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Below is a problem worked out in the Introduction to Algorithms by Cormen et. al.

(I am not having problem with the proof but only I want to clarify the meaning conveyed by few statements in the text while solving the recurrence and the statements are given as ordered list at the end. Simply because I want to master the text.)

$$T(n) = 3T(\lfloor n/4 \rfloor) + \Theta(n^2)$$

Now the authors attempt to first find a good guess of the recurrence relation using the recursion tree method and for that they allow sloppiness and assumes $T(n)=3T(n/4) + cn^2$.

Recursion Tree

Though the above recursion tree is not quite required for my question but I felt like including it to make the background a bit clearer.

The guessed candidate is $T(n)=O(n^2)$. Then the authors proof the same using the substitution method.

In fact, if $O(n^2)$ is indeed an upper bound for the recurrence (as we shall verify in a moment), then it must be a tight bound. Why? The first recursive call contributes a cost of $\Theta(n^2)$ , and so $\Omega(n^2)$ must be a lower bound for the recurrence. Now we can use the substitution method to verify that our guess was correct, that is, $T(n)=O(n^2)$ is an upper bound for the recurrence $T(n) = 3T(\lfloor n/4 \rfloor) + cn^2$ We want to show that $T(n)\leq d n^2$ for some constant $d > 0$.

Now there are a few things which I want to get clarified...

(1) if $O(n^2)$ is indeed an upper bound for the recurrence. Here the sentence means (probably) $\exists$ a function $f(n) \in O(n^2)$ such that $T(n)\in O(f(n))$

(2) $\Omega(n^2)$ must be a lower bound for the recurrence Here the sentence means probably $\exists$ a function $f(n) \in \Omega(n^2)$ such that $T(n)\in \Omega(f(n))$

(3) $T(n)=O(n^2)$ is an upper bound for the recurrence $T(n) = 3T(\lfloor n/4 \rfloor) + cn^2$ This sentence can be interpreted as follows assume that $T'(n) = 3T'(\lfloor n/4 \rfloor) + cn^2$ and $\exists$ a function $T(n) \in O(n^2)$ such that $T'(n)\in O(T(n))$

(4) $T(n)\leq d n^2$ for some constant $d > 0$ We are using induction to verify to the definition of Big Oh...

I feel that the author could simply have written the $T(n)$ is Upper Bounded by $n^2$ and Lower Bounded by $n^2$ or the author could have simply written $T(n) = O(n^2)$ and $T(n)=\Omega(n^2)$, did the author just use the above style of statements as pointed out in $(1),(2),(3)$ just for more clearer explanation or there are some extra meaning conveyed which I am missing out.

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  1. "$O(n^2)$ is indeed an upper bound for the recurrence" means $T(n) \in O(n^2)$. That is, $\exists n_0 \ge 1, c>0, \forall n \ge n_0, T(n) \le c n^2$.

  2. "$\Omega(n^2)$ must be a lower bound for the recurrence" means $T(n) \in \Omega(n^2)$. That is $\exists n_0 \ge 1, c>0, \forall n \ge n_0, T(n) \ge c n^2$.

  3. This is exactly the same as 1).

  4. Yes. In this step you use induction to show that $T(n) \le d n^2$ for some choice of a constant $d > 0$.

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    $\begingroup$ I was a bit confused with exact meaning as to whether they are using the asymptotic notation in the usual set sense as $O(n^2)$ is a set after all, or as an anonymous function. Thanks for the help. The statement is quite odd to interpret as the first meant a "set is an upper bound bound for the recurrence" and the second interpretation (as a function) as I thought which ultimately leads to what you said if we use the transitivity property for the Big Oh or Big Omega... $\endgroup$ – Abhishek Ghosh Jun 19 at 18:16

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