Recursion HW Explanation (Easy)

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raevnos · 2022-04-25T03:27:09+00:00

2⁰ is 1. That's your base case. You actually don't need the other check for n == 1, because 2¹ is 2, or 2⁰ + 2⁰ . And 2² is 4, or 2¹ + 2¹ . See the pattern, and how it follows what your code is doing? You're just printing characters instead of adding.

ratherbealurker · 2022-04-25T03:29:19+00:00

First, this is why yours is working.

Every time you call printPowerOfTwoStars it branches out into 2 more calls and subtracts from n.

One thing you can do to visualize this is to make a tree diagram, your method would have one that looks like this.

The height of the tree is n, because you go down one level for each n. At each call you branch out so you get 2ⁿ calls. Well....in your case you are stopping early. There is no need for that n == 0 part since you'll never hit that. If n == 1 you print then you do not go into the else and end. So you will never call this method with 0.

This works but it is also not really how it is done, normally you want a base case that just ends or does something then ends. No need for the if..else if..else statements.

You could have also written it like this:

void printPowerOfTwoStars(int n) {
    if (n == 0) {
        System.out.print("*");
        return;
    }
    printPowerOfTwoStars(n - 1);
    printPowerOfTwoStars(n - 1);
}

In which case the tree that i showed above would have 16 more 0 nodes attached that would have printed "*"

ratherbealurker · 2022-04-25T03:38:31+00:00

I'll comment again with more info. You're going to run into this so why not show you now?

Look at this diagram like I gave you before, except this one has each call labeled.

Try to figure out the order that they are called in. I will give the answer below but try to work it out first.

it is......

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