Let’s write a simple program to add numbers from n to 0. But instead of using iterative approach, why not try recursive approach?
We call this program sum
. We know sum(0) == 0
, so this is our base case. How do we arrive at the base case? sum(n) == n sum(n-1)
, until finally reaching sum(0)
. The Java code is as follows:
<code class="language-java">int sum(int n) { if (n == 0) { return 0; } return n + sum(n - 1); }</code>
Recursion has an inherent flaw when the base case is far away from the input value... In most languages, function calls use the program's stack to store function call information, so very large recursions can cause a stack overflow.
But, is there a way to avoid this? Actually, there is. This is an old strategy called trampoline.
The basic idea of the springboard strategy is that part of the program returns a "value" or a "continuation". What is continuation? A function that will continue processing.
It’s roughly as follows:
<code class="language-java">let trampolim = primeiraChamada(input); while (trampolim is continuation) { trampolim = trampolim.continue(); } return trampolim;</code>
sum
? Let us model sum
the program as: Instead of simply recursing, use continuations. One way is to use acc
as an object passed via a continuation. So when sum_trampoline(0, acc)
is reached, we return acc
. How to proceed?
Let’s go from sum_trampoline(n, acc)
to sum_trampoline(n-1, acc n)
. The first input is sum_trampoline(n, 0)
.
So, the code is as follows:
<code class="language-java">Object sum_trampoline_bootstrap(int n) { return sum_trampoline(n, 0); } Object sum_trampoline(int n, int acc) { if (n == 0) { return acc; } return (Supplier<object>) () -> sum(n - 1, acc + n); }</code>
The springboard needs to be roughly of the following form:
<code class="language-java">let trampolim = primeiraChamada(input); while (trampolim is continuation) { trampolim = trampolim.continue(); } return trampolim;</code>
But this gives a lot of coding freedom and is not very intuitive for the Java world. We can check if it is a continuation by asking the object. What if we asked "Was the value found?" Another thing is that since Java doesn't have sum-types, return trampolim
will actually return the trampolim
type instead of returning the value. We can go back to trampolim.value()
.
Finally, a key point is the bootstrapping of the springboard. To do this, we can use a function to convert the input into the appropriate pogo return value. Inputs and results can be generalized for better use:
<code class="language-java">public static <R> R trampoline(IN input, Function<IN, TrampolineStep<R>> trampolinebootStrap) { TrampolineStep<R> nextStep = trampolinebootStrap.apply(input); while (!nextStep.gotValue()) { nextStep = nextStep.runNextStep(); } return nextStep.value(); }</code>
TrampolineStep<R>
What about the interface?
It defines three methods:
gotValue
: Asks if the value has been found value
: Returns the found value runNextStep
: Returns a value or a continuation It basically has two states:
Therefore, we can use static methods to initialize it. For cases where the value has been found, the value needs to be passed:
<code class="language-java">int sum(int n) { if (n == 0) { return 0; } return n + sum(n - 1); }</code>
For the case of continuation, you need to pass how to get the next item of the continuation:
<code class="language-java">let trampolim = primeiraChamada(input); while (trampolim is continuation) { trampolim = trampolim.continue(); } return trampolim;</code>
sum_trampoline
How will this be achieved?
<code class="language-java">Object sum_trampoline_bootstrap(int n) { return sum_trampoline(n, 0); } Object sum_trampoline(int n, int acc) { if (n == 0) { return acc; } return (Supplier<object>) () -> sum(n - 1, acc + n); }</code>
The classic implementation of Fibonacci follows the recursive definition:
<code class="language-java">let trampolim = primeiraChamada(input); while (trampolim is continuation) { trampolim = trampolim.continue(); } return trampolim;</code>
There is also an iterative version that expands the Fibonacci definition not recursively, but forward: starting from 0 and 1 until the corresponding values are reached:
<code class="language-java">public static <R> R trampoline(IN input, Function<IN, TrampolineStep<R>> trampolinebootStrap) { TrampolineStep<R> nextStep = trampolinebootStrap.apply(input); while (!nextStep.gotValue()) { nextStep = nextStep.runNextStep(); } return nextStep.value(); }</code>
There is a forward version of this implementation, using "tail call recursion":
<code class="language-java">static <X> TrampolineStep<X> valueFound(X value) { return new TrampolineStep() { @Override public boolean gotValue() { return true; } @Override public X value() { return value; } @Override public TrampolineStep<X> runNextStep() { return this; } }; }</code>
Here I separate the input interface, which prepares the numbers that will be used in the tail call recursive Fibonacci. As it moves forward, we start with the mapping fib[0] => 0
, fib[1] => 1
and navigate from index 0 until we reach index n.
Fibonacci: From Tail Call to Springboard
The example offib_tc
illustrates the Fibonacci springboard well:
<code class="language-java">static <X> TrampolineStep<X> goonStep(Supplier<TrampolineStep<X>> x) { return new TrampolineStep() { @Override public boolean gotValue() { return false; } @Override public X value() { throw new RuntimeException("dont call this"); } @Override public TrampolineStep<X> runNextStep() { return x.get(); } }; }</code>
Please note that this is just a skeleton and requires a complete implementation of the TrampolineStep
interface and a complete implementation of the trampoline
functions to compile and run. Additionally, IN
needs to be replaced with a specific input type.
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