Once upon a time, programming interviews would include a fizzbuzz problem to weed out the folks who couldn’t string together a few lines of code. We can debate when and how such things are appropriate interview questions, but one thing that is always appropriate is to use them as inspiration for practising our own skills.^{1}
There are various common problems offered in such a vein, including fizzbuzz itself, computing certain prime numbers, and computing Fibonacci number. A few years back, I had a go at writing my own Fibonacci function. When I started researching approaches, I discovered an intriguing bit of matrix math, so I learned something while practicing my skills.^{2}
enter the matrix
One problem with calculating a Fibonacci number is that naïve algorithms require n additions. This is obviously expensive for large values of n. But of course, there are some interesting things we can do to improve on this.
In this solution, we observe that we can express the Fibonacci number F(n)
using a 2x2 matrix that is raised to the power of n:
[ 1 1 ] n [ F(n+1) F(n) ]
[ 1 0 ] = [ F(n) F(n1) ]
On the face of it, raising someting to the power of n turns n additions into n multiplications. n multiplications sounds worse than n additions, however there is a trick about raising something to a power that we can exploit. Let’s start by writing some code to multiply matrices:
Multiplying two matrices is a little interesting if you have never seen it before:
[ a b ] [ e f ] [ ae + bg af + bh ]
[ c d ] times [ g h ] = [ ce + dg cf + dh ]
Our matrices always have diagonal symmetry, so we can simplify the calculation because c is always equal to b:
[ a b ] [ d e ] [ ad + be ae + bf ]
[ b c ] times [ e f ] = [ bd + ce be + cf ]
Now we are given that we are multiplying two matrices with diagonal symmetry. Will the result have diagonal symmetry? In other words, will ae + bf
always be equal to bd + ce
? Remember that a = b + c
at all times and d = e + f
provided that each is a power of [[1,1], [1,0]]
. Therefore:
ae + bf = (b + c)e + bf
= be + ce + bf
bd + ce = b(e + f) + ce
= be + bf + ce
That simplifies things for us, we can say:
[ a b ] [ d e ] [ ad + be ae + bf ]
[ b c ] times [ e f ] = [ ae + bf be + cf ]
And thus, we can always work with three elements instead of four. Let’s express this as operations on arrays:
[a, b, c] times [d, e, f] = [ad + be, ae + bf, be + cf]
Which we can code in JavaScript, using array destructuring:
let times = (...matrices) =>
matrices.reduce(
([a,b,c], [d,e,f]) => [a*d + b*e, a*e + b*f, b*e + c*f]
);
times([1, 1, 0]) // => [1, 1, 0]
times([1, 1, 0], [1, 1, 0]) // => [2, 1, 1]
times([1, 1, 0], [1, 1, 0], [1, 1, 0]) // => [3, 2, 1]
times([1, 1, 0], [1, 1, 0], [1, 1, 0], [1, 1, 0]) // => [5, 3, 2]
times([1, 1, 0], [1, 1, 0], [1, 1, 0], [1, 1, 0], [1, 1, 0]) // => [8, 5, 3]
To get exponentiation from multiplication, we could write out a naive implementation that constructs a long array of copies of [1, 1, 0]
and then calls times
:
let naive_power = (matrix, n) =>
times(...new Array(n).fill([1, 1, 0]));
naive_power([1, 1, 0], 1) // => [1, 1, 0]
naive_power([1, 1, 0], 2) // => [2, 1, 1]
naive_power([1, 1, 0], 3) // => [3, 2, 1]
naive_power([1, 1, 0], 4) // => [5, 3, 2]
naive_power([1, 1, 0], 5) // => [8, 5, 3]
Very interesting, and less expensive than multiplying any two arbitrary matrices, but we are still performing n multiplications when we raise a matrix to the nth power. What can we do about that?
exponentiation with matrices
Now let’s make an observation: instead of accumulating a product by iterating over the list, let’s Divide and Conquer. Let’s take the easy case: Don’t you agree that times([1, 1, 0], [1, 1, 0], [1, 1, 0], [1, 1, 0])
is equal to times(times([1, 1, 0], [1, 1, 0]), times([1, 1, 0], [1, 1, 0]))
?
This saves us an operation, since times([1, 1, 0], [1, 1, 0], [1, 1, 0], [1, 1, 0])
is implemented as:
times([1, 1, 0],
times([1, 1, 0],
times([1, 1, 0], [1, 1, 0]))
Whereas times(times([1, 1, 0], [1, 1, 0]), times([1, 1, 0], [1, 1, 0]))
can be implemented as:
let double = times([1, 1, 0], [1, 1, 0]),
quadruple = times(double, double);
This only requires two operations rather than three. Furthermore, this pattern is recursive. For example, naive_power([1, 1, 0], 8)
requires seven operations:
times([1, 1, 0],
times([1, 1, 0],
times([1, 1, 0],
times([1, 1, 0],
times([1, 1, 0],
times([1, 1, 0],
times([1, 1, 0], [1, 1, 0])))))))
However, it can be formulated with just three operations:
let double = times([1, 1, 0], [1, 1, 0]),
quadruple = times(double, double),
octuple = times(quadruple, quadruple);
Of course, we left out how to deal with odd numbers. Fixing that also fixes how to deal with even numbers that aren’t neat powers of two:
let power = (matrix, n) => {
if (n === 1) return matrix;
let halves = power(matrix, Math.floor(n / 2));
return n % 2 === 0
? times(halves, halves)
: times(halves, halves, matrix);
}
power([1, 1, 0], 1) // => [1, 1, 0]
power([1, 1, 0], 2) // => [2, 1, 1]
power([1, 1, 0], 3) // => [3, 2, 1]
power([1, 1, 0], 4) // => [5, 3, 2]
power([1, 1, 0], 5) // => [8, 5, 3]
Now we can perform exponentiation of our matrices, and we take advantage of the symmetry to perform log2n multiplications.
and thus to fibonacci
We can now write our complete fibonacci function:
let times = (...matrices) =>
matrices.reduce(
([a,b,c], [d,e,f]) => [a*d + b*e, a*e + b*f, b*e + c*f]
);
let power = (matrix, n) => {
if (n === 1) return matrix;
let halves = power(matrix, Math.floor(n / 2));
return n % 2 === 0
? times(halves, halves)
: times(halves, halves, matrix);
}
let fibonacci = (n) =>
n < 2
? n
: power([1, 1, 0], n  1)[0];
fibonacci(62)
// => 4052739537881
If we’d like to work with very large numbers, JavaScript’s integers are insufficient. Using a library like BigInteger.js, our solution becomes:
import { zero, one } from 'biginteger';
let times = (...matrices) =>
matrices.reduce(
([a, b, c], [d, e, f]) => [
a.times(d).plus(b.times(e)),
a.times(e).plus(b.times(f)),
b.times(e).plus(c.times(f))
]
);
let power = (matrix, n) => {
if (n === 1) return matrix;
let halves = power(matrix, Math.floor(n / 2));
return n % 2 === 0
? times(halves, halves)
: times(halves, halves, matrix);
}
let fibonacci = (n) =>
n < 2
? n
: power([one, one, zero], n  1)[0];
Let’s stretch our wings and calculate the 19,620,614th Fibonacci number:^{3}
fibonacci(19620614).toString()
// =>
29554981652302145421961363135286189884298419359021591207414
94029404508891979849589048890639433083583865137532017734839
03976989846431379560920380384049354648973349793444866743699
77583527866834756211404857224578913159290369224744375346007
69568427064684727742279727974589543524504574566687346018118
// ...
// ...69,490 lines elided...
// ...
44917243010555501044826365112091652635017254277919365055752
92134790638460565443537453870610661665070132289987927432062
35114452925784094009088430159367013806505734580798084002841
21219542644844237050682927035326929321204843947060841278604
7707601726389614978163177
(full result, as calculated in Safari)
We’re done. And this is a win over the typical recursive or even iterative solution for large numbers, because while each operation is more expensive, we only perform log2n operations.^{4}
(This is a translation of a blog post written in 2008. It feels cleaner than the Ruby original, possibly because of the destructuring, and possibly because writing functions is idiomatic JavaScript, whereas refining core classes is idiomatic Ruby. What do you think? Discuss on Hacker News.)
notes:

Actually, let me be candid: I just like programming, and I find it’s fun, even if I don’t magically transform myself into a 10x programming ninja through putting in 10,000 hours of practice. But practice certainly doesn’t hurt. ↩

There is a closedform solution to the function
fib
, but floating point math has some limitations you should be aware of before using it in an interview. Naturally, if you’re running into some of those limits, you would use a BigInt library such as BigInteger.js. ↩ 
19620614 is a number near and dear to me :) ↩

No, this isn’t the fastest implementation by far. But it beats the pants off of a naïve iterative implementation. ↩