In the last chapter we discussed the derivation of how to represent numbers in our machine diagrams. However, the representing something is meaningless if you don’t have any means of interacting with that thing. In this chapter we will come up with a way of working with numbers, and implement it as a machine diagram. This is going to be the “heaviest” chapter yet, but the material is surprisingly easy to follow as long as you Don’t Panic!

Takeaway: Take complicated-seeming things slow and make sure you Don’t Panic!

Addition as a Syntactic Operation

Now that we have a notion of numbers, there are two natural things we might want to use them for: adding two numbers together, and describing how many times we want some “action” done. Because we don’t yet have a notion of what an “action” might be (remember, our machine diagrams never change, once they’ve been given their inputs), it seems more prudent to look into adding two numbers together.

Just like when we started looking for a way of representing numbers, our approach will be to study how adding numbers works in a symbolic sense. Asking about some kind of meaning behind adding two numbers can’t possibly be a fruitful endeavor, since we can’t teach our diagrams about meaning.

This approach is know as being syntactic in nature, which means it follows the structure of what the problem “looks like”. Syntax is in strict dichotomy with semantics, which describe the meaning and interpretation of something.

A useful intuition behind syntax and semantics is that syntax exists in the world, and semantics exist in our minds. Obviously, syntax is going to be the easier thing to work with in most cases.

Takeaway: Syntax is the factual structure of something, while semantics is the interpretation we put on that structure.

The divide between syntax and semantics requires a much deeper philosophical dive than we will devote to it here today, but we’ll discuss it in the next interlude. In particular, the interplay between syntax and semantics goes a long way to answering some of our earlier philosophical questions behind “why does any of this work?”

But I digress. The question we’re asking is “how does addition between numbers work?” and furthermore, “what does addition look like on a syntactic level?”

The good news is that you already know the answer to both of these things. Pop quiz: using a pen and paper, compute the answer to \(1243+412\). I’ll give you a second.

Done? Good. Unless you’re a math wizard, your approach to computing this sum probably looked something like filling in this table, one column at a time:

1 2 4 3
+ 0 4 1 2

The approach is this: starting from the right, add the two symbols together (which you’ve memorized), fill that symbol into the bottom line in the same column, and repeat with the column to the left of it.

Make no mistake, this computation we’ve performed was purely syntactic. You didn’t have to know anything about numbers to do it. In fact, if we completely jumbled up all symbols with ones you’d never seen before, and gave you a book about how to add any two of these new symbols, you’d still do just fine.

Let’s do one more example, just to make sure we’ve got it down:

2 4 7 7
+ 0 4 2 6

Keep the syntactic process we just discussed in mind as you work through this example.

Uh oh! Notice anything wrong? Our original strategy forgot to take into account what happens when we add up two numbers and they’re larger than ten. We forgot we sometimes needed to carry a one between columns!

Takeaway: Always check a couple of examples before assuming you got it right.

Let’s fill in the table from the last example together.

1 1
2 4 7 7
+ 0 4 2 6
2 9 0 3

Whenever one column overflows ten, we need to do our old shuffle again, where we move the 1 over to the next column. Furthermore, we realize that each column is potentially a sum of three numbers – the two we’re adding, and a carry from the last column, if necessary.

Great! So we have some idea of what we need to do. Again, like with our number representation, all we need to do to translate this knowledge to binary is to replace the word “ten” with “two” in the entire analysis. It might not be immediately obvious why this is a safe thing to do, but it is because we never did any analysis that depended on the number ten itself. If we had made a claim like “because 10 is 2 more than 8”, then we might be in hot water, but we didn’t so we’re cool.

Because our analysis of how to perform addition is on a per-column basis, our machine diagram of addition should itself be in terms of how to add a column. We said that addition of a column depends on three things: the two numbers being added, and a value being carried in from the previous column. The output is the sum to put in this column, and the carry to move to the next column.

In short, we want a machine that looks like this:

(where Cin is short for “carry-in”, Cout is likewise “carry-out”, and S is the sum for the column)

Pack Up and Carry Out

What we need now is to determine the function table for such a machine. We’ll start with Cout, since it’s easier to reason about. We need to carry a 1 when our column has overflown. Our column can store a maximum of \(1\text{b}\), but we’re adding three wires together, which is potentially as large as \(11\text[b] = 3\). Therefore, we want to output a 1 on Cout whenever two-or-more of our inputs are 1.

Our function table for Cout must then look like this:

A B Cin Cout
0 0 0 0
0 0 1 0
0 1 1 1
0 1 0 0
1 1 0 1
1 1 1 1
1 0 1 1
1 0 0 0

Turning function tables into combinations of gates is a bit of an art, but thankfully it’s pretty easy to reason our way through this one. We want to output a 1 whenever two-or-more of our inputs are 1.

We know how to output a 1 if two inputs are 1 (an and gate), and we know how to output a 1 if any of our inputs is a 1 (an or gate), so we’ve already done most of the work.

If we make an and gate for every combination of our inputs, each and gate will tell us if both of them are set. If we then combine each of the output of those and gates with an or gate, we’ll learn if any two of them are set. Makes sense?

Armed with this knowledge, we draw out our machine to calculate Cout:

It’s kind of crazy, I know, but Don’t Panic! While there’s a lot going on here, we don’t need to pay attention to all of it simultaneously. Let’s trace through this diagram together.

Start on the left, at input A. Track its wire, and see that it feeds into both the first and second and gates. There’s nowhere else to go here, because we don’t know where the other inputs to those and gates are coming from.

So we go and we look at input B. We follow its wire, and see that it goes into the first and third and gates. Now, the first and gate is fully connected, so we know its output must be 1 only when A and B are both 1. The result of this and gate flows to the top-most or gate, but we don’t know where the other one comes from, so we stop this line of inquiry.

All the way back to the beginning, where we look at input C. Its wire connects to the second and third and gates, both of which are now fully connected. The second and gate computes an “and” of A and C, while the third and computes an “and” of B and C.

So far, so good. We’ve computed the “and” of every possible pair of A, B, and C. If we follow the outputs of these, we see the top-most or gate now answers the “or” of A and B or A and C. The bottom-most or gate answers A and C or B and C.

Finally, we get to the last or gate, which computes A and B or A and C or B and C. Which is to say, it computes whether any two of A, B, and C are 1. Exactly what we wanted.

Wow! What an ordeal. Give yourself a pat on the back if you made it through that entire line of reasoning. Give yourself one even if you just managed to read all of it without skimming through.

You might have noticed that we did strictly more work than was necessary; in particular, we computed an or with A and C twice. You can actually see it in the diagram – it’s the big left-half of a rectangle of wires coming out of the A and C gate. Doing more work than necessary isn’t a problem, and it helped make the diagram a little prettier.

Being More Exclusive

Before we get into figuring out how to compute the S output of our Add machine, let’s take a short intermission and build a gate that will turn out to be very helpful in our endeavors.

When you think about it, the semantics of the or gate are kinda funky. Recall that or is 1 if either or both of its inputs is 1. That’s generally not how humans think about “or”. If your friend asked you whether you wanted to go skiing this weekend or next, and you replied “yes”, they’d probably be rather surprised if you meant that you wanted to go both weekends.

When humans say “or”, we usually mean “one or the other, but not both”. This is what’s known as an exclusive or, or xor for short. Let’s take a peek at its function table.

A B A xor B
0 0 0
0 1 1
1 1 0
1 0 1

(We’ve labeled the Output column to be A xor B here as a shorthand – since xor (and the other gates we’ve looked at) only ever give us back one output, it’s unambiguous to refer to this output as the result of the gate on the inputs.)

Study the function table above – notice that A xor B is 1 only when A and B have different values.

We can use the same trick we did for Cout to write this function table out in gates: we want A and (not B) or (not A) and B. The phrasing of that sentence should be very suggestive of what the implementation of this machine will be.

Which, of course, we give a new symbol due to popularity:

Takeaway: xor determines whether two inputs have different values.

Some (Sum) Output

Back, finally, to the task at hand: defining the column sum of A, B, and Cin. We wanted to do that a long time ago, though you’ve probably forgotten since we’ve been through such a journey to get here. Don’t give up, the finish line is in sight!

We built all of that xor stuff because xor has an interesting property: it’s output looks a lot like the sum of two binary numbers, if you ignore carrying.

No! It’s not just crazy talk! Let’s take a look at a few examples:

  • 0 xor 0 is 0, which is the same as \(0\text{b}+0\text{b}=0\text{b}\)
  • 0 xor 1 is 1, which is the same as \(0\text{b}+1\text{b}=1\text{b}\)
  • 1 xor 1 is 0, which is the same as \(1\text{b}+1\text{b}=10\text{b}\) if we focus only on the rightmost digit

At first blush, that “focusing only on the rightmost digit” is a little worrisome, until you realize that that is exactly the semantics we want for the sum. Why? Because the sum is supposed to be the result we leave in a column after addition completely ignoring any carries. Another way to think of xor is addition that doesn’t care about carries.

Which, of course, makes our implementation of the adder sum very simple – we just need to xor “add” our three inputs together, and the result is what we want.

Let’s do it. Last one for today, I promise.

Easy. Nothin’ to it. All that’s left is to implement Add in terms of Cout and Sum by routing A, B, and Cin to the same input wires in both of our smaller, “helper” machines. If you want to try your hand at it, go ahead, but I’ll assume you know how to (literally) connect the dots.

In the next chapter we’ll hook up several of these column adders in a chain, and tie all of the pieces together. We’ll then take a break from numbers for a while and discuss some ways of making these messes of machine diagrams a little easier to work with.


None. Take a break. You’ve earned it.