Home Previous Next

CSC205::Lecture Note::Week 15
Assignments | Code | Handouts | Resources | Email Thurman | Tweets::@compufoo
GDT::Bits:: Time  |  Weather  |  Populations  |  Special Dates

Overview Assignment(s):  All assignments have been assigned.

Code BST.java {output} | QuickSort.java

RPN calculator

algorithm for a RPN (Reverse Polish Notation) calculator
   // binary operators take two operands (left and right)
   // let l = left and r = right
   // + * commutative
   // - / not commutative
   // stack-based; pop() order matters
   // r = pop() and l = pop()

   // evaluate:  (2 + 3) * 4
   //  postfix:  2 3 + 4 *
   op = +
   r = pop()
   l = pop()
   x = eval(l op r)
   op = *
   r = pop()
   l = pop()
   x = eval(l op r)

{TopOfPage} {Oracle.com::API Specification | Tutorial} {Derek Banas Videos} {Data Structure Visualizations}


QuickSort  was first developed by C.A.R. Hoare. It requires the picking of a partition element (pivot) and then partially sorting the array about the partition. You can then sort each of the two partitions by recursive application of the same technique. The algorithm can sort quite rapidly, but it can also sort very slowly [O(n^2) worst case].

QuickSort is a fast divide-an-conquer algorithm when comparison sorting arrays in place (frequently O(n log n)).

The basic algorithm for QuickSort is recursive and consists of the following steps:

  1. If the number of elements is 0 or 1, return.
  2. Pick any element and call it the pivot.
  3. Partition remaining elements into two disjoint groups called L and R.
  4. Return the result of QuickSort(L), followed by the pivot, followed by QuickSort(R).
The pivot divides array elements into two groups: those smaller than the pivot and those larger than the pivot.

The partition step places every element except the pivot in one of two groups.

   13  81 43  92  31  65  57  26  75  0

   Assume 65 is selected as the pivot.  Partition the elements.

    Left:  13  43  31  57  26  0
   Right:  81  92  75

   Quicksort the left items.  Quicksort the right items.
   0  13  26  31  43  57      75  81  92
   Original list:
      8  1  4  9  6  3  5  2  7  0

   Select the pivot  6  and swap the pivot with the last element.

      8  1  4  9  0  3  5  2  7  6

   Run   i  from left-to-right, and  j  from right-to-left.
   When  i  sees a large element,  i  stops.  When  j  sees
   a small element,  j  stops.  If  i  and  j  have not
   crossed, swap their items and continue; otherwise, stop
   this loop.

      i  stops at element 8; j stops at element 2
      the 8 and 2 are swapped

   2  1  4  9  0  3  5  8  7  6

      i  stops at element 9; j stops at element 5
      the 9 and 5 are swapped

   2  1  4  5  0  3  9  8  7  6

      i  stops at element 9; j stops at element 3
      i and j have crossed -- swap pivot with
      i'th element (6 with 9)

   2  1  4  5  0  3  6  8  7  9
Picking the pivot is important to good performance. Never choose the first element as the pivot (you should also avoid the last element). The middle element is usually a reasonable choice. The perfect pivot is one that results in equally sized partitions. Picking a random pivot has proven efficient.

The  median-of-three  is a popular technique for picking a pivot. Sort the first, middle and last elements of the list and use the median (in this case middle) value. Using this technique is good because while in the process of determining the pivot, you are also getting started on the sorting of the list (element 0 will be less than the pivot and element N-1 will be greater than the pivot).

QuickSort pseudo-code:

   quickSort(T a[], int left, int right) 

      if (left >= right) return;
      SWAP (a, left, (left+right)/2);
      last = left;
      for (i = left+1; i <= right; i++) {
         if (COMPARE (a, i,  left) {
            SWAP (a, last, i);

      SWAP (a, left, last);

      quickSort(a, left, last-1);
      quickSort(a, last+1, right);

{TopOfPage} {Oracle.com::API Specification | Tutorial} {Derek Banas Videos} {Data Structure Visualizations}


A tree is a "nonlinear" structure that consists of nodes and edges (or branches). A tree is a set of nodes and edges that connect the nodes. There is exactly one path between two nodes. A path is a connected sequence of edges.

Arrays, stacks, queues, and linked-lists define collections of objects that are sequentially accessed. These are linear lists: they have first and last elements, and each interior element has a unique successor.

In a nonlinear structure, an element may have multiple successors.

Tree structures are used to represent a "hierarchy" of information. A commonly used hierarchy is an operating system's file system.

A tree structure consists of a set of nodes that originate from a unique starting node called the root.

Some nodes may be considered a parent node which implies they have zero or more children nodes to which they are connected. Every node except the root node has a parent.

The children of a node and children of those children are called descendants, and parents and grandparents of a node are its ancestors.

Nodes are siblings when they have the same parent.

A node that doesn't have any children is called a leaf.

Each node in a tree is the root of a subtree, which is defined by the node and all descendants of the node.

            |          |           |
         father       uncle       aunt
      |           |
   brother      sister

      grandfather  is the  root  node.  grandfather is the parent
      node to father, uncle and aunt.  father, uncle, aunt, brother
      and sister are all descendants of grandfather.  father and
      grandfather are ancestors of brother and sister.  brother and
      sister are children of father.  brother are sister are leaf
      nodes.  brother and sister are siblings.  father is the root
      of the subtree that contains the father, brother, sister nodes.

A single-inheritance class hierarchy can be represented by a tree.

             |                  |            |            |
         Container            Button       Canvas    TextComponent
             |                                            |
   +---------+----------+                            +----+----+
   |         |          |                            |         |
 Panel  ScrollPane   Window                      TextArea   Textfield
                  |          |
               Dialog      Frame

Movement from a parent node to its child and other descendants is done along a path.

The level of a node is the length of the path from the root to the node.

The depth of a tree is the maximum level of any node in the tree. The depth of a node is the length of its path from the node to the root.

            A               level 0
          / | \
         B  C  D            level 1
          \  \
           E  F             level 2
          / \
         G   H              level 3

   The depth of this tree is 3 (the longest path from the root
   to a node).  The path length for node E is 2.  The path
   length for node D is 1.  The path for node H is:  A->B->E->H.


A general tree is a set of nodes that is either empty, or has a designated node called the root from which descend zero or more subtrees. Each node is not an ancestor of itself, and each subtree itself satisfies the definition of a treee.

Note that a tree is defined in a recursive fashion; consequently, most tree-processing algorithms are recursive.

YouTube.com::CS 61B Lecture 23: Trees and Traversals

{TopOfPage} {Oracle.com::API Specification | Tutorial} {Derek Banas Videos} {Data Structure Visualizations}

Binary Trees

A binary tree is a popularly used tree in which each node has at most two (2) subtrees. These subtrees are designated as the left and right, respectively. Either or both of these subtrees may be empty.

The following operations are performed on a binary tree:

[Definition] An algorithm that systematically "visits" all items in a tree is called a tree traversal.

Here is a binary tree and its respective (depth-first) traversals:

                     /     \
                  B           C
               /            /   \
            D             E        F
                        /   \
                      G       H

       Preorder traversal: A (root); B,D (left); C,E,G,H,F (right)
        Inorder traversal: D,B (left); A (root); G,E,H,C,F (right)
      Postorder traversal: D,B (left); G,H,E,F,C (right); A (root)

The following tree takes on a heap property:

                         /    \
                        /      \
                      84        63
                     /  \         \
                   68    79        12
                   /      \       /  \
                 32       67     6   10
                                    /  \
                                   8    9

   The data in any given node of the tree is greater than or 
   equal to the data in its left and right subtrees.

A binary tree can have a hierarchy that exhibits the property known as the ordering property.

                         /    \
                        /      \
                      84        103
                     /  \       / \
                   68    86    90  109
                  /  \        /  \
                32   74      88   97
                    /  \
                  70   80

   The data in each node of the tree are greater than all of the
   data in that node's left subtree and less than or equal
   to all the data in the right subtree.

A binary tree with the ordering property is often called a binary search tree (BST).

A complete binary tree of depth N is a tree in which each level 0 to N-1 has a full set of nodes and all leaf nodes at level N occupy the leftmost positions in the tree.

               A          Complete tree with depth 3.
             /   \
            B     C
           / \   / \
         D    E F   G
        / \   /\
       H   I J  K         Leaf nodes occupy leftmost positions.

A full binary tree is a complete tree that contains 2N nodes at level N.

               A           Full tree with depth 2.
             /   \
            B     C        Every node has two children.
           / \   / \
         D    E F   G      22 is 4 and there are 4 nodes.

A degenerate tree is one that has a single leaf node and each non-leaf node has only one child. [Note: a degenerate tree is equivalent to a linked-list.]


Many compilers make use of tree structures in obtaining forms of an arithmetic expression for evaluation.

   Operators are non-leaf nodes and operands are leaf nodes.

                    /   \
                  -       *
                 / \     / \
                A   B   C   ÷
                           / \
                          E   F

      (A - B) + C * (E / F)       infix
      + - A B * C / E F           prefix
      A B - C E F / * +           postfix

Binary Tree Structure

A binary tree structure is built with nodes. A tree node contains a data field and two pointer (i.e. link) fields.

Given the non-linear structure of trees, a linked-list type of searching (i.e. sequential) is not available; therefore, a variety of traversal methods (inorder, preorder, postorder) are commonly used.

Tree traversal methods rely heavily on recursion. Each traversal method performs three actions at a node:

The descent terminates when an empty tree pointer is encountered.

A node visit depends on the application (i.e. it is application-dependent).

Inorder Traversal

An inorder traversal begins its action at a node by first descending to its left subtree. After recursively descending through the nodes in that subtree, the inorder traversal takes the second action at the node and uses the data value. The traversal completes its action at the node by performing a recursive scan of the right subtree. In the recursive descent through the subtrees, the actions of the algorithm are repeated at each new node.

  1. traverse the left subtree
  2. visit the node
  3. traverse the right subtree
Postorder Traversal

The postorder traversal delays a visit to a node until after a recursive descent of the left subtree and a recursive descent of the right subtree.

  1. traverse the left subtree
  2. traverse the right subtree
  3. visit the node
Preorder Traversal

The preorder traversal visits the node first, then it does recursive descents of the left and right subtrees, respectively.

  1. visit the node
  2. traverse the left subtree
  3. traverse the right subtree
Breadth First Traversal (or Level Traversal)

A breath first (or level) traversal scans the tree nodes level-by-level starting with the root at level 0.

A common implementation of this algorithm is queue-based.

   create queue  Q
   insert root into  Q
   while Q is not empty
      delete front node  p  from  Q
      use node  p  to identify its children at next level
         if (p.left() != null)
         if (p.right() != NULL)
      Using the queue helps ensure that nodes on the same
      level are visited in the correct order (in this case


Trees perform inserts, deletes, and finds in O(log N) time (on average). A degenerate tree can be O(N).


The following inputs inserted in the given order produce a good binary tree.


What the binary tree look like? [answer]

What order do these state abbreviations need to be inserted in order to produce a degenerate tree? [answer]

WolframAlpha.com::Binary Tree

{TopOfPage} {Oracle.com::API Specification | Tutorial} {Derek Banas Videos} {Data Structure Visualizations}

Binary Search Tree

A general binary tree can hold a large collection of data and yet provide fast access.

Recall, creating a "container" or "collection" class using an array or linked-list, requires those collections to be searched using a linear search. This can be inefficient when dealing with large amounts of data [O(N)].

Trees are efficient for searching because the path to any data value is no greater than the depth of the tree. Searching is maximized with a complete binary tree O(log2N).

To store elements in a tree for efficient access, a  binary search tree  (BST) can be used.

   For each node, the data values in the left subtree
   are less than the value of the node and the data
   values in the right subtree are greater than or equal
   to the value of the node.
Using a BST implies that the data found in each node contain a  key . In many cases, the data of a node is a record and the key is a field in the record.

Operations performed on a BST are:

   find -- search a tree of a specific data value
   insert -- find appropriate insertion spot and add new
             data value to the tree
   delete -- search a tree and remove the first occurrence
             of a given data value; reattach all subtrees
             to maintain BST structure
   update -- if key at current position matches key for
             the data item, update data value; otherwise,
             insert data value into the tree

Insertion on a Binary Search Tree

Search the tree and find the location where the new node is to reside. This is done be scanning the left and right subtrees until the insertion point is found. For each step in the path, maintain a record of the current node and the parent of the current node. The process terminates when an empty subtree is encountered. At this location, the new node is inserted as a child of the parent.
        /  \
      20    35
     /        \
   12          40

   Insert the value 32 into this tree.
   Step 1:  compare 32 with 25 -- traverse the right subtree
   Step 2:  comapre 32 with 35 -- traverse the left subtree
   Step 3:  insert 32 as left child of parent (i.e. 35).

        /  \
      20    35
     /     /  \
   12    32    40 

BST Pseudo-code For find(), min(), max(), delete()

BST is a collection of Node objects

class BST                class Node
   Node root                int key
   int nodecnt              Node parent
                            Node left
                            Node right

   if root is null
      return null
   return find(key, root)

find(key, node)
   if key equals node.key
      return node
   if key less than node.key
      if node.left null
         return null
      return find(key, node.left)
   if node.right null
      return null
   return find(key, node.right)

   if node.left null
      return node
   return min(node.left)

max(node)  // left to the reader
   node = find(key)
   if node null
   decrement nodecnt

   // case 0: delete childless root node
   if node.parent null    // or nodecnt equals 0
      if node.left null and node.right null
         set root to null

   // case 1:  delete leaf node
   if node.left null and node.right null
      if node.parent !null
         if node.parent.left equals node
            set node.parent.left to null
            set node.parent.right to null

   // case 2:  delete node having two kids
   if node.left !null and node.right !null
      minnode = min(node.right)
      set minnode.left to node.left
      set node.left.parent to minnode
      if minnode.right !null and minnode.parent not equals node
         set minnode.parent.left to minnode.right
         set minnode.right.parent to minnode.parent
      else if minnode equals minnode.parent.left
         set minnode.parent.left to null
      if minnode.parent not equal node
         set minmode.right to node.right
         set node.right.parent to minnode
      set minnode.parent to node.parent
      if node.parent !null
         if node.parent.left equals node
            set node.parent.left to minnode
            set node.parent.right to minnode
         set root = minnode
         set node.parent = null

   // case 3: delete node having left child only
   if node.left !null
      if node.parent !null
         if node.parent.left equals node
            set node.parent.left to node.left
            set node.parent.right to node.left
         set node.left.parent to node.parent
      set root to node.left
      set root.parent to null

   // case 4:  delete node having right child only
   if node.parent !null) 
      if node.parent.right equals node
         set node.parent.right to node.right
         set node.parent.left to node.right
      set node.right.parent to node.parent
   set root to node.right
   set root.parent to null

Update on a Binary Search Tree

Find the node that needs updating and copy in new data. If node note not found, then insert it. What if key being modified? Find node, delete it and then insert new node containing the updated key.

{TopOfPage} {Oracle.com::API Specification | Tutorial} {Derek Banas Videos} {Data Structure Visualizations}

2-3-4 Trees

2-3-4 Trees are trees that have nodes which may have more than two children.

A benefit of a 2-3-4 tree is that it contains shorter search paths because there are fewer levels in the tree.

When searching a binary tree, you either go left or right at each node. In a 2-3-4 tree, there can be several different directions that a search path may follow.


   An  m-node  in a search tree stores  m-1  data values
   k1 < k2 < ... < km-1,
   and has links to  m  subtrees  T1, ..., Tm,
   where for each value  i, all data values in  
   Ti < ki <= all data values in Ti+1

Here is an example 2-3-4 tree.

                            /  \
            +---------------    ------------+
            |                               |
        27     38                       60     70
       /    |    \                     /    |    \
   16,25  33,36  41,46,48          55,59  65,68  73,75,79

      This tree has 9 nodes.  The root node contains one value.
      Both nodes at level 1 have two values.  On level two,
      there are 4 nodes containing 2 values, and 2 nodes 
      containing 3 values.

   Search for the value 36.

      36 is less than 53, traverse the left subtree.
      36 greater than 27 and less than 38, traverse 
      the center subtree.  36 less than 33, compare
      to the next number.  36 equals 36 -- found.

2-3-4 trees satisfy the following properties:

In a nutshell, 2-3-4 tree is constructed as follows:

   Construct a 2-3-4 tree using the inputs:  10 3 9 15 21 44 18 1 12 5 7

   insert 10:       10
   insert  3:       3,10
   insert  9:       3,9,10
   insert 15:       split
               3              10

         15 greater than 9 -- search right subtree

               3              10,15

   insert 21:
               3              10,15,21

   insert 44:
      44 greater than 9 -- search right subtree
      internal node is full -- split
      median value moved into parent

            3        10       21,44

   insert 18:
      18 greater than 15 -- search right subtree
            3         10       18,21,44

   insert 1:
      1 less than 9 -- search left subtree
         1,3         10       18,21,44

   insert 12:
      12 greater than 9 and less than 15 -- search middle subtree
         1,3        10,12       18,21,44

   insert 5:
      5 less than 9 -- search left subtree
         1,3,5     10,12       18,21,44

   insert 7:  
      7 less than 9 -- search left subtree
      internal node is full; split
      median value 3 moved into parent
         1,5,7      10,12       18,21,44

2-3-4 trees stay balanced during insertions and deletions.

Think about top-down insertion versus the demonstrated bottom-up insertion. [Don't allow any parent nodes to become 4-nodes.]

2-3-4 trees have numerous unused pointers. Explore Red-Black trees for an alternative.

   2-3-4 tree with N nodes has 4*N pointers
   each node (excluding the root) has only one pointer to it
   4*N - (N-1) = 3N+1 of the pointers are null 
      (i.e. approximately 75% of the pointers are null)

Example 2-3-4 tree node class.

   class Node234 {
      Object data[3];
      Node234 links[4];

azrael.digipen.edu::Deleting Elements From a 2-3-4 Tree

YouTube.com::UCBerkeley::CS61B::Lecture 26: Balanced Search Trees

{TopOfPage} {Oracle.com::API Specification | Tutorial} {Derek Banas Videos} {Data Structure Visualizations}

Big-O Notation

The run-time efficiency of a program has two main ingredients: space and time.

Space efficiency is a measure of the amount of memory a program requires.

Time efficiency is a measure of how long a program takes to execute.

In general, there is a trade-off between space and time. If your algorithm uses more space, then typically it will execute faster and vice versa.

The relationship between N (where N is a variable representing the number of inputs on which your algorithm must operate) and the running time of an algorithm as N becomes large is called the computational complexity of that algorithm.

Computational complexity is denoted using BigO notation. The letter O stands for the word order as it is used in the phrase on the order of. It signifies approximation.

Given two functions f and g, the function g is said to dominate function f if there is some positive constant c such that:

   c * g(x) >= f(x), for all possible values of  x

Asymptotic dominance is a variation of dominance such that:

   c * g(x) >= f(x), for all  x >= x'

In other words, function g asymptotically dominates function f for all "large" values of x. Big-O notation is represented using asymptotic dominance.

          5N   versus     N2
   N=1:   5    versus     1
   N=2:   10   versus     4
   N=3:   15   versus     9
   N=4:   20   versus     16
   N=5:   25   versus     25
   N=6:   30   versus     36
   N=7:   35   versus     49

      In this example, x'  is equal to 6 and the constant is 1.

The following are some examples of how dominance is used to determine BigO values:

   O(N3) + O(N2)  =>  O(N3)
      As  N  gets large, N cubed grows at a faster rate than
      does  N squared.  For large values of N, N cubed dominates.
   O(10 * N) + O(1000 * N) + O(1)   =>  O(N)
      Constant values are ignored.  The significance of the
      constant values are minimized for large values of N.
   O(N log N) + O(log N) + O(N2)  =>  O(N2)
      N squared dominates  N log N  which in turn dominates  log N.

[About logarithms.]

Let's review:

BigO-notation is used for three distinct purposes.

  1. To bound the error that is made when we ignore small terms in mathematical formulas.
  2. To bound the error that is made when we ignore parts of a program that contribute a small amount to the total being analyzed.
  3. To allow us to classify algorithms according to upper bounds on their total running times.

Categories of Common Running Times
O(1) constant time (most efficient)
O(N) linear time (polynomial time)
O(N log N) "linearithmic" time (polynomial time)
O(N2) quadratic time (polynomial time)
O(N3) cubic time (polynomial time)
O(log N) logarithmic time
O(2N) exponential time (impractical for large values of N)

The following table presents growth rates for some of the common functions (i.e running times):

    N     log N         N   N log N      N**2          N**3          2**N
    1         0         1         0         1             1             2
    2         1         2         2         4             8             4
    4         2         4         8        16            64            16
    8         3         8        24        64           512           256
   16         4        16        64       256          4096         65536
   32         5        32       160      1024         32768   4.29497e+09
   64         6        64       384      4096        262144   1.84467e+19
  128         7       128       896     16384       2097152   3.40282e+38

Typically constant factors are ignored in BigO analysis (although in reality constants frequently matter).

To compare the speed of different algorithms, computer scientists use bigO notation. The bigO notation doesn't exactly predict performance times. Instead, what the bigO notation tells us is that given two algorithms with the same order, then both will slow down at roughly the same rate as the number of items being sorted gets larger.

Search algorithms also have bigO values. Linear search is O(n), binary search is O(log n). Once again, these differences are most important for large data sets. If you have 1 million items in an array, a linear search will require about 1 million tests, but a binary search will only require 20 tests (because 2 to the 20 power is about 1 million -- the 'log' in bigO notation is always log-base-2 -- when you have a logarithmic algorithm, the analysis need not be concerned with the base of the logarithm, since this can change the total running time only by a constant factor (and constant factors are ignored).

   for (i = SOME_CONSTANT; i <= ANOTHER_CONSTANT; i++)      // O(1)
      // loop body requiring constant time
   for (i = SOME_CONSTANT; i <= N; i++)                     // O(N)
      // loop body requiring constant time
   for (i = SOME_CONSTANT; i <= N; i++)                     // O(N*N)
      for (j = ANOTHER_CONSTANT; j <= N; j++)
         // loop body requiring constant time

   for (i = SOME_CONSTANT; i <= N; i++)                     // O(N*M)
      // loop body requiring time O(M), where M is a variable

The following is example of how one can go about determining the approximate running time of an algorithm.

    // the following logic initializes a 'n' element array
    index = 0;                // assignment is 1 step
    while (index < n) {       // EXPR  index < n  is 1 step
       array[index] = -1;     // assignment is 1 step
       index++;               // increment is 1 step

             Si + Sc*(n+1) + Sb*n
   n = 0:    1  + 1*1      + 2*0    =  2
   n = 1;    1  * 1*2      + 2*1    =  5
   n = 2;    1  * 1*3      + 2*2    =  8
   n = 3:    1  + 1*4      + 2*3    =  11
   n = 6:    1  + 1*7      + 2*6    =  20
   n = 9:    1  + 1*10     + 2*9    =  29
   n = 300:  1  + 1*301    + 300*2  =  902
      Si => # of initialization steps
      Sc => # of steps in the condition EXPR
      sb => # of steps in the loop body
   Conclusion:  linear -- O(n)

Analysis of a Couple of Code Snippets

   for (k = 1; k <= n / 2; k++) {
      for (j = 1; j <= n * n; j++) {

Since these loops are nested, the number of times statements within the innermost loop are executed is the product of the number of repetitions of the two individual loops. Hence the efficiency is n3, or O(N3) in BigO terms, with a constant of proportionality equal to 1/2.

   for (k = 1; k <= n * 10; k++) {

   for (j = 1; j <= n * n * n; j++) {

Since one loop follows the other, the number of operations executed by both loops is the sum of the individual loop efficencies. Hence the efficiency is 10n + n3, or O(N3) in BigO terms.

Analysis of Exchange Sort

Assume we have array A with length N and we want to find the minimum element in the array.

   Pass 1:  compare the  n-1  elements A[1]...A[n-1] 
            with A[0] and, if necessary, exchange elements
            so that A[0] always has the smallest value

   Pass 2:  compare the  n-2  elements A[2]...A[n-1] 
            with A[1] and, if necessary, exchange elements
            so that A[1] less than all elements to the right

   Pass i:  compare the  n-i  elements A[i]...A[n-1]
            with A[i-1] and, if necessary, exchange elements 

   The total number of comparisons is given by the
   arithmetic series  f(n) from 1 to n-1:
      f(n) = (n-1) + (n-2) + ... + 3 + 2 + 1 =  n * (n-1) / 2

   The number of comparisons depends on  n2.

Using an Array for a List

Append an item: O(1). Insert an item at the front of the array: O(N).

Simple Sorts

Bubble, selection, insertion, and exchange are all O(N2).

Shell sort can give much better performance than O(N2) in the worst case.

Quick sort in the worst case can be O(N2), but on average it is O(N log N).

Merge sort is an O(N log N) sort.

Order of a Function: Mathematical Definition

The order of a function is defined as follows:

Given two non-negative functions f and g, the order of f is g if and only if g asymptotially dominates f. Using BigO notation, this can said as: f = O(g) (f is of order g).

Rules for Comparing BigO Estimates

Some rules to help calculate and compare BigO estimates:

   Let  f  and  g  be functions, and  C  a constant:
      O(C * f) = O(f)
      O(f * g) = O(f) * O(g)
      O(f / g) = O(f) / O(g)
      O(f) > O(g), if and only if  f  dominates  g
      O(f + g) = Max[O(f), O(g)]

About Logarithms

A logarithm of base b for the value y is the power to which you must raise b to get y. Normally, this is written as: logby = x.

   logby = x  <=>  bx = y  <=>  blogby = y

      where <=>  means "is equivalent to"

Logarithms have the following properties:

    For any positive values of  m,  n, and  r, and any positive integers
    a  and  b:
      log nm = log n + log m
      log n/m = log n - log m
      log nr = r log n
      logan = logb n / logba

YouTube.com::Berkeley.edu::CS61B::Lecture 19: Asymptotic Analysis

{TopOfPage} {Oracle.com::API Specification | Tutorial} {Derek Banas Videos} {Data Structure Visualizations}

Home Previous Next