## Two's Complement and negative Numbers

### Introduction

I am in the progress of reading the excellent book Code: The Hidden Language of Computer Hardware and Software. The book walks you through the chronology of inventions that made the computer possible. One chapter is dedicated to number representation and arithmentic, which is where the two's complement is naturally explained.

This is where I realized that I knew the "how it works", but not the "why it works". The explanation in the book didn't get through to me, either. As I had to piece together the information and couldn't find a source, which did the job for me, I decided to write it down in this article.

### What Problem are we solving?

Imagine you already know how to build a machinery, which can add two positive numbers. It doesn't matter whether you use transistors, tubes, relays or gears. Just something that "mechanically" takes two positive numbers and outputs the result. Like this black box: To visualize the operation of an adder machinery, we can look at the number line of whole numbers (natural numbers plus 0). The first operand, which is "4" in our example above, may be seen as the starting point and the second operand, which is "3" in the same example, may be seen as movement into the direction of increasing numbers: Now the goal is to modify this machinery to enable subtraction and not only addition. Using the two's complement achieves this, by actually turning subtraction into addition. How this works will be explained in the remainder of this article.

### Wrapping Around

In terms of our number line we can only move into the direction of increasing numbers using our adder machinery. Subtraction in this analogy is moving into the opposite direction. Let's walk through the steps to achieve this.

For this, observe a limitation of any real-world adder machinery: the number of digits accepted as input and produced as output are limited, because its a physical thing, which cannot accept or output unbounded values. In our simple example above this limit is one digit. The numbers are represented in the decimal system for now, so the digits 0-9 are accepted and produced.

But even when adding two single digits, the result may have two digits as in 8+3=11. But since the output was wired to only have one digit, the result will be 8+3=1. The carry digit is simply discarded, because it has no place to go. The machinery will signal this overflow on a dedicated output. So if we keep adding one to our result and we cross 9, we will in a sense wrap around to 0, then 1, then 2 etc. The key observation to make here now is that with such a limitation, when we add 3 to 8 we arrive at the same number as if we had subtracted 7 from 8. Let's visualize this on our number line: Now there seems to be a way to turn a subtraction into an addition, which arrives at the same result. We need to understand the systematic behind this observation, to add it to our machinery.

### The Method of Complements

To make the wrap-around more apparent, We will turn the line into a circle: If you look at our circle above, we see that starting at 8 and either adding 3 (the grey segment) or subtracting 7 (the blueish segment) will both arrive at 1 as its result. Both distances taken together describe the full circle. This is no coincidence.

A full circle in our decimal system has 10 digits. So if I want to reach the same number moving into either direction, then the two distances sum up to 10. So I either go 3 steps into the negative direction, or 10-3=7 into the positive direction. And this is where the term complement stems from: I take the complement to 10 to change direction.

This is the basic algorithm then:

1. Check if we are adding or subtracting two numbers
2. If we add, just input the two numbers and you're done
3. If we subtract, complement the second operand to 10, then add

And here is the adder extended with the complementer showing a subtraction of 8-7=1 in action: In case we built our adder to accomodate for more digits, let's say 3, then the wrap-around would happen at 999 and our ring had 1000 representable numbers (0 to 999). Thus, we change direction by taking the complement with respect to 1000 (103).

Before we can pad ourselves on our shoulder, there is one thing we need to fix: when we're subtracting, we deliberately cross nine, which falsly triggers the overflow signal like in the example above. It shouldn't.

And since we are subtracting, we will eventually run into underflows, which actually should be singaled, as in 4-6. If we follow through with our algorithm above, we complement 6 to 10, yielding 4 and than add it to the first operand: 4+4=8. The result has no carry, so no underflow is signalled.

You may have noticed that our adder behaves exactly oposing to our desired signaling behavior, so inverting it's behavior will do the trick.

In the below diagrams, I added an inverter to the overflow signal, which is only active when we're subtracting. The diagrams show two cases: one with an underflow and another which stays in range. I'll walk you through it.  The first diagram shows 4-6=8, i.e. a larger number subtractred from a smaller one. The second operand 6 is complemented to 10 to result in 4 and the addition will have a carry of 0, which will be inverted to 1, so an underflow is correctly signalled.

The second one shows 4-3=1, i.e. a smaller number subtracted from a larger one. The second operand 3 is complemented to 10 to result in 7, resulting in 11. The carry will be discarded in the output and inverted for the underflow signal, resulting in no underflow. Nice!

Addition remains unchanged, as the complementer will not complement and the over-/underflow signal will not be inverted. Compare the following two additions, one triggering an overflow and the other that stays in range:  ### The Complementer

We need to talk about the complementer. I admit that I fooled you a bit: the complementer subtracts the operand from 10, so we actually need a subtractor! Sorry I wasn't honest.

But my betrayal wasn't that bad. At least we only need to subtract from a fixed number, which is 10n, with n being the number of digits. Still, high-school subtraction will involve borrowing a one from the next digit, if the digit we are subtracting (the subtrahend) is bigger than the one being subtracted from (the minuend).

We can do at least one easier with a trick: if we don't subtract from 10n, but from 10n-1 and add the missing one back at the end, the result will stay the same, but the complement is taken to a series of nines. No borrows involved anymore. An example for complementing with 1000:

1000-1-777+1=223
is equivalent to
999-777+1=223

Now we can at least regard each pair of digits in isolation and just deal with the problem of subtracting from 9, as 9 is always larger or equally large as any other digit we are subtracting from it. Once we get to binary, this trick will resolve all our problems, so keep it in mind.

A note on notation that drove me mad. In the literature I studied, the complement to 10n is called the ten's complement. The complement to ten minus one is called the nines' complement. It drove me crazy, because the nines' complement is just a series of nines, but not 9n.

What I didn't see for a long time was the position of the apostrophe. The complement to 10n is the ten's complement (genitive), so the complement of ten. The nines' complement (plural) is the complement to a series of nines.

I call that bad word usability.

### Signed Numbers

Our little machine can now subtract positive numbers by complementing one operand, but we can use the complement also to represent negative numbers. We wanted to move from 8 to 1. Not being able to move 7 steps into the negative direction, we moved 3 into the positive, because its effect was the same.

So the numbers -7 and 3 behave the same under addition. As do all the others when complemented to 10. So we might as well regard each number as its negative complemented partner: An explanation of why this works and why I sneakily left out the complement to zero is in order.

Firstly, the ten's complement (mind the genivite ;-) of zero is 10. As we only keep one digit it will fall back to zero. This is deemed as one of the advantages of this method: 0 and -0 have the same representation. No wasted digits.

Secondly, let's take another look at our adder/subtractor machinery, which we painstainkingly designed until now and see how it subtracts the positive numbers 3 from 4: Remember? We assert the subtraction signal to complement the 3, so instead of going back 3 steps in our number circle, we go forward 7 steps, arriving at 1. It wouldn't be any different, if we input 7 directly and didn't assert the subtraction signal: So in short: 4-3 is the same as 4+7. I hope it makes sense now that we could also say: 4-3 is the same as 4+(-3), reinterpreting the 7 as (-3). Only that pesky over-/underflow signal is acting up again. We'll fix that in a second.

We gained the power of expressing negative numbers. In fact, we can just choose any number of digits to be their negative counterparts. Let's pick the digits 4 to 9 to actually mean -6 to -1. I purposely picked an uneven number of digits to be interpreted as negative, because most literature will tell you to cut the set of digits in half and interpret one half to be negative. This approach has benefits in binary as we will see, but it is by no means mandatory to get the complement method to work: Let's test this new interpretation for a subtraction that stays in range: 1+(-4)=-3. So the first operand is 1, the second is the ten's complement to 4, which is 6. But we won't use our complementer, but input the 6 ourselves and we don't assert the subtraction signal, as we are adding a negative number: Effectively we move 6 steps into the positive direction, which is the same as moving 4 steps into the negative direction. The result is 7. Looking at our number circle above, 7 is interpreted as -3, so 1+(-4)=-3. Bingo.

As promised, we need to fix our over-/underflow signal, because the machinery will still signal an overflow for crossing the 0 from the negative direction, e.g. for -2+3=1 (where instead of -2 we would use 8).

It turns out that there is no elegant solution in our decimal system. We would really have to look at it case by case:

1. A positive number added to a positive number: If we started at 0, 1, 2 or 3 and added a positive number, but ended up at 4 (-6), 5 (-5) ..., then we overflowed
2. A negative number added to a negative number: If we started at one of the negative numbers and added a negative number or subtracted a positive one, but ended up positive, we underflowed
3. A negative number added to a positive or vice versa: This combination can never over- or underflow. Just check out the number circle and try it.

I know that it is hard to imagine how to put this logic into our machinery, but as with our complementer I have to ask for your patience, because in binary it will be trivial to implement.

### What we've learned so far

The good news is that you now know every trick there is to know to understand why the complement method works. In the remainder of this article we will merely transfer this knowledge to the binary system. We will do this by first generalizing to arbitrary number bases and then go to binary. But first we sum up what we now know:

• The fixed width digits make the adder wrap-around
• Thus instead of subtracting, we can add the complement to come around from the other side
• When we subtract, we deliberately overflow, so we invert the overflow signal
• The complementer is still subtraction, but a bit easier when we subtract from 10n-1. Will become trivial in binary
• We can input the complement of numbers directly, interpreting the complements as negative numbers, e.g. 1+(-3)
• The over- and underflow need to be checked with a weird if/else statement, which again becomes trivial in binary

Ok, now for some math to generalize this knowledge to arbitary bases.

As you may know, 10 is the radix of our decimal system, which means we have 10 uniqe digits. The complement method works with any radix of any number system, giving this method it's name the radix complement.

In general we can formalize the radix complement as follows :

ac = bn-a,

where ac is the complement to a, b is the radix or base of the number system and n is the number of digits to which we constrain our machinery to.

We also need the complement to the radix minus one, which is called the dimminished radix complement :

where adc is the dimminished radix complement to a, while the rest has the same definition as above. And from the formulas above it is easy to see that the radix complement equals the diminished radix complement plus one.

One property that also translates to the generalized case is that the number we subtract from in the diminished radix complement is always a series of the digit with the highest value in the number system at hand, e.g. 999 in decimal or 444 in quinary or 111 in binary.

Let's translate this knowledge to the quinary system, which has base 5. A quick recap on numbering systems: if we have n digits to our disposal, we count up until we run out of digits, after which we will add a carry to the next position and start from zero again. The quinary system on a number line thus looks like this: where 10 is not the number of fingers on our hands, but the number of stars a very good hotel has. We just don't have this nice symbol 5 in our quinary system to express it, so we're using a combination of the symbols we have to our disposal.

The diagram below shows the number circle for one quinary digit: If we take the example of 2-1=1 and consider our formula above, we can achieve the same result by taking the five's complement of 1, which is 4. It follows that 2+4=11 (look at our number line above, if you don't belive me). Now we discard the second 1 and arrive at the correct result of 1. We just exercised 2+(-1)=1 in quinary.

The complement method works. What about the overflow suppression? Nothing special in quinary, just invert the carry again, if we are subtracting.

What about signed numbers? Still works! Let's do a quick reinterpretation: and test it against 1+(-2). -2 becomes 5-2=3, so instead of going two steps negative, we go 3 steps positive and arrive at 4, which we interpret as -1. 1+(-2)=-1 checks out.

Also in quinary we need subtraction for our complementer module. And also here we can avoid borrows and at least concern ourselves with each digit in isolation, if we complement to the dimished radix, which is 5-1=4 or 53-1=444, when dealing with three digits.

As I said, the dimished radix is always a series of digits with the highest value. Think about it: in any number system, the base itself is always a 1 followed by zeros (like 5 on the number line above is written as 10). 6 in the senary number system (base 6) is also 10. So it should be clear that if we subtract one from 10 or 100 or 1000 etc. that the result will be a series of the digits with the highest value (100-1=99 in decimal, 100-1=44 in quinary etc.).

And like in the decimal case with signed numbers, we don't care about crossing zero anymore when trying to detect over- and underflows. Also in quinary, the logic is still weird. We need to check all cases. As promised, this will resolve in binary.

Finally we can go to binary and the two's complement, which should be a piece of cake as we already know everything there is to know and the two remaining mysteries (complementer and over-/underflow detection) will take care of itself in binary.

### The Two's Complement

We seem to have established a certain procedure, so let's stick to it. First the binary system on a number line, but instead of allowing only one digit, we exand to three. Otherwise we'd already wrap-around after two digits, which looks stupid: Again, we can display the line as a circle to make the wrap-around more obvious: By now, we are experts in changing direction to express subtraction as addition. And we also can use fancy mathematical terms for this act: if we want to execute the subtraction 011-010=001, we turn it into 011+(-010) by taking the radix complement in binary, i.e. the two's complement, from 010:

23-010
expressed in the binary system will become
10 3-010
is equivalent to
1000-010=110

This results in 011+110=1001. Cross out the leading one because of our constraint to only support three digits and we see that it works.

I promised that complementing in binary will become trivial to implement in hardware, so here it goes: I asked you to remember the diminished radix complement, which was

ac=bn-1-a,

which will always lead to a series of the numeral with the highest value in the respective number system. So in binary for three digits, the diminished radix complement looks as follows:

ac=103-1-a,
which is equivalent to
ac=1000-1-a,
which is equivalent to
ac=111-a.

So in binary the dimished radix complement is taken with respect to a series of ones and thus called the ones' complement (mind the stupid apostrophe).

Subtracting any binary number from a series of ones is just flipping all the bits from the subtrahend, e.g. 111-010=101 or 111-001=110. It shouldn't be too hard to imagine that it is easy to implement a bit-flipper in hardware. For those who forgot their digital logic class, it's the behavior of a XOR gate, when one input is asserted.

There you go! To change directions in the number circle, we take the two's complement, which is easily done by first taking the ones' complement (flipping all the bits) and then adding the one back. The adder now looks as follows: That's it for unsigned numbers. We also already suppressed the overflow signal, when deliberately crossing zero on subtraction.

The last and final step is to reinterpret a set of binary numbers as being negative numbers. But before we write the negative numbers to the circle, let's take another look at the unmodified circle with unsigned numbers: Remember the second problem I promised would dissolve in binary? When reinterpreting a set of numbers as negative, we had this weird logic to check if we went from positive to negative numbers etc. That logic was weird, because it was not obvious how to detect if we passed our imaginary over- or underflow threshold.

But in binary, as you can verify above, this becomes easy if we interpret all numbers with a leading one as negative and those with a leading zero as positive. Then we can use the leading bit to check, if two positive numbers were added and resulted in a negative.

Ok, remember we will now reinterpret all values with a leading one to a negative number, but it is just reinterpretation, so the actual values that go into the adder as operands are the ones from the above circle. Here we go: Now the over-/underflow happens when stepping from 011 to 100 (-100) or from 100 (-100) to 011. Let's reiterate our conditions for over-/underflow detection:

1. A positive number added to a positive number: if the leading bits of both operands are zero and the result it one, then we overflowed. This is easy to implement as a logic expression and thus as a logic circuit.
2. A negative number added to a negative number: If both leading bits are one and the leading bit of the result is zero, we underflowed.
3. A negative number added to a positive or vice versa: This combination can never over- or underflow. Just check out the number circle and try it.

And we're done. I never thought that such an everyday seemingly simple concept would take me so long to write about. This was really a surprise, but I am happy I made it this far.

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