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Dear Students
Read latest announcement from CS619.
It is to inform you about the assignments interface on VULMS. All the students who are registering in the project for the first time in Fall 2011
Semester should ignore the assignments interface on VULMS.
The submission method is already explained in this file. After the completion of project and groups, you will get further information about
starting the project from same announcements page.
Remember Me in Your Prayers
Best regard's
Ch. Muhammad Afaaq (Arrien)
MBA (Finance)
Islamabad
Afaaq_Tariq@yahoo.com
0346-5329264
For latest assignments solved quizzes files GDB, Solve & Unsolved Past Papers come join us in
http://groups.google.com/group/vustudymania
If u like me than raise your hand with me
If not then raise ur standard
That's about me … !
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Remember Me in Your Prayers
Best regard's
Ch. Muhammad Afaaq (Arrien)
MBA (Finance)
Islamabad
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0346-5329264
For latest assignments solved quizzes files GDB, Solve & Unsolved Past Papers come join us in
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If u like me than raise your hand with me
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Friends now a day's viva & presentations start
and fellows need help am making a threat in ma site www.vubaba.com and 2470+ fellows visit that threat in few days means fellows can get help from there as per requirement and help purpose more viva & presentation experience share its now century complete 108+ fellows current viva experience its v useful for viva students links are updated daily
Given below in 10 different pages :)
http://www.vubaba.com/showthread.php?t=4701
http://www.vubaba.com/showthread.php?t=4701&page=2
http://www.vubaba.com/showthread.php?t=4701&page=3
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http://www.vubaba.com/showthread.php?t=4701&page=10
http://www.vubaba.com/showthread.php?t=4701&page=11
Remember Me in Your Prayers
Best regard's
Ch. Muhammad Afaaq (Arrien)
MBA (Finance)
Islamabad
Afaaq_Tariq@yahoo.com
0346-5329264
For latest assignments solved quizzes files GDB, Solve & Unsolved Past Papers come join us in
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If u like me than raise your hand with me
If not then raise ur standard
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Data Structures
Data Structures and Algorithm Analysis in C++ Chapter. 3
3.1, 3.2, 3.2.1, 3.2.2
1) List Implementation
· add Method
· next Method
· remove Method
· find Method
· Other Methods
2) Analysis Of Array List
3) List Using Linked Memory
4) Linked List
Today, we will discuss the concept of list operations. You may have a fair idea of ' start operation' that sets the current pointer to the first element of the list while the tail operation moves the current pointer to the last element of the list. In the previous lecture, we discussed the operation next that moves the current pointer one element forward. Similarly, there is the 'back operation' which moves the current pointer one element backward.
List Implementation
Now we will see what the implementation of the list is and how one can create a list in C++. After designing the interface for the list, it is advisable to know how to implement that interface. Suppose we want to create a list of integers. For this purpose, the methods of the list can be implemented with the use of an array inside. For example, the list of integers (2, 6, 8, 7, 1) can be represented in the following manner where the current position is 3.
A | 2 | 6 | 8 | 7 | 1 |
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| current | size |
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| 1 | 2 | 3 | 4 | 5 |
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| 3 | 5 |
In this case, we start the index of the array from 1 just for simplification against the usual practice in which the index of an array starts from zero in C++. It is not necessary to always start the indexing from zero. Sometimes, it is required to start the indexing from 1. For this, we leave the zeroth position and start using the array from index 1 that is actually the second position. Suppose we have to store the numbers from 1 to 6 in the array. We take an array of 7 elements and put the numbers from the index 1. Thus there is a correspondence between index and the numbers stored in it. This is not very useful. So, it does not justify the non-use of zeroth position of the array out-rightly. However for simplification purposes, it is good to use the index from 1.
add Method
Now we will talk about adding an element to the list. Suppose there is a call to add an element in the list i.e. add(9). As we said earlier that the current position is 3, so by adding the element 9 to the list, the new list will be (2, 6, 8, 9, 7, 1).
To add the new element (9) to the list at the current position, at first, we have to make space for this element. For this purpose, we shift every element on the right of 8 (the current position) to one place on the right. Thus after creating the space for new element at position 4, the array can be represented as
A | 2 | 6 | 8 |
| 7 | 1 |
|
| current | size |
|
| 1 | 2 | 3 | 4 | 5 |
|
|
| 3 | 5 |
Now in the second step, we put the element 9 at the empty space i.e. position 4. Thus the array will attain the following shape. The figure shows the elements in the array in the same order as stored in the list.
| A | 2 | 6 | 8 | 9 | 7 | 1 |
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| current | size |
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| 1 | 2 | 3 | 4 | 5 | 6 |
|
| 4 | 6 |
We have moved the current position to 4 while increasing the size to 6. The size shows that the elements in the list. Where as the size of the array is different that we have defined already to a fixed length, which may be 100, 200 or even greater.
next Method
Now let's see another method, called 'next'. We have talked that the next method moves the current position one position forward. In this method, we do not add a new element to the list but simply move the pointer one element ahead. This method is required while employing the list in our program and manipulating it according to the requirement. There is also an array to store the list in it. We also have two variables- current and size to store the position of current pointer and the number of elements in the list. By looking on the values of these variables, we can find the state of the list i.e. how many elements are in the list and at what position the current pointer is.
The method next is used to know about the boundary conditions of the list i.e. the array being used by us to implement the list. To understand the boundary conditions, we can take the example of an array of size 100 to implement the list. Here, 100 elements are added to the array. Let's see what happens when we want to add 101st element to the array? We used to move the current position by next method and reached the 100th position. Now, in case of moving the pointer to the next position (i.e. 101st), there will be an error as the size of the array is 100, having no position after this point. Similarly if we move the pointer backward and reach at the first position regardless that the index is 0 or 1. But what will happen if we want to move backward from the first position? These situations are known as boundary conditions and need attention during the process of writing programs when we write the code to use the list. We will take care of these things while implementing the list in C++ programs.
remove Method
We have seen that the add method adds an element in the list. Now we are going to discuss the remove method. The remove method removes the element residing at the current position. The removal of the element will be carried out as follows. Suppose there are 6 elements (2, 6, 8, 9, 7, 1) in the list. The current pointer is pointing to the position 5 that has the value 7. We remove the element, making the current position empty. The size of the list will become 5. This is represented in the following figure.
| A | 2 | 6 | 8 | 9 |
| 1 |
|
| current | size |
| | 1 | 2 | 3 | 4 | 5 | 6 |
|
| 5 | 6 5 |
We fill in the blank position left by the removal of 7 by shifting the values on the right of position 5 to the left by one space. This means that we shift the remaining elements on the right hand side of the current position one place to the left so that the element next to the removed element (i.e. 1) takes its place (the fifth position) and becomes the current position element. We do not change the current pointer that is still pointing to the position 5. Thus the current pointer remains pointing to the position 5 despite the fact that there is now element 1 at this place instead of 7. Thus in the remove method, when we remove an element, the element next to it on the right hand side comes at its place and the remaining are also shifted one place to the right. This step is represented by the following figure.
| A | 2 | 6 | 8 | 9 | 1 |
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| current | size |
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| 1 | 2 | 3 | 4 | 5 |
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| 5 | 5 |
find Method
Now lets talk about a function, used to find a specific element in the array. The find (x) function is used to find a specific element in the array. We pass the element, which is to be found, as an argument to the find function. This function then traverses the array until the specific element is found. If the element is found, this function sets the current position to it and returns 1 i.e. true. On the other hand, if the element is not found, the function returns 0 i.e. false. This indicates that the element was not found. Following is the code of this find(x) function in C++.
int find (int x)
{
int j ;
for (j = 1; j < size + 1; j++ )
if (A[j] == x )
break ;
if ( j < size + 1) // x is found
{
current = j ; //current points to the position where x found
return 1 ; // return true
}
return 0 ; //return false, x is not found
}
In the above code, we execute a for loop to traverse the array. The number of execution of this loop is equal to the size of the list. This for loop gets terminated when the value of loop variable (j) increases from the size of the list. However we terminate the loop with the break statement if the element is found at a position. When the control comes out from the loop, we check the value of j. If the value of j is less than the size of the array, it means that the loop was terminated by the break statement. We use the break statement when we find the required element (x) in the list. The execution of break statement shows that the required element was found at the position equal to the value of j. So the program sets the current position to j and comes out the function by returning 1 (i.e. true). If the value of j is greater than the size of the array, it means that the whole array has traversed and the required element is not found. So we simply return 0 (i.e. false) and come out of the function.
There are some other methods to implement the list using an array. These methods are very simple, which perform their task just in one step (i.e. in one statement). There is a get() method , used to get the element from the current position in the array. The syntax of this function is of one line and is as under
return A[current] ;
This statement returns the element to which the current is pointing to (i.e. the current position) in the list A.
Another function is update(x). This method is used to change (set) the value at the current position. A value is passed to this method as an argument. It puts that value at the current position. The following statement in this method carries out this process.
A [current] = x ;
Then there is a method length( ).This method returns the size of the list. The syntax of this method is
return size ;
You may notice here that we are returning the size of the list and not the size of the array being used internally to implement the list. This size is the number of the elements of the list, stored in the array.
The back() method decreases the value of variable current by 1. In other words, it moves the current position one element backward. This is done by writing the statement.
current -- ;
The -- is a decrement operator in C++ that decreases the value of the operand by one. The above statement can also be written as
current = current -1 ;
The start() method sets the current position to the first element of the list. We know that the index of the array starts from 0 but we use the index 1 for the starting position. We do not use the index zero. So we set the current position to the first element by writing
current = 1 ;
Similarly, the end() method sets the current position to the last element of the list i.e. size. So we write
current = size ;
Analysis of Array List
Now we analyze the implementation of the list while using an array internally. We analyze different methods used for the implementation of the list. We try to see the level upto which these are efficient in terms of CPU's time consumption. Time is the major factor to see the efficiency of a program.
First of all, we have talked about the add method. When we add an element to the list, every element is moved to the right of the current position to make space for the new element. So, if the current position is the start of the list and we want to add an element in the beginning, we have to shift all the elements of the list to the right one place. This is the worst case of adding an element to the list. Suppose if the size of the list is 10000 or 20000, we have to do the shift operation for all of these 10000 or 20000 elements. Normally, it is done by shifting of elements with the use of a for loop. This operation takes much time of the CPU and thus it is not a good practice to add an element at the beginning of a list. On the other hand, if we add an element at the end of the list, it can be done by carrying out 'no shift operation'. It is the best case of adding an element to the list. However, normally we may have to move half of the elements. The usage of add method is the matter warranting special care at the time of implementation of the list in our program. To provide the interface of the list, we just define these methods.
When we remove an element at the current position in the list, its space gets empty. The current pointer remains at the same position. To fill this space, we shift the elements on the right of this empty space one place to the left. If we remove an element from the beginning of the list, then we have to shift the entire remaining elements one place to the left. Suppose there is a large number of elements, say 10000 or 20000, in the list. We remove the first element from the list. Now to fill this space, the remaining elements are shifted (that is a large number). Shifting such a large number of elements is time consuming process. The CPU takes time to execute the for loop that performs this shift operation. Thus to remove an element at the beginning of the list is the worst case of remove method. However it is very easy to remove an element at the end of the list. In average cases of the remove method we expect to shift half of the elements. This average does not mean that in most of the cases, you will have to shift half the elements. It is just the average. We may have to shift all the elements in one operation (if we remove at the beginning) and in the second operation, we have to shift no element (if we remove at the end). Similarly, in certain operations, we have to shift just 10, 15 elements.
We have discussed that the find method takes an element and traverses the list to find that element. The worst case of the find method is that it has to search the entire list from beginning to end. So, it finds the element at the end of the array or the element is not found. On average the find method searches at most half the list.
The other methods get (), length () etc are one-step methods. They carry out their operation in one instruction. There is no need of any loop or other programming structures to perform the task. The get() method gets the value from the specified position just in one step. Similarly the update() method sets a value at the specific position just in one-step. The length () method returns the value of the size of the list. The other methods back(), start() and end() also perform their tasks only in one step.
List using Linked Memory
We have seen the implementation of the list with the use of an array. Now we will discuss the implementation of the list while using linked memory. In an array, the memory cells of the array are linked with each other. It means that the memory of the array is contiguous. In an array, it is impossible that one element of the array is located at a memory location while the other element is located somewhere far from it in the memory. It is located in very next location in the memory. It is a property of the array that its elements are placed together with one another in the memory. Moreover, when we have declared the size of the array, it is not possible to increase or decrease it during the execution of the program. If we need more elements to store in the array, there is need of changing its size in the declaration. We have to compile the program again before executing it. Now array will be of the new size. But what happens if we again need to store more elements? We will change the code of our program to change the declaration of the array while recompiling it.
Suppose we have used the dynamic memory allocation and created an array of 100 elements with the use of new operator. In case of need of 200 elements, we will release this array and allocate a new array of 200 elements. Before releasing the previous array, it will wise to copy its elements to the new array so that it does not lose any information. Now this new array is in 'ready for use' position. Thus the procedure of creating a new array is not an easy task.
To avoid such problems, usually faced by the programmers while using an array, there is need of using linked memory in which the various cells of memory, are not located continuously. In this process, each cell of the memory not only contains the value of the element but also the information where the next element of the list is residing in the memory. It is not necessary that the next element is at the next location in the memory. It may be anywhere in the memory. We have to keep a track of it. Thus, in this way, the first element must explicitly have the information about the location of the second element. Similarly, the second element must know where the third element is located and the third should know the position of the fourth element and so on. Thus, each cell (space) of the list will provide the value of the element along with the information about where the next element is in the memory. This information of the next element is accomplished by holding the memory address of the next element. The memory address can be understood as the index of the array. As in case of an array, we can access an element in the array by its index. Similarly, we can access a memory location by using its address, normally called memory address.
Linked List
For the utilization of the concept of linked memory, we usually define a structure, called linked list. To form a linked list, at first, we define a node. A node comprises two fields. i.e. the object field that holds the actual list element and the next that holds the starting location of the next node.
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object
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next |
A chain of these nodes forms a linked list. Now let's consider our previous list, used with an array i.e. 2, 6, 8, 7, 1. Following is the figure which represents the list stored as a linked list.
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| 2 |
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| 6 |
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| 8 |
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| 7 |
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| 1 |
| size = 5 |
current
This diagram just represents the linked list. In the memory, different nodes may occur at different locations but the next part of each node contains the address of the next node. Thus it forms a chain of nodes which we call a linked list.
While using an array we knew that the array started from index 1that means the first element of the list is at index 1. Similarly in the linked list we need to know the starting point of the list. For this purpose, we have a pointer head that points to the first node of the list. If we don't use head, it will not be possible to know the starting position of the list. We also have a pointer current to point to the current node of the list. We need this pointer to add or remove current node from the list. Here in the linked list, the current is a pointer and not an index as we used while using an array. The next field of the last node points to nothing .It is the end of the list. We place the memory address NULL in the last node. NULL is an invalid address and is inaccessible.
Now again consider the list 2, 6, 8, 7, 1. The previous figure represents this list as a linked list. In this linked list, the head points to 2, 2 points to 6, 6 points to 8, 8 points to 7 and 7 points to 1. Moreover we have the current position at element 8.
This linked list is stored in the memory. The following diagram depicts the process through which this linked list is stored in the memory.
| | 1051 | 6 |
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| 1052 | 1063 |
| current | 1053 |
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| 1054 | 2 |
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| 1055 | 1051 |
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| 1056 |
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| 1057 | 7 |
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| 1058 | 1060 |
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| 1059 |
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| 1060 | 1 |
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| 1061 | 0 |
| head | 1062 | 1054 |
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| 1063 | 8 |
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| 1064 | 1057 |
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| 1065 |
|
We can see in the figure that each memory location has an address. Normally in programming, we access the memory locations by some variable names. These variable names are alias for these locations and are like labels that are put to these memory locations. We use head and current variable names instead of using the memory address in numbers for starting and the current nodes. In the figure, we see that head is the name of the memory location 1062 and the name current is used for the memory address 1053. The head holds the address 1054 and the element 2, the first one in the list, is stored in the location 1054. Similarly current holds the address 1063 where the element 8 is stored which is our current position in the list. In the diagram, two memory locations comprise a node. So we see that the location 1054 holds the element 2 while the next location 1055 holds the address of the memory location (1051) where the next element of the list (i.e. 6) is stored. Similarly the next part of the node that has value 6 holds the memory address of the location occupied by the next element (i.e. 8) of the list. The other nodes are structured in a similar fashion. Thus, by knowing the address of the next element we can traverse the whole list.
