) / log(2)) ) instead. But that was not the question (…how can I find the next number k > n…)

[endolith]

@Wauzl Yeah, but the log method is inaccurate for higher numbers anyway. if x = 536870912 it returns 1073741824 instead of 536870912.

[primo]

Fastest method here, why no upvotes? Although, I changed to while(t > (t & -t)).

[v01pe]

Ha, so simple! Didn't think about that one :)

[oHo]

See also similar answer in another question: stackoverflow.com/a/10143264/938111 (answered by ydroneaud on 2012). See Wikipedia builtin bitwise functions for other compilers: en.wikipedia.org/wiki/Find_first_set#Tool_and_library_suppor‌​t             This a C++ 64-bits version:        constexpr uint64_t nextPowerOfTwo64 (uint64_t x) { return 1ULL << (sizeof(uint64_t) * 8 - __builtin_clzll(x)); }

[NetMage]

@primo At least in C# the accepted answer is over 2.5 times faster on average for 32-bit integers.

[Martin Kealey]

nitpick: The logic is correct, but the description of the Z flag is inverted: jz jumps when the Z flag is set - which is its state after testing a 0 value. (This goes all the way back to Intel's 8008 CPU.)

[Martin Kealey]

By definition uint64_t (if it exists) has 64 data bits and no padding bits, therefore sizeof(uint64_t)*CHAR_BIT must be 64. Perhaps you were thinking of long long or uint_fast64_t?