408/912-notes
1
0
Fork 0
Code Issues Pull Requests Projects Releases Wiki Activity

update the display of latex equations

Browse Source
This commit is contained in:
Shine wOng
2019-05-15 09:14:25 +08:00
parent 0ab96e17f7
commit f17fc05662
1 changed files with 5 additions and 5 deletions
Download Patch File Download Diff File Expand all files Collapse all files

10
thu_dsa/chp1/master theorem.md
View File

@@ -9,9 +9,9 @@ We all know that for `divide-and-conquer` algorithms, there are two way to analy
## Introduction
First we'll make some abstractions. Lets' consider an algorithm implemented in the form of a recursion. Generally, we can assume that to solve a problem of scale `n`, we can divide it into `a` subproblems, whose scale would be `n/b`, with f(n) being the time to create the subproblems and combine their results in the above procedure.
First we'll make some abstractions. Lets' consider an algorithm implemented in the form of a recursion. Generally, we can assume that to solve a problem of scale `n`, we can divide it into `a` subproblems, whose scale would be `n/b`, with $f(n)$ being the time to create the subproblems and combine their results in the above procedure.
The runtime of subck an algorithm on an input of size 'n', usually denoted T(n), can be expressed by the recurrence relation
The runtime of subck an algorithm on an input of size 'n', usually denoted $T(n)$, can be expressed by the recurrence relation
$$
T(n) = aT(\frac{n}{b}) + f(n)
@@ -19,7 +19,7 @@ $$
## Generic solution
First we'll assume f(n) being in the form of $f(n) = n^d$. Generally, this is indeed the most common form of f(n).In this way, the equation should look like this:
First we'll assume $f(n)$ being in the form of $f(n) = n^d$. Generally, this is indeed the most common form of $f(n)$.In this way, the equation should look like this:
$$
T(n) = aT(\frac{n}{b}) + n^d
@@ -58,11 +58,11 @@ $$
> Case 1: q = 1
$T(n) = K * n^d = log_{b}^{n} * n^d = O(logn · n^d) $
$T(n) = k * n^d = log_{b}^{n} * n^d = O(logn · n^d) $
> Case 2: q < 1
Then T(n) is a convergent geometric progression. So T(n) can be expressed as:
Then $T(n)$ is a convergent geometric progression. So $T(n)$ can be expressed as:
$$
T(n) = O(n^d)