I am aware that to get a running time by recursion tree method, we need to draw a tree and find:

a) number of levels in tree.

Since left side of tree decreases by 1 in size, so it's longest path from root. Subproblem size at level i is n-i . setting n - i = 1 when it hits a size of 1, we get number of levels, i = n - 1.

b) cost per node in tree : cn

c) Number of nodes at level i: This is where i am stuck. Not able to find nodes at level i since left side decreases by 1, right side by half. Naturally, tree is more dense towards left side. Not every node will have two children.

if i can get answer to c, i can calculate T(n) = cost of level 0 + cost of level 1 + cost of level 2 + ... cost of level n-1. if y1 is number of nodes at level 1, y2 at level 2, etc... then
=> T(n) = cn + y1 * cn + y2 * cn + y3 * cn + .... yn-1 * cn to get total cost.

Can anyone guide me to the approach i am taking ? is it correct ? can i take an assumption that for sufficiently large n, we can ignore T(n/2) and then proceed ? .

Online searching confused me. Problem is 4.4-5 from CLRS.

Please see here

This solution says T(n) = O(2^n) and T(n) = omega(n^2) and does not explain how.

Also see here

This solution says T(n) = O(n^2). but contradicts with above solution

  • $\begingroup$ It's a lovely problem. The T(n/2) is what makes it so nice, and it can absolutely not be ignored. Assume T(n) ≈ c n^k. Then T(n) ≈ c (n-1)^k + c (n/2)^k ≈ c (1 +2^-k)n^k - c n^(k-1) >> T(n). So the solution is not polynomial. $\endgroup$
    – gnasher729
    Commented Apr 23, 2020 at 23:19
  • $\begingroup$ The problems you quote are all different. It's very difficult to get an idea by looking at the formula only. (Well, Yuval can... ) Using a spreadsheet it looks like T(n) = n ^ f(n), where f(n) is a quite slowly growing function. If T(1) = 1 then T(n) ≈ n^3 for n around 130 or so, T(n) = n^2.5 for n around 38. $\endgroup$
    – gnasher729
    Commented Apr 23, 2020 at 23:24

2 Answers 2


Let $S(n) = T(n) - 2n - 2$. You can check that $S(n) = S(n-1) + S(n/2)$ (ignoring the fact that $n/2$ need not be an integer). This shows that the additive $n$ term doesn't make a big difference.

For large $n$, we have roughly $S(n) - S(n-1) \approx S'(n)$, and so we are led to solve the differential equation $$ S'(n) = S(n/2). $$ Consider $f(n) = \exp (\tfrac{1}{2}\log_2^2 n)$. Then $$ f'(n) = \exp (\tfrac{1}{2}\log_2^2 n) \cdot \frac{\ln n}{(\ln 4)n}, $$ whereas $$ f(n/2) = \exp(\tfrac{1}{2} (\log_2 n - 1)^2) \approx \exp(\tfrac{1}{2}\log^2 n) \exp(-\log n) = \exp(\tfrac{1}{2} \log^2 n) \cdot \frac{1}{n}. $$ This suggests that, at the very least, $\ln S(n) = \Theta(\log^2 n)$.

Where does this come from? You can think of $S(n)$ (with an appropriate base case!) as the number of ways to go from $n$ to zero by applying two operations: subtract 1 and divide by 2. A "typical" such sequence will contain roughly $\log_2n$ many operations of the second type, out of $\Theta(n)$ operations in total, leading to the very rough estimate $\binom{\Theta(n)}{\log_2 n}$, which is also of the form $\exp \Theta(\log^2 n)$.

Consider for concreteness the following precise definition of $S(n)$: the base case is $S(0) = 1$, and for $n > 0$, $$ S(n) = S(n-1) + S(\lfloor n/2 \rfloor). $$ This is sequence A000123. Knuth, An almost linear recurrence, showed that $$ \log_4 S(n) \sim \log_4^2 n, $$ that is, the ratio of the two terms tends to 1 as $n \to \infty$. The OEIS entry contains even more precise asymptotics.

  • $\begingroup$ I find exp (log^2 n) hard to imagine, but its the same as (exp (log n)) ^ log n = n^(log n). So it's a power of n, with a slowly growing exponent. $\endgroup$
    – gnasher729
    Commented Apr 23, 2020 at 23:29
  • $\begingroup$ @Yuval Thanks for the response. how do you conclude S(n) - S(n-1) = S'(n) ?. Sorry if it's lame, I am a beginner in CS theory. Also how did you come up with f(n) = exp(1/2 log^n ) ? . why this particular function ? $\endgroup$
    – sherelock
    Commented May 2, 2020 at 8:42
  • $\begingroup$ The mean value theorem states that for a nice enough function, $S(n) - S(n-1) = S'(n+\theta)$ for some $\theta \in [0,1]$. If $n$ is large then $S'(n+\theta) \approx S'(n)$, for some meaning of $\approx$. $\endgroup$ Commented May 2, 2020 at 8:51
  • $\begingroup$ I came up with $f(n)$ by trial-and-error. With more trial-and-error, it should be possible to come up with an even better solution. $\endgroup$ Commented May 2, 2020 at 8:52

I disagree with the Fibonacci hypnosis, but I'm not hundred percent sure of my answer

You see $ T(n)= T(n-1) +c*n = O(n ^2) $

$ T(n) = T(n/2) + c*n =O(n log n) $

However, u don't get ur T(n) by straightforward adding.

If u go one level in the recursion (u can draw a tree)

T(n) =T(n-2) +T((n-1)/2) +T(n/2-1)+ T(n/4) + [n + (n-1) + (n-1)/2 + (n/2-1) +n/4]

This to show u it is also not just $ O(n^2) $

I have to complete it, but I suspect either $ O((n^2) * log n) $ or $ O((n^2) * n ^ {log n} ) = O(n^3) $

{There is a term n(1+1/2+1/4+..1/n) that decays in log n steps, I have to check again}

  • $\begingroup$ You are right, I'm very sorry. I'll edit my answer, or maybe delete it temporarily. $\endgroup$
    – ShAr
    Commented Apr 23, 2020 at 21:21

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