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Binary Search Algorithm: Implementation and Sorting, Slides of Artificial Intelligence

Bidhan Chandra Krishi ViswavidyalayaArtificial Intelligence

An explanation of the binary search algorithm, including its implementation using javascript code and its application to sort a list using selection sort. The document also discusses the limitations of binary search and introduces insertion sort as an alternative.

Typology: Slides

2012/2013

Uploaded on 04/24/2013

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CSCI 100
Think Like Computers
Lecture 23
Fall 2008
Last Time …
•Lists (arrays)
•Implementing algorithms
•Sequential search
•Binary search
Binary Search
•binarySearch(list, target) returns index (if found) or -1
•[“Charlie”, “Dave”, “Jack”, “Jerald”, “Jerroll”, “Laura”,
“Lois”, “Winnie”]
•Use 3 index variables
left: the leftmost range index (inclusive)
right: the rightmost range index (exclusive)
mid: the middle index for the range (left, right)
•Some things to remember:
left < right : the range contains value
Otherwise, nothing Æsearch fail
How Do We Implement It?
•[“Charlie”, “Dave”, “Jack”, “Jerald”, “Jerroll”,
“Laura”, “Lois”, “Winnie”]
•Compare array[mid] with target
If equal, we found it Æreturn mid
Otherwise, we shrink the range and starts with the
new left, right, mid values (another iteration of the
loop)
array[mid] > target : right = mid
array[mid] < target : left = mid+1;
New mid : Math.floor(( left + right )/2), as always
Example
•[“Charlie”, “Dave”, “Jack”, “Jerald”,
“Jerroll”, “Laura”, “Lois”, “Winnie”]
•Search for “Zoe”:
•left right mid list[mid]
•1st try 0 8 4 Jerroll
•2nd try 5 8 6 Lois
•3rd try 7 8 7 Winnie
•4th try 8 8 Æno values, return -1
Exercise
•[“Charlie”, “Dave”, “Jack”, “Jerald”,
“Jerroll”, “Laura”, “Lois”, “Winnie”]
•Search for “Jones”:
•left right mid list[mid]
•1st try
•2nd try
•3rd try
•…
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CSCI 100

Think Like Computers

Lecture 23

Fall 2008

Last Time …

  • Lists (arrays)
  • Implementing algorithms
  • Sequential search
  • Binary search

Binary Search

  • binarySearch(list, target) returns index (if found) or -
  • [“Charlie”, “Dave”, “Jack”, “Jerald”, “Jerroll”, “Laura”, “Lois”, “Winnie”]
  • Use 3 index variables Š left: the leftmost range index (inclusive) Š right: the rightmost range index (exclusive) Š mid: the middle index for the range (left, right)
  • Some things to remember: Š left < right : the range contains value Š Otherwise, nothing Æ search fail

How Do We Implement It?

  • [“Charlie”, “Dave”, “Jack”, “Jerald”, “Jerroll”, “Laura”, “Lois”, “Winnie”]
  • Compare array[mid] with target Š If equal, we found it Æ return mid Š Otherwise, we shrink the range and starts with the new left, right, mid values (another iteration of the loop) ƒ array[mid] > target : right = mid ƒ array[mid] < target : left = mid+1; ƒ New mid : Math.floor(( left + right )/2), as always

Example

  • [“Charlie”, “Dave”, “Jack”, “Jerald”, “Jerroll”, “Laura”, “Lois”, “Winnie”]
  • Search for “Zoe”:
  • left right mid list[mid]
  • 1 st^ try 0 8 4 Jerroll
  • 2 nd^ try 5 8 6 Lois
  • 3 rd^ try 7 8 7 Winnie
  • 4 th^ try 8 8 Æ no values, return -

Exercise

  • [“Charlie”, “Dave”, “Jack”, “Jerald”, “Jerroll”, “Laura”, “Lois”, “Winnie”]
  • Search for “Jones”:
  • left right mid list[mid]
  • 1 st^ try
  • 2 nd^ try
  • 3 rd^ try
  • …

Writing the Code

  • Begin:
  • var left, right, mid;
  • left = 0; right = list.length;
  • mid = Math.floor( (left+right)/2 );
  • Loop:
  • while ( left < right ) { // items in range
  • …
  • }

Writing the Code

  • Inside the loop:
  • (1) compare list[mid] and target
  • if ( list[mid] == target ) return mid;
  • (2) // otherwise, update the range
  • else if ( list[mid] > target ) {
  • right = mid;
  • } else {
  • left = mid+1;
  • }
  • (3) don’t forget the mid:
  • mid = Math.floor( (left+right)/2 );

Writing the Code

  • If loop is finished (did not return during the loop)
  • Then target is not in the list!
  • return -1;
  • That’s it!

How to Sort?

  • Any idea?
  • list = [ 3, 20, 11, 13, 8, 10, 6, 2];

First, How to Swap?

  • You’ll need a temporary variable:
  • var temp;
  • temp = x;
  • x = y;
  • y = temp;
  • Notice the sequence: first save/write, then you can modify (and lose its original value)

Swap List Elements

  • Swap elements at index i and j:
  • var temp;
  • temp = list[i];
  • list[i] = list[j];
  • list[j] = temp;

Put it Together

  • k = 0;
  • while (! k == list.length-1 ) {
  • sequential search the min of list range (k, list.length) Æ index i
  • swap list elements at positions k and i.
  • k = k+1;
  • }

Use Functions for

Frequent Operations

  • Sequential search for min in list range:
  • minIndex(list, begin, end)
  • Swap list elements
  • swap(list, k, i)

Exercise: minIndex(…)

  • function minIndex(list, begin, end) {
  • // list: the input array
  • // begin: the first index of the range (inclusive)
  • // end: the last index of the range (exclusive)
  • // Assumes: end > begin
  • // Returns: the index of the min value
  • }

Better Sorting Algorithms?

  • Selection sort is quite slow.
  • Insertion sort is better…
  • Similar to arranging a hand of cards Š Pick up cards one by one Š Insert the current card at the right position Š The (partial) hand is always sorted Š The whole hand is sorted after inserting the last card

Insertion Sort

  • Partial hand: list range (0, k) Š Start with k= Š Want to grow k, until k==list.length
  • The next card to be picked: list[k]
  • Insert at the right position: Š Compare list[k] with list[k-1]: if less, swap k & k-1. Otherwise, done Š Compare list[k-1] with list[k-2]: if less, swap k-1 & k-2. Otherwise, done Š … Š This would also be a loop!

Insertion Sort

  • k=1;
  • while ( k < list.length ) {
  • j = k;
  • while ( j > 0 ) {
  • if ( list[j] < list [j-1] ) swap(list, j, j-1);
  • else break; // abort the loop since we are done!
  • j = j-1;
  • } // end of inner while loop
  • k = k+1;
  • } // end of outer while

Š This can be improved, because we don’t really need full swaps