Programming assignments at the master’s level can be quite challenging. Students are often tasked with complex problems that require a deep understanding of algorithms, data structures, and advanced programming techniques. At programminghomeworkhelp.com, we specialize in providing expert assistance with such assignments, ensuring that students in Australia and beyond can excel in their academic pursuits. In this blog post, we will explore two master-level C programming questions, complete with detailed solutions, showcasing how our expert help can make a difference.
Question 1: Advanced Data Structures – Implementing a Balanced Binary Search Tree
Problem Statement:
Design and implement a balanced binary search tree (BST) in C. Your implementation should include the following features:
- Insertion of nodes while maintaining the balance of the tree.
- Deletion of nodes with rebalancing.
- An in-order traversal method to display the tree’s contents.
- A method to search for a specific value.
The balanced BST should ensure optimal performance for insertion, deletion, and search operations, aiming for a time complexity of O(log n).
Solution:
To address this problem, we will use an AVL Tree, which is a type of self-balancing binary search tree. The AVL Tree automatically maintains balance through rotations during insertion and deletion operations. Below is a C implementation of an AVL Tree with the required features:
#include <stdio.h>
#include <stdlib.h>
// Node structure
typedef struct Node {
int key;
struct Node *left, *right;
int height;
} Node;
// Utility function to get the height of the tree
int height(Node *N) {
if (N == NULL)
return 0;
return N->height;
}
// Utility function to get the maximum of two integers
int max(int a, int b) {
return (a > b) ? a : b;
}
// Utility function to create a new node
Node* newNode(int key) {
Node* node = (Node*) malloc(sizeof(Node));
node->key = key;
node->left = NULL;
node->right = NULL;
node->height = 1;
return(node);
}
// Right rotate utility function
Node* rightRotate(Node *y) {
Node *x = y->left;
Node *T2 = x->right;
// Perform rotation
x->right = y;
y->left = T2;
// Update heights
y->height = max(height(y->left), height(y->right)) + 1;
x->height = max(height(x->left), height(x->right)) + 1;
// Return new root
return x;
}
// Left rotate utility function
Node* leftRotate(Node *x) {
Node *y = x->right;
Node *T2 = y->left;
// Perform rotation
y->left = x;
x->right = T2;
// Update heights
x->height = max(height(x->left), height(x->right)) + 1;
y->height = max(height(y->left), height(y->right)) + 1;
// Return new root
return y;
}
// Get balance factor of node
int getBalance(Node *N) {
if (N == NULL)
return 0;
return height(N->left) - height(N->right);
}
// Insert a node and balance the tree
Node* insert(Node* node, int key) {
// Perform the normal BST insert
if (node == NULL)
return(newNode(key));
if (key < node->key)
node->left = insert(node->left, key);
else if (key > node->key)
node->right = insert(node->right, key);
else
return node; // Duplicates are not allowed
// Update height of this ancestor node
node->height = 1 + max(height(node->left), height(node->right));
// Get the balance factor of this ancestor node
int balance = getBalance(node);
// If this node becomes unbalanced, then there are 4 cases
// Left Left Case
if (balance > 1 && key < node->left->key)
return rightRotate(node);
// Right Right Case
if (balance < -1 && key > node->right->key)
return leftRotate(node);
// Left Right Case
if (balance > 1 && key > node->left->key) {
node->left = leftRotate(node->left);
return rightRotate(node);
}
// Right Left Case
if (balance < -1 && key < node->right->key) {
node->right = rightRotate(node->right);
return leftRotate(node);
}
// Return the (unchanged) node pointer
return node;
}
// In-order traversal of the tree
void inOrder(Node *root) {
if (root != NULL) {
inOrder(root->left);
printf("%d ", root->key);
inOrder(root->right);
}
}
// Search for a node in the tree
Node* search(Node* root, int key) {
if (root == NULL || root->key == key)
return root;
if (root->key < key)
return search(root->right, key);
return search(root->left, key);
}
// Driver program
int main() {
Node *root = NULL;
// Constructing the AVL tree
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 30);
root = insert(root, 40);
root = insert(root, 50);
root = insert(root, 25);
// Print the in-order traversal of the AVL tree
printf("In-order traversal of the AVL tree: ");
inOrder(root);
printf("\");
// Search for a value
Node *result = search(root, 30);
if (result != NULL)
printf("Found node with key %d\", result->key);
else
printf("Node not found\");
return 0;
}
This implementation covers insertion, balancing, traversal, and searching within an AVL Tree. It ensures efficient operations, demonstrating the practical application of advanced data structures. For students needing C assignment help in Australia, our expert writers can offer customized solutions similar to the one above, tailored to your specific needs.
Question 2: Complex Algorithm Design – Solving the Traveling Salesman Problem (TSP) with Dynamic Programming
Problem Statement:
The Traveling Salesman Problem (TSP) is a classic problem in combinatorial optimization. Given a list of cities and the distances between each pair of cities, determine the shortest possible route that visits each city exactly once and returns to the origin city.
Implement a dynamic programming approach to solve TSP in C. Your solution should include:
- A function to compute the minimum cost path visiting all cities.
- A function to print the optimal route.
Solution:
The dynamic programming approach to solve the TSP involves using a bitmask to represent subsets of cities and an array to store the minimum costs. Below is a C implementation of the dynamic programming solution for TSP:
#include <stdio.h>
#include <limits.h>
// Define the number of cities
#define N 4
// Initialize distance matrix
int dist[N][N] = {
{0, 10, 15, 20},
{10, 0, 35, 25},
{15, 35, 0, 30},
{20, 25, 30, 0}
};
// Initialize memoization table
int dp[1 << N][N];
int path[1 << N][N];
// Function to solve the TSP problem using dynamic programming
int tsp(int mask, int pos) {
if (mask == (1 << N) - 1) // All cities visited
return dist[pos][0]; // Return to the starting city
if (dp[mask][pos] != -1)
return dp[mask][pos];
int ans = INT_MAX;
for (int city = 0; city < N; city++) {
if ((mask & (1 << city)) == 0) { // If city not visited
int newAns = dist[pos][city] + tsp(mask | (1 << city), city);
if (newAns < ans) {
ans = newAns;
path[mask][pos] = city;
}
}
}
return dp[mask][pos] = ans;
}
// Function to print the optimal path
void printPath() {
int mask = 0;
int pos = 0;
printf("Optimal path: 0 ");
for (int i = 1; i < N; i++) {
int nextCity = path[mask][pos];
printf("-> %d ", nextCity);
mask |= (1 << nextCity);
pos = nextCity;
}
printf("-> 0\");
}
// Driver program
int main() {
// Initialize dp and path tables
for (int i = 0; i < (1 << N); i++) {
for (int j = 0; j < N; j++) {
dp[i][j] = -1;
path[i][j] = -1;
}
}
// Calculate minimum cost
int minCost = tsp(1, 0);
printf("Minimum cost: %d\", minCost);
// Print optimal path
printPath();
return 0;
}
This solution uses dynamic programming to efficiently compute the shortest path for the TSP problem, optimizing the route through memoization. Our C assignment help in Australia covers such advanced topics, offering detailed explanations and customized code that aligns with academic requirements.
Conclusion
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