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.
> Without loss of generality, let the number be xx...xx100...000,
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.
- 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.
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.
In the "two's complement" representation, there is a sign bit, but the meaning isn't "invert this number", it's "subtract a large, fixed power of 2 from this number".
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.
You're thinking about a sign and magnitude representation, which is not how integers are represented in a modern computer.
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.
Signedness doesn't affect anything here, as long as the representation is in two's complement. The negation of x (i.e. -x) in two's complement is ~x + 1. The interpretation of the bits as signed or unsigned doesn't change any of the following steps.
There's still a sign bit. The sign bit tells you whether to interpret the rest of the number as positive or treat as negative and undo the 2s complement operation.
Some people do see it that way, and I think it's fine if you do, but I think I disagree. To me, the definition of N-bit two's complement is that -x (i.e. 0 - x) is represented by 2^N - x. The fact that that is equal to ~x + 1 is not obvious, or (as I see it) part of the definition.
Or briefly, copied from my StackOverflow answer[1]:
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)
The linked article is wrong in only mentioning 2 signed integer representations. Old versions of C allowed 3 representations for integers, floats use one of those (sign-magnitude, also used widedly by bignum libraries) and one other (offset binary), and base negative-2 is also possible in theory (not sure if practical for anything), for a total of 5.
I find this a pretty confusing explanation because the actual mechanism is buried in the middle and you can't understand the first diagram without it.
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.
Grammar nazi correction, but one I think is both interesting and a useful way to remember how these work: ones' complement has the apostrophe at the end (but two's complement is correct).
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.
Note that this gives 0 when x == O. For applications where you need the largest power of 2 that divides x because you are going to actually divide x by that you may want to check for 0 first.
It is a simple trick, but in practice you probably want to know the position of the last nonzero bit instead of the 2^n pattern. And most CPUs have instructions for that, which are more efficient and as a bonus also document better what is happening.
Nice article, which is putatively about the question posed, but really is about how fun number theory can be. This trick and the "reverse the order of bits with XOR" kinds of things are gateways into the properties of numbers and their bases. Base 2 happens to be "easy" because you can use simple boolean functions to move it around but once you get the hang of it you can do this with other bases as well.
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!
This was a real joy to read. Especially kudos for the (I guess already oldschool?) plain text figures. I was reading this with lynx (as I stll frequently use it) and totally enjoyed the flawless accessibility of this article. Whatever the motivation, thanks for putting out content like this, it is totally appreciated!
Aww thanks! Actually we both should thank Hugo for converting the MarkDown into a very simple HTML structure. I also try to follow semantic web whenever possible (article, header, footer etc. instead of divsoup).
It's easy to see when a number is divisible by 2 in binary,
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.
Readit News
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.
https://en.wikipedia.org/wiki/Two%27s_complement
https://en.wikipedia.org/wiki/Modular_arithmetic
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.
Deleted Comment
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.
[1]: https://stackoverflow.com/a/63552117/1405588
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.