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.
next = pow(2, ceil(log(x)/log(2))); This works by finding the number you'd have raise 2 by to get x (take the log of the number, and divide by the log of the desired base, see wikipedia for more). Then round that up with ceil to get the nearest whole number power.
The most straightforward way to convert a positive power of two into the form 2n is to count the number n of divisions by 2 that it takes to reach a quotient of 1. For example, the number 524,288 requires 19 divisions to reach 1, giving 219: 524,288/2 = 262,144. 262,144/2 = 131,072.
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).
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