Linear and Binary search

Prepared by Nisha Soms
Reference:
(i) Kruse and Ryba Ch 7.1-7.3 and 9.6
(ii) https://cs-eb.bu.edu/fac/gkollios/cs113/Slides/Searching.ppt

Problem: Search
 We are given a list of records.
 Each record has an associated key.
 Give efficient algorithm for searching for a record
containing a particular key.
 Efficiency is quantified in terms of average time
analysis (number of comparisons) to retrieve an item.

Search
[ 0 ] [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 700 ]
Number 506643548
Number 233667136
Number 281942902
Number 155778322
Number 580625685
Number 701466868
…
Number 580625685
Each record in list has an associated key.
In this example, the keys are ID numbers.
Given a particular key, how can we
efficiently retrieve the record from the list?

Serial Search
 Step through array of records, one at a time.
 Look for record with matching key.
 Search stops when
 record with matching key is found
 or when search has examined all records without
success.

Pseudocode for Serial Search
// Search for a desired item in the n array elements
// starting at a[first].
// Returns pointer to desired record if found.
// Otherwise, return NULL
…
for(i = first; i < n; ++i )
if(a[first+i] is desired item)
return &a[first+i];
// if we drop through loop, then desired item was not found
return NULL;

Serial Search Analysis
 What are the worst and average case running times for
serial search?
 We must determine the O-notation for the number of
operations required in search.
 Number of operations depends on n, the number of
entries in the list.

Worst Case Time for Serial Search
 For an array of n elements, the worst case time
for serial search requires n array accesses:
O(n).
 Consider cases where we must loop over all n
records:
 desired record appears in the last position of the
array
 desired record does not appear in the array at all

Average Case for Serial Search
Assumptions:
1. All keys are equally likely in a search
2. We always search for a key that is in the array
Example:
 We have an array of 10 records.
 If search for the first record, then it requires 1
array access; if the second, then 2 array accesses.
etc.
The average of all these searches is:
(1+2+3+4+5+6+7+8+9+10)/10 = 5.5

Average Case Time for Serial Search
Generalize for array size n.
Expression for average-case running time:
(1+2+…+n)/n = n(n+1)/2n = (n+1)/2
Therefore, average case time complexity for serial
search is O(n).

Binary Search
 Perhaps we can do better than O(n) in the average
case?
 Assume that we are give an array of records that is
sorted. For instance:
 an array of records with integer keys sorted from
smallest to largest (e.g., ID numbers), or
 an array of records with string keys sorted in
alphabetical order (e.g., names).

Binary Search Pseudocode
…
if(size == 0)
found = false;
else {
middle = index of approximate midpoint of array segment;
if(target == a[middle])
target has been found!
else if(target < a[middle])
search for target in area before midpoint;
else
search for target in area after midpoint;
}
…

Binary Search
[ 0 ] [ 1 ]
Example: sorted array of integer keys. Target=7.
3 6 7 11 32 33 53
[ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ]

Binary Search
[ 0 ] [ 1 ]
3 6 7 11 32 33 53
[ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ]
Find approximate midpoint

Binary Search
[ 0 ] [ 1 ]
3 6 7 11 32 33 53
[ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ]
Is 7 = midpoint key? NO.

Binary Search
[ 0 ] [ 1 ]
3 6 7 11 32 33 53
[ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ]
Is 7 < midpoint key? YES.

Binary Search
[ 0 ] [ 1 ]
3 6 7 11 32 33 53
[ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ]
Search for the target in the area before midpoint.

Binary Search
[ 0 ] [ 1 ]
3 6 7 11 32 33 53
[ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ]
Target = key of midpoint? NO.

Binary Search
[ 0 ] [ 1 ]
3 6 7 11 32 33 53
[ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ]
Target < key of midpoint? NO.

Binary Search
[ 0 ] [ 1 ]
3 6 7 11 32 33 53
[ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ]
Target > key of midpoint? YES.

Binary Search
[ 0 ] [ 1 ]
3 6 7 11 32 33 53
[ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ]
Search for the target in the area after midpoint.

Binary Search
[ 0 ] [ 1 ]
3 6 7 11 32 33 53
[ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ]
Find approximate midpoint.
Is target = midpoint key? YES.

Binary Search Implementation
void search(const int a[ ], size_t first, size_t size, int target, bool& found, size_t& location)
{
size_t middle;
if(size == 0) found = false;
else {
middle = first + size/2;
if(target == a[middle]){
location = middle;
found = true;
}
else if (target < a[middle])
// target is less than middle, so search subarray before middle
search(a, first, size/2, target, found, location);
else
// target is greater than middle, so search subarray after middle
search(a, middle+1, (size-1)/2, target, found, location);
}
}

Relation to Binary Search Tree
Corresponding complete binary search tree
3 6 7 11 32 33 53
3
6
7
11
32
33
53
Array of previous example:

Search for target = 7
Start at root:
Find midpoint:
3 6 7 11 32 33 53
3
6
7
11
32
33
53

Search left subarray:
Search left subtree:
3 6 7 11 32 33 53
3
6
7
11
32
33
53

Find approximate midpoint of subarray:
Visit root of subtree:
3 6 7 11 32 33 53
3
6
7
11
32
33
53

Search right subarray:
Search right subtree:
3 6 7 11 32 33 53
3
6
7
11
32
33
53

Binary Search: Analysis
 Worst case complexity?
 What is the maximum depth of recursive calls in
binary search as function of n?
 Each level in the recursion, we split the array in half
(divide by two).
 Therefore maximum recursion depth is floor(log2n)
and worst case = O(log2n).
 Average case is also = O(log2n).

Can we do better than O(log2n)?
 Average and worst case of serial search = O(n)
 Average and worst case of binary search =
O(log2n)
 Can we do better than this?
YES. Use a hash table!

Linear and Binary search

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