Submission details: Make a directory and name it, using your first and last names, applying the camel notation, for example haniGirgis. Please name the main file, where execution starts, as hw2.cpp. Compress your directory by the zip utility, and submit the compressed file using Blackboard Introduction Suppose you had a document (perhaps a newspaper article, or an unattributed manuscript for a book), and you were interested in knowing who wrote it. One way to try to determine the authorship of the anonymous document is by comparing properties of the anonymous document with properties of known documents, and seeing if there is enough similarity to make a judgment of authorship. Some simple properties one might use to distinguish different authors include: Vocabulary (ie. the set of words an author uses). Word frequencies (ie. the frequencies with which an author uses words). Bigram frequencies (ie the frequencies of two consecutive words). Bigram probabilities (ie. the probability that one word follows another word). Terminology A unigram is a sequence of words of length one (ie. a single word). A bigram is a sequence of words of length two. • The conditional probability of an event E2 given another event E1, written P(E2|E1), is the probability that E2 will occur given that event El has already occurred. We write p(w(k)]w[k-1)) for the conditional probability of a word w in position k, w(k), given the immediately preceding word, w(k-1). You determine the conditional probabilities by determining unigram counts (the number of times each word appears, written c(w{k}), bigram counts (the number of times each pair of words appears, written c(w(k-1) w(k)), and then dividing each bigram count by the unigram count of the first word in the bigram: P(WORD(k)|WORD(k-1) = c(WORD(k-1) WORD(k)) / c(WORD(k-1)) For example, if the word "time" occurs seven times in a text, and "time of occurs threetimes, then the probability of "of occurring after "time" is 3/7. Project description In this project, you will be determining conditional probabilities of bigrams. To do this, you will write a C program, which reads in a file of text and produces three output files, as described below.

Based on the requirements in the image make sure the code below matches the description and run, i have already done some code Make sure implement  INCORPORATE CODE BELOW

#include <stdio.h>
#include <stdlib.h>

//Represent a node of the singly linked list
struct node{
    int data;
    struct node *next;
};

//Represent the head and tail of the singly linked list
struct node *head, *tail = NULL;

//addNode() will add a new node to the list
void addNode(int data) {
    //Create a new node
    struct node *newNode = (struct node*)malloc(sizeof(struct node));
    newNode->data = data;
    newNode->next = NULL;

    //Checks if the list is empty
    if(head == NULL) {
        //If list is empty, both head and tail will point to new node
        head = newNode;
        tail = newNode;
    }
    else {
        //newNode will be added after tail such that tail's next will point to newNode
        tail->next = newNode;
        //newNode will become new tail of the list
        tail = newNode;
    }
}

//removeDuplicate() will remove duplicate nodes from the list
void removeDuplicate() {
    //Node current will point to head
    struct node *current = head, *index = NULL, *temp = NULL;

    if(head == NULL) {
        return;
    }
    else {
        while(current != NULL){
            //Node temp will point to previous node to index.
            temp = current;
            //Index will point to node next to current
            index = current->next;

            while(index != NULL) {
                //If current node's data is equal to index node's data
                if(current->data == index->data) {
                    //Here, index node is pointing to the node which is duplicate of current node
                    //Skips the duplicate node by pointing to next node
                    temp->next = index->next;
                }
                else {
                    //Temp will point to previous node of index.
                    temp = index;
                }
                index = index->next;
            }
            current = current->next;
        }
    }
}

//display() will display all the nodes present in the list
void display() {
    //Node current will point to head
    struct node *current = head;
    if(head == NULL) {
        printf("List is empty \n");
        return;
    }
    while(current != NULL) {
        //Prints each node by incrementing pointer
        printf("%d ", current->data);
        current = current->next;
    }
    printf("\n");
}

struct node* findMiddle(struct node* b, struct node* e)
{
    if(b == NULL)
        return NULL;
    struct node* slow = b;
    struct node* fast = b->next;
    while(fast != e)
    {
        fast=fast->next;
        if(fast != e)
        {
            slow = slow->next;
            fast = fast->next;
        }
    }
    return slow;
};

struct node* binarysearch(int searchItem)
{
    struct node* beg = head;
    struct node* end = NULL;
    do
    {
        struct node* middle = findMiddle(beg, end);
        if(middle == NULL)
            return middle;
        else if(middle->data == searchItem)
            return middle;
        else if(middle->data < searchItem)
            beg = middle->next;
        else
            end = middle;
    }while(end==NULL || end != beg);
    return NULL;
}

int main()
{
    //Adds data to the list
    addNode(1);
    addNode(2);
    addNode(3);
    addNode(2);
    addNode(2);
    addNode(4);
    addNode(1);

    printf("Originals list: \n");
    display();

    //Removes duplicate nodes
    removeDuplicate();

    printf("List after removing duplicates: \n");
    display();

    printf("Search for 3 in the linked list:\n");
    struct node* n = binarysearch(3);
    if(n==NULL)
    {
        printf("Item not Found\n");
    }
    else
    {
        printf("Item Found\n");
        printf("%d\n",n->data);
    }
    printf("Search for 7 in the linked list:\n");
    n = binarysearch(7);
    if(n==NULL)
    {
        printf("Item not Found\n");
    }
    else
    {
        printf("Item Found\n");
        printf("%d",n->data);
    }
    return 0;
}