Playing with the Ackermann function: a tour of computational complexity
Once in a while, I like to play around with Oak to test new mathematical ideas I come across. Today, I was reminded of the Ackermann function, and thought I'd write a small Oak program to get a feel for how the function behaved.
The Ackermann function is quite simple to write. It takes two integers and returns an integer.
fn ackermann(m, n) if {
m = 0 -> n + 1
n = 0 -> ackermann(m - 1, 1)
_ -> ackermann(m - 1, ackermann(m, n - 1))
}
It's best known in mathematics and computer science for two properties:
- It increases super-exponentially, even for very small input values.
- It's not a primitive recursive function, which means that it isn't expressible in terms of simple, finite
for
loops.
We'll take a look at both of those properties here.
The script
To quickly play around with values of the Ackermann function, I wrote a simple script that let me type in input values into an interactive prompt and measure the run-time of the function:
Cli := cli.parse()
Time? := Cli.opts.t != ? | Cli.opts.time != ?
std.println('Ackermann function calculator.')
with std.loop() fn {
print('.> ')
// input is of the form "M N, M N, M N"
// assume input never fails b/c it's just a toy calculator
args := input().data |> str.split(',') |> std.map(fn(pair) {
pair |>
str.split(' ') |>
std.map(int) |>
std.compact() |>
std.take(2)
})
args |> with std.each() fn(pair) {
[m, n] := pair
prefix := if len(args) {
1 -> ''
_ -> 'ack({{0}}, {{1}}) = ' |> fmt.format(m, n)
}
if m != ? & n != ? {
true -> {
start := nanotime()
a := ackermann(m, n)
elapsed := nanotime() - start
(prefix + if {
Time? -> '{{0}} ({{1}}ms)'
_ -> '{{0}}'
}) |> fmt.printf(a, math.round(elapsed / 1000000, 3))
}
_ -> std.println('Invalid input. Try again.')
}
}
}
It works like this:
$ oak ackermann.oak --time
Ackermann function calculator.
.> 1 1
3 (0.008ms)
.> 2 2
7 (0.068ms)
.> 3 4
125 (17.101ms)
.> 3 5, 3 6, 3 7
ack(3, 5) = 253 (54.885ms)
ack(3, 6) = 509 (220.24ms)
ack(3, 7) = 1021 (973.239ms)
.> 3 10
8189 (157760.616ms)
.>
Big numbers are big
The Ackermann function stays quite tame for m = 1
and m = 2
.
.> 1 1, 1 2, 1 3, 1 4, 1 5, 1 6, 1 7, 1 8, 1 9, 1 10
ack(1, 1) = 3 (0.009ms)
ack(1, 2) = 4 (0.009ms)
ack(1, 3) = 5 (0.012ms)
ack(1, 4) = 6 (0.015ms)
ack(1, 5) = 7 (0.019ms)
ack(1, 6) = 8 (0.022ms)
ack(1, 7) = 9 (0.024ms)
ack(1, 8) = 10 (0.023ms)
ack(1, 9) = 11 (0.024ms)
ack(1, 10) = 12 (0.034ms)
.> 2 1, 2 2, 2 3, 2 4, 2 5, 2 6, 2 7, 2 8, 2 9, 2 10
ack(2, 1) = 5 (0.025ms)
ack(2, 2) = 7 (0.04ms)
ack(2, 3) = 9 (0.067ms)
ack(2, 4) = 11 (0.095ms)
ack(2, 5) = 13 (0.131ms)
ack(2, 6) = 15 (0.187ms)
ack(2, 7) = 17 (0.208ms)
ack(2, 8) = 19 (0.24ms)
ack(2, 9) = 21 (0.29ms)
ack(2, 10) = 23 (0.351ms)
Both sequences increase linearly by 1 and 2, and are quick to compute. For m = 3
, though, things start looking different:
.> 3 1, 3 2, 3 3, 3 4, 3 5, 3 6, 3 7, 3 8, 3 9, 3 10
ack(3, 1) = 13 (0.159ms)
ack(3, 2) = 29 (0.759ms)
ack(3, 3) = 61 (3.57ms)
ack(3, 4) = 125 (12.222ms)
ack(3, 5) = 253 (50.915ms)
ack(3, 6) = 509 (212.928ms)
ack(3, 7) = 1021 (980.088ms)
ack(3, 8) = 2045 (4745.548ms)
ack(3, 9) = 4093 (26737.925ms)
ack(3, 10) = 8189 (157760.616ms)
Not only do the numbers increase rapidly, the runtime increases even faster -- ack(3, 10)
took over 2 minutes to compute. The runtime of the Ackermann function increases so quickly because the function changes its return value by 1 each iteration, and only sometimes. That means every increment of 1 in the function's return value corresponds to several invocations of the Ackermann function during its computation.
Some facts about the next sequence, ack(4, _)
:
- I tried to compute a sequence for
ack(4, _)
but only got as far asack(4, 0) = 13 (0.574ms)
. ack(4, 1)
took too long to compute, and I had to stop the program, but according to Wikipedia, its value should be65533
.ack(4, 2)
is too large to even write down here, and is best described as2^65536 − 3
.
I expected the function itself to increase quickly, from its reputation, but I was surprised the run-time increased much faster than even the function itself.
To recurse or not to recurse
After getting these results, I tried to compile the Ackermann function to JavaScript and run the script on Node.js, where Oak often runs faster especially for numerical computations. A naive attempt, compiling the same file using oak build --web
and running on Node.js, resulted in a stack overflow at ack(3, 10)
.
That's fine, I thought. Oak's recursion limit must be higher than JavaScript's, so the stack overflowed deep in the recursive call. Usually, in this situation, I'd rewrite the function so that the function is tail-recursive, so that Oak's compiler could optimize the recursion down to a loop. But for the Ackermann function, there isn't a straightforward way to unroll the recursion into a simple tail recursion, because as noted above, the Ackermann function isn't primitive recursive!
The most common way to rewrite the Ackerman function using loops (or basic tail recursion) is actually a kind of a cheat: rather than using the programming language's stack, which can overflow, we can simulate a stack manually, using a growable list of numbers. The Ackermann function then operates not on two parameters to a function, but the top two numbers of this manually-managed stack.
I wrote this new loop-based variant of the Ackermann function (which I affectionately called stackermann
) in Oak using JavaScript's arrays:
fn stackermann(m, n) {
stack := [m, n] // begin with the stack [m, n]
with std.loop() fn(_, break) if len(stack) {
// if stack has only 1 value, return it
1 -> break(stack.0)
_ -> {
// Ackermann function operating on
// the top 2 values of the stack
n := stack.pop()
m := stack.pop()
if {
m = 0 -> stack.push(n + 1)
n = 0 -> {
stack.push(m - 1)
stack.push(1)
}
_ -> {
stack.push(m - 1)
stack.push(m)
stack.push(n - 1)
}
}
}
}
}
This function, based on a loop and a manually managed stack, runs slightly faster on Node.js than the original Oak version runs natively:
.> 3 1, 3 2, 3 3, 3 4, 3 5, 3 6, 3 7, 3 8, 3 9, 3 10
ack(3, 1) = 13 (1ms)
ack(3, 2) = 29 (1ms)
ack(3, 3) = 61 (3ms)
ack(3, 4) = 125 (6ms)
ack(3, 5) = 253 (5ms)
ack(3, 6) = 509 (19ms)
ack(3, 7) = 1021 (73ms)
ack(3, 8) = 2045 (301ms)
ack(3, 9) = 4093 (1095ms)
ack(3, 10) = 8189 (4375ms)
Armed with this slightly faster implementation of the algorithm, we can now attack ackermann(4, 0)
and ackermann(4, 1)
.
ack(4, 0) = 13 (0ms)
ack(4, 1) = 65533 (275487ms)
Even with the speed improvements, the run-time of ackermann(4, 1)
still dwarfs that of any in the ackermann(3, _)
sequence. ackermann(4, 2)
and beyond are, unfortunately, still out of our reach. It would take many days of compute on my lowly laptop to find those answers, and I wasn't about to put mine through that today.
As a last step in my journey to make the function run even faster, I explored the possibility of memoizing the Ackermann function. But this Stack Overflow answer confirmed my suspicions that memoizing the Ackermann function doesn't yield much speed-up, because the domain of the function is two-dimensional, and memoization yields minor speed improvements at the cost of much more heap memory usage. Even computing ackermann(4, 2)
, even with more efficient programming languages and memoization, can result in out-of-memory errors.
At this point, I decided to put my own exploration to a halt. Looking at the two different implementations of the Ackermann function -- the recursion-based and loop/stack-based -- gave me a good sense of how the function achieves its massive computational complexity, and why simplifying it isn't trivial.
If you're interested in digging deeper, both the Wikipedia page and Computerphile's video on the subject seem like great starting points.