A simple rule to get the hypothesis.
Assume $T(n)\ge0$ for all $n$. Suppose the recurrence relation is
$$T(n) = a_1T(b_1n) + a_2T(b_2n) + cn$$
where $a_i, b_i, c$ are constants, $a_i\ge0$ and $0\lt b_i\lt 1$ for all $i$, $c>0$.
The asymptotic approximation depends on the value $a_1b_1+a_2b_2$.
- If $a_1b_1+a_2b_2<1$, then $T(n)=\Theta(n)$;
- If $a_1b_1+a_2b_2=1$, then $T(n)=\Theta(n\log n)$;
- If $a_1b_1+a_2b_2>1$, then $T(n)=\Theta(n^p)$ for some $p>1$.
Some examples
| Recurrence relation |
$a_1b_1$ + $a_2b_2$ |
$T(n)$ |
| $T(n)=2T(n/2)+n$ |
$2\times\frac12$ = $1$ $\ \ \ $($a_2=0$) |
$\Theta(n\log n)$ |
| $T(n)=T(2n/3)+2T(n/6)+5n$ |
$\frac23+$ $2\times$ $\frac16=$ $1$ |
$\Theta(n\log n)$ |
| $T(n)= 3T(\frac{n}{5}) + T(\frac{n}{6}) + 3n$ |
$3\times\frac15$ + $\frac16$ = $\frac{23}{30}<1$ |
$\Theta(n)$ |
| $T(n)= 3T(\frac{n}{5}) + 3T(\frac{n}{6}) + \frac n7$ |
$3\times\frac15$ + $3\times\frac36$ = $\frac{11}{10}>1$ |
$\Theta(n^p)$ for some $p>1$ |
What about more general situations?
The simple rule above should be sufficient for most of the basic exercises or tests in an introductory course.
The rule stays the same basically when there are more terms like $af(bn)$ in the recurrence relation. Or $cn$ is not changed much, such as $cn+d$ for some constant $d$ or $cn+\log n$.
For more general situations, you can take a look at the Akra–Bazzi method. It is also explained here. You can find how the exponent $p$ in the case of $a_1b_1+a_2b_2>1$ is defined.
