answer:Here's the completed Dictionary class with the `put`, `get`, and `remove` methods: ```cpp #include<iostream> #include<string> using namespace std; // Node class representing each key-value pair in the dictionary class Node{ public: string key; int value; Node *next; Node(string k, int v){ key = k; value = v; next = nullptr; } }; // Dictionary class with separate chaining (linked lists) for collision resolution class Dictionary{ private: Node *table[128]; // Array of linked lists int size; // Current size of the dictionary // Simple hash function that takes a string key and returns an index in the range of the size of the array int hash(string key){ int sum = 0; for(int i = 0; i < key.length(); i++){ sum += key[i]; } return sum % 128; } public: // Constructor to initialize the dictionary Dictionary(){ size = 0; for(int i = 0; i < 128; i++){ table[i] = nullptr; } } // Destructor to free the memory allocated by the dictionary ~Dictionary(){ for(int i = 0; i < 128; i++){ Node* curr = table[i]; while (curr) { Node* next = curr->next; delete curr; curr = next; } } } // Adds a new key-value pair to the dictionary. If the key already exists, update the value. void put(string key, int value){ int index = hash(key); Node* curr = table[index]; while (curr) { if (curr->key == key) { // Key already exists, update the value curr->value = value; return; } curr = curr->next; } // Key does not exist, create a new node and add it to the linked list Node* newNode = new Node(key, value); newNode->next = table[index]; table[index] = newNode; size++; } // Returns the value associated with the given key, or -1 if the key is not in the dictionary. int get(string key){ int index = hash(key); Node* curr = table[index]; while (curr) { if (curr->key == key) { // Key found, return the value return curr->value; } curr = curr->next; } // Key not found, return -1 return -1; } // Removes the key-value pair from the dictionary. If the key is not in the dictionary, do nothing. void remove(string key){ int index = hash(key); Node* curr = table[index]; Node* prev = nullptr; while (curr) { if (curr->key == key) { // Key found, remove the node from the linked list if (prev) { prev->next = curr->next; } else { table[index] = curr->next; } delete curr; size--; return; } prev = curr; curr = curr->next; } // Key not found, do nothing } }; // Example usage of the Dictionary class int main(){ Dictionary dict; dict.put(apple, 5); dict.put(banana, 10); dict.put(orange, 15); cout << dict.get(apple) << endl; // Output: 5 cout << dict.get(banana) << endl; // Output: 10 cout << dict.get(orange) << endl; // Output: 15 cout << dict.get(grape) << endl; // Output: -1 dict.remove(banana); cout << dict.get(banana) << endl; // Output: -1 return 0; } ``` **Explanation**: * The `put` method adds a new key-value pair to the dictionary. If the key already exists, it updates the value. * The `get` method returns the value associated with the given key, or -1 if the key is not in the dictionary. * The `remove` method removes the key-value pair from the dictionary. If the key is not in the dictionary, it does nothing. * The dictionary uses a simple hash function that takes a string key and returns an index in the range of the size of the array. * The dictionary handles collisions using separate chaining (linked lists), where each index of the array contains the head of a linked list that stores all key-value pairs that hash to that index. **Time Complexity**: * `put`: O(1) (

answer:Let's break down the relationship between statements A and B. We're given that if A is true, then B is true. This is a one-way implication, often denoted as A implies B or if A, then B. We don't have information about the reverse implication, so we can't assume that if B is true, then A must be true as well. Now, let's examine each case: 1. If we know that A is true, we can conclude that B is also true. This follows directly from the given implication: if A is true, then B is true. 2. If we know that B is true, we can't make any definitive conclusion about A. We know that A being true would make B true, but we don't know if B being true requires A to be true. There could be other reasons why B is true, unrelated to A. So, in this case, we can't infer the truth value of A. 3. If we know that B is false, we can conclude that A is false as well. This might seem counterintuitive at first, but it follows from the original implication. If A were true, then B would have to be true, but we know that B is false. So, our initial assumption that A is true must be incorrect, meaning A is actually false. By carefully considering the one-way implication between A and B, we can make these conclusions based on the given information.

answer:When we know that B is false, we can conclude that A must also be false. Here's a more detailed explanation of the reasoning: We're given that if A is true, then B is true. If we assume that A is true, then B would have to be true according to this implication. However, we're given that B is actually false. This creates a contradiction, because our assumption that A is true would force B to be true, but we know that's not the case. To resolve this contradiction, we must conclude that our initial assumption – that A is true – is incorrect. If A were true, B would be true, but since B is false, A cannot be true. Therefore, A must be false. Think of it like a chain of events: if A happens, B must happen. If B didn't happen (i.e., B is false), it means the chain of events was never triggered, which means A didn't happen (i.e., A is false).

answer:That's the crucial distinction. When B is true, we can't infer the truth value of A with certainty. There could be other reasons why B is true, unrelated to A. The implication if A, then B only tells us that A being true guarantees B being true, but it doesn't guarantee that A is the only possible reason for B being true. On the other hand, when B is false, we can make a conclusive inference about A – namely, that A must be false, as we discussed earlier. So, to summarize: knowing B is true doesn't give us conclusive information about A, but knowing B is false does.