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

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