Construct a PDA that recognizes {w|w in {0,1|* where w = xy
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Construct a PDA that recognizes {w|w in {0,1|* where w = xy , with 2|y| < |x| < 3|y|, and x includes 0101 as a substring.
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- 2 pts 2/5 Animo Space Support 2D-DP-7 void mysteryFunction(int myList[][5], int r, int c) { int i, j, k, curr, temp; for ( (1) ) for (i = 0; i < c; i++) curr = i; for (j = i; j < c; j++) if(_(2)_) curr = j; temp = myList[k][i]; myList[k][i] = myList[k][curr]; myList[k][curr] = temp; int main() int i,j; int input[2][5] = { {7, 2, 3, 15, 8}, {56, 4, 12, 5, 9}}; mysteryFunction(input, 2, 5); for (i = 0; i < 2; i++) { for (j = 0; j < 5; j++) printf("%d ", input[i][j]); printf("\n"); return 0; Output: 2 378 15 459 12 56 Given the code segment above, what should be the content of _(2)__ in order to achieve the provided output? Assume the necessary libraries and functions are included and implemented.In this project, you will develop algorithms that find road routes through the bridges to travel between islands. The input is a text file containing data about the given map. Each file begins with the number of rows and columns in the map considered as maximum latitudes and maximum longitudes respectively on the map. The character "X" in the file represents the water that means if a cell contains "X" then the traveler is not allowed to occupy that cell as this car is not drivable on water. The character "0" in the file represents the road connected island. That means if a cell contains "0" then the traveler is allowed to occupy that cell as this car can drive on roads. The traveler starts at the island located at latitude = 0 and longitude = 0 (i.e., (0,0)) in the upper left comer, and the goal is to drive to the island located at (MaxLattitude-1, MaxLongitudes-1) in the lower right corner. A legal move from an island is to move left, right, up, or down to an immediately adjacent cell…1. Using the above class map, write function void map::mapToGraph(graph &g){...} to create a graph g that represents the legal moves in the map m. Each vertex should represent a cell, and each edge should represent a legal move between adjacent cells. 2. Write a recursive function findPathRecursive(graph &g, stack &moves) that looks for a path from the start island to the destination island. If a path from the start to the destination exists, your function should call the map::printPath() function that should print a sequence of correct moves (Go left, Go right, Go down, Go up, etc.). If no path from the start to the destination exists, the program should print, "No path exists". If a solution exists the solver should also simulate the solution to each map by calling the map::print() function. The map::print() function prints out a map visualization, with the goal and current position of the car in the map at each move, marked to show the progress. Hint: consider recursive-DFS. 3.…
- Q2: Design a PDA for accepting a language {0^1m0n | m, n>=1} give the ID form as we explain in the lecture. Apply the string "00011000" on this PDA using ID transitions.C++ Use the driver code typed out below to answer the question. // { Driver Code Starts #include <bits/stdc++.h>using namespace std; // } Driver Code Endsclass Solution { public: // Function to return a path vector consisting of vertex ids from vertex 0 to target vector <int> shortestPath(int V, vector adj[], int target) { // Enter code here! }}; // { Driver Code Starts.int main() { int tc; cin >> tc; while (tc--) { int V, E, target; cin >> V >> E >> target; vector adj[V]; for (int i = 0; i < E; i++) { int u, v; cin >> u >> v; adj[u].push_back(v); // adj[v].push_back(u); } // string s1; // cin>>s1; Solution obj; vector ans = obj. shortestPath(V, adj, target); for (int i = 0; i < ans.size(); i++) { cout << ans[i] << " "; } cout << endl; } return 0;} // } Driver Code Ends Given a directed graph. The task is complete the function vector <int> shortestPath(int V, vector…3. The class definition for sparse matrix is shown below. struct Triple{int row, col, value;};class Matrix; // forward declarationclass MatrixNode{friend class Matrix;friend istream& operator>>(istream&, Matrix&); // for reading in a matrixprivate:MatrixNode *down, *right;bool head; // a head node or notunion{ // anonymous unionMatrixNode *next;Triple triple;};MatrixNode(bool, Triple*); // constructor};MatrixNode::MatrixNode(bool b, Triple *t) // constructor{head = b;if (b){right = down = this; // row/column header node}elsetriple = *t; // element node or header node for list of header nodes}class Matrix{friend istream& operator>>(istream&, Matrix&);public:~Matrix(); // destructorprivate:MatrixNode *headnode;}; Based on this class, do the following tasks. Write the C++ function, operator+(const Matrix& b) const, which returns the matrix *this + b. Write the C++ function, operator*(const Matrix& b) const, which returns the matrix *this *…
- In questions 4-10 estimate the Big O value by analyzing the code. Note the algorithms are written in English. Hint: you are interested in the number of operations for each algorithm.∑ = {C,A,G,T}, L = { w : w = CAjGnTmC, m = j + n }. For example, CAGTTC ∈ L; CTAGTC ∉ L because the symbols are not in the order specified by the characteristic function; CAGTT ∉ L because it does not end with C; and CAGGTTC ∉ L because the number of T's do not equal the number of A's plus the number of G's. Prove that L ∉ RLs using the RL pumping theorem.IN C LANGUAGE I HAD AN ERROR. URGENT!! struct Graph* createGraph(int V, int E){ struct Graph* graph = new Graph; graph->V = V; graph->E = E; graph->edge = new Edge[E]; return graph;} // A structure to represent a subset for union-findstruct subset { int parent; int rank;}; // A utility function to find set of an element i// (uses path compression technique)int find(struct subset subsets[], int i){ // find root and make root as parent of i // (path compression) if (subsets[i].parent != i) subsets[i].parent = find(subsets, subsets[i].parent); return subsets[i].parent;} // A function that does the union of two sets of x and yvoid Union(struct subset subsets[], int x, int y){ int xroot = find(subsets, x); int yroot = find(subsets, y); // Attach smaller rank tree under root of high // rank tree (Union by Rank) if…
- Define multiFold which folds an element of MkType returning a tuple of type (a,b). You should define this as a right fold . multiFold:: MkType a b -> (c,d) -> (a -> c -> c) -> (b -> d -> d) -> (c,d)multiFold = undefined answer in haskell form. thank you.C PROGRAMMING Implement dijkstras alorithm Check that the Graph graph, and starting node, id, are valid• Create the set S containing all the networks (vertices) except the source node (you might wantto use an array for this.• Create an array to represent the table D and initialise it with the weights of the edges from thesource node, or infinity if no edge exists. You should use the constant DBL_MAX to representinfinity.• Create an array to represent the table R and initialise it with the next hops if an edge existsfrom the source, or 0 otherwise.• Then repeatedly follow the remaining rules of Dijkstra’s algorithm, updating the values in D andR until S is empty.• Each of the values required to complete the above can be found by calling the variousfunctions (get_vertices(), get_edge(), edge_destination(), edge_weight(), etc.)in the supplied graph library.• Once Dijkstra’s algorithm has run, you will need to create the routing table to be returned byallocating enough memory for the…Please convert the code to C language //separate chaining#include <bits/stdc++.h>using namespace std; class node {public: int data; node* next; node() { data = 0; next = NULL; } node(int x) { data = x; next = NULL; }}; node* add(node* head, int data){ if (head == NULL) { head = new node(data); return head; } node* temp = head; while (temp->next) { temp = temp->next; } temp->next = new node(data); return head;} void print(node* head){ if (!head) { cout << "NULL\n"; return; } node* temp = head; while (temp) { cout << temp->data << "->"; temp = temp->next; } cout << "NULL\n";} int main(){ //set of input numbers vector<int> arr{ 123, 456, 763, 656, 908, 238, 231 }; //initialize the hash table //each entry of the hash table is a linkedlist vector<node*> hash(10); //size of…