What is tail recursion?

Question

Consider a simple function that adds the first N natural numbers. (e.g. sum(5) = 0 + 1 + 2 + 3 + 4 + 5 = 15).

Here is a simple JavaScript implementation that uses recursion:

function recsum(x) {
    if (x === 0) {
        return 0;
    } else {
        return x + recsum(x - 1);
    }
}

If you called recsum(5), this is what the JavaScript interpreter would evaluate:

recsum(5)
5 + recsum(4)
5 + (4 + recsum(3))
5 + (4 + (3 + recsum(2)))
5 + (4 + (3 + (2 + recsum(1))))
5 + (4 + (3 + (2 + (1 + recsum(0)))))
5 + (4 + (3 + (2 + (1 + 0))))
5 + (4 + (3 + (2 + 1)))
5 + (4 + (3 + 3))
5 + (4 + 6)
5 + 10
15

Note how every recursive call has to complete before the JavaScript interpreter begins to actually do the work of calculating the sum.

Here's a tail-recursive version of the same function:

function tailrecsum(x, running_total = 0) {
    if (x === 0) {
        return running_total;
    } else {
        return tailrecsum(x - 1, running_total + x);
    }
}

Here's the sequence of events that would occur if you called tailrecsum(5), (which would effectively be tailrecsum(5, 0), because of the default second argument).

tailrecsum(5, 0)
tailrecsum(4, 5)
tailrecsum(3, 9)
tailrecsum(2, 12)
tailrecsum(1, 14)
tailrecsum(0, 15)
15

In the tail-recursive case, with each evaluation of the recursive call, the running_total is updated.

Note: The original answer used examples from Python. These have been changed to JavaScript, since Python interpreters don't support tail call optimization. However, while tail call optimization is part of the ECMAScript 2015 spec, most JavaScript interpreters don't support it.

In traditional recursion, the typical model is that you perform your recursive calls first, and then you take the return value of the recursive call and calculate the result. In this manner, you don't get the result of your calculation until you have returned from every recursive call.

In tail recursion, you perform your calculations first, and then you execute the recursive call, passing the results of your current step to the next recursive step. This results in the last statement being in the form of (return (recursive-function params)). Basically, the return value of any given recursive step is the same as the return value of the next recursive call.

The consequence of this is that once you are ready to perform your next recursive step, you don't need the current stack frame any more. This allows for some optimization. In fact, with an appropriately written compiler, you should never have a stack overflow snicker with a tail recursive call. Simply reuse the current stack frame for the next recursive step. I'm pretty sure Lisp does this.

An important point is that tail recursion is essentially equivalent to looping. It's not just a matter of compiler optimization, but a fundamental fact about expressiveness. This goes both ways: you can take any loop of the form

while(E) { S }; return Q

where E and Q are expressions and S is a sequence of statements, and turn it into a tail recursive function

f() = if E then { S; return f() } else { return Q }

Of course, E, S, and Q have to be defined to compute some interesting value over some variables. For example, the looping function

sum(n) {
  int i = 1, k = 0;
  while( i <= n ) {
    k += i;
    ++i;
  }
  return k;
}

is equivalent to the tail-recursive function(s)

sum_aux(n,i,k) {
  if( i <= n ) {
    return sum_aux(n,i+1,k+i);
  } else {
    return k;
  }
}

sum(n) {
  return sum_aux(n,1,0);
}

(This "wrapping" of the tail-recursive function with a function with fewer parameters is a common functional idiom.)

This excerpt from the book Programming in Lua shows how to make a proper tail recursion (in Lua, but should apply to Lisp too) and why it's better.

A tail call [tail recursion] is a kind of goto dressed as a call. A tail call happens when a function calls another as its last action, so it has nothing else to do. For instance, in the following code, the call to g is a tail call:

function f (x)
  return g(x)
end

After f calls g, it has nothing else to do. In such situations, the program does not need to return to the calling function when the called function ends. Therefore, after the tail call, the program does not need to keep any information about the calling function in the stack. ...

Because a proper tail call uses no stack space, there is no limit on the number of "nested" tail calls that a program can make. For instance, we can call the following function with any number as argument; it will never overflow the stack:

function foo (n)
  if n > 0 then return foo(n - 1) end
end

... As I said earlier, a tail call is a kind of goto. As such, a quite useful application of proper tail calls in Lua is for programming state machines. Such applications can represent each state by a function; to change state is to go to (or to call) a specific function. As an example, let us consider a simple maze game. The maze has several rooms, each with up to four doors: north, south, east, and west. At each step, the user enters a movement direction. If there is a door in that direction, the user goes to the corresponding room; otherwise, the program prints a warning. The goal is to go from an initial room to a final room.

This game is a typical state machine, where the current room is the state. We can implement such maze with one function for each room. We use tail calls to move from one room to another. A small maze with four rooms could look like this:

function room1 ()
  local move = io.read()
  if move == "south" then return room3()
  elseif move == "east" then return room2()
  else print("invalid move")
       return room1()   -- stay in the same room
  end
end

function room2 ()
  local move = io.read()
  if move == "south" then return room4()
  elseif move == "west" then return room1()
  else print("invalid move")
       return room2()
  end
end

function room3 ()
  local move = io.read()
  if move == "north" then return room1()
  elseif move == "east" then return room4()
  else print("invalid move")
       return room3()
  end
end

function room4 ()
  print("congratulations!")
end

So you see, when you make a recursive call like:

function x(n)
  if n==0 then return 0
  n= n-2
  return x(n) + 1
end

This is not tail recursive because you still have things to do (add 1) in that function after the recursive call is made. If you input a very high number it will probably cause a stack overflow.