2

I am trying to analyze the Time Complexity of a recursive algorithm that solves the Generate all sequences of bits within Hamming distance t problem. The algorithm is this:

// str is the bitstring, i the current length, and changesLeft the
// desired Hamming distance (see linked question for more)
void magic(char* str, int i, int changesLeft) {
        if (changesLeft == 0) {
                // assume that this is constant
                printf("%s\n", str);
                return;
        }
        if (i < 0) return;
        // flip current bit
        str[i] = str[i] == '0' ? '1' : '0';
        magic(str, i-1, changesLeft-1);
        // or don't flip it (flip it again to undo)
        str[i] = str[i] == '0' ? '1' : '0';
        magic(str, i-1, changesLeft);
}

What is the time complexity of this algorithm?

I fond myself pretty rusty when it comes to this and here is my attempt, which I feel is no where near the truth:

t(0) = 1
t(n) = 2t(n - 1) + c
t(n) = t(n - 1) + c
     = t(n - 2) + c + c
     = ...
     = (n - 1) * c + 1
    ~= O(n)

where n is the length of the bit string.

Related questions: 1, 2.

Community
  • 1
  • 1
gsamaras
  • 71,951
  • 46
  • 188
  • 305

1 Answers1

5

It's exponential:

t(0) = 1
t(n) = 2 t(n - 1) + c
t(n) = 2 (2 t(n - 2) + c) + c          = 4 t (n - 2) + 3 c
     = 2 (2 (2 t(n - 3) + c) + c) + c  = 8 t (n - 3) + 7 c
     = ...
     = 2^i t(n-i) + (2^i - 1) c         [at any step i]
     = ...
     = 2^n t(0) + (2^n - 1) c          = 2^n + (2^n - 1) c
    ~= O(2^n)

Or, using WolframAlpha: https://www.wolframalpha.com/input/?i=t(0)%3D1,+t(n)%3D2+t(n-1)+%2B+c

The reason it's exponential is that your recursive calls are reducing the problem size by 1, but you're making two recursive calls. Your recursive calls are forming a binary tree.

Aziz
  • 20,065
  • 8
  • 63
  • 69
  • Result feels like correct Aziz, but could you please explain the intuition? I mean the way you crafted out this answer. Learn the man to fish, don't just give him the fish! :D – gsamaras Dec 10 '16 at 16:56
  • This method of solving the recurrence relation is called "telescoping". You basically substitute the function t(n) over and over, until you are able to determine a pattern in the relation (for a given step `i`). Then you'll be able to determine the final step with t(0), which is the solution. – Aziz Dec 10 '16 at 16:58
  • Thanks Aziz very much, shouldn't we introduce the `changesLeft` in the equation somehow? I mean in [its iterative version, we did](http://stackoverflow.com/questions/41086928/time-complexity-of-an-iterative-algorithm), it's called `dist` there (I am trying to compare them theoretically). I know that for a given `changesLeft`, (`n` choose `changesLeft`) times, we will reach this line of code: `printf("%s\n", str);`. I posted a [relevant question](http://stackoverflow.com/questions/41105621/trying-to-compare-a-recursive-and-iterative-algorithm); if you have some time, please take a look! :) – gsamaras Dec 12 '16 at 16:51