I'm having a hard time with the following question:

Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 4T(n/2 + 2) + n$. Use the substitution method to verify your answer.

This is not homework, I'm just practicing myself for an upcoming exam.

The thing I'm having a hard time with is the $n/2 + 2$, what will the height be of this recursion tree?

I came up with the following formula to calculate the cost of each level after a lot of labor: $2^i n + 2^{i+2}(2^i -1)$ not 100% sure this is correct either.

Any help appreciated, really looking forward to the answer :D
I often make stupid mistakes and I just started doing algo's for my first time.

  • 2
    $\begingroup$ Where did that formula come from? Looks like it came out of thin air. More generally, what do you understand? Do you understand how to build a recursion tree? If so, have you tried applying those methods to this problem? What did you get? What happens if you try to solve $U(n) = 4U(n/2) + n$? Can you find a solution to that? If you can, have you tried checking whether that solution also happens to solve the original recurrence (asymptotically)? The $n/2+2$ is indeed a pain, but the first thing I would try would be to analyze $U(n)$. $\endgroup$
    – D.W.
    Commented Dec 28, 2013 at 20:00
  • $\begingroup$ Hi @D.W. yea I understand how to do it for U(n) = 4U(n/2) + n. The only thing that was giving me trouble is this n/2 + 2, all other exercises I did successfully. The formula came from the book. Just read your answer, thanks a lot. I think the substitution method is indeed the method I should learn for this question. I'm gonna work on it now :D $\endgroup$ Commented Dec 29, 2013 at 2:11

1 Answer 1


I'll suggest two methods, but since this is your exercise, you'll have to work out some of the details. Important: study them both.

Method 1 (the dumb method)

We're going to use guess-and-check. In other words, we're going to guess a solution to the recurrence $T$, and then we'll check whether our guess is correct.

How do we come up with a reasonable guess? Well, we could try stuff blindly, but I have a better suggestion. The reason this recurrence is difficult is because of the nasty $+2$ in $T(n/2+2)$. So if that's inconvenient, let's throw it away: let's look at what happens with the related recurrence where that isn't present. In other words, define a new recurrence $U(\cdot)$ by

$$U(n) = 4 U(n/2) + n.$$

Now use your methods to find a solution to $U(n)$ (e.g., recursion tree, etc.). Finally, use that formulate for $U(n)$ and let's use that as our guess for $T(n)$: let's check whether it also provides an asympotically valid solution to $T(n)$. If it is, ta-da, you are done!

I call this a "dumb" method because, while it might happen to work on this example, there's no guarantee it will work in every situation. So that's why it is helpful to know a more powerful method:

Method 2 (transform of variables)

My suggestion is to first apply a change of variables to make that nasty $+2$ go away, then solve the transformed recurrence using standard methods you already know.

Here's an example. Suppose we define a new recurrence for $S(n)$ by making the definition $S(n) = T(n+3)$. (This corresponds to the change of variables $n \mapsto n+3$.) Can you derive a recurrence relation for $S(n)$? Sure, with some simple manipulation of the definitions of $S$ and $T$, you ought to be able to derive a recurrence of the form

$$S(n) = 4 S(\text{something}) + \text{stuff}.$$

In particular, I think you'll find that the $\text{something}$ has the form $n/2+\text{blah}$. So, go do it. Sit down with a pencil and work out the exact recurrence for $S(n)$ (fill in the $\text{something}$ and $\text{stuff}$ parts with their exact expressions).

Now if we were lucky, we'd get a recurrence of the form $S(n) = 4 S(n/2) + \text{stuff}$. That'd be lucky, because then you could use your standard methods to solve that kind of recurrence (now that the nasty $+2$ is gone). Unfortunately, when you work this out, you will discover that you were not lucky. The nasty $+2$ got replaced by some other $+\text{blah}$, which is still nasty (since $\text{blah}\ne 0$).

But don't give up. It turns out that $S(n)=T(n+3)$ was not lucky. So try a different definition. Try replacing the $3$ with some other number. If you play around with it a bit, I think you will discover that there is another number you can use instead of $3$ that will be lucky: it'll make the recurrence look like

$$S(n) = 4S(n/2) + \text{stuff},$$

and that's a recurrence you should know how to solve (e.g., recursion trees nad such). Once you find a solution for $S(n)$, you can plug back into the definition $S(n)=T(n+3)$ to get a solution for $T(n)$, and you're good.

This will work. It takes a little more work, but it is more general, so it is worth knowing. I suspect it is the method that your instructor wanted you to learn.

  • $\begingroup$ I got $S(n) = T(n-4)$ , and then $S(n) = O(n^2)$ so $T(n) = S(n+4) = O((n+4)^2) = O(n^2)$ thats why the dummy method also worked. Were this correct conclusions? I wrote a total of 4 A4's for this question already lol. Fun question. Learned a lot. $T(n-4) = 4T((n-4)/2 + 2) + n-4 = 4T(n/2 -2 +2) + n-4$ $\endgroup$ Commented Dec 29, 2013 at 7:02
  • 1
    $\begingroup$ We get here T(n - 4) = 4T(n/2) + n-4, and if we write this in S(n) form, then the equation becomes, S(n) = 4S(n/2 + 4) + n - 4, so what I don't understand is how is the above equation simplified, to solve the recurrence. Please help $\endgroup$ Commented May 11, 2017 at 3:38
  • $\begingroup$ @D.W. what you said will not lead to any change. I don't know how people upvoted your answer. $\endgroup$ Commented Apr 7, 2019 at 19:04
  • $\begingroup$ @SamStoelinga how did you know that S(n)=O(n^2) check what Debasish said $\endgroup$ Commented Apr 7, 2019 at 19:05
  • 1
    $\begingroup$ That's because what Sam Stoelinga wrote isn't correct (as Debasish Mohapatra correctly explains). You'll need to set $S(n) = T(n+c)$ for some constant $c$, and you'll need to choose $c$ appropriately to make things work. Contrary to what Sam Stoelinga wrote, $c=-4$ doesn't work, but there is another value of $c$ that does work. If you pick the right $c$ it will lead to a change that simplifies things. I'm going to let you discover what value does work, because it is a good exercise. $\endgroup$
    – D.W.
    Commented Apr 7, 2019 at 19:14

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