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# A program to compute the nth Fibonacci number

Since I'm looking for a job and people often like to ask for a fizzbuzz program to weed out the folks who can't string together a few lines of code, I thought I'd write up a program to compute the nth Fibonacci number. There's an intriguing bit of matrix math involved, so I learned something while implementing it.

Reminder: This was written in 2008, it does not necessarily reflect my employment situation today.

Recently I was having lunch with some of Toronto's most interesting Ruby developers, and the subject of interview questions came up. Specifically, writing a program to compute the nth Fibonacci number. Unsurprisingly, we agreed it might be useful as a screener for weeding out the people who were completely delusional about their prospects as a professional programmer. You know, the type of person who stares blankly at the screen and has no idea where to start, even when told that all we wanted was a program that correctly prints an answer. No tests, no specs, no shouldas, no passengers, no CSS, just prove you actually can write something, anything.

We also agreed that any Guess the answer I'm thinking of problem is terrible, including Fibonacci. Should it iterate? Recurse? Memoize? Is it overkill to use advanced math to be really fast? Or will you get tossed out of the interview for writing a naive recursive solution? Unless the interviewer is very specific that they just want you to prove that you actually have written ten or more lines of working Ruby code in your life, there could be any number of reasons to disqualify any solution.

But we batted around a few ideas for how to gild the lily, and since I have just found myself contemplating the prospect of being asked to write a program to compute the nth Fibonacci number, I thought I'd get some practice and write one in advance. Since I'm doing it for myself, I'm going to optimize for maximum personal growth.

Here we go...

## Enter The Matrix

One problem with calculating a Finonacci number is that naive algorithms require n addition operations. There are some interesting things we can do to improve on this. One way is to transform n additions into raising something to the power of n. This turns n additions into n multiplications. That seems retrograde, but hold on to your disbelief.

This is actually nice, because there is a trick about raising something to a power that we can exploit. But first things first. As explained in Sum of Fibonacci numbers?, we can express the Fibonacci number F(n) using a 2x2 matrix:

[ 1 1 ] n      [ F(n+1) F(n)   ]
[ 1 0 ]    =   [ F(n)   F(n-1) ]

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 Ruby:

def times(*ems)
ems.inject do |product, matrix|
a,b,c = product; d,e,f = matrix
[a*d + b*e, a*e + b*f, b*e + c*f]
end
end

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]

Very interesting. We could write out a naive implementation that constructs a long array of copies of [1,1,0] and then calls times:

def naive_power(m, n)
times(*(1..n).map { |n| m })
end

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]

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]))? And that 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:

begin
double = times([1,1,0], [1,1,0])
times(double, double)
end

This only requires two operations rather than three. Furthermore, it is recursive. naive_power([1,1,0], 8) requires seven operations. However, it can be formulated as:

begin
double = times([1,1,0], [1,1,0])
times(double, double)
end
end

Now we only need three operations compared to seven. 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:

def power(m, n)
if n == 1
m
else
halves = power(m, n / 2)
if n % 2 == 0
times(halves, halves)
else
times(halves, halves, m)
end
end
end

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 write our complete fibonacci function:

def fib(n)
return n if n < 2
power([1,1,0], n - 1).first
end

And dress things up in idiomatic Ruby using the anonymous module pattern:

class Integer

include(Module.new do

times = lambda do |*ems|
ems.inject do |product, matrix|
a,b,c = product; d,e,f = matrix
[a*d + b*e, a*e + b*f, b*e + c*f]
end
end

power = lambda do |m, n|
if n == 1
m
else
halves = power.call(m, n / 2)
if n % 2 == 0
times.call(halves, halves)
else
times.call(halves, halves, m)
end
end
end

define_method :matrix_fib do
return self if self < 2
power.call([1,1,0], self - 1).first
end

end)

end

(0..20).map { |n| n.matrix_fib }
# => [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765]

We're done!

p.s. No we're not done: Another program to compute the nth Fibonacci number.

p.p.s. No, this isn't the fastest implementation by far. But it beats the pants off of a naive iterative implementation. See fibonacci.rb for details.

Christoph Temmel forked homoiconic and added these observations:

*Another option would be to decompose *

A = [ 1 1 ]
[ 1 0 ]

into along its eigenspace (eigenvalues \lambda_{1,2}=-\frac{1\pm\sqrt{5}}{2} = -\frac{1}{2}\mp\frac{\sqrt{5}{2}}) to get

A = Q^t D

and

D = [ \lambda_1 0         ]
[ 0         \lambda_2 ]

where Q is the orthonormal matrix of the normated eigenvectors with QQ^t = I and D is the above diagonal matrix with the eigenvalues \lambda_{1,2} (see also Wikipedia). Now it's easy to take powers of A

A^n = Q^t D^n D

*The only difficulty is, that in order to avoid floating point arithmetic one would have to do the powers in the field \mathbb{Q}[\sqrt{5}] - this would ask for a custom datatype with overloading of addition and multiplication (the part needed for this exercise). *

### More fun with Fib

As noted on news.ycombinator.com, There is a closed-form solution to the function fib.

http://en.wikipedia.org/wiki/Fibonacci_number#Closed-form_expression

In the context of a hypothetical job interview, of course, the correct answer is whatever produces the most interesting discussion. If asked in a real interview, I might ask "Do you want the closed-form or are you trying to get me to correctly write a recursive function?" If you write out the closed-form, once in a very long while you will probably encounter someone who tries to out-clever your cleverness and demands that you write it without Math.sqrt or some other arbitrary way of forcing you to demonstrate that your brain has the right shape for handling recursion.

Recent work:

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