# Writing programs that solve the Programming Projects helps to solidify your understanding

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Programming Projects

Writing programs that solve the Programming Projects helps to solidify your understanding

of the material and demonstrates how the chapter’s concepts are applied.

(As noted in the Introduction, qualified instructors may obtain completed solutions

to the Programming Projects on the publisher’s Web site.)

3.1 In the bubbleSort.java program (Listing 3.1) and the BubbleSort Workshop

applet, the in index always goes from left to right, finding the largest item and

carrying it toward out on the right. Modify the bubbleSort() method so that it’s

bidirectional. This means the in index will first carry the largest item from left

to right as before, but when it reaches out, it will reverse and carry the smallest

item from right to left. You’ll need two outer indexes, one on the right (the old

out) and another on the left.

3.2 Add a method called median() to the ArrayIns class in the insertSort.java

program (Listing 3.3). This method should return the median value in the

array. (Recall that in a group of numbers half are larger than the median and

half are smaller.) Do it the easy way.

3.3 To the insertSort.java program (Listing 3.3), add a method called noDups() that

removes duplicates from a previously sorted array without disrupting the order.

(You can use the insertionSort() method to sort the data, or you can simply

use main() to insert the data in sorted order.) One can imagine schemes in

which all the items from the place where a duplicate was discovered to the end

of the array would be shifted down one space every time a duplicate was

discovered, but this would lead to slow O(N2) time, at least when there were a

lot of duplicates. In your algorithm, make sure no item is moved more than

once, no matter how many duplicates there are. This will give you an algorithm

with O(N) time.

3.4 Another simple sort is the odd-even sort. The idea is to repeatedly make two

passes through the array. On the first pass you look at all the pairs of items,

a[j] and a[j+1], where j is odd (j = 1, 3, 5, …). If their key values are out of

order, you swap them. On the second pass you do the same for all the even

values (j = 2, 4, 6, …). You do these two passes repeatedly until the array is

sorted. Replace the bubbleSort() method in bubbleSort.java (Listing 3.1) with

an oddEvenSort() method. Make sure it works for varying amounts of data.

You’ll need to figure out how many times to do the two passes.

The odd-even sort is actually useful in a multiprocessing environment, where a

separate processor can operate on each odd pair simultaneously and then on

each even pair. Because the odd pairs are independent of each other, each pair

can be checked—and swapped, if necessary—by a different processor. This

makes for a very fast sort.

112 CHAPTER 3 Simple Sorting

3.5 Modify the insertionSort() method in insertSort.java (Listing 3.3) so it counts

the number of copies and the number of comparisons it makes during a sort

and displays the totals. To count comparisons, you’ll need to break up the

double condition in the inner while loop. Use this program to measure the

number of copies and comparisons for different amounts of inversely sorted

data. Do the results verify O(N2) efficiency? Do the same for almost-sorted data

(only a few items out of place). What can you deduce about the efficiency of

this algorithm for almost-sorted data?

3.6 Here’s an interesting way to remove duplicates from an array. The insertion sort

uses a loop-within-a-loop algorithm that compares every item in the array with

every other item. If you want to remove duplicates, this is one way to start.

(See also Exercise 2.6 in Chapter 2.) Modify the insertionSort() method in the

insertSort.java program so that it removes duplicates as it sorts. Here’s one

approach: When a duplicate is found, write over one of the duplicated items

with a key value less than any normally used (such as –1, if all the normal keys

are positive). Then the normal insertion sort algorithm, treating this new key

like any other item, will put it at index 0. From now on the algorithm can

ignore this item. The next duplicate will go at index 1, and so on. When the

sort is finished, all the removed dups (now represented by –1 values) will be

found at the beginning of the array. The array can then be resized and shifted

down so it starts at 0.

And here are the respective instruction given by teacher to follow

3_1

Develop bidirectional method bidiBubbleSort( ).

The main( ) may look like:

public static void main(String[] args)

{

int maxSize = 100;           // array size

ArrayBub arr;                // reference to array

arr = new ArrayBub(maxSize); // create the array

arr.insert(7);               // insert 7 items

arr.insert(6);

arr.insert(5);

arr.insert(4);

arr.insert(3);

arr.insert(2);

arr.insert(1);

arr.display();               // display items

arr.bidiBubbleSort();        // bidirectional bubble sort

arr.display();               // display them again

}  // end main()

The output may look like:

7 6 5 4 3 2 1

1 2 3 4 5 6 7

3_2

It is a simple project and will give you minimum points.

Define the median( ) method.

The main( ) may look like:

public static void main(String[] args)

{

int maxSize = 100;            // array size

ArrayIns arr;                 // reference to array

arr = new ArrayIns(maxSize);  // create the array

arr.insert(77);               // insert 10 items

arr.insert(99);

arr.insert(44);

arr.insert(55);

arr.insert(22);

arr.insert(88);

arr.insert(11);

arr.insert(00);

arr.insert(66);

arr.insert(33);               // even number of elems

arr.display();                // display items

long med = arr.median();      // find median

System.out.println(“Median is ” + med);  // show median

arr.insert(109);              // odd number of elems

med = arr.median();           // find median

arr.display();                // display items

System.out.println(“Median is ” + med);  // show median

}  // end main()

The output may look like:

77 99 44 55 22 88 11 0 66 33

Median is 55

0 11 22 33 44 55 66 77 88 99 109

Median is 55

3_3

Define the method noDups( ).

The main( ) may look like:

public static void main(String[] args)

{

int maxSize = 100;            // array size

ArrayIns arr;                 // reference to array

arr = new ArrayIns(maxSize);  // create the array

arr.insert(12);

arr.insert(12);

arr.insert(13);

arr.insert(13);

arr.insert(15);

arr.insert(27);

arr.insert(27);

arr.insert(27);

arr.insert(27);

arr.insert(32);

arr.insert(33);

arr.insert(34);

arr.insert(44);

arr.insert(44);

arr.insert(55);

arr.insert(56);

arr.insert(57);

arr.insert(57);

arr.display();                // display array

arr.noDups();                 // remove duplicates

arr.display();                // display it again

}  // end main()

The output may look like:

12 12 13 13 15 27 27 27 27 32 33 34 44 44 55 56 57 57

12 13 15 27 32 33 34 44 55 56 57

3_4

Develop the method oddEvenSort( ).  In the textbook in the description there is a small mistake (slip pen): even indexes are 2, 4, … .  But should be 0, 2, 4, … .

You may use an outer loop. The limit on the outer loop is related to nElems like this:

nElems 1 2 3 4 5 6

k <      1 1 2 2 3 3

The main( ) method may look like:

public static void main(String[] args)

{

int maxSize = 14;             // array size

ArrayIns arr;                 // reference to array

arr = new ArrayIns(maxSize);  // create the array

arr.insert(81);

arr.insert(77);

arr.insert(99);

arr.insert(44);

arr.insert(55);

arr.insert(22);

arr.insert(88);

arr.insert(77);

arr.insert(11);

arr.insert(00);

arr.insert(44);

arr.insert(66);

arr.insert(33);

arr.insert(33);

arr.display();                // display items

arr.oddEvenSort();            // sort them

arr.display();                // display them again

}  // end main()

The output may look like:

81 77 99 44 55 22 88 77 11 0 44 66 33 33

0 11 22 33 33 44 44 55 66 77 77 81 88 99

See additionally the output that traces odd-even code:

81 77 99 44 55 22 88 77 11 0 44 66 33 -1 -2

81 77 99 44 55 22 88 11 77 0 44 33 66 -2 -1

77 81 44 99 22 55 11 88 0 77 33 44 -2 66 -1

77 44 81 22 99 11 55 0 88 33 77 -2 44 -1 66

44 77 22 81 11 99 0 55 33 88 -2 77 -1 44 66

44 22 77 11 81 0 99 33 55 -2 88 -1 77 44 66

22 44 11 77 0 81 33 99 -2 55 -1 88 44 77 66

22 11 44 0 77 33 81 -2 99 -1 55 44 88 66 77

11 22 0 44 33 77 -2 81 -1 99 44 55 66 88 77

11 0 22 33 44 -2 77 -1 81 44 99 55 66 77 88

0 11 22 33 -2 44 -1 77 44 81 55 99 66 77 88

0 11 22 -2 33 -1 44 44 77 55 81 66 99 77 88

0 11 -2 22 -1 33 44 44 55 77 66 81 77 99 88

0 -2 11 -1 22 33 44 44 55 66 77 77 81 88 99

-2 0 -1 11 22 33 44 44 55 66 77 77 81 88 99

-2 -1 0 11 22 33 44 44 55 66 77 77 81 88 99

-2 -1 0 11 22 33 44 44 55 66 77 77 81 88 99

You can write a similar code, but it is optional.

3_5

public void put(int index, long value)  // insert at index

and modify the method insertionSort( ).

The main( ) method may look like:

public static void main(String[] args)

{

int maxSize = 25;             // array size

ArrayIns arr;                 // reference to array

arr = new ArrayIns(maxSize);  // create the array

for(int j=0; j<maxSize; j++)  // insert in-order items

arr.insert(j);

arr.put(10, 7);               // out of order item at 10

arr.put(20, 13);              // out of order item at 20

arr.display();                // display items

arr.insertionSort();          // insertion-sort them

arr.display();

}  // end main()

The output may look like:

0 1 2 3 4 5 6 7 8 9 7 11 12 13 14 15 16 17 18 19 13 21 22 23 24

Copies=58, comparisons=34

0 1 2 3 4 5 6 7 7 8 9 11 12 13 13 14 15 16 17 18 19 21 22 23 24

3_6

Modify insertion sort to remove duplicates.

/* Just after the normal comparison (top of while loop), the two

elements are also compared to see if they’re duplicates. If so,

the one in ‘temp’ is replaced by a sentinal, which has a value

lower than any key. On exit from the while loop, this

sentinal will end up inserted at the lowest index, following

the shifts of all the larger items to the right.

Now we no longer end the while loop when ‘in’ gets to 1, but when

it gets to ‘start’, which records the number of duplicates found

so far. At the buttom of the for loop, we check if there is

a sentinal at the lowest index (‘start’). If so, we bump

up ‘start’, which makes the sentinal invisible to the rest

of the algorithm. This will happen whenever there’s a duplicate,

so the array will shrink from the bottom up. The smaller the

array, the faster the sorting algorithm runs.

At the end of this method we shift the cells left (writing over

the sentinals), so the array again starts at 0. (One can imagine

situations where this shift would not be necessary).

*/

The main( ) may look like:

public static void main(String[] args)

{

int maxSize = 100;            // array size

ArrayIns arr;                 // reference to array

arr = new ArrayIns(maxSize);  // create the array

arr.insert(77);

arr.insert(77);

arr.insert(99);

arr.insert(44);

arr.insert(55);

arr.insert(22);

arr.insert(88);

arr.insert(77);

arr.insert(11);

arr.insert(00);

arr.insert(44);

arr.insert(66);

arr.insert(33);

arr.insert(33);

arr.display();                // display array

arr.insertionSort();          // sort and de-dup it

arr.display();                // display it again

}  // end main()

The output may look like:

77 77 99 44 55 22 88 77 11 0 44 66 33 33

0 11 22 33 44 55

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