Time complexity of recursive algorithm

Question

I have a grid with x-sided field in it. Every field contains a link to it's x surrounding fields. [x is constant]

I have an algorithm which is implemented in this field, (which can probably be optimized):

[java like pseudocode]

public ArrayList getAllFields(ArrayList list) {

  list.addToList(this);

  for each side {
    if ( ! list.contains(neighbour) && constantTimeConditionsAreMet()) {
      neighbour.getAllFields(list) //Recursive call
    }
  }

  return list;

}

I'm having trouble finding the time complexity.

ArrayList#contains(Object) runs in linear time

How do i find the time complexity? My approach is this:

T(n) = O(1) + T(n-1) +
c(nbOfFieldsInArray - n) [The time to check the ever filling ArrayList]

T(n) = O(1) + T(n-1) + c*nbOfFieldsInArray - cn

Does this give me T(n) = T(n-1) + O(n)?

Where's the recursive call meant to happen? Is 'getContinent' meant to be 'getAllFields'? — Gian
– Gian, Commented Apr 26, 2011 at 21:19
Change the list to a hash table and you have an O(n) algorithm where n = number of fields :) — Antti Huima
– Antti Huima, Commented Apr 27, 2011 at 21:14
I was planning on doing that. Would you agree the algorithm is O(n^2) now? — Samuel
– Samuel, Commented Apr 28, 2011 at 9:28

Jim Mischel · Accepted Answer · 2011-04-27 18:25:05Z

2

The comment you added to your code is not helpful. What does getContinent do?

In any case, since you're using a linear search (ArrayList.contains) for every potential addition to the list, then it looks like the complexity will be Omega(n^2).

edited Apr 27, 2011 at 18:25

answered Apr 26, 2011 at 21:51

Jim Mischel

135k25 gold badges197 silver badges377 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

Samuel Over a year ago

Very sorry, i had named it differently before, i forgot i renamed it later and thanks; That's a hint already :)

Aryabhatta Over a year ago

+1: But to nitpick: 'at least O(n^2)' is not too meaningful. Perhaps you mean to say Omega(n^2)?

Jim Mischel Over a year ago

@Moron: noted. Although my reading of en.wikipedia.org/wiki/… indicates that "Omega(n^2)" means "O larger than n^2", which (I thought) was the same as "at least O(n^2)". I can see, though, how my term is less informative.

Aryabhatta Over a year ago

@JIm: It is not 'less informative'. Technically speaking, it is actually meaningless.

Aryabhatta Over a year ago

To be more clear: O(n^2) means not worse than cn^2: i.e. BigOh is used to state upper bounds. So what you are saying is "at least not worse than cn^2". Omega is used to state lower bounds, which is exactly what you need here...

|

pgpb.padilla · Accepted Answer · 2015-12-03 16:39:49Z

1

You recurrence seems correct T(n) = T(n-1) + theta(1).

If you draw the recursion tree you'll notice you have a single branch with the values theta(n-1), theta(n-2), ..., theta(2), theta(1), if you add up all the levels you get the arithmetic series 1+2+3+...+n

S1 = 1+2+3+...+n

If you define

S2 = n+...+3+2+1

and then calculate S1+S2 you get

S1 + S2 = 2*S1 = (n+1) + (n+1) + ... + (n+1) = n(n+1)

therefore

2*S1 = n(n-1) => S1 = n(n-1)/2

which means T(n) = 1/2 theta(n(n-1)) = 1/2 theta(n^2) = theta(n^2)

edited Dec 3, 2015 at 16:39

answered Dec 3, 2015 at 6:39

pgpb.padilla

2,4362 gold badges21 silver badges47 bronze badges

Collectives™ on Stack Overflow

Time complexity of recursive algorithm

2 Answers 2

6 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

6 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related