COMP3010: Algorithm Theory and Design
Daniel Sutantyo, Department of Computing, Macquarie University
6.1 - Recursion-Tree Method
Recursion-Tree Method
6.1 - Recursion-Tree Method
- In the previous lecture, we learnt about recurrence equations
- In this lecture, we're going to derive the complexity of the algorithm based on this recurrence equations
- The steps:
- Draw out the recursion tree until you can see a pattern (2-3 levels should be enough)
- put the cost of each subproblem in the nodes
- Count the number of operations on each level
- Count the number of levels (work out the height of the tree)
- Count the total number of operations
- Draw out the recursion tree until you can see a pattern (2-3 levels should be enough)
Mergesort
6.1 - Recursion-Tree Method
\(\text{mergeSort}(0,n)\)
\(\text{mergeSort}(0,\lfloor\frac{n}{2}\rfloor)\)
\(\text{mergeSort}(\lceil\frac{n}{2}\rceil,n)\)
\(\text{mergeSort}(0,\lfloor\frac{n}{4}\rfloor)\)
\(\text{mergeSort}(\lceil\frac{n}{4}\rceil,\lfloor\frac{n}{2}\rfloor)\)
\(\text{mergeSort}(\lceil\frac{n}{2}\rceil,\lfloor\frac{3n}{4}\rfloor)\)
\(\text{mergeSort}(\lceil\frac{3n}{4}\rceil,n)\)
Mergesort
6.1 - Recursion-Tree Method
\(\text{mergeSort}(n)\)
\(\text{mergeSort}(\frac{n}{2})\)
\(\text{mergeSort}(\frac{n}{2})\)
\(\text{mergeSort}(\frac{n}{4})\)
\(\text{mergeSort}(\frac{n}{4})\)
\(\text{mergeSort}(\frac{n}{4})\)
\(\text{mergeSort}(\frac{n}{4})\)
\(T(n) = \begin{cases} \Theta(1) &\text{if $n = 1$}\\ 2T(n/2) + cn &\text{if $n > 1$} \end{cases}\)
Mergesort
6.1 - Recursion-Tree Method
\(\text{mergeSort}(n)\)
\(\text{mergeSort}(\frac{n}{2})\)
\(\text{mergeSort}(\frac{n}{2})\)
\(\text{mergeSort}(\frac{n}{4})\)
\(\text{mergeSort}(\frac{n}{4})\)
\(\text{mergeSort}(\frac{n}{4})\)
\(\text{mergeSort}(\frac{n}{4})\)
\(T(n) = \begin{cases} \Theta(1) &\text{if $n = 1$}\\ 2T(n/2) + cn &\text{if $n > 1$} \end{cases}\)
- assume:
- \(n\) is a power of 2
- we are doing \(cn\) operations on each subproblem of size \(n\)
- we are doing 1 operation on the leaves of the tree
Mergesort
6.1 - Recursion-Tree Method
- Step 1: Draw the recursion tree
\(T(n)\)
\[T\left(\frac{n}{2}\right)\]
\[T\left(\frac{n}{4}\right)\]
1
\[T\left(\frac{n}{2}\right)\]
1
1
1
1
1
1
1
. . .
. . .
. . .
\(T(n) = \begin{cases} \Theta(1) &\text{if $n = 1$}\\ 2T(n/2) + cn &\text{if $n > 1$} \end{cases}\)
\[T\left(\frac{n}{4}\right)\]
\[T\left(\frac{n}{4}\right)\]
\[T\left(\frac{n}{4}\right)\]
Mergesort
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n = 1$}\\ 2T(n/2) + cn &\text{if $n > 1$} \end{cases}\)
- Step 2: Count the number of operations on each level
\(cn\)
\[\frac{cn}{2}\]
\[\frac{cn}{4}\]
1
\[\frac{cn}{4}\]
\[\frac{cn}{2}\]
\[\frac{cn}{4}\]
\[\frac{cn}{4}\]
1
1
1
1
1
1
1
. . .
. . .
. . .
\[cn\]
\[cn\]
\[cn\]
\[\Theta(n)\]
Mergesort
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n = 1$}\\ 2T(n/2) + cn &\text{if $n > 1$} \end{cases}\)
- Step 3: Count the number of levels (height of the tree)
\(cn\)
\[\frac{cn}{2}\]
\[\frac{cn}{4}\]
1
\[\frac{cn}{4}\]
\[\frac{cn}{2}\]
\[\frac{cn}{4}\]
\[\frac{cn}{4}\]
1
1
1
1
1
1
1
. . .
. . .
. . .
\[cn\]
\[cn\]
\[cn\]
\[\Theta(n)\]
Mergesort
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n = 1$}\\ 2T(n/2) + cn &\text{if $n > 1$} \end{cases}\)
- Step 3: Count the number of levels (height of the tree)
- on each level we divide the problem size by 2
- sizes: \(\quad n \quad\rightarrow \quad n/2 \quad\rightarrow\quad n/4 \quad\rightarrow\quad\cdots\quad\rightarrow\quad n/2^i\)
- remember, we assume that \(n\) is a power of 2
- the final level is when \(n /2^i = 1\)
- i.e. when \(\quad 2^i = n \quad\rightarrow\quad i = \log_2n\)
- so there are \(\log_2 n+1\) levels
- just to check: level \(i\) has \(2^i\) nodes, so the final level has \(2^{\log_2 n}\) nodes = \(n\) nodes (as we expected)
Mergesort
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n = 1$}\\ 2T(n/2) + cn &\text{if $n > 1$} \end{cases}\)
- Step 4: Count the total number of operations
- number of levels: \(\log_2 n\)
- number of operations on each level: \(n\)
- total number of operations: \(n \log_2 n\)
\(T(n) = \Theta(n \log n)\)
Mergesort
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n = 1$}\\ 2T(n/2) + cn &\text{if $n > 1$} \end{cases}\)
- Step 4: Count the total number of operations
- number of levels: \(\log_2 n\)
- number of operations on each level: \(n\)
- total number of operations: \(n \log_2 n\)
\(T(n) = \Theta(n \log n)\)
\[\underbrace{cn + cn+ \cdots + cn}_{\text{$(\log_2 n)$ terms}} + \Theta (n) = \Theta(n \log n)\]
Linear Search
6.1 - Recursion-Tree Method
public static int linearSearch(int[] a, int i, int j, int target) {
if (i > j)
return -1;
int m = (i+j)/2;
if (a[m] == target)
return m;
return Math.max(linearSearch(a,i,m-1,target), linearSearch(a,m+1,j,target));
}\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 2T(n/2) + \Theta(1)&\text{otherwise} \end{cases}\)
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 2T(n/2) + \Theta(1)&\text{otherwise} \end{cases}\)
- Step 1: Draw the recursion tree
\(T(n)\)
\[T\left(\frac{n}{2}\right)\]
1
1
1
1
1
1
1
1
. . .
. . .
. . .
- assume:
- \(n\) is a power of 2
- we are doing \(1\) operation on each node (including the leaves)
\[T\left(\frac{n}{2}\right)\]
\[T\left(\frac{n}{4}\right)\]
\[T\left(\frac{n}{4}\right)\]
\[T\left(\frac{n}{4}\right)\]
\[T\left(\frac{n}{4}\right)\]
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 2T(n/2) + \Theta(1)&\text{otherwise} \end{cases}\)
- Step 2: Count the number of operations on each level
\(cn\)
\[\frac{cn}{2}\]
\[\frac{cn}{4}\]
1
\[\frac{cn}{4}\]
\[\frac{cn}{2}\]
\[\frac{cn}{4}\]
\[\frac{cn}{4}\]
1
1
1
1
1
1
1
. . .
. . .
. . .
\[cn\]
\[cn\]
\[cn\]
\[\Theta(n)\]
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 2T(n/2) + \Theta(1)&\text{otherwise} \end{cases}\)
- Step 2: Count the number of operations on each level
\(1\)
\[1\]
\[1\]
1
\[1\]
\[1\]
\[1\]
\[1\]
1
1
1
1
1
1
1
. . .
. . .
. . .
\[4\]
\[2\]
\[1\]
\[???\]
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 2T(n/2) + \Theta(1)&\text{otherwise} \end{cases}\)
- Step 2: Count the number of operations on each level
- as before, the final level is when \(n /2^i = 1\)
- i.e. when \(\quad 2^i = n \quad\rightarrow\quad i = \log_2n\)
- level \(i\) has \(2^i\) nodes, so the final level has \(2^{\log_2 n}\) nodes = \(n\) nodes
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 2T(n/2) + \Theta(1)&\text{otherwise} \end{cases}\)
- Step 2: Count the number of operations on each level
\[4\]
\[2\]
\[1\]
\[\Theta(2^{\log_2 n}) = \Theta(n)\]
\(1\)
\[1\]
\[1\]
1
\[1\]
\[1\]
\[1\]
\[1\]
1
1
1
1
1
1
1
. . .
. . .
. . .
- Step 3: Count the number of levels (height of the tree)
- as in the case of mergesort, there are \(\log_2 n\) levels
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 2T(n/2) + \Theta(1)&\text{otherwise} \end{cases}\)
- Step 4: Count the total number of operations
- unfortunately, we cannot just multiply the number of levels with the number of operations on each level (because it changes)
- we need to find the sum of
- \(1 + 2 + 4 + \dots + \Theta(n) = 2^0 + 2^1 + 2^2 + \dots + 2^{\log_2 n-1} + \Theta(n)= \displaystyle\sum_{k=0}^{\log_2 n-1} 2^k + \Theta(n)\)
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 2T(n/2) + \Theta(1)&\text{otherwise} \end{cases}\)
- Step 4: Count the total number of operations
- we need to find the sum of
- \(1 + 2 + 4 + \dots + n = 2^0 + 2^1 + 2^2 + \dots + 2^{\log_2 n-1} = \displaystyle\sum_{k=0}^{\log_2 n-1} 2^k\)
- see CLRS Appendix A for the formula for geometric series
\[\sum_{k=0}^m x^k = 1 + x + x^2 + \cdots + x^m = \frac{x^{m+1}-1}{x-1}\]
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 2T(n/2) + \Theta(1)&\text{otherwise} \end{cases}\)
- Step 4: Count the total number of operations
\[\sum_{k=0}^{\log_2 n-1} 2^k = 1 + 2 + 2^2 + \cdots + 2^{\log_2 n-1}= \frac{2^{\log_2 n}-1}{2-1} = n - 1\]
\[\sum_{k=0}^m x^k = 1 + x + x^2 + \cdots + x^m = \frac{x^{m+1}-1}{x-1}\]
so linear search is \(\Theta(n)\)
- Total number of operation is
- \(\displaystyle\sum_{k=0}^{\log_2 n-1} 2^k + \Theta(n) = (n-1) + \Theta(n) = \Theta(n)\)
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/3) + \Theta(1)&\text{otherwise} \end{cases}\)
- assume:
- \(n\) is a power of 3
- we are doing \(\Theta(1)\) operation on each node
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/3) + \Theta(1)&\text{otherwise} \end{cases}\)
\(1\)
1
1
1
1
1
1
. . .
. . .
\[9\]
\[3\]
\[1\]
\[\Theta(3^{\log_3 n}) = \Theta(n)\]
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/3) + \Theta(1)&\text{otherwise} \end{cases}\)
- there are \(\log_3 n\) levels
- we need to find the sum of
- \(3^0 + 3^1 + 3^2 + \dots + 3^{\log_3 n-1} = \displaystyle\sum_{k=0}^{\log_3 n-1} 3^k\)
- the formula for geometric series:
\[\sum_{k=0}^m x^k = 1 + x + x^2 + \cdots + x^m = \frac{x^{m+1}-1}{x-1}\]
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/3) + \Theta(1)&\text{otherwise} \end{cases}\)
- Step 4: Count the total number of operations
\[\sum_{k=0}^m x^k = 1 + x + x^2 + \cdots + x^m = \frac{x^{m+1}-1}{x-1}\]
\[\sum_{k=0}^{\log_3 n-1} x^k = 1 + 3 + 3^2 + \cdots + 3^{\log_3 n -1} = \frac{3^{\log_3 n}-1}{3-1} = \frac{n-1}{2}\]
- so linear search is again \(\Theta(n)\)
- but can you see why we need less than \(n\) operations?
- (it's a mistake!)
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/3) + \Theta(1)&\text{otherwise} \end{cases}\)
\(2\)
2
2
2
2
2
2
. . .
. . .
\[9\]
\[3\]
\[1\]
\[\Theta(3^{\log_3 n}) = \Theta(n)\]
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/3) + \Theta(1)&\text{otherwise} \end{cases}\)
\[\sum_{k=0}^{\log_3 n-1} x^k = 1 + 3 + 3^2 + \cdots + 3^{\log_3 n -1} = \frac{3^{\log_3 n}-1}{3-1} = \frac{n-1}{2}\]
\[\sum_{k=0}^{\log_3 n-1} x^k = 2(1) + 2(3) + 2(3^2) + \cdots + 2(3^{\log_3 n -1}) = \frac{2(3^{\log_3 n}-1)}{3-1} =n-1\]
Linear Search
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/3) + \Theta(1)&\text{otherwise} \end{cases}\)
- Total number of operation is
- \(\displaystyle\sum_{k=0}^{\log_3 n-1} 3^k + \Theta(n) = (n-1) + \Theta(n) = \Theta(n)\)
Textbook Example
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/4) + cn^2&\text{otherwise} \end{cases}\)
- CLRS page 89-90
\(T(n)\)
\[T\left(\frac{n}{4}\right)\]
\[T\left(\frac{n}{4}\right)\]
\[T\left(\frac{n}{4}\right)\]
- Step 1: Draw the recursion tree
we assume \(n\) is a power of 4
Textbook Example
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/4) + cn^2&\text{otherwise} \end{cases}\)
\(cn^2\)
\[\frac{cn^2}{16}\]
\[\frac{cn^2}{16}\]
\[\frac{cn^2}{16}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
1
1
1
1
. . . . . . . . . . . . . . . . . . . . .
1
1
1
1
1
1
1
1
1
1
1
1
1
1
- Step 2: Count the number of operations on each level
Textbook Example
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/4) + cn^2&\text{otherwise} \end{cases}\)
\(cn^2\)
\[\frac{cn^2}{16}\]
\[\frac{cn^2}{16}\]
\[\frac{cn^2}{16}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
1
1
1
. . . . . . . . . . . . . . . . . . . . .
1
1
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
1
1
1
1
1
1
\(cn^2\)
\[\frac{3}{16}cn^2\]
\[\frac{9}{256}cn^2\]
???
- Step 2: Count the number of operations on each level
Textbook Example
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/4) + cn^2&\text{otherwise} \end{cases}\)
- Step 3: Count the number of levels
- on each level we divide the problem size by 4
- the subproblem sizes:
- \(n \quad\rightarrow \quad n/4 \quad\rightarrow\quad n/16 \quad\rightarrow\quad\cdots\quad\rightarrow\quad n/4^i\)
- the final level is when \(n /4^i = 1\)
- i.e. when \(\quad 4^i = n \quad\rightarrow\quad i = \log_4 n\)
- level \(i\) has \(3^i\) nodes, so the final level has \(3^{\log_4 n}\) nodes
Textbook Example
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/4) + cn^2&\text{otherwise} \end{cases}\)
- Step 3: Count the number of levels
\(cn^2\)
\[\frac{cn^2}{16}\]
\[\frac{cn^2}{16}\]
\[\frac{cn^2}{16}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
1
1
1
. . . . . . . . . . . . . . . . . . . . .
1
1
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
\[\frac{cn^2}{256}\]
1
1
1
1
1
1
\(cn^2\)
\[\frac{3}{16}cn^2\]
\[\frac{9}{256}cn^2\]
\(\Theta(3^{\log_4 n})\)
Textbook Example
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/4) + cn^2&\text{otherwise} \end{cases}\)
- Step 4: Count the number of operations
\(cn^2\)
\[\frac{3}{16}cn^2\]
\[\frac{9}{256}cn^2\]
\(\Theta(3^{\log_4 n})\)
Textbook Example
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/4) + cn^2&\text{otherwise} \end{cases}\)
- Step 4: Count the number of operations
\(cn^2\)
\[\frac{3}{16}cn^2\]
\[\frac{9}{256}cn^2\]
\(\Theta(3^{\log_4 n})\)
\(+\)
\(+\)
\(+ \dots +\)
\[cn^2 + \frac{3}{16}cn^2 + \left(\frac{3}{16}\right)^2cn^2+\cdots+ \left(\frac{3}{16}\right) ^{\log_4 n - 1}cn^2+ \Theta(3^{\log_4 n})\]
\[cn^2\sum_{i=0}^{\log_4n-1}\left(\frac{3}{16}\right)^i+ \Theta(3^{\log_4 n})\]
\[= \frac{(3/16)^{\log_4 n}-1}{(3/16)-1}cn^2+ \Theta(3^{\log_4 n})\]
Textbook Example
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/4) + cn^2&\text{otherwise} \end{cases}\)
- Step 4: Count the number of operations
\[cn^2\sum_{i=0}^{\log_4n-1}\left(\frac{3}{16}\right)^i+ \Theta(3^{\log_4 n})\]
\[= \frac{(3/16)^{\log_4 n}-1}{(3/16)-1}cn^2+ \Theta(3^{\log_4 n})\]
- this is ugly ...
- but here we can be a bit sloppy again:
- use geometric sum of an infinite series, with \(|x| < 1\)
\[\sum_{k=0}^\infty x^k = \frac{1}{1-x}\]
\[\sum_{k=0}^m x^k = 1 + x + x^2 + \cdots + x^m = \frac{x^{m+1}-1}{x-1}\]
Textbook Example
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/4) + cn^2&\text{otherwise} \end{cases}\)
- Step 4: Count the number of operations
\[\sum_{k=0}^n x^k <\sum_{k=0}^\infty x^k = \frac{1}{1-x}\]
- if \(|x| < 1\) and each term is positive:
- so:
\[\sum_{i=0}^{\log_4n-1}\left(\frac{3}{16}\right)^i cn^2+ \Theta(3^{\log_4 n})\]
\[< \sum_{i=0}^{\infty}\left(\frac{3}{16}\right)^i cn^2+ \Theta(3^{\log_4 n})\]
\[=\frac{1}{1-(3/16)} cn^2+ \Theta(3^{\log_4 n})\]
Textbook Example
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/4) + cn^2&\text{otherwise} \end{cases}\)
- Step 4: Count the number of operations
- One more rule: \(3^{\log_4 n} = n^{\log_4 3}\)
\[\frac{1}{1-(3/16)} cn^2+ \Theta(3^{\log_4 n})\]
\[=\frac{1}{1-(3/16)} cn^2+ \Theta(n^{\log_4 3})\]
- \(\log_4 3\) is about 0.79248125, so we have
\[\frac{1}{1-(3/16)} cn^2+ \Theta(n^{0.79}) = O(n^2)\]
Textbook Example
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/4) + cn^2&\text{otherwise} \end{cases}\)
- Step 4: Count the number of operations
\[cn^2 + \frac{3}{16}cn^2 + \left(\frac{3}{16}\right)^2cn^2+\cdots+ \left(\frac{3}{16}\right) ^{\log_4 n - 1}cn^2+ \Theta(3^{\log_4 n})\]
\[\left(\frac{3}{16}\right)^{\log_4 n} cn^2 = \frac{3^{\log_4 n}}{16^{\log_4 n}}cn^2\]
\[\left(\frac{3}{16}\right)^{\log_4 n} cn^2\]
\(16^{\log_4 n} = (4^2)^{\log_4 n} = (4^{\log_4 n})^2 = n^2\)
\[\left(\frac{3}{16}\right)^{\log_4 n} cn^2 = \frac{3^{\log_4 n}}{n^2}cn^2 = c\cdot 3^{\log_4 n}\]
Textbook Example
6.1 - Recursion-Tree Method
\(T(n) = \begin{cases} \Theta(1) &\text{if $n \le 1$}\\ 3T(n/4) + cn^2&\text{otherwise} \end{cases}\)
- Step 4: Count the number of operations
\[cn^2 + \frac{3}{16}cn^2 + \left(\frac{3}{16}\right)^2cn^2+\cdots+ \left(\frac{3}{16}\right) ^{\log_4 n - 1}cn^2+ \Theta(3^{\log_4 n})\]
\[cn^2 + \frac{3}{16}cn^2 + \left(\frac{3}{16}\right)^2cn^2+\cdots+ \left(\frac{3}{16}\right) ^{\log_4 n - 1}cn^2+ \left(\frac{3}{16}\right)^{\log_4 n}cn^2 \]
\[cn^2\sum_{i=0}^{\log_4n-1}\left(\frac{3}{16}\right)^i+ \Theta(3^{\log_4 n})\]
\[cn^2\sum_{i=0}^{\log_4n}\left(\frac{3}{16}\right)^i\]
or
The End
6.1 - Recursion-Tree Method
- we will do more in the workshop and there will be a submission question as well for this week
- it is going to show up in the exam, but the formulas for geometric sums will be provided
6.1 - Recursion-Tree Method
I told you that recursion-tree will be in the exam
I just want a pass
COMP3010 - 6.1 - Recursion Tree Method
By Daniel Sutantyo
COMP3010 - 6.1 - Recursion Tree Method
Recursion-tree method for finding the running time of a divide and conquer algorithm
