Turning a recursive function into an iterative function

Question

I have written the following recursive function, but am incurring a runtime error due to maximum recursion depth. I was wondering is it possible to write an iterative function to overcome this:

def finaldistance(n):

   if n%2 == 0:
       return 1 + finaldistance(n//2) 

   elif n != 1:
       a = finaldistance(n-1)+1
       b = distance(n)
       return min(a,b)

   else:
       return 0

What I have tried is this but it does not seem to be working,

def finaldistance(n, acc):

   while n > 1:

     if n%2 == 0:
        (n, acc) = (n//2, acc+1)

     else:
        a = finaldistance(n-1, acc) + 1
        b = distance(n)
        if a < b:
          (n, acc) = (n-1, acc+1)

        else:
          (n, acc) =(1, acc + distance(n))

   return acc

Answer 1

Johnbot's solution shows you how to solve your specific problem. How in general can we remove this recursion? Let me show you how, by making a series of small, clearly correct, clearly safe refactorings.

First, here's a slightly rewritten version of your function. I hope you agree it is the same:

def f(n):
  if n % 2 == 0:
    return 1 + f(n // 2) 
  elif n != 1:
    a = f(n - 1) + 1
    b = d(n)
    return min(a, b)
  else:
    return 0

I want the base case to be first. This function is logically the same:

def f(n):
  if n == 1:
    return 0
  if n % 2 == 0:
    return 1 + f(n // 2) 
  a = f(n - 1) + 1
  b = d(n)
  return min(a, b)

I want the code that comes after each recursive call to be a method call and nothing else. These functions are logically the same:

def add_one(n, x):
    return 1 + x

def min_distance(n, x):
    a = x + 1
    b = d(n)
    return min(a, b)

def f(n):
    if n == 1:
        return 0
    if n % 2 == 0:
        return add_one(n, f(n // 2))
    return min_distance(n, f(n - 1))

Similarly, we add helper functions that compute the recursive argument:

def half(n):
    return n // 2

def less_one(n):
    return n - 1

def f(n):
    if n == 1:
        return 0
    if n % 2 == 0:
        return add_one(n, f(half(n))
    return min_distance(n, f(less_one(n))

Again, make sure you agree that this program is logically the same. Now I'm going to simplify the computation of the argument:

def get_argument(n):
    return half if n % 2 == 0 else less_one    

def f(n):
    if n == 1:
        return 0
    argument = get_argument(n) # argument is a function!
    if n % 2 == 0:
        return add_one(n, f(argument(n)))
    return min_distance(n, f(argument(n)))

Now I'm going to do the same thing to the code after the recursion, and we'll get down to a single recursion:

def get_after(n):
    return add_one if n % 2 == 0 else min_distance

def f(n):
    if n == 1:
        return 0
    argument = get_argument(n)
    after = get_after(n) # this is also a function!
    return after(n, f(argument(n)))  

Now I'm noticing that we're passing n to get_after, and then passing it right along to "after" again. I'm going to curry these functions to eliminate that problem. This step is tricky. Make sure you understand it!

def add_one(n):
    return lambda x: x + 1

def min_distance(n):
    def nested(x):
        a = x + 1
        b = d(n)
        return min(a, b)
    return nested

These functions did take two arguments. Now they take one argument, and return a function that takes one argument! So we refactor the use site:

def get_after(n):
    return add_one(n) if n % 2 == 0 else min_distance(n)

and here:

def f(n):
    if n == 1:
        return 0
    argument = get_argument(n)
    after = get_after(n) # now this is a function of one argument, not two
    return after(f(argument(n)))  

Similarly we notice that we are calling get_argument(n)(n) to get the argument. Let's simplify that:

def get_argument(n):
    return half(n) if n % 2 == 0 else less_one(n)   

And let's make it just slightly more general:

base_case_value = 0

def is_base_case(n):
  return n == 1

def f(n):
    if is_base_case(n):
        return base_case_value
    argument = get_argument(n)
    after = get_after(n)
    return after(f(argument))  

OK, we now have our program in an extremely compact form. The logic has been spread out into multiple functions, and some of them are curried, to be sure. But now that the function is in this form we can easily remove the recursion. This is the bit that is really tricky is turning the whole thing into an explicit stack:

def f(n):
    # Let's make a stack of afters. 
    afters = [ ]
    while not is_base_case(n) :
        argument = get_argument(n)
        after = get_after(n)
        afters.append(after)
        n = argument
    # Now we have a stack of afters:
    x = base_case_value
    while len(afters) != 0:
        after = afters.pop()
        x = after(x)
    return x

Study this implementation very carefully. You will learn a lot from it. Remember, when you do a recursive call:

after(f(something))

you are saying that after is the continuation -- the thing that comes next -- of the call to f. We typically implement continuations by putting information about the location in the callers code onto the "call stack". What we're doing in this removal of recursion is simply moving continuation information off of the call stack and onto a stack data structure. But the information is exactly the same.

The important thing to realize here is that we typically think of the call stack as "what is the thing that happened in the past that got me here?". That is exactly backwards. The call stack tells you what you have to do after this call is finished! So that's the information that we encode in the explicit stack. Nowhere do we encode what we did before each step as we "unwind the stack", because we don't need that information.

As I said in my initial comment: there is always a way to turn a recursive algorithm into an iterative one but it is not always easy. I've shown you here how to do it: carefully refactor the recursive method until it is extremely simple. Get it down to a single recursion by refactoring it. Then, and only then, apply this transformation to get it into an explicit stack form. Practice that until you are comfortable with this program transformation. You can then move on to more advanced techniques for removing recursions.

Note that of course this is almost certainly not the "pythonic" way to solve this problem; you could likely build a much more compact, understandable method using lazily evaluated list comprehensions. This answer was intended to answer the specific question that was asked: how in general do we turn recursive methods into iterative methods?

I mentioned in a comment that a standard technique for removing a recursion is to build an explicit list as a stack. This shows that technique. There are other techniques: tail recursion, continuation passing style and trampolines. This answer is already too long, so I'll cover those in a follow-up answer.

Answer 2

Read this answer after you read my first answer.

Again, we are answering the question in general of "how do you turn a recursive algorithm into an iterative algorithm", in this case in Python. As noted previously, this is about exploring the general idea of transforming a program; this is not the "pythonic" way to solve the specific problem.

In my first answer I started by rewriting the program into this form:

def f(n):
    if is_base_case(n):
        return base_case_value
    argument = get_argument(n)
    after = get_after(n)
    return after(f(argument))  

And then transformed it into this form:

def f(n):
    # Let's make a stack of afters. 
    afters = [ ]
    while not is_base_case(n) :
        argument = get_argument(n)
        after = get_after(n)
        afters.append(after)
        n = argument
    # Now we have a stack of afters:
    x = base_case_value
    while len(afters) != 0:
        after = afters.pop()
        x = after(x)
    return x

The technique here is to construct an explicit stack of "after" calls for a particular input, and then once we have it, run down the whole stack. We are essentially simulating what the runtime already does: constructs a stack of "continuations" that say what to do next.

A different technique is to let the function itself decide what to do with its continuation; this is called "continuation passing style". Let's explore it.

This time, we're going to add a parameter c to the recursive method f. c is a function that takes what would normally be the return value of f, and does whatever was suppose to happen after the call to f. That is, it is explicitly the continuation of f. The method f then becomes "void returning".

The base case is easy. What do we do if we're in the base case? We call the continuation with the value we would have returned:

def f(n, c):
    if is_base_case(n):
        c(base_case_value)
        return

Easy peasy. What about the non-base case? Well, what were we going to do in the original program? We were going to (1) get the arguments, (2) get the "after" -- the continuation of the recursive call, (3) do the recursive call, (4) call "after", its continuation, and (5) return the computed value to whatever the continuation of f is.

We're going to do all the same things, except that when we do step (3) now we need to pass in a continuation that does steps 4 and 5:

    argument = get_argument(n)
    after = get_after(n)
    f(argument, lambda x: c(after(x)))

Hey, that is so easy! What do we do after the recursive call? Well, we call after with the value returned by the recursive call. But now that value is going to be passed to the recursive call's continuation function, so it just goes into x. What happens after that? Well, whatever was going to happen next, and that's in c, so it needs to be called, and we're done.

Let's try it out. Previously we would have said

print(f(100))

but now we have to pass in what happens after f(100). Well, what happens is, the value gets printed!

f(100, print)

and we're done.

So... big deal. The function is still recursive. Why is this interesting? Because the function is now tail recursive! That is, the last thing it does in the non-base case is call itself. Consider a silly case:

def tailcall(x, sum):
  if x <= 0:
    return sum
  return tailcall(x - 1, sum + x)

If we call tailcall(10, 0) it calls tailcall(9, 10), which calls (8, 19), and so on. But any tail-recursive method we can rewrite into a loop very, very easily:

def tailcall(x, sum):
  while True:
    if x <= 0:
      return sum
    x = x - 1
    sum = sum + x

So can we do the same thing with our general case?

# This is wrong!
def f(n, c):
    while True:
        if is_base_case(n):
            c(base_case_value)
            return
        argument = get_argument(n)
        after = get_after(n)
        n = argument
        c = lambda x: c(after(x))

Do you see what is wrong? the lambda is closed over c and after, which means that every lambda will use the current value of c and after, not the value it had when the lambda was created. So this is broken, but we can fix it easily by creating a scope which introduces new variables every time it is invoked:

def continuation_factory(c, after)
    return lambda x: c(after(x))


def f(n, c):
    while True:
        if is_base_case(n):
            c(base_case_value)
            return
        argument = get_argument(n)
        after = get_after(n)
        n = argument
        c = continuation_factory(c, after)

And we're done! We've turned this recursive algorithm into an iterative algorithm.

Or... have we?

Think about this really carefully before you read on. Your spider sense should be telling you that something is wrong here.

---

The problem we started with was that a recursive algorithm is blowing the stack. We've turned this into an iterative algorithm -- there's no recursive call at all here! We just sit in a loop updating local variables.

The question though is -- what happens when the final continuation is called, in the base case? What does that continuation do? Well, it calls its after, and then it calls its continuation. What does that continuation do? Same thing.

All we've done here is moved the recursive control flow into a collection of function objects that we've built up iteratively, and calling that thing is still going to blow the stack. So we haven't actually solved the problem.

Or... have we?

What we can do here is add one more level of indirection, and that will solve the problem. (This solves every problem in computer programming except one problem; do you know what that problem is?)

What we'll do is we'll change the contract of f so that it is no longer "I am void-returning and will call my continuation when I'm done". We will change it to "I will return a function that, when it is called, calls my continuation. And furthermore, my continuation will do the same."

That sounds a little tricky but really its not. Again, let's reason it through. What does the base case have to do? It has to return a function which, when called, calls my continuation. But my continuation already meets that requirement:

def f(n, c):
    if is_base_case(n):
        return c(base_case_value)

What about the recursive case? We need to return a function, which when called, executes the recursion. The continuation of that call needs to be a function that takes a value and returns a function that when called executes the continuation on that value. We know how to do that:

    argument = get_argument(n)
    after = get_after(n)
    return lambda : f(argument, lambda x: lambda: c(after(x)))

OK, so how does this help? We can now move the loop into a helper function:

def trampoline(f, n, c):
    t = f(n, c)
    while t != None:
        t = t()

And call it:

trampoline(f, 3, print)

And holy goodness it works.

Follow along what happens here. Here's the call sequence with indentation showing stack depth:

trampoline(f, 3, print)
  f(3, print)

What does this call return? It effectively returns lambda : f(2, lambda x: lambda : print(min_distance(x)), so that's the new value of t.

That's not None, so we call t(), which calls:

  f(2, lambda x: lambda : print(min_distance(x))

What does that thing do? It immediately returns

lambda : f(1,
  lambda x:
    lambda:
      (lambda x: lambda : print(min_distance(x)))(add_one(x))

So that's the new value of t. It's not None, so we invoke it. That calls:

  f(1,
    lambda x:
      lambda:
        (lambda x: lambda : print(min_distance(x)))(add_one(x))

Now we're in the base case, so we *call the continuation, substituting 0 for x. It returns:

      lambda: (lambda x: lambda : print(min_distance(x)))(add_one(0))

So that's the new value of t. It's not None, so we invoke it.

That calls add_one(0) and gets 1. It then passes 1 for x in the middle lambda. That thing returns:

lambda : print(min_distance(1))

So that's the new value of t. It's not None, so we invoke it. And that calls

  print(min_distance(1))

Which prints out the correct answer, print returns None, and the loop stops.

Notice what happened there. The stack never got more than two deep because every call returned a function that said what to do next to the loop, rather than calling the function.

If this sounds familiar, it should. Basically what we're doing here is making a very simple work queue. Every time we "enqueue" a job, it is immediately dequeued, and the only thing the job does is enqueues the next job by returning a lambda to the trampoline, which sticks it in its "queue", the variable t.

We break the problem up into little pieces, and make each piece responsible for saying what the next piece is.

Now, you'll notice that we end up with arbitrarily deep nested lambdas, just as we ended up in the previous technique with an arbitrarily deep queue. Essentially what we've done here is moved the workflow description from an explicit list into a network of nested lambdas, but unlike before, this time we've done a little trick to avoid those lambdas ever calling each other in a manner that increases the stack depth.

Once you see this pattern of "break it up into pieces and describe a workflow that coordinates execution of the pieces", you start to see it everywhere. This is how Windows works; each window has a queue of messages, and messages can represent portions of a workflow. When a portion of a workflow wishes to say what the next portion is, it posts a message to the queue, and it runs later. This is how async await works -- again, we break up the workflow into pieces, and each await is the boundary of a piece. It's how generators work, where each yield is the boundary, and so on. Of course they don't actually use trampolines like this, but they could.

The key thing to understand here is the notion of continuation. Once you realize that you can treat continuations as objects that can be manipulated by the program, any control flow becomes possible. Want to implement your own try-catch? try-catch is just a workflow where every step has two continuations: the normal continuation and the exceptional continuation. When there's an exception, you branch to the exceptional continuation instead of the regular continuation. And so on.

The question here was again, how do we eliminate an out-of-stack caused by a deep recursion in general. I've shown that any recursive method of the form

def f(n):
    if is_base_case(n):
        return base_case_value
    argument = get_argument(n)
    after = get_after(n)
    return after(f(argument))
...
print(f(10))

can be rewritten as:

def f(n, c):
    if is_base_case(n):
        return c(base_case_value)
    argument = get_argument(n)
    after = get_after(n)
    return lambda : f(argument, lambda x: lambda: c(after(x)))
...
trampoline(f, 10, print)

and that the "recursive" method will now use only a very small, fixed amount of stack.

Answer 3

First you need to find all the values of n, luckily your sequence is strictly descending and only depends on the next distance:

values = []
while n > 1:
  values.append(n)
  n = n // 2 if n % 2 == 0 else n - 1

Next you need to calculate the distance at each value. To do that we need to start from the buttom:

values.reverse()

And now we can easily keep track of the previous distance if we need it to calculate the next distance.

distance_so_far = 0
for v in values:
  if v % 2 == 0:
    distance_so_far += 1
  else:
    distance_so_far = min(distance(v), distance_so_far + 1)

return distance_so_far

Stick it all together:

def finaldistance(n):
    values = []
    while n > 1:
      values.append(n)
      n = n // 2 if n % 2 == 0 else n - 1

    values.reverse()
    distance_so_far = 0
    for v in values:
      if v % 2 == 0:
        distance_so_far += 1
      else:
        distance_so_far = min(distance(v), distance_so_far + 1)

    return distance_so_far

And now you're using memory instead of stack.

(I don't program in Python so this is probably not be idiomatic Python)