CODEWITHSUNDEEP
javaalgorithmrecursion
Abwatts
Abwatts

Reputation: 31

Stuck with understanding recursion

For the past couple of hours, I have tried to understand why the output for this program is behaving in such a weird manner. The first part of the recursive functions makes sense to me, however, the rest I, unfortunately, find extremely confusing.

I was wondering how come that the program is going backward, and even adds to a variable that should be subtracted instead. This does not make any sense to me. I would appreciate any help, thanks.

Here is my recursive function:

public static void threeWays(int id, int n) {
    if (n >= 0) {
        System.out.println(id + "=" + n);
        threeWays(id + 1, n - 1);
        threeWays(id + 1, n - 2);
        threeWays(id + 1, n - 3);
    }
}

public static void main(String[] args) {
    threeWays(1, 4);
}

And here's the output:

1=4

2=3

3=2

4=1

5=0

4=0

3=1

4=0

3=0

2=2

3=1

4=0

3=0

2=1

3=0

Upvotes: 1

Views: 143

Answers (2)

ghchoi
ghchoi

Reputation: 5166

Python3 version:

  def threeWays(f, i, n):
      if n >= 0:
          print(f, i, '=', n)
          threeWays('1st', i + 1, n - 1)
          threeWays('2nd', i + 1, n - 2)
          threeWays('3rd', i + 1, n - 3)

  threeWays('    ', 1, 4)

At the first sight, it could, could look like DFS but actually not...

When threeWays is recursively called from 2nd and 3rd threeWays, 1st or 1st and 2nd threeWay(s) is/are also called...!

threeWays('   ', 1, 4)            ->     1 = 4
  threeWays('1st', 2, 3)          -> 1st 2 = 3
    threeWays('1st', 3, 2)        -> 1st 3 = 2
      threeWays('1st', 4, 1)      -> 1st 4 = 1
        threeWays('1st', 5, 0)    -> 1st 5 = 0
          threeWays('1st', 6, -1)
        threeWays('2nd', 5, -2)
        threeWays('3rd', 5, -3)
      threeWays('2nd', 4, 0)      -> 2nd 4 = 0
      threeWays('3rd', 4, -1)
    threeWays('2nd', 3, 1)        -> 2nd 3 = 1
      threeWays('1st', 4, 0)      -> 1st 4 = 0
      threeWays('2nd', 4, -1)
      threeWays('3rd', 4, -2)
    threeWays('3rd', 3, 0)        -> 3rd 3 = 0
...

Upvotes: 1

sprinter
sprinter

Reputation: 27986

You might find it more useful if you indent the output to show how deep the call stack is at each point. I replaced your print statement with System.out.println(" ".repeat(id) + n); and received the output:

  4
    3
      2
        1
          0
        0
      1
        0
      0
    2
      1
        0
      0
    1
      0

The structure here is that every time a number is printed, the next level shows the three numbers below in descending order, until it gets to zero. That is exactly what your function does.

Upvotes: 0

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