Recursion tree for recursive equation

Question

I have this recursive equation: T(n) = T(n/2) + T(n/5) + T(n/9) + Θ(n)

I draw my recursion tree like this:

          cn
     /     |     \
   n/2    n/5    n/9
  / | \  / | \  / | \
  ..................

The tree has log(n) + 1 levels, each level has 3 times more nodes than the level above, and subproblem sizes decrease by a factor of 2 each time. Now this is how I see the total cost is:

I forgot to put this: Is my solution correct?

@templatetypedef So I think the longest path is n -> n/2 -> n/4... So the number nodes at depth i is 3^i, and each node at depth i, has a cost of c(n / (2^i)). So the total cost is 3^i * c(n / (2^i)), or (3/2)^i * cn — PTN, Oct 18 '17 at 18:47
Hint: the time complexity's upper bound is given by the depth of the *longest* branch of the recursion tree. — meowgoesthedog, Oct 18 '17 at 18:48
@PTN correct. Now think about what the "effective" recurrence relation would be, i.e. a simpler one which gives the same upper bound. — meowgoesthedog, Oct 18 '17 at 18:51
@PTN I did not downvote this. Just put a comment. Although I do find it hilarious that you downvoted one of my questions because you thought I downvoted yours. Would appreciate it if you removed your downvote. — kkaosninja, Oct 18 '17 at 20:11

meowgoesthedog · Accepted Answer · 2017-12-10T02:59:11.903

Suppose we have the slightly more general relation:

Where f(n) is some function, e.g. cn.

If we re-substitute this relation into itself, i.e. each of the recursive calls, the next layer of recursive calls produced will have its parameter multiplied by the corresponding factor:

If we continue, the pattern is given by this expression:

... i.e. a Trinomial expansion. The coefficients of each T term is given by the trinomial coefficients, and the argument by the different powers of λμν:

As you can see from the expansion, the f-terms are one level of recursion behind the T-terms. So, the sum of all f-terms, taking into consideration that they must be accumulated unlike the T-terms:

All of the above can be proven by induction, and are generalizable to any number of recursive calls.

When do we terminate, i.e. get rid of the T-terms? As I said in my comment, when recursion reaches the end of the longest path. This is given by considering the slowest decreasing term (as you correctly deduced):

Thus the time complexity function's tightest closed-form bound is given by:

This is very easy to evaluate if f(n) is some power of n.

For the example, f(n) = Θ(n), α = β = γ = 1, λ = 2, μ = 5, ν = 9. Thus:

The term inside the brackets is exactly the trinomial expansion from earlier. Thus:

Since τ < 1, the exponential term vanishes. Therefore T(n) = O(n).

Numerical tests to confirm this:

n         T(n)
-----------------------------
10        21.86111111
100       328.1442901
1000      3967.333431
10000     44150.87621
100000    471262.9357
1000000   4910520.041
10000000  50415530.84

log-log plot of T(n) against n:

The gradient is 1.054 ± 0.01166, which is very close to the theoretical value of 1, thus strongly supporting the result of T(n) = O(n).

Recursion tree for recursive equation

1 Answers1