Question PLEASE READ MY POST AND QUESTION BEFORE POSTING. ONLY C++ NOT JAVA. Also I need this program to run and output all valid words. Program NEEDS TO ATTACH TEXT FILE WORDS.TXT, that part of the question. You can make up a word.txt with 10 random words to test program, but please ADD WORDS.TXT. Thank you. 13. A popular word game involves finding words from a grid of randomly generated letters. Words must be at least three letters long and formed from adjoining letters. Letters may not be reused and it is valid to move across diagonals. As an example, consider the following 4×4 grid of letters: ER 13 Recursion The word "FAB" is valid (letters in the upper left corner) and the word "KNIFE" is valid. The word "BABE" is not valid because the "B" may not be reused. The word "MINE" is not valid because the " E " is not adjacent to the "N". Write a program that uses a 4∗4 two-dimensional array to represent the game board. The program should randomly select letters for the board. You may wish to select vowels with a higher probability than consonants. You may also wish to always place a " U " next to a " Q " or to treat " QU " as a single letter. The program should read the words from the text file words. txt (included on the website with this book) and then use a recursive algorithm to determine if the word may be formed from the letters on the game board. The program should output all valid words from the file that are on the game board.

Here is your answer Run the code below   // C++ program for Boggle game #include <cstring> #include <iostream> using namespace std;    #define M 3 #define N 3    // Let the given dictionary be following string dictionary[] = { "GEEKS", "FOR", "QUIZ", "GO" }; int n = sizeof(dictionary) / sizeof(dictionary[0]);    // A given function to check if a given string is present in // dictionary. The implementation is naive for simplicity. As // per the question dictionary is given to us. bool isWord(string& str) {     // Linearly search all words     for (int i = 0; i < n; i++)         if (str.compare(dictionary[i]) == 0)             return true;     return false; }    // A recursive function to print all words present on boggle void findWordsUtil(char boggle[M][N], bool visited[M][N], int i,                    int j, string& str) {     // Mark current cell as visited and append current character     // to str     visited[i][j] = true;     str = str + boggle[i][j];        // If str is present in dictionary, then print it     if (isWord(str))         cout << str << endl;        // Traverse 8 adjacent cells of boggle[i][j]     for (int row = i - 1; row <= i + 1 && row < M; row++)         for (int col = j - 1; col <= j + 1 && col < N; col++)             if (row >= 0 && col >= 0 && !visited[row][col])                 findWordsUtil(boggle, visited, row, col, str);        // Erase current character from string and mark visited     // of current cell as false     str.erase(str.length() - 1);     visited[i][j] = false; }    // Prints all words present in dictionary. void findWords(char boggle[M][N]) {     // Mark all characters as not visited     bool visited[M][N] = { { false } };        // Initialize current string     string str = "";        // Consider every character and look for all words     // starting with this character     for (int i = 0; i < M; i++)         for (int j = 0; j < N; j++)             findWordsUtil(boggle, visited, i, j, str); }    // Driver program to test above function int main() {     char boggle[M][N] = { { 'G', 'I', 'Z' },                           { 'U', 'E', 'K' },                           { 'Q', 'S', 'E' } };        cout << "Following words of dictionary are presentn";     findWords(boggle);     return 0; }       Note that the above solution may print the same word multiple times. For example, if we add “SEEK” to the dictionary, it is printed multiple times. To avoid this, we can use hashing to keep track of all printed words. In below set 2, I changed this Trie based optimized solution:       // C++ program for Boggle game #include <bits/stdc++.h> using namespace std;    // Converts key current character into index // use only 'A' through 'Z' #define char_int(c) ((int)c - (int)'A')    // Alphabet size #define SIZE (26)    #define M 3 #define N 3    // trie Node struct TrieNode {     TrieNode* Child[SIZE];        // isLeaf is true if the node represents     // end of a word     bool leaf; };    // Returns new trie node (initialized to NULLs) TrieNode* getNode() {     TrieNode* newNode = new TrieNode;     newNode->leaf = false;     for (int i = 0; i < SIZE; i++)         newNode->Child[i] = NULL;     return newNode; }    // If not present, inserts a key into the trie // If the key is a prefix of trie node, just // marks leaf node void insert(TrieNode* root, char* Key) {     int n = strlen(Key);     TrieNode* pChild = root;        for (int i = 0; i < n; i++) {         int index = char_int(Key[i]);            if (pChild->Child[index] == NULL)             pChild->Child[index] = getNode();            pChild = pChild->Child[index];     }        // make last node as leaf node     pChild->leaf = true; }    // function to check that current location // (i and j) is in matrix range bool isSafe(int i, int j, bool visited[M][N]) {     return (i >= 0 && i < M && j >= 0 && j < N && !visited[i][j]); }    // A recursive function to print all words present on boggle void searchWord(TrieNode* root, char boggle[M][N], int i,                 int j, bool visited[][N], string str) {     // if we found word in trie / dictionary     if (root->leaf == true)         cout << str << endl;        // If both I and j in  range and we visited     // that element of matrix first time     if (isSafe(i, j, visited)) {         // make it visited         visited[i][j] = true;            // traverse all childs of current root         for (int K = 0; K < SIZE; K++) {             if (root->Child[K] != NULL) {                 // current character                 char ch = (char)K + (char)'A';                    // Recursively search reaming character of word                 // in trie for all 8 adjacent cells of boggle[i][j]                 if (isSafe(i + 1, j + 1, visited)                     && boggle[i + 1][j + 1] == ch)                     searchWord(root->Child[K], boggle,                                i + 1, j + 1, visited, str + ch);                 if (isSafe(i, j + 1, visited)                     && boggle[i][j + 1] == ch)                     searchWord(root->Child[K], boggle,                                i, j + 1, visited, str + ch);                 if (isSafe(i - 1, j + 1, visited)                     && boggle[i - 1][j + 1] == ch)                     searchWord(root->Child[K], boggle,                                i - 1, j + 1, visited, str + ch);                 if (isSafe(i + 1, j, visited)                     && boggle[i + 1][j] == ch)                     searchWord(root->Child[K], boggle,                                i + 1, j, visited, str + ch);                 if (isSafe(i + 1, j - 1, visited)                     && boggle[i + 1][j - 1] == ch)                     searchWord(root->Child[K], boggle,                                i + 1, j - 1, visited, str + ch);                 if (isSafe(i, j - 1, visited)                     && boggle[i][j - 1] == ch)                     searchWord(root->Child[K], boggle,                                i, j - 1, visited, str + ch);                 if (isSafe(i - 1, j - 1, visited)                     && boggle[i - 1][j - 1] == ch)                     searchWord(root->Child[K], boggle,                                i - 1, j - 1, visited, str + ch);                 if (isSafe(i - 1, j, visited)                     && boggle[i - 1][j] == ch)                     searchWord(root->Child[K], boggle,                                i - 1, j, visited, str + ch);             }         }            // make current element unvisited         visited[i][j] = false;     } }    // Prints all words present in dictionary. void findWords(char boggle[M][N], TrieNode* root) {     // Mark all characters as not visited     bool visited[M][N];     memset(visited, false, sizeof(visited));        TrieNode* pChild = root;        string str = "";        // traverse all matrix elements     for (int i = 0; i < M; i++) {         for (int j = 0; j < N; j++) {             // we start searching for word in dictionary             // if we found a character which is child             // of Trie root             if (pChild->Child[char_int(boggle[i][j])]) {                 str = str + boggle[i][j];                 searchWord(pChild->Child[char_int(boggle[i][j])],                            boggle, i, j, visited, str);                 str = "";             }         }     } }    // Driver program to test above function int main() {     // Let the given dictionary be following     char* dictionary[] = { "GEEKS", "FOR", "QUIZ", "GEE" };        // root Node of trie     TrieNode* root = getNode();        // insert all words of dictionary into trie     int n = sizeof(dictionary) / sizeof(dictionary[0]);     for (int i = 0; i < n; i++)         insert(root, dictionary[i]);        char boggle[M][N] = { { 'G', 'I', 'Z' },                           { 'U', 'E', 'K' },                           { 'Q', 'S', 'E' } };        findWords(boggle, root);        return 0; } tHANK YOU! ...