Changing variables in recurrence relations

Question

Currently, I am self-studying Intro to Algorithms (CLRS) and there is one particular method they outline in the book to solve recurrence relations.

The following method can be illustrated with this example. Suppose we have the recurrence

$$T(n) = 2T(\sqrt n) + \log n$$

Initially they make the substitution m = lg(n), and then plug it back in to the recurrence and get:

$$T(2^m) = 2T(2^{\frac{m}{2}}) + m$$

Up to this point I understand perfectly. This next step is the one that's confusing to me.

They now "rename" the recurrence $S(m)$ and let $S(m) = T(2^m)$, which apparently produces

$$S(m) = 2S(m/2) + m$$

For some reason it's not clear to me why this renaming works, and it just seems like cheating. Can anyone explain this better?

Answer 1

It's certainly not cheating. Think in calculus how substitution may be used to solve a tricky integral. The substitution makes the equation more manageable for manipulation. Additionally, substitution may transform somewhat complex recurrences into familiar ones.

This is exactly what occurred in your example. We define a new recurrence $S(m)=T(2^m)$. Remember that $T(2^m) = 2T(2^\frac{m}{2}) + m$. Notice that, $S(m/2) = T(2^\frac{m}{2})$. If this particular point is still unclear, let $k =m/2$ and notice all we are doing is this $S(k) = T(2^k)$. Now, we can express $S(m)$ by expanding it to: $$ S(m) = 2S(m/2) + m.$$ Solving for $S$ we see that it resolves to our familiar friend $O(m \log m)$. Now that we've solved $S$ we want to express this in terms of $T(n)$. To do this, merely plug back in our original value for $m$ and we have $T \in O(\log n \log \log n)$.

Answer 2

What $S(m) = T(2^m)$ means is that $S$ and $T$ are two different functions which produce the same result while taking inputs as $m$ and $2^m$ respectively.

Function $S$ can be considered as an operator with two internal steps (otherwise, composition of functions):

1. $S'$ operator: Input:$m$, Output:$2^m$

2. $T$ operator(original function): Input:output of first part, Output: As defined originally.

Therefore then transitions are:

$$m\to 2^m\to T(2^m) = S(m)$$ $$\tfrac{m}2\to2^{m/2}\to T(2^{m/2}) = S(\tfrac{m}2)\,.$$