Way to go from recursion to iteration

Question

I've used recursion quite a lot on my many years of programming to solve simple problems, but I'm fully aware that sometimes you need iteration due to memory/speed problems.

So, sometime in the very far past I went to try and find if there existed any "pattern" or text-book way of transforming a common recursion approach to iteration and found nothing. Or at least nothing that I can remember it would help.

• Are there general rules?
• Is there a "pattern"?

Answer 1

Strive to make your recursive call Tail Recursion (recursion where the last statement is the recursive call). Once you have that, converting it to iteration is generally pretty easy.

Answer 2

Well, in general, recursion can be mimicked as iteration by simply using a storage variable. Note that recursion and iteraction are generally equivalent; one can almost always be converted to the other. A tail-recursive function is very easily converted to an iterative one. Just make the accumulator variable a local one, and iterate instead of recurse. Here's an example in C++ (C were it not for the use of a default argument):

// tail-recursive
int factorial (int n, int acc = 1)
{
  if (n == 1)
    return acc;
  else
    return factorial(n - 1, acc * n);
}

// iterative
int factorial (int n)
{
  int acc = 1;
  for (; n > 1; --n)
    acc *= n;
  return acc;
}

Knowing me, I probably made a mistake in the code, but the idea is there.

Answer 3

A question that had been closed as a duplicate of this one had a very specific data structure:

The node had the following structure:

typedef struct {
    int32_t type;
    int32_t valueint;
    double  valuedouble;
    struct  cNODE *next;
    struct  cNODE *prev;
    struct  cNODE *child;
} cNODE;

The recursive deletion function looked like:

void cNODE_Delete(cNODE *c) {
    cNODE*next;
    while (c) {
        next=c->next;
        if (c->child) { 
          cNODE_Delete(c->child)
        }
        free(c);
        c=next;
    }
}

In general, it is not always possible to avoid a stack for recursive functions that invoke itself more than one time (or even once). However, for this particular structure, it is possible. The idea is to flatten all the nodes into a single list. This is accomplished by putting the current node's child at the end of the top row's list.

void cNODE_Delete (cNODE *c) {
    cNODE *tmp, *last = c;
    while (c) {
        while (last->next) {
            last = last->next;   /* find last */
        }
        if ((tmp = c->child)) {
            c->child = NULL;     /* append child to last */
            last->next = tmp;
            tmp->prev = last;
        }
        tmp = c->next;           /* remove current */
        free(c);
        c = tmp;
    }
}

This technique can be applied to any data linked structure that can be reduce to a DAG with a deterministic topological ordering. The current nodes children are rearranged so that the last child adopts all of the other children. Then the current node can be deleted and traversal can then iterate to the remaining child.

Answer 4

Recursion is nothing but the process of calling of one function from the other only this process is done by calling of a function by itself. As we know when one function calls the other function the first function saves its state(its variables) and then passes the control to the called function. The called function can be called by using the same name of variables ex fun1(a) can call fun2(a). When we do recursive call nothing new happens. One function calls itself by passing the same type and similar in name variables(but obviously the values stored in variables are different,only the name remains same.)to itself. But before every call the function saves its state and this process of saving continues. The SAVING IS DONE ON A STACK.

NOW THE STACK COMES INTO PLAY.

So if you write an iterative program and save the state on a stack each time and then pop out the values from stack when needed, you have successfully converted a recursive program into an iterative one!

The proof is simple and analytical.

In recursion the computer maintains a stack and in iterative version you will have to manually maintain the stack.

Think over it, just convert a depth first search(on graphs) recursive program into a dfs iterative program.

All the best!

Answer 5

The stacks and recursion elimination article captures the idea of externalizing the stack frame on heap, but does not provide a straightforward and repeatable way to convert. Below is one.

While converting to iterative code, one must be aware that the recursive call may happen from an arbitrarily deep code block. Its not just the parameters, but also the point to return to the logic that remains to be executed and the state of variables which participate in subsequent conditionals, which matter. Below is a very simple way to convert to iterative code with least changes.

Consider this recursive code:

struct tnode
{
    tnode(int n) : data(n), left(0), right(0) {}
    tnode *left, *right;
    int data;
};

void insertnode_recur(tnode *node, int num)
{
    if(node->data <= num)
    {
        if(node->right == NULL)
            node->right = new tnode(num);
        else
            insertnode(node->right, num);
    }
    else
    {
        if(node->left == NULL)
            node->left = new tnode(num);
        else
            insertnode(node->left, num);
    }    
}

Iterative code:

// Identify the stack variables that need to be preserved across stack 
// invocations, that is, across iterations and wrap them in an object
struct stackitem 
{ 
    stackitem(tnode *t, int n) : node(t), num(n), ra(0) {}
    tnode *node; int num;
    int ra; //to point of return
};

void insertnode_iter(tnode *node, int num) 
{
    vector<stackitem> v;
    //pushing a stackitem is equivalent to making a recursive call.
    v.push_back(stackitem(node, num));

    while(v.size()) 
    {
        // taking a modifiable reference to the stack item makes prepending 
        // 'si.' to auto variables in recursive logic suffice
        // e.g., instead of num, replace with si.num.
        stackitem &si = v.back(); 
        switch(si.ra)
        {
        // this jump simulates resuming execution after return from recursive 
        // call 
            case 1: goto ra1;
            case 2: goto ra2;
            default: break;
        } 

        if(si.node->data <= si.num)
        {
            if(si.node->right == NULL)
                si.node->right = new tnode(si.num);
            else
            {
                // replace a recursive call with below statements
                // (a) save return point, 
                // (b) push stack item with new stackitem, 
                // (c) continue statement to make loop pick up and start 
                //    processing new stack item, 
                // (d) a return point label
                // (e) optional semi-colon, if resume point is an end 
                // of a block.

                si.ra=1;
                v.push_back(stackitem(si.node->right, si.num));
                continue; 
ra1:            ;         
            }
        }
        else
        {
            if(si.node->left == NULL)
                si.node->left = new tnode(si.num);
            else
            {
                si.ra=2;                
                v.push_back(stackitem(si.node->left, si.num));
                continue;
ra2:            ;
            }
        }

        v.pop_back();
    }
}

Notice how the structure of the code still remains true to the recursive logic and modifications are minimal, resulting in less number of bugs. For comparison, I have marked the changes with ++ and --. Most of the new inserted blocks except v.push_back, are common to any converted iterative logic

void insertnode_iter(tnode *node, int num) 
{

+++++++++++++++++++++++++

    vector<stackitem> v;
    v.push_back(stackitem(node, num));

    while(v.size())
    {
        stackitem &si = v.back(); 
        switch(si.ra)
        {
            case 1: goto ra1;
            case 2: goto ra2;
            default: break;
        } 

------------------------

        if(si.node->data <= si.num)
        {
            if(si.node->right == NULL)
                si.node->right = new tnode(si.num);
            else
            {

+++++++++++++++++++++++++

                si.ra=1;
                v.push_back(stackitem(si.node->right, si.num));
                continue; 
ra1:            ;    

-------------------------

            }
        }
        else
        {
            if(si.node->left == NULL)
                si.node->left = new tnode(si.num);
            else
            {

+++++++++++++++++++++++++

                si.ra=2;                
                v.push_back(stackitem(si.node->left, si.num));
                continue;
ra2:            ;

-------------------------

            }
        }

+++++++++++++++++++++++++

        v.pop_back();
    }

-------------------------

}

Answer 6

It seems nobody has addressed where the recursive function calls itself more than once in the body, and handles returning to a specific point in the recursion (i.e. not primitive-recursive). It is said that every recursion can be turned into iteration, so it appears that this should be possible.

I just came up with a C# example of how to do this. Suppose you have the following recursive function, which acts like a postorder traversal, and that AbcTreeNode is a 3-ary tree with pointers a, b, c.

public static void AbcRecursiveTraversal(this AbcTreeNode x, List<int> list) {
        if (x != null) {
            AbcRecursiveTraversal(x.a, list);
            AbcRecursiveTraversal(x.b, list);
            AbcRecursiveTraversal(x.c, list);
            list.Add(x.key);//finally visit root
        }
}

The iterative solution:

        int? address = null;
        AbcTreeNode x = null;
        x = root;
        address = A;
        stack.Push(x);
        stack.Push(null)    

        while (stack.Count > 0) {
            bool @return = x == null;

            if (@return == false) {

                switch (address) {
                    case A://   
                        stack.Push(x);
                        stack.Push(B);
                        x = x.a;
                        address = A;
                        break;
                    case B:
                        stack.Push(x);
                        stack.Push(C);
                        x = x.b;
                        address = A;
                        break;
                    case C:
                        stack.Push(x);
                        stack.Push(null);
                        x = x.c;
                        address = A;
                        break;
                    case null:
                        list_iterative.Add(x.key);
                        @return = true;
                        break;
                }

            }


            if (@return == true) {
                address = (int?)stack.Pop();
                x = (AbcTreeNode)stack.Pop();
            }


        }