When data items are stored in a collection such as a list, we say that they have a linear or sequential relationship. Each data item is stored in a position relative to the others. In Java arrays and ArrayLists, these relative positions are the index values of the individual items. Since these index values are ordered, it is possible for us to visit them in sequence. This process gives rise to our first search technique, the sequential search.

Figure 5.3.1 shows that if we start at the first item in the list and move from item to item, we follow the underlying sequential order until we either find what we are looking for or run out of items, which would mean the item we were searching for was not present.

The Java implementation for this algorithm is shown in the following program. Listing 5.3.2 The function needs two items–the array and the item we are looking for–and returns a boolean value as to whether it is present.

Subsection5.3.1Analysis 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.

If the item is not in the list, the only way to know that 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.

What about the average case? On average, we will find the item about half way 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 5.3.3 summarizes these results.

Table5.3.3.Comparisons Used in a Sequential Search of an Unordered List

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?

Assume that the list of items was constructed so that the items are 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 5.3.4 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 5.3.5 The following program shows this variation of the sequential search function.

Table 5.3.6 summarizes these results. Note that in the best case we might discover that the item is not in the list 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 list only in the case where we do not find the item.

Table5.3.6.Comparisons Used in Sequential Search of an Ordered List

Case

Best Case

Worst Case

Average Case

item is present

\(1\)

\(n\)

\(\frac{n}{2}\)

item not present

\(1\)

\(n\)

\(\frac{n}{2}\)

ExercisesSelf Check

1.

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?

5

Five comparisons would get the second 18 in the list.

10

You do not need to search the entire list, only until you find the key you are looking for.

4

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

In this case only 2 comparisons were needed to find the key.

2.

Suppose you are doing a sequential search of the ordered list [3, 5, 6, 8, 11, 12, 14, 15, 17, 18]. How many comparisons would you need to do in order to find the key 13?

10

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

Since 11 is less than the key value 13 you need to keep searching.

7

Since 14 is greater than the key value 13 you can stop.

6

Because 12 is less than the key value 13 you need to keep going.