Long division

Tavian Barnes

I have a confession to make: I never learned how to do long division. That might seem strange for someone whose early childhood identity was basically, “the kid who's good at math.” In fact, that's kind of why I didn't learn it—in grade three, I was doing well enough that they bumped me up to the fifth grade math class, skipping right over fourth grade where (I assume) it would have been taught. (Funnily enough, my dad, a math professor, thought the jump ahead was a bad idea.)

I did know how to divide by one-digit numbers, with a process that's apparently called short division. Since teachers were usually not cruel enough to assign things like 39483÷123, I fumbled my way through the rest of elementary and high school without really having to admit that I couldn't do one of the four basic arithmetic operations. Along the way, I started picking up computer programming, gradually shifting into my new identity, “the kid who's good at computers.” One thing I liked about computers is they would do division for you.

As I got better at programming, I stumbled on a way to cash in my talents for a kind of nerd clout. A teacher suggested I enter one of my programs in the local science fair, which some people took more seriously than I would have guessed. Not just your stereotypical baking-soda-and-vinegar volcanoes, there were students experimenting with stem cells and studying black holes. There were even science fair nationals, as I was surprised to learn when my project won me a trip there.

My project simulated the orbits of planets according to Newtonian gravity. Initially I was just happy to work on it for its own sake (and to procratinate my actual schoolwork), but nationals were super fun and I wanted to make sure I went back the next year. I figured that meant adding more kinds of physics, and making it more accurate. I didn't quite know enough math to improve its accuracy the good way, so I did it the bad way: arbitrary-precision arithmetic.

Loooonnnngggg division

In order to show off as much as possible, I implemented arbitrary-precision arithmetic myself, without those fancy libraries that do everything for you. Unfortunately, one of the things I had to implement was (you guessed it) division. At this point I had already finished all of high school math (which even included polynomial long division), and I still didn't really know how to divide two numbers. Sadly, my first attempt is lost to time, as I hadn't yet heard of revision control, but it looked something like this:

# My original implementation was in C++, but I'll use Python for exposition
def slow_divide(num, denom):
    guess = 0
    while (guess + 1) * denom <= num:
        guess += 1
    return guess

This is an algorithm you might use if long division just doesn't take long enough for you. I knew it took “exponential time,” but I didn't really have a feeling for what that meant until I ran this code for the first time. My project sat there for hours without making any progress. I had to find a faster way.

The division algorithm

Like any reasonable programmer might do to find an algorithm for division, I typed “division algorithm” into Google. It led me straight to the Division Algorithm, which is famously not an algorithm. Rather, it's a common name for the mathematical theorem that the quotient and remainer are unique when dividing two integers. Specifically, given two integers nn (numerator) and dd (denominator), one can find the unique integers qq (quotient) and rr (remainder) such that n=qd+rn = q d + r and 0r<d0 \le r < |d|.

That's great and all, but it doesn't really tell you how to find them. I remember that at the time, this unfortunate name gave me a lot of trouble finding actual algorithms for division. Luckily Wikipedia has since renamed some articles so that division algorithm now refers to actual algorithms.

Short divison

As lost as I was implementing division in full generality, I did know how to divide by one-digit numbers well enough to implement that. Short division is pretty easy, enough that you can even do it in your head with some practice. On paper it looks like this:

Say we want to compute 61320÷761320 \div 7:

7)613207 \overline{\smash{)}61320}

We work from left to right, starting with 6÷7=06 \div 7 = 0, remainder 66.

  07)661320\begin{aligned} & \;\, 0 \\[-4pt] 7 & \overline{\smash{)}\cancel{6}_{6}1320} \end{aligned}

Take the remainder from the last step and write it in front of the next digit. In this case not much changed, and we calculate 61÷7=861 \div 7 = 8, remainder 55.

  087)6615320\begin{aligned} & \;\, 08 \\[-4pt] 7 & \overline{\smash{)}\cancel{6}_{6}\cancel{1}_{5}320} \end{aligned}

53÷7=753 \div 7 = 7, remainder 44.

  0877)66153420\begin{aligned} & \;\, 087 \\[-4pt] 7 & \overline{\smash{)}\cancel{6}_{6}\cancel{1}_{5}\cancel{3}_{4}20} \end{aligned}


  08767)661534200\begin{aligned} & \;\, 0876 \\[-4pt] 7 & \overline{\smash{)}\cancel{6}_{6}\cancel{1}_{5}\cancel{3}_{4}\cancel{2}_{0}0} \end{aligned}

And, finally,

  087607)661534200\begin{aligned} & \;\, 08760 \\[-4pt] 7 & \overline{\smash{)}\cancel{6}_{6}\cancel{1}_{5}\cancel{3}_{4}\cancel{2}_{0}\cancel{0}} \end{aligned}

Indeed, we can check that 8760×7=613208760 \times 7 = 61320.

Why does this work? Let's start with a single-digit numerator n1n_1 divided by a single digit denominator dd. The “division algorithm” tells us that there is a unique quotient q1q_1 and remainder r1r_1 such that n1=q1d+r1n_1 = q_1 d + r_1. (In the example above, 6=0×7+66 = 0 \times 7 + 6.)

Now let's extend it to a two-digit numerator 10n1+n210 n_1 + n_2. We can plug in the above expression for n1n_1 to get

10n1+n2=10(q1d+r1)+n2=(10q1)d+10r1+n2.\begin{aligned} 10 n_1 + n_2 & = 10 (q_1 d + r_1) + n_2 \\ & = (10 q_1) d + 10 r_1 + n_2. \end{aligned}

Again, the division algorithm tells us we can find q2q_2 and r2r_2 such that 10r1+n2=q2d+r210 r_1 + n_2 = q_2 d + r_2. (In the above example, 61=8×7+561 = 8 \times 7 + 5.) It follows that

10n1+n2=(10q1)d+q2d+r2=(10q1+q2)d+r2.\begin{aligned} 10 n_1 + n_2 & = (10 q_1) d + q_2 d + r_2 \\ & = (10 q_1 + q_2) d + r_2. \end{aligned}

We can continue this process to extract the digits of the quotient and remainder for longer and longer numerators:

100n1+10n2+n3=(100q1+10q2+q3)d+r3,1000n1+100n2+10n3+n4=(1000q1+100q2+10q3+q4)d+r4,\begin{aligned} 100 n_1 + 10 n_2 + n_3 & = (100 q_1 + 10 q_2 + q_3) d + r_3, \\ 1000 n_1 + 100 n_2 + 10 n_3 + n_4 & = (1000 q_1 + 100 q_2 + 10 q_3 + q_4) d + r_4, \\ \end{aligned}


In code, it could look something like this:

def short_divide(num, denom):
    # Initialize the quotient and remainder to zero
    quot = 0
    rem = 0

    # Loop over the digits of the numerator
    for digit in str(num):
        # Write the remainder in front of the next digit of the numerator
        chunk = 10 * rem + int(digit)

        # Perform a two-digit by one-digit division to find the next quotient
        # digit.  If using the built-in division operator `//` feels like
        # cheating, imagine we used a look-up table or something instead.
        quot *= 10
        quot += chunk // denom

        # Compute the remainder for the next iteration
        rem = chunk % denom

    return quot

>>> short_divide(61320, 7)

Long division

Now that we have an algorithm for dividing by one-digit numbers, what do we have to change to make it work for bigger denominators? Embarassingly enough, nothing:

>>> short_divide(61320, 73)
>>> 840 * 73

None of the correctness of the math above relied on the denominator having only one digit. Long division is just short division with extra writing! All this time I'd thought I didn't know long division, I actually did, if I'd thought about it for a few minutes.

I will say that the division in chunk // denom feels more like cheating now that our denominators are larger. We can't really use a look-up table any more to evaluate it. But we could use a different algorithm for it—say, the terrible, exponential, count-up-from-zero algorithm I started with:

def long_divide(num, denom):
    quot = 0
    rem = 0

    for digit in str(num):
        chunk = 10*rem + int(digit)
        temp = slow_divide(chunk, denom)

        quot *= 10
        quot += temp

        rem = chunk - temp * denom

    return quot

>>> long_divide(61320, 73)

slow_divide() is fine in this case: since we're only using it to compute a single digit, it won't ever have to loop more than 10 times.

If we were serious numericists implementing division for a real arbitrary-precision arithmetic package, we'd be using a much larger base than 10, maybe 2322^{32} or 2642^{64} or something. Suddenly slow_divide() seems pretty slow again. But as usual in computer science, we could speed up the linear scan using binary search:

def binsearch_divide(num, denom):
    low = 0
    high = num

    while low < high:
        mid = (low + high) // 2
        if (mid + 1) * denom <= num:
            low = mid + 1
            high = mid

    return low

Actually, this is a perfectly fine general-purpose division algorithm (as long as I got the edge cases right):

>>> binsearch_divide(61320, 73)

It is slightly slower than long division since we have to do full-width multiplications, but on the other hand, it's easier to see why the approach is correct and efficient. I'm always hearing about plans to teach computer science concepts to kids in grade school, and I'm normally hesitant about such things, but maybe teaching “binary search division” alongside long division would be a good way to introduce students to algorithmic thinking.