1. The largest power of 2 that divides x is just 2^(number of trailing zeros in x)
2. Crucial observation: -x == ~x + 1
3. ~x flips all the bits of x bits, so none of the bits of ~x match those of x. (i.e. (x & ~x) == 0)
4. When you do +1, all the trailing 1's flip AGAIN, becoming zero like they were in x. The next highest 0 (say it was the n'th) also flips, becoming 1... like it was in x.
5. Crucial observation: The n'th 0 did NOT match the corresponding bit in x prior to the increment, therefore it MUST match after the increment. All higher bits stay as-is.
6. This leaves only the n'th bits matching in x and ~x + 1, isolating the highest power-of-2 divisor of x when you AND them.
Without loss of generality, let the number be xx...xx100...000, where x stands for any bit and there are n trailing zeros.
1. The largest power of 2 that divides it is 2^n.
2. The twos-complement representation is xx...xx011...111, where the x's are inverted from before, plus one.
3. After adding one, we get xx...xx100...000. Notice that the carryover stopped exactly where the zero was (which is where our first 1 from the right originally was).
4. All the "x" bits are now inverted from their original values, so their & is zero. All the original trailing zero bits are still zero, so their & is zero. The middle 1 has stayed 1, so its & is 1.
5. Thus, the overall & is 00...00100...000 = 2^n, as required.
Note that this is with loss of generality, since there is no guarantee of a 1 bit anywhere in the number ;) but you can handle the zero case specially.
I battled to understand the post. Your explanation got me there very quickly.
- x is 2^b*k for some odd k < 2^(n-b)
- -x is 2^n - 2^b*k = 2^b*(2^(n-b)-k)
- k is odd by definition, and (2^(n-b)-k) + k = 0 (mod 2^(n-b)). This means that the LSB must be 1 in both operands, which results in a carry out, and in the following ith bits we have that the sum is 0 mod 2^i if and only if the ith bits of (2^(n-b)-k) and k are distinct. Thus x & -x = 2^b.
I’m not trying to dunk on you, but I can’t help to note that the denseness of math is too much for an idiot like me.
Erm no. -x is - 2^b * k and it's the "Two's complement" notation which is 2^n - 2 * b * k
This is because the way we generate the "Two's complement" number. Given A in binary representation, let's call A1 the number you get by flipping every bit in A also called one's complement. Observe how every bit in A+A1 is 1 because if the bit was 0 in A then the bit is 1 in A1 and vica versa. So then that's 2^N-1. Two's complement is created by adding 1 to A1 so then A + A1 is 2^N or A1 = 2^N-A. But you do need to prove this before you can use it.
The reason this has become the most common representation for signed numbers in computer hardware is that it makes the difference between signed and unsigned numbers basically irrelevant as far as the hardware is concerned. When the difference does matter (which it does in division, conversions between different word sizes, conversions to/from text, and sometimes multiplication), there are two different instructions for signed and unsigned, and it's up to the programmer or compiler to pick the right one.
The modern version is two's complement. It still has a sign bit, but negating a number involves more than just changing the sign bit since the representation is modular.
0000 is 0
0001 is 1
0010 is 2
0011 is 3
0100 is 4
0101 is 5
0110 is 6
0111 is 7
1111 is -1
1110 is -2
1101 is -3
1100 is -4
1011 is -5
1010 is -6
1001 is -7
1000 is -8
The highest bit is always 1 when the sign is negative.
x = y | 1 | z
-x = ~x + 1
-x = (~y | 0 | ~z) + 1
-x = ~y | 1 | z
x & -x = (y & ~y) | (1 & 1) | (z & z)
x & -x = 0 | 1 | z
To be annoyingly pedantic, this is not simply an observation, this is essentially the definition of 2's complement arithmetic.
Your explanation is more thorough.
v 000...00010100
~v 111...11101011 (not used directly, all bits opposite)
-v 111...11101100 (-v == ~v + 1; this causes all low 1 bits to overflow and carry)
v&-v 000...00000100 (has a single 1 bit, from the carry)
TIL we have 5 such representations! :)
The goal is to produce the number `0...010...0` with the same number of trailing zeroes as x.
If you flip all the bits in x, then `...10...0` turns into `...01...1`. Adding one cascades the tail into `...10...0`.
You can do this entirely in unsigned math and you don't need to drag two's complement into it. The fact that it corresponds to `x & -x` is a curiosity, and then you have to explain why it works for e.g. negative numbers, where it counts trailing ones instead but still produces a positive.
There are plenty of other tricks like this, e.g. `x & (x - 1) == 0` tells you if a (non-zero) x is a power of two, but it doesn't work for negatives.
Ones' complement is the "complement of ones (plural)" in that if you take an n-bit number and the representation of its additive inverse in that scheme and add them as ordinary binary numbers you get a result that is n ones. Eg. using 4 bits, 6 is 0110, -6 is 1001, adding them with ordinary binary addition gives 1111.
Two's complement is the "complement of two (to the n)" in that if you do the same thing you get 2^n. Eg. 6 is 0110, -6 is 1010, adding them with ordinary binary addition gives 10000, or 2^4.
“Some people, notably Donald Knuth, recommend using the placement of the apostrophe to distinguish between the radix complement and the diminished radix complement. In this usage, the four's complement refers to the radix complement of a number in base four while fours' complement is the diminished radix complement of a number in base 5. However, the distinction is not important when the radix is apparent (nearly always), and the subtle difference in apostrophe placement is not common practice. Most writers use one's and nine's complement, and many style manuals leave out the apostrophe, recommending ones and nines complement.”
I.e., the distinction isn’t as well established as some claim, and in the context of binary presentations it doesn’t really matter.
I really didn't appreciate number theory until I started writing cryptography code and trying to understand some of the 'tricks' used to optimize the algorithms. Fun stuff!
Very glad to know the site works well with Lynx!
just like in decimal: 70 is divisible by 10, 500 is divisible by 100,
so in binary 10 is divisible by 2 10100 is divisible by 4 101010111000 is divisible by 8 etc. (We also know 101010111001 can't be divisible by 8, cause it's only 1 more than a number divisible by 8)
Knowing this we can look at the question backwards take 24 and -24, because 24 is divisible by 8 it must end with a 1 and 3 0s ...0011000
when we take the compliment we get 1100111 which for the same reason must end in a 0 and a run of 1s,
now when we add one to this, all the ones will roll over and we end up with the same tail "1000" as in 24, and since this anything to the left of the tail is a compliment of the corresponding bit in x a bitwise and will just be the tail.
01100 (12, 2 bit even)
10011 (inv 12, & = 0)
10100 (inv 12 +1 aka -12)
00100 &, 2 bit even> On the other hand, if you prefer the result be either -1, 0, or +1, then use:
> sign = (v != 0) | -(int)((unsigned int)((int)v) >> (sizeof(int) * CHAR_BIT - 1));
> // Or, for more speed but less portability:
> sign = (v != 0) | (v >> (sizeof(int) * CHAR_BIT - 1)); // -1, 0, or +1
> // Or, for portability, brevity, and (perhaps) speed:
> sign = (v > 0) - (v < 0); // -1, 0, or +1
Or if you prefer portability, speed, and readability at the same time:
int sign(int i) {
if(i > 0)
return 1;
else if(i < 0)
return -1;
else
return 0;
}
mov ecx, edi
sar ecx, 31
test edi, edi
mov eax, 1
cmovle eax, ecx
ret
(still useful as a reference for implementing an optimizer, to be read as pseudo code)
For example:
#include <stdlib.h>
#include <stdio.h>
void decompose_bits(unsigned x) {
while (x) {
unsigned bit = x & -x;
printf("0x%x (%u)\n", bit, bit);
x -= bit;
}
}
int main(int argc, char *argv[]) {
unsigned x = atoi(argv[1]);
decompose_bits(x);
}And then after that: what use can this be put to?
It (using the address of the nodes as arguments) can serve as a tiebreaker in a Cartesian tree (such as one implementing a first-fit memory allocator) or even to replace the random priority value in a treap (meaning you need neither storage nor computation for the priority node of the treap).
It's that carry that ripples through, making bits flip until it gets to that 'largest power of 2'