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Tail call optimization

Some code reformatted to fit better on this page.

Let’s see what GPT knows about tail call optimization. We’ll start with some code that was explicitly written to be ammenable to TCO. Note that this is a pretty famous problem, used in the early chapters of the classic text Structure and Interpretation of Computer Programs.

Can this code be compiled with tail-call-optimizaton:

const change3 = (amt, coins) => {
  const rows = Math.max(...coins) + 1;
  const table = Array(rows)
    .fill()
    .map((_, i) => Array(coins.length).fill(i === 0 ? 1 : 0));

  const helper = (i, c) => {
    if (c === coins.length) {
      if (i === amt) {
        return table[i % table.length][c - 1];
      } else {
        return helper(i + 1, 0);
      }
    } else {
      const smallerAmount =
        i - coins[c] < 0
          ? 0
          : table[(i - coins[c]) % table.length][c];
      const fewerCoins = c === 0 ? 0 : table[i % table.length][c - 1];
      table[i % table.length][c] = smallerAmount + fewerCoins;
      return helper(i, c + 1);
    }
  };

  return helper(1, 0);
};

The provided code is written in JavaScript, which does not have guaranteed tail-call optimization support across all implementations. The ECMAScript 2015 (ES6) specification included proper tail calls (PTC) as part of the standard, but not all JavaScript engines have implemented this feature. In fact, some major engines, like V8 (used in Chrome and Node.js), have opted not to implement it due to concerns over performance and debugging complexity.

As a result, while the provided code may work correctly and efficiently in an engine that supports tail-call optimization, there is no guarantee that it will do so in all JavaScript environments.

If you need to ensure that your code can benefit from tail-call optimization, consider using a language that guarantees support for this feature, such as Scheme or Haskell, or use an iterative approach instead of recursion in JavaScript.

That fact about Javascript engines not actually supporting TCO even though the spec requires it, was news to me. So that’s useful! But I’m actually more interested in analyzing the code. In this case it got it right.

Now let’s try one that is not eligible for TCO:

Can this code be compiled with tail-call-optimization:

const change = (amt, coins) => {
  if (amt < 0 || coins.length === 0) {
    return 0;
  } else if (amt === 0) {
    return 1;
  } else {
    return change(amt - coins[0], coins) + change(amt, coins.slice(1));
  }
};

As mentioned in my previous response, JavaScript does not have guaranteed tail-call optimization support across all implementations. However, let's analyze the provided code in terms of whether it is tail-recursive, which is a prerequisite for tail-call optimization.

The provided change function is not tail-recursive because in the last return statement:

return change(amt - coins[0], coins) + change(amt, coins.slice(1));

The function calls itself twice and then adds the results together. The addition operation prevents the function from being tail-recursive, as the function needs to store the intermediate state between the two recursive calls in order to perform the addition.

For a function to be tail-recursive, the last operation in the function should be the recursive call itself, with no other operations performed after the call. In this case, you may want to refactor the code to use an iterative approach or use helper functions with accumulators to achieve a tail-recursive structure, if using a language that supports tail-call optimization.

That’s correct and the analysis seems good.

Since actual Javascript engines may not support TCO we might want to translate into actual iterative form. Let’s see if GPT can do it for us:

Can you rewrite the tail-recursive change3 from before in iterative form?

Certainly! Here's an iterative version of the change3 function provided earlier:

const change3 = (amt, coins) => {
  const rows = Math.max(...coins) + 1;
  const table = Array(rows)
    .fill()
    .map((_, i) => Array(coins.length).fill(i === 0 ? 1 : 0));

  for (let i = 1; i <= amt; i++) {
    for (let c = 0; c < coins.length; c++) {
      const smallerAmount =
        i - coins[c] < 0
          ? 0
          : table[(i - coins[c]) % table.length][c];
      const fewerCoins = c === 0 ? 0 : table[i % table.length][c - 1];
      table[i % table.length][c] = smallerAmount + fewerCoins;
    }
  }

  return table[amt % table.length][coins.length - 1];
};

In this version, I've replaced the recursive helper function with two nested for loops. The outer loop iterates over the amounts from 1 to amt, and the inner loop iterates over the coin denominations. The rest of the logic remains the same as in the original version.

That’s pretty darn good! It’s almost identical to the code I wrote myself. And it’s obviously not just spitting out a memorized iterative solution to this famous problem, insofar as it’s using the same variable names and other parts of the original function.

Some time, later I asked it to help me analyze some code. Sadly, this time it got it confidently wrong:

In a Javascript implementation that performs tail call optimization, can this function be optimized with TCO:

const loop = (f) => f ** 2 > n || (n % f !== 0 && loop(f + 1));

First, it's important to note that not all JavaScript environments support tail call optimization (TCO). As mentioned in previous responses, TCO is part of the ECMAScript 2015 (ES6) specification, but some major JavaScript engines have not implemented it.

Now, let's analyze the given function to see if it's tail-recursive. The provided function loop has a logical OR (||) in the return statement:

const loop = (f) => f ** 2 > n || (n % f !== 0 && loop(f + 1));

In this function, the recursive call loop(f + 1) appears in the right-hand side of the OR operation. Unfortunately, this function is not tail-recursive because the OR operation requires the function to store the intermediate state before making the recursive call. For a function to be tail-recursive, the last operation in the function must be the recursive call itself, with no other operations performed afterward.

To make this function tail-recursive, you can refactor it to use a helper function with an accumulator, which is often the pattern when converting a non-tail-recursive function to a tail-recursive one.

First, off, I was really trying to head it off at the pass about telling me that not all Javascript engines support TCO. Oh well. But more importantly, I’m pretty sure that’s wrong. In this case it’s not true that the or operation “requires the function to store the intermediate state before making the call to g()”.

Let’s try a simpler example, taken from https://exploringjs.com/es6/ch_tail-calls.html#sec_checking-for-tail-calls

Is the call to g in this function in tail position:

const a = () => f() || g();

In the given function a, the call to g is not in tail position. The reason is that the return statement includes a logical OR (||) operation:

const a = () => f() || g();

In this case, the JavaScript engine first evaluates the result of f(). If the result of f() is truthy, it will be returned directly; otherwise, the engine will call g(). The logical OR operation requires the function to store the intermediate state before making the call to g(), which prevents the function from being tail-recursive. For a function to be tail-recursive, the last operation in the function must be the recursive call itself, with no other operations performed afterward.

However, it's important to note that tail-call optimization is relevant to recursive functions, and the provided function a does not appear to be recursive. If you meant to ask whether the call to g can benefit from tail-call optimization in a language that supports it, the answer would still be no, due to the reasons explained above.

It’s still not true that the “logical OR operation requires the function to store the intermediate state before making the call to g()”. Let’s try with what should be equivalent:

Is the call to g in tail position in this function?

const a = () => {
  const fResult = f();
  if (fResult) {
      return fResult;
  } else {
      return g();
  }
};

In the given function a, the call to g is indeed in tail position. The function has been refactored to separate the calls to f() and g() in conditional branches:

const a = () => {
    const fResult = f();
    if (fResult) {
        return fResult;
    } else {
        return g();
    }
};

Now, when the function reaches the else branch, the last operation is the call to g(), and there are no other operations performed afterward. This makes the call to g in tail position.

However, it's important to note that tail-call optimization is relevant to recursive functions, and the provided function a does not appear to be recursive. If you meant to ask whether the call to g can benefit from tail-call optimization in a language that supports it, the answer would be yes, assuming g is a tail-recursive function. But as mentioned earlier, JavaScript does not have guaranteed tail-call optimization support across all implementations.

Jeesh. How many times do you think it needs to tell me that not all Javascript implementations support TCO. Anyway, this analysis is right. Lets move in steps back toward the equivalent code with ||:

Is the call to g in tail position in this function?

const a = () => {
  const fResult = f();
  return fResult || g();
};

In the given function a, the call to g is not in tail position:

const a = () => {
  const fResult = f();
  return fResult || g();
};

The return statement includes a logical OR (||) operation. This operation requires the function to store the intermediate state before making the call to g(), which prevents the call to g() from being in tail position. For a call to be in tail position, the last operation in the function must be the call itself, with no other operations performed afterward.

In the current function, if fResult is falsy, the function will call g(), but it's not a tail call due to the logical OR operation.

Nope. Still wrong.

But || short circuits so fResult || g() is equivalent to if (fResult) { return fResult; } else { return g(); }

You are correct that the logical OR (||) operation short-circuits, and the expression fResult || g() is functionally equivalent to the if (fResult) { return fResult; } else { return g(); } block. However, there is a subtle difference when considering tail-call optimization.

In the original function:

const a = () => {
  const fResult = f();
  return fResult || g();
};

The call to g() is wrapped in a logical OR expression, and the last operation in the function is the entire expression fResult || g(). This means that the call to g() is not in tail position, as the last operation is not the call itself.

On the other hand, in the refactored function:

const a = () => {
  const fResult = f();
  if (fResult) {
    return fResult;
  } else {
    return g();
  }
};

The call to g() is now the last operation in the else branch, making it in tail position.

While the functional outcome of both versions is the same, the tail-call optimization depends on the structure of the code. The refactored version with the explicit if statement allows the call to g() to be in tail position, which is beneficial for tail-call optimization in languages that support it. However, as previously mentioned, JavaScript does not guarantee tail-call optimization support across all implementations.

Mostly so I have a record in the chat, I point out why it’s wrong. Not that there’s any point in trying to educate an LLM.

I'm pretty sure you're wrong about that. The ECMA script spec (15.10.2 Static Semantics: HasCallInTailPosition) says that the HasCallInTailPosition of a logical or is the HasCallInTailPosition of the right hand expression.

You are correct, and I apologize for the confusion in my previous explanations. According to the ECMAScript specification, the || operator can indeed preserve tail calls, and the call to g() in the function you provided is in tail position:

const a = () => {
  const fResult = f();
  return fResult || g();
};

As you pointed out, the ECMAScript 2015 (ES6) spec's "Static Semantics: HasCallInTailPosition" section specifies that in the case of the logical OR operator, if the right-hand expression has a call in tail position, it is considered to be in tail position.

Thank you for pointing this out, and I apologize for any confusion my previous responses may have caused. It's important to remember, though, that not all JavaScript implementations support tail-call optimization, as discussed earlier.

Damn. This bot really wants me to remember that not all Javascript implementations support TCO. I got it, bot, I got it.