DS The Sequential Search

Section 6.3 The Sequential Search

When data items are stored in a container type such as a Python list or a C++ array/vector, we say that they have a linear or sequential relationship. Each data item is stored in a position relative to the others. In Python lists, these relative positions are the index values of the individual items. In C++ arrays, these are simply adjacent memory locations each equally sized to fit the data type of the container. Since these index values are ordered, it is possible for us to visit them in sequence. This process gives rise to our first searching technique, the sequential search.

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Figure 6.3.1 shows how this search works. Starting at the first item in the list, we simply move from item to item, following the underlying sequential ordering until we either find what we are looking for or run out of items. If we run out of items, we have discovered that the item we were searching for was not present.

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described in detail following the image — Figure 6.3.1. Sequential Search of a List of Integers.
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Both the Python and C++ implementations for this algorithm are shown in Task 6.3.1.b and Task 6.3.1.a> respectively. The function needs the list and the item we are looking for and returns a boolean value as to whether it is present. The boolean variable found is initialized to False and is assigned the value True if we discover the item in the list (or vector, in the case of C++).

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Exploration 6.3.1. Sequential Search.

(a) C++ Implementation.

#include <iostream>
#include <vector>
using namespace std;

// Checks to see if item is in a vector
// returns true or false (1 or 0)
// using sequential Search
bool sequentialSearch(vector<int> avector, int item) {
    unsigned int pos = 0;
    bool found = false;

    while (pos < avector.size() && !found) {
        if (avector[pos] == item) {
            found = true;
        } else {
            pos++;
        }
    }

    return found;
}

int main() {
    // Vector initialized using an array
    int arr[] = {1, 2, 32, 8, 17, 19, 42, 13, 0};
    vector<int> testvector(arr, arr + (sizeof(arr) / sizeof(arr[0])));

    cout << sequentialSearch(testvector, 3) << endl;
    cout << sequentialSearch(testvector, 13) << endl;

    return 0;
}

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(b) Python Implementation.

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Subsection 6.3.1 Analysis of Sequential Search

To analyze searching algorithms, we need to decide on a basic unit of computation. Recall that this is typically the common step that must be repeated in order to solve the problem. For searching, it makes sense to count the number of comparisons performed. Each comparison may or may not discover the item we are looking for. In addition, we make another assumption here: the list of items is not ordered in any way. The items have been placed randomly into the list. In other words, the probability that the item we are looking for is in any particular position is exactly the same for each position of the list.

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If the item is not in the list, the only way to know it is to compare it against every item present. If there are \(n\) items, then the sequential search requires \(n\) comparisons to discover that the item is not there. In the case where the item is in the list, the analysis is not so straightforward. There are actually three different scenarios that can occur. In the best case we will find the item in the first place we look, at the beginning of the list. We will need only one comparison. In the worst case, we will not discover the item until the very last comparison, the \(n^{th}\) comparison.

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What about the average case? On average, we will find the item about halfway into the list; that is, we will compare against \(\frac{n}{2}\) items. Recall, however, that as n gets large, the coefficients, no matter what they are, become insignificant in our approximation, so the complexity of the sequential search, is \(O(n)\text{.}\) Table 6.3.2 summarizes these results.

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Table 6.3.2. Comparisons Used in a Sequential Search of an Unordered List

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Case	Best Case	Worst Case	Average Case
item is present	\(1\)	\(n\)	\(\frac{n}{2}\)
item is not present	\(n\)	\(n\)	\(n\)

We assumed earlier that the items in our collection had been randomly placed so that there is no relative order between the items. What would happen to the sequential search if the items were ordered in some way? Would we be able to gain any efficiency in our search technique?

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Assume that the list of items was constructed so that the items were in ascending order, from low to high. If the item we are looking for is present in the list, the chance of it being in any one of the n positions is still the same as before. We will still have the same number of comparisons to find the item. However, if the item is not present there is a slight advantage. Figure 6.3.3 shows this process as the algorithm looks for the item 50. Notice that items are still compared in sequence until 54. At this point, however, we know something extra. Not only is 54 not the item we are looking for, but no other elements beyond 54 can work either since the list is sorted. In this case, the algorithm does not have to continue looking through all of the items to report that the item was not found. It can stop immediately. Listing 6.3.4 shows this variation of the sequential search function.

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Listing 6.3.4. Ordered Sequential Search

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#include <iostream>
#include <vector>
using namespace std;

// Checks to see if item is in a vector
// returns true or false (1 or 0)
// using ordered sequential Search
bool orderedSequentialSearch(vector<int> avector, int item) {
    unsigned int pos = 0;
    bool found = false;
    bool stop = false;
    while (pos < avector.size() && !found && !stop) {
        if (avector[pos] == item) {
            found = true;
        } else {
            if (avector[pos] > item) {
                stop = true;
            } else {
                pos++;
            }
        }
    }

    return found;
}

int main() {
    // Vector initialized using an array
    int arr[] = {0, 1, 2, 8, 13, 17, 19, 32, 42};
    vector<int> testvector(arr,arr+(sizeof(arr)/sizeof(arr[0])));

    cout << orderedSequentialSearch(testvector, 3) << endl;
    cout << orderedSequentialSearch(testvector, 13) << endl;

    return 0;
}

Table 6.3.5 summarizes these results. Note that in the best case we might discover that the item is not in the vector by looking at only one item. On average, we will know after looking through only \(\frac {n}{2}\) items. However, this technique is still \(O(n)\text{.}\) In summary, a sequential search is improved by ordering the vector only in the case where we do not find the item.

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Table 6.3.5. Comparisons Used in Sequential Search of an Ordered List

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Case	Best Case	Worst Case	Average Case
item is present	\(1\)	\(n\)	\(\frac{n}{2}\)
item not present	\(1\)	\(n\)	\(\frac{n}{2}\)

Reading Questions Reading Questions

1.

For the following unordered sequential list: {0, 1, 2, 13, 8, 19, 17, 32, 42}, searching for would produce the best case scenario, and searching for would produce the worst case scenario.

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2.

Suppose you are doing a sequential search of the list [15, 18, 2, 19, 18, 0, 8, 14, 19, 14]. How many comparisons would you need to do in order to find the key 18?

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5
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Five comparisons would get the second 18 in the list.
10
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You do not need to search the entire list, only until you find the key you are looking for.
4
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No, remember in a sequential search you start at the beginning and check each key until you find what you are looking for or exhaust the list.
2
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In this case only 2 comparisons were needed to find the key.

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3.

Suppose you are doing a sequential search using a program that is enhanced to handle ordered lists more efficiently. When passing the list [3, 5, 6, 8, 11, 12, 14, 15, 17, 18] as a parameter, how many comparisons would you need to do in order to find the key 13?

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10
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You do not need to search the entire list, since it is ordered you can stop searching when you have compared with a value larger than the key.
5
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Since 11 is less than the key value 13 you need to keep searching.
7
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Since 14 is greater than the key value 13 you can stop.
6
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Because 12 is less than the key value 13 you need to keep going.

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