# Clarification of the proof involving the regularity condition in Master Theorem

I was going the text Introduction to Algorithms by Cormen et al. Where I came across the following statement in the proof of the third case of the Master's Theorem.

(The Statement of Master theorem) Let $$a \geqslant 1$$ and $$b > 1$$ be constants, let $$f(n)$$ be a function, and let $$T (n)$$ be deﬁned on the nonnegative integers by the recurrence( the recursion divides a problem of size $$n$$ into $$a$$ problems of size $$n/b$$ each and takes $$f(n)$$ for the divide and combine)

$$T(n) = aT(n/b)+ f (n)$$ ;

where we interpret $$n/b$$ to mean either $$\lceil b/n \rceil$$ or $$\lfloor b/n \rfloor$$. Then $$T(n)$$ has the following asymptotic bounds:

1. If $$f(n)=O (n^{\log_ba - \epsilon})$$ for some constant $$\epsilon > 0$$, then $$T(n)=\Theta (n^{\log_ba})$$.

2. If $$f(n)=\Theta (n^{\log_ba})$$, then $$T(n)=\Theta (n^{\log_ba}\lg n)$$

3. If $$f(n)=\Omega (n^{\log_ba + \epsilon})$$ for some constant $$\epsilon > 0$$, and if $$af(n/b) \leqslant cf(n)$$ for some constant $$c < 1$$ and all sufﬁciently large n, then $$T(n)=\Theta (f(n))$$.

For $$n$$ as exact powers of $$b$$ we restrict the domain of T(n) as follows:

$$T(n)= \Theta(1), n=1$$ $$T(n)=aT(n/b)+f(n) ,n=b^i$$

Now in the proof of Master's Theorem with $$n$$ as exact power of $$b$$ the expression for $$T(n)$$ reduces to :

$$T(n)=\Theta(n^{\log_ba})+\sum_{j=0}^{\log_bn -1} a^jf(n/b^j)$$ Then the authors assume the following,

$$g(n)=\sum_{j=0}^{\log_bn -1} a^jf(n/b^j)$$

Then for the proof of the 3rd case of the Master's Theorem the authors prove the following lemma,

Lemma 1 : If $$a\cdot f(n/b)\leqslant c\cdot f(n)$$ for some constant $$c<1$$ and for all $$n\geqslant b$$ then $$g(n)=\Theta(f(n))$$

They say that:

under their assumption that $$c<1$$ and $$n \geqslant b$$,they have $$a \cdot f(n/b)\leqslant c \cdot f(n) \implies f(n/b)\leqslant (c/a) \cdot f(n)$$

then iterating $$j$$ times yields, $$f(n/b^j)\leqslant (c/a)^j \cdot f(n)$$

I could not quite get the mathematics used behind iterating $$j$$ times.

Moreover I could not quite get the logic behind the assumption of $$n\geqslant b$$ for the situation that $$n$$ should be sufficiently large. (As the third case of the Master's Theorem says.)

The proof of the lemma continues as follows:

$$f(n/b^j)\leqslant (c/a)^j\cdot f(n) \iff a^j\cdot f(n/b^j)\leqslant c^j\cdot f(n)$$ So, $$g(n)=\sum_{j=0}^{\log_bn -1} a^jf(n/b^j)$$ $$\leqslant \sum_{j=0}^{\log_bn -1} c^jf(n)$$ $$\leqslant f(n)\sum_{j=0}^{\infty} c^j,$$ as $$c<1$$ we have an infinite geometric series $$= f(n) \left(\frac{1}{1-c}\right)$$ $$=O(f(n))$$ as $$c$$ is a constant. (Note that $$T(n)=\Omega(f(n))$$ from the recursion diagram.)

Then the authors proof the third case of the Masters Theorem for $$n$$ as exact power of $$b$$:

Lemma 2: Let $$a \geqslant 1$$ and $$b>1$$, if $$f(n)=\Omega (n^{\log_ba + \epsilon})$$ for some constant $$\epsilon > 0$$, and if $$af(n/b) \leqslant cf(n)$$ for some constant $$c < 1$$ and all sufﬁciently large n, then $$T(n)=\Theta (f(n))$$.

So $$T(n) = \Theta(n^{\log_ba}) + g(n) = \Theta(n^{\log_ba}) + \Theta(f(n)) =\Theta(f(n))$$ as $$f(n)=\Omega (n^{\log_ba + \epsilon})$$

Moreover in the similar proof for the third case of the general master theorem (not assuming $$n$$ as exact powers of $$b$$) there again the book assumes that $$n\geqslant b+b/(b-1)$$ to go with the situation of sufficiently large $$n$$.

I do not quite understand what the specific value has to do and why such is assumed as sufficiently large $$n$$

(I did not give the details of the second situation as I feel that it shall be something similar to the first situation but however it can be found here)

Let's start with the issue of iteration. Suppose that a function $$f$$ satisfies $$f(n/b) \leq (c/a)f(n).$$ Then it also satisfies $$f(n/b^2) \leq (c/a)f(n/b) \leq (c/a)^2 f(n).$$ You can prove by induction that for all integer $$t \geq 0$$, $$f(n/b^t) \leq (c/a)^t f(n).$$
As for your second question, about assuming that $$n$$ is large enough: the proof is just sloppy. You cannot assume that $$f(n/b) \leq (c/a) f(n)$$ holds for all $$n \geq b$$. Indeed, in Introduction to Algorithms, third edition, they do not make such an assumption for the case where $$n$$ is a power of $$b$$.
They do seem to make such as assumption in the case of general $$n$$, but what they are really saying is that the inequality $$f(\lceil n/b \rceil) \leq (c/a) f(n)$$ only makes sense for $$n \ge b + b/(b-1)$$. Using the idea of the proof of the special case where $$n$$ is a power of $$b$$, you can complete the proof of the general case. I would, however, strongly suggest ignoring such technicalities at present. The master theorem is essentially a calculation, and you can trust the authors that it works out. Nothing interesting is hidden under the rug.
• @YuvalFimus I have modified the question with the details of the proof included, I could understand your explanation of the iteration part... Please can you help me out with the $n\geqslant b$ part as you said that answering it required more details... Jun 20, 2020 at 6:23
• This is of no help. Crucial context is missing. Their assumption doesn't imply that $f(n/b) \leq (c/a) f(n)$ for all $n \geq b$, and they must be saying something about this. Jun 20, 2020 at 9:09
• Nothing more is written about their assumption and implication. May be it was weird and that's why they dealt away with it in 3rd edition(as you pointed out). I thought that they might have got the assumption of $n\geq b+ \frac{b}{b-1}$ in same manner as they made the assumption for $n \geq b$. I was quite disturbed as I was unable to make out those two technicalities(which are sloppy and not quite clear). Thanks a lot for the help... Jun 20, 2020 at 9:56