1.4.1 Walkthrough: Recursion
Recursion is a common pattern for solving certain problems in most programming languages, and Hoon is no exception. One of the classically recursive problems is that of factorial. The factorial of n is the product of all positive integers less than or equal to N. Thus the factorial of 5, denoted as 5!, would be:
5 * 4 * 3 * 2 * 1 = 120
Factorial In Hoon
Let's implement a factorial calculator in Hoon. To do so, we'll write a
recursive gate
(Hoon's equivalent of a function) to perform the relevant
computation. Save the code below as factorial.hoon
in your ship's /gen
directory.
= n=@ud ?: =(n 1) 1 (mul n $(n (dec n)))
Our gate
takes one sample (argument) n
that must nest inside @ud
, the
unsignedinteger type.
Next we check to see if n
is 1
. If so, the result is just 1
, since
1 * 1 = 1
.
If, however, n
is not 1
, then we branch to the final line of the code,
(mul n $(n (dec n)))
, where the recursion logic lives. Here, we multiply n
by the recursion of n
minus 1
. $
initiates recursion: it calls the gate
that we're already in, but replaces its sample.
In our example, we multiply n
with the product of this entire gate
all over again with $(n (dec n))
, except when that new gate has its sample
decremented by one. This works recursively because each new gate will, of
course, itself contain the code to call a furtherdecremented gate. Gates will
continue to call new, furtherdecremented gates until n
is 1
, and that 1
will be the final number to be multiplied by.
Let's run the program in the Dojo:
> +factorial 5
120
It may help to visualize the operation of our example gate. The pseudoHoon below illustrates what happens when we use it to find the factorial of 5:
(factorial 5)
(mul 5 (factorial 4))
(mul 5 (mul 4 (factorial 3)))
(mul 5 (mul 4 (mul 3 (factorial 2))))
(mul 5 (mul 4 (mul 3 (mul 2 (factorial 1)))))
(mul 5 (mul 4 (mul 3 (mul 2 1))))
(mul 5 (mul 4 (mul 3 2)))
(mul 5 (mul 4 6))
(mul 5 24)
120
It's easy to see how we're "floating" gate calls until we reach the final
iteration of such calls that only produces a value. The mul n
component of the
gate leaves something like mul 5
, waiting for the final series of terms
to be operated upon. The $(n (dec n))
component is what expands out the
expression, as illustrated by (factorial 4)
. Once the expression cannot be
expanded out further, the operations work backwards, successively feeding values
into the mul
functions behind them.
TailCall Optimization
Our last example isn't a very efficient use computing resources. The pyramidshaped illustration approximates what's happening on the call stack, a memory structure that tracks the instructions of the program. In our example code, every time a parent gate calls another gate, the gate being called is "pushed" to the top of the stack in the form of a frame. This process continues until a value is produced instead of a function, completing the stack.
Push order Pop order
(fifth frame) ^ 
(fourth frame)  
(third frame)  
(second frame)  
(first frame)  V
Once this stack of frames is completed, frames "pop" off the stack starting at the top. When a frame is popped, it executes the contained gate and passes produced data to the frame below it. This process continues until the stack is empty, giving us the gate's output.
When a program's final expression uses the stack in this way, it's considered to
be not tailrecursive. This usually happens when the last line of executable
code calls more than one gate, our example code's (mul n $(n (dec n)))
being
such a case. That's because such an expression needs to hold each iteration of
$(n (dec n)
in memory so that it can know what to run against the mul
function every time.
To reiterate: if you have to manipulate the result of a recursion as the last
expression of your gate, as we did in our example, the function is not
tailrecursive, and therefore not very efficient with memory. A problem arises
when we try to recurse more times that we have space on the stack. This will
result in our computation failing and producing a stack overflow. If we tried
to find the factorial of 5.000.000
, for example, we would almost certainly
run out of stack space.
But the Hoon compiler, like most compilers, is smart enough to notice when the last statement of a parent can reuse the same frame instead of needing to add new ones onto the stack. If we write our code properly, we can use a single frame that simply has its values replaced with each recursion.
A TailRecursive Gate
With a bit of refactoring, we can write a version of our factorial gate that is tailrecursive and can take advantage of this feature:
= n=@ud =/ t=@ud 1  ?: =(n 1) t $(n (dec n), t (mul t n))
The above code should look familiar. We are still building a gate that
takes one argument n
. This time, however, we are also putting a face on a
@ud
and setting its initial value to 1. The 
here is used to create a new
gate with one arm $
and immediately call it. Think of 
as the recursion
point.
We then evaluate n
to see if it is 1. If it is we return the value of t
. In
the case that n
is anything other than 1, we perform our recursion:
$(n (dec n), t (mul t n))
All we are doing here is recursing our new gate and modifying the values of n
and t
. t
is used as an accumulator variable that we use to keep a running
total for the factorial computation.
Let's use pseudoHoon to illustrate how the stack is working in this example for factorial 5.
(factorial 5)
( 5 1)
( 4 5)
( 3 20)
( 2 60)
( 1 120)
120
We simply multiply t
and n
to produce the new value of t
, and then
decrement n
before repeating. Since this $
call is the final and solitary
thing that is run in the default case and since we are doing all computation
before the call, this version is properly tailrecursive. We don't need to do
anything to the result of the recursion except recurse it again. That means that
each iteration can be replaced instead of held in memory.
A Note on $
$
(pronounced "buc") is, in its use with recursion, a reference to the gate that we are inside
of. That's because a gate is just a core with a single arm named $
. The
subject is searched depthfirst, head before tail, with faces skipped, and
stopping on the first result. In other words, the first match found in the head
will be returned. If you wished to refer to the outer $
in this context, the
idiomatic way would be to use ^$
. The ^
operator
skips the first match of a name.
Exercises

Write a recursive gate that produces the first n Fibonacci numbers

Write a recursive gate that produces a list of moves to solve the Tower of Hanoi problem. Disks are stacked on a pole by decreasing order of size. Move all of the disks from one pole to another with a third pole as a spare, moving one disc at a time, without putting a larger disk on top of a smaller disk.