recursive to iterative (java)

Question

as a beginner in programming I am trying to convert the following recursive method to an iterative one but I just don't get the hang of it. The method or function has a binary tree like recursion and I would like to use an array for the iterative solution.. unfortunately I am very confused how to do it.

I have already checked the way of converting the fibonnaci recursive method to an iterative one. But I think this is not the same here. Also I am not sure if a tree search method is useful?! Any help, hint, idea would be appreciated. Thanks.

public static int funct(int n) {

  if (n == 0) return 1;
  if (n == 1) return 2;
  if n > 1 return funct(n-2)*funct(n/2);

}

funct(n/2)? What if n is odd? That is a strange function you have here — fge
– fge, Commented Sep 8, 2016 at 10:26
@fge it seems it's related to A242634 where the answer is 2^A242634(n) — Sylwester
– Sylwester, Commented Sep 8, 2016 at 15:21

Liviu Stirb · Accepted Answer · 2016-09-08 11:01:53Z

8

Since every n-th member is computed by others before if you can cache all in a list. You start by adding the first 2 known members. Fibonacci its easier because you always need only previous value.

private static int func(int n) {
    List<Integer> values = new ArrayList<>(n+1);
    values.add(1);
    values.add(2);
    for (int i = 2; i <= n; i++) {
        values.add(values.get(i - 2) * values.get(i / 2));
    }

    return values.get(n);
}

edited Sep 8, 2016 at 11:01

answered Sep 8, 2016 at 10:29

Liviu Stirb

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3 Comments

xenteros Over a year ago

That's the answer.

Raskayu Over a year ago

I was thinking using a while with a Stack, but yours looks btter

Jekin Kalariya Over a year ago

super solution with super logic

Joop Eggen · Accepted Answer · 2016-09-08 14:57:40Z

2

Now the real function is without last if:

public static int funct(int n) {
    if (n == 0) return 1;
    if (n == 1) return 2;
    return funct(n-2) * funct(n/2);
}

As the recursive calls refer to smaller parameters one can cache all return values upto n.

Unfortunately this already spoils the pleasure, as the resulting code is complete:

public static int funct(int n) {
    int[] results = new int[n+1];
    results[0] = 1;
    results[1] = 2;
    int i = 2;
    while (i <= n) {
        results[i] = results[i-2] * results[i/2];
        ++i;
    }
    return results[n];
}

It indeed looks like fibonacci.

In general one would not need to fill all items of results. like probably results[n - 1]. Unfortunately you should have learnt prior to this problem:

Solving tail recursion.
Using a stack (like here) to use inner results of a recurive call.

You might look into those topics.

Math afterlude

The initial values are powers of 2. As the result is a product of earlier results, all results will be powers of 2.

f(0) = 1 = 2^0
f(1) = 2 = 2^1
f(n) = f(n - 2) * f(n / 2)

Hence you can introduce:

g(0) = 0
g(1) = 1
g(n) = g(n - 2) + g(n / 2)
f(n) = 2^g(n)

This will enlarge the range you can calculate as say 2¹⁰⁰.

You will also see:

g(2k + 1) = g(2k) + 1

So you will only need a domain of even numbers:

g(2k) = g(2(k-1)) + g(k - k%2) + k%2

edited Sep 8, 2016 at 14:57

answered Sep 8, 2016 at 10:44

Joop Eggen

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6 Comments

Liviu Stirb Over a year ago

you should also increment i in the while loop

Joop Eggen Over a year ago

@user1121883 now it is infinitely better.

Jekin Kalariya Over a year ago

@joop just miss the link :)

Joop Eggen Over a year ago

@JekinKalariya tail recursion in wikipedia was not very convincing for me, and stack is more or less what I did. We both know that simple math will yield some power of 2 formula f(n) = 2^{n-2 + [(n-8):2].[(n-6):2] / 2} given n > 5with integer division and a sum 1+2+3+...+k. (Not sure whether I got it right.) It would indeed be nice to show that math derivation, but the OP has to do that work himself.

MainStack Over a year ago

@Joop: Thanks.. that's really great.. I very much like your implementation! I definitely will have a look at the mentioned topics of tail recursion and use of stack. I am also thinking to optimize the size of the array. I don't know if this has any benefit but one can set the length to max(n-2, n/2) + 1. What do you think?

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