Labor of Division (Episode IV): Algorithm D
April 28th, 2021

Algorithm D is Knuth's celebrated multiword integer division algorithm. This post tries to make Algorithm D approachable, and also has an idea for an improvement, in the last section. A followup post provides code to take advantage of that improvement.

To build intuition, we'll use concrete base-10 values, instead of a general base and digit count like Knuth does. It should be straightforward to generalize to arbitrary bases and digit counts.

### Background

The problem is dividing one big number by another big number. The numbers are big enough to require multiple "digits", where a digit is a number in some comfortable base. The examples will be base 10, but in practical applications the base is the width of a hardware divider, or sometimes half that. For example, when dividing a 256-bit number by a 128-bit number, we may pick 64-bit digits, for a base of $2^{64}$, which is huge.

It's hard to beat grade-school long division. In long division, we crank out the quotient digits from left to right by peeling off a prefix of the numerator, dividing that, and then using the remainder in the next step. For example:

1. Compute 1793 ÷ 25. Start with the first three digits of the dividend, i.e. 179.
2. 179 ÷ 25 = 7, remainder 4. Our first quotient digit is 7.
3. Take the remainder and append ("bring down") the next digit 3 to form a new dividend 43.
4. 43 ÷ 25 = 1, remainder of 18.
5. Full quotient is 71 with remainder of 18.

Our dividend was four digits, but through long division we only had to divide a three digit number. This is the general irreducible step in long division: if our divisor is N digits, we must work in N+1 digit chunks from the dividend. So our simplified problem is to compute $q = \lfloor \tfrac n d \rfloor$, where n has exactly one more digit than d, and q is a single digit. (If q were not a single digit, then we peeled off too much of the dividend; back up one digit and try again.)

Now we switch to a sub-algorithm for this simpler problem.

### Estimating the quotient digit

We have an N digit denominator and an N+1 digit numerator; the quotient q is a single digit. How do we find it?

The big idea is to "zero out" the low digits: keep top two from the numerator, and top one from the denominator. For example, instead of $\lfloor \tfrac {2356} {395} \rfloor$, compute $\lfloor \tfrac {2300} {300} \rfloor$, which is the same as $\lfloor \tfrac {23} {3} \rfloor$, which is easy. We hope that this new "estimate" quotient, q̂ will be close to q, and so help us find q:

$$\require{enclose} q = \lfloor \tfrac {2356} {395} \rfloor \\ \hat{q} = \lfloor \tfrac {23\enclose{downdiagonalstrike}{56}} {3\enclose{downdiagonalstrike}{95}} \rfloor = \lfloor \tfrac {2300} {300} \rfloor = \lfloor \tfrac {23} {3} \rfloor = 7\\ q \stackrel{?}{\approx} 7$$

How close is q̂ to q? We'll try to bound the error: let's make q̂ as far off as possible, in both directions.

### Is q̂ too small?

We'll try to make q̂ smaller than q. (Spoiler: we can't.)

If we want q̂ < q, then this zeroing process must make the quotient smaller. Let's arrange for the numerator to be reduced as much as possible (lots of 9s) and the denominator not at all (lots of zeros). For example $\lfloor \tfrac {2399} {300} \rfloor$ has a good chance of making q̂ too small. Apply some algebra:

$\lfloor \tfrac {2399} {300} \rfloor = \lfloor \tfrac {2300 + 99} {300} \rfloor = \lfloor \tfrac {2300} {300} + \tfrac {99} {300} \rfloor = \lfloor \tfrac {23} {3} + \tfrac {99} {300} \rfloor$

Now what is the fractional contribution of the two terms in the last expression? The first term can contribute at most 2/3, and the second at most 99/300, which is less than 1/3. So their sum is always less than 1, so it is always wiped out by the floor, and we have:

$\lfloor \tfrac {2300} {300} \rfloor = \lfloor \tfrac {2399} {300} \rfloor$

Which is kind of neat. So q̂ is never too small.

(This is surprising because it's not true with ordinary division: obviously for example $\tfrac {9999} {100} \gt \tfrac {9900} {100}$. It's true in integer division because of the floor.)

### Is q̂ too big?

Now let's reverse it: make this zeroing process increase the quotient by as much as possible. What if the numerator ends in 0s (so it doesn't shrink) and the denominator end in lots of 9s (so it gets much smaller)? Like:

$q = \lfloor \tfrac {10000} {1999} \rfloor = 5\\ \hat{q} = \lfloor \tfrac {10000} {1000} \rfloor = 10 \not\approx 5$

q̂ is a terrible estimate. When we zeroed out the denominator, we reduced it by nearly half, because its leading digit was 1. If the leading digit were larger, then zeroing out the rest wouldn't affect its value so much.

Let's bound the error using that leading digit. We will just call the numerator $n$, and the denominator has leading digit $v$ (following Knuth), where $0 \lt v \lt 10$. We hope to maximize the error, so arrange for the numerator to have trailing 0s and the denominator trailing 9s. Examine the difference:

$\hat{q} - q = \lfloor \tfrac n {v000} \rfloor - \lfloor \tfrac n {v999} \rfloor$

If we eliminate the floors, the fractional difference is always less than 1 (perhaps less than 0), so we have:

$\hat{q} - q \lt \tfrac n {v000} - \tfrac n {v999} + 1$

Apply some algebra to the difference of fractions:

\begin{align} \tfrac n {v000} - \tfrac n {v999} & = \tfrac {n \times {v999} - n \times {v000}} {v000 \times v999} \\[2ex] & = \tfrac { n \times ( v999 - v000 ) } {v000 \times v999} \\[2ex] & = \tfrac { n \times 999 } {v000 \times v999} \\[2ex] & = \tfrac {999} {v000} \times \tfrac n {v999} \\[2ex] & \lt \tfrac {1} {v} \times \tfrac n {v999} \\[2ex] & \lt \tfrac {1} {v} \times 10 \\[2ex] \end{align} The last step used the fact that the quotient q is a single digit. Inserting this, we get the relation:

$\hat{q} - q \lt \tfrac n {v000} - \tfrac n {v999} + 1 \lt \tfrac {10} v + 1$

So when our denominator's leading digit is 1, q̂ may be too big by as much as 11! But if the leading digit were, say, 8, then q̂ is at most 2 too big.

### Normalization

Now for the trick: we can always arrange for the denominator to have a leading digit of 5 or more, by just multiplying it. If our denominator has a leading digit of, say, 4, just multiply the whole denominator by 2. And then multiply the numerator by 2 also, to keep the quotient the same.

Let's work an example: $$q = \lfloor \tfrac {192} {29} \rfloor = 6 \\[2ex] \hat{q} = \lfloor \tfrac {190} {20} \rfloor = 9$$ q̂ is a bad estimate because the denominator has a low leading digit of 2. So multiply top and bottom by 3 first: $$q = \lfloor \tfrac {192} {29} \rfloor = \lfloor \tfrac {192 \times 3} {29 \times 3} \rfloor = \lfloor \tfrac {576} {87} \rfloor = 6 \\ \hat{q} = \lfloor \tfrac {570} {80} \rfloor = \lfloor \tfrac {57} {8} \rfloor = 7$$ and our estimate q̂ is closer.

Increasing the leading digit of the divisor is the "normalization" step. After normalizing, the leading digit of at least 5 implies that $\hat{q} - q \lt \tfrac {10} 5 + 1 = 3$. And as q̂ and q are integral, we get the awesome result that $\hat{q} - q \le 2$. This normalization step ensures our estimate is off by at most 2!

Note that normalizing increases the leading digit of the divisor but does not add a new digit. Alas, it may add a new digit for the dividend, if you normalize before entering the long division algorithm (and you typically do). Also note that while normalizing does not affect the quotient, it does affect the remainder, which must be scaled down by the normalization factor.

### Multiply and subtract

Now that we have an estimated quotient digit, we proceed prosaically as in long division. Multiply the divisor by the quotient digit and subtract to produce a partial remainder. Slap on the next digit of the dividend ("bring down the...") and that is your new dividend.

If we over-estimated the quotient digit, we may get a negative remainder. For example, when the divisor is "just barely" larger than some factor of the dividend: $$q = \lfloor \tfrac {8000} {4001} \rfloor = 1 \\ \hat{q} = \lfloor \tfrac {8} {4} \rfloor = 2 \\ r = 8000 - \hat{q} \times {4001} \lt 0$$

Whoops, but it's no problem: just decrement our quotient, and add back the divisor, until the remainder is no longer negative. But as we are performing multiword arithmetic, we should try to avoid these expensive extra computations. It would be better to correct the estimated quotient digit immediately.

Now we turn to techniques to do that.

### Correcting the estimate

Here we'll restrict ourselves to dividing a 3 digit dividend by a 2 digit divisor, producing a 1 digit quotient. We will see how to rapidly correct the error in q̂.

In this section, think of a "digit" as half of a machine word. Working with a two digit number is fast, but three or more digits is slow. "Overflow" means "requires more than two digits."

As proven above, we know that (after normalization) q̂ may exceed q by at most 2. Let's return to our previous example, and then extend it by computing a remainder r̂:

$$q = \lfloor \tfrac {576} {87} \rfloor = 6 \\[1ex] \hat{q} = \lfloor \tfrac {570} {80} \rfloor = 7 \\[1ex] \hat{r} = {570} \bmod {80} = 10 \\[1ex] \hat{q} \times 80 + \hat{r} = {570}$$

r̂ is "the remainder of the estimated quotient". It is NOT an estimate of the actual remainder; it has nothing whatsoever to do with the actual remainder, because we ignored the low digit of the divisor. This is true even if the estimated quotient digit is precise. In this sense "r-hat" is a poor choice of symbol, but we'll keep it for continuity with Knuth.

Now our quotient digit 7 is too high, but let's pretend we didn't know that and see what happens: $$\begin{array}{r}7\\ 87\enclose{longdiv}{576}\\ -\underline{609}\\ {\color{red}{?}} \end{array}$$

The actual remainder would be negative. But to discover that, we had to perform an expensive full multiply (7 x 87) which produces 3 digits. We can avoid that full multiply with r̂:

$$\hat{q} \times 87 \gt 576 \implies \\ \hat{q} \times 80 + \hat{q} \times 7 \gt 576 \implies \\ \hat{q} \times 80 + \hat{r} + \hat{q} \times 7 \gt 576 + \hat{r} \implies \\ 570 + \hat{q} \times 7 \gt 576 + \hat{r} \implies \\ \hat{q} \times 7 \gt 6 + \hat{r}$$ So take our estimated quotient digit, multiply it by the last digit in the divisor, and compare that to the sum of the the last digit of the dividend and the "remainder of the estimated quotient". If it's larger, subtract one from the estimated quotient and repeat. But note we only have to test this two times.

This explains the (otherwise baffling) comparison step in Algorithm D.

One last question: now that we have the correct quotient digit, don't we have to perform that widening multiply anyways to compute the remainder? The answer is no: the remainder is guaranteed to fit in two digits, so we can just use ordinary wrapping arithmetic to find it. For example, the first quotient digit is 6, so the first remainder is $576 - 6 \times 87 = 54$; let's wrap at two digits so mod 100:

\begin{align} r &= (576 - 6 \times 87) \bmod 100 \\ &= 76 - (6 \times 87) \bmod 100 \\ &= 76 - 22 \\ &= 54 \end{align}

Once the quotient digit is correct, we may just compute the remainder directly. Modulo overflow is expected and fine.

### Optimizing the estimate correction

I am optimistic this idea is original; I have not seen it published. This section will be more rigorous for that reason.

Let's restate what we know using variables. We work in some digit base $b$. We have a three digit dividend $u_2 u_1 u_0$, a two digit divisor $v_1 v_0$, and we know their quotient $q = \lfloor \frac {u_2 u_1 u_0} {v_1 v_0} \rfloor \lt b$ , i.e. fits in a single digit. We compute an estimate $\hat{q} = \lfloor \tfrac {u_2 u_1 0} {v_1 0} \rfloor$, and the corresponding remainder $\hat{r}$. We know that $\hat{q}$ is too big iff $\hat{q} \times v_0 \gt u_0 + \hat{r}$. If so, decrement $\hat{q}$, recompute $\hat{r}$, and repeat the check exactly once: if it still too big decrement $\hat{q}$ once more.

Of course when we decrement the quotient, we may easily compute the new remainder by just adding back the divisor. With this idea we get the following two checks:

1. $\hat{q} \times v_0 \gt u_0 + \hat{r} \\$
2. $(\hat{q} - 1) \times v_0 \gt u_0 + (\hat{r} + v_10)$

If neither are true, then q̂ is precise. If only the first is true, q̂ is too big by 1. If both are true, q̂ is too big by 2.

One non-obvious risk is that the last sum may overflow. This is mitigated by the $\hat{r} \lt b$ check in both Knuth and Warren. However it is possible to avoid the overflow through some algebra:

$$(\hat{q} - 1) \times v_0 \gt u_0 + (\hat{r} + v_10) \implies \\ \hat{q} \times v_0 - v_0 - u_0 \gt \hat{r} + v_10 \implies \\ \hat{q} \times v_0 - u_0 - \hat{r} \gt v_0 + v_10 \implies \\ \hat{q} \times v_0 - (u_0 + \hat{r}) \gt v_1 v_0 \\$$

How nice is this last expression! $v_1 v_0$ is just the original divisor. And the difference $\hat{q} \times v_0 - (u_0 + \hat{r})$ may also be used for the first check: if it is not positive, then the inequality in check 1 was not satisifed.

What about overflow? The product $\hat{q} \times v_0$ cannot overflow, because $\hat{q} \le q + 2 \implies \hat{q} \le b + 1$, therefore $\hat{q} \times v_0 \le v_0 * b + v_0$ which clearly does not overflow.

Also the sum $u_0 + \hat{r}$ does not overflow, because $\hat{r} \lt v_1 0$, therefore $\hat{r} + u_0 \lt v_1 u_0$.

This idea more efficiently finds $q$ in the 3 ÷ 2 case, avoiding the loop and some other checks from Knuth. See the next post for an updated narrowing division algorithm that exploits this idea.