Main theorem (general formula)

The Master Theorem is an important tool for analyzing the time complexity of recursive algorithms. It is applicable to a class of formally defined recursive relations, where a divide-and-conquer strategy is usually used to solve the problem.

Limitations of the simplified version of the main theorem

Three cases of the simplified version of the main theorem:

1. $IF$ $f(n)=O(n^{lo{g}_b(a-\epsilon )})$，and $e>0$，Then $T(n)=\Theta (n^{lo{g}_b(a)})$
2. $IF$ $f\left(n\right)=\Theta (n^{lo{g}_b(a)}\cdot lo{g}^{k}n)$，and $k\ge 0$，Then $T\left(n\right)=\Theta (n^{lo{g}_b\left(a\right)}\cdot lo{g}^{k+1}n)$
3. $IF$ $f\left(n\right)=\Omega (n^{lo{g}_b(a+\epsilon )})$，and $e>0$，$a\cdot f(\frac{n}{b})\le c\cdot f\left(n\right)$ A certain constant < /span>$c<1$ 和own Foothold $n$ 成立，Then $T\left(n\right)=\Theta (f(n))$

The simplified form has certain restrictions. For example, the form must be: $$T\left(n\right)=a\cdot T(\frac{n}{b})+f\left(n\right)$$

并且 $f\left(n\right)=n^c$

Three counterexamples:

1. The number of subproblems is not constant

$$T\left(n\right)=n\cdot T\left(\frac{n}{2}\right)+n^2$$

2. The number of sub-problems is less than 1

$$T\left(n\right)=\frac{1}{2}T\left(\frac{n}{2}\right)+n^2$$

3. The time to decompose the problem and merge the solutions is not $n^c$

$$T\left(n\right)=2T(\frac{n}{2})+nlogn$$

---

General form of the main theorem

$$T(n)=a\cdot T(\frac{n}{b})+f(n),a>0,b>1$$

1. $IF$ $\exists \epsilon >0$ Differentiation $f(n)=O(n^{lo{g}_b(a-\epsilon )})$ ，Then $T\left(n\right)=\Theta (n^{lo{g}_b(a)})$
2. $IF$ $\exists k\ge 0$ so that $f\left(n\right)=\Theta (n^{lo{g}_b(a)}\cdot lo{g}^{k}n)$，Then $T\left(n\right)=\Theta (n^{lo{g}_b\left(a\right)}\cdot lo{g}^{k+1}n)$
3. $IF$ $\exists \epsilon >0$ so that $f\left(n\right)=\Omega (n^{lo{g}_b(a+\epsilon )})$，
   且对于certain constant $c<1$ 和own Foothold $n$ 有 $a\cdot f(\frac{n}{b})\le c\cdot f(n)$ ，Then $T\left(n\right)=\Theta (f(n))$

Main considerations Function $n^{lo{g}_{b}a}$ given $f\left(n\right)$'s growth rate system< /span>

Context 1:$n^{lo{g}_{b}a}$ 比 $f\left(n\right)$ Increased pleasure

$$T\left(n\right)=9T(\frac{n}{3})+n$$

• $n^{lo{g}_{b}a}=n^2$,
• $f\left(n\right)=n=O(n^{2-ϵ}),ϵ\le 1$
• $\Rightarrow T\left(n\right)=O(n^2)$

---

Context 2:$n^{lo{g}_{b}a}$ given $f\left(n\right)$ Growth rate similar

$$T\left(n\right)=T\left(\frac{2n}{3}\right)+1$$

• $n^{lo{g}_{b}a}=n^{lo{g}_{3/2}1}=n^0=1$,
• $f(n)=1=\Theta (n^{lo{g}_{b}a}lo{g}^{0}n)$
• $\Rightarrow T(n)=O(logn)$

---

Context 3:$n^{lo{g}_{b}a}$ 比 $f\left(n\right)$ Increased arrogance

• $f(n)$比$n^{lo{g}_{b}a}$ Grow faster, at least faster $\Theta \left(n^{ϵ}\right)$ 倍，且 $af(\frac{n}{b})\le cf(n)$

$$T\left(n\right)=3T(\frac{n}{4})+nlogn$$

• $n^{lo{g}_{b}a}=n^{lo{g}_{4}3=n^{0.793}}$,
• $f\left(n\right)=nlogn=\Omega (n^{lo{g}_{4}3+ϵ}),ϵ\le 0.207$
• $af\left(\frac{n}{b}\right)=3(\frac{n}{4})log(\frac{n}{4})\le \frac{3}{4}nlogn=cf(n),c=\frac{3}{4}$
• $\Rightarrow T\left(n\right)=O(nlogn)$

---

Cases where the main theorem does not apply

1. $n^{lo{g}_{b}a}$ given $f\left(n\right)$
2. $n^{lo{g}_{b}a}$ 比 $f\left(n\right)$ Increased pleasure, however Not happy $O(n^{ϵ})$ 倍
3. $n^{lo{g}_{b}a}$ 比 $f\left(n\right)$ Increased arrogance, however Conceited $O\left(n^{ϵ}\right)$ 倍

---