The Word Ladder problem is a classic algorithmic challenge that involves transforming one word into another by changing a single letter at a time, with each intermediate word being a valid word from a given dictionary. This problem not only tests one’s understanding of graph traversal algorithms but also has practical applications in fields like natural language processing and computational linguistics.

Program Overview

This C++ program solves the Word Ladder problem using the Breadth-First Search (BFS) algorithm. The goal is to find the shortest transformation sequence from a starting word to a target word, ensuring that each intermediate word exists in a provided dictionary.

Program Structure

1. Graph Representation: The words are treated as nodes in a graph, where an edge exists between two words if they differ by exactly one letter.
2. BFS Implementation: BFS is employed to explore the graph level by level, ensuring the shortest path is found.
3. Word Transformation: The program generates all possible one-letter transformations of the current word and checks their validity against the dictionary.
4. Path Reconstruction: Once the target word is found, the program reconstructs the path from the starting word to the target word.

Code Explanation

The program begins by defining the word list (dictionary) and the starting and ending words. It uses BFS to traverse through all possible valid transformations, keeping track of the path taken to reach each word. When the target word is found, the program outputs the sequence of transformations.

Complete C++ Program

// C++ Program to solve the Word Ladder problem using BFS

#include <iostream>
#include <vector>
#include <queue>
#include <unordered_set>
#include <unordered_map>
#include <string>
using namespace std;

// Function to generate all possible one-letter transformations of a word
vector<string> generateTransformations(const string &word, const unordered_set<string> &dict) {
    vector<string> transformations;
    string temp = word;
    for(int i = 0; i < word.length(); i++) {
        char original_char = temp[i];
        for(char c = 'a'; c <= 'z'; c++) {
            if(c == original_char) continue;
            temp[i] = c;
            if(dict.find(temp) != dict.end()) {
                transformations.push_back(temp);
            }
        }
        temp[i] = original_char;
    }
    return transformations;
}

// Function to perform BFS and find the shortest transformation sequence
vector<string> findWordLadder(const string &start, const string &end, const vector<string> &wordList) {
    unordered_set<string> dict(wordList.begin(), wordList.end());
    vector<string> result;

    if(dict.find(end) == dict.end()) {
        return result; // No possible transformation
    }

    queue<string> q;
    unordered_map<string, string> parent; // To reconstruct the path

    q.push(start);
    dict.erase(start); // Mark as visited

    while(!q.empty()) {
        string current = q.front();
        q.pop();

        vector<string> neighbors = generateTransformations(current, dict);

        for(const auto &neighbor : neighbors) {
            if(parent.find(neighbor) == parent.end()) {
                parent[neighbor] = current;
                if(neighbor == end) {
                    // Reconstruct the path
                    string word = end;
                    while(word != start) {
                        result.push_back(word);
                        word = parent[word];
                    }
                    result.push_back(start);
                    reverse(result.begin(), result.end());
                    return result;
                }
                q.push(neighbor);
                dict.erase(neighbor); // Mark as visited
            }
        }
    }

    return result; // No transformation found
}

int main() {
    string start = "hit";
    string end = "cog";
    vector<string> wordList = {"hot", "dot", "dog", "lot", "log", "cog"};

    vector<string> ladder = findWordLadder(start, end, wordList);

    if(!ladder.empty()) {
        cout << "Shortest transformation sequence:\n";
        for(const auto &word : ladder) {
            cout << word << " ";
        }
        cout << "\n";
    }
    else {
        cout << "No transformation sequence exists from " << start << " to " << end << ".\n";
    }

    return 0;
}
    

Program Documentation

Function: generateTransformations

Purpose: Generates all valid one-letter transformations of a given word by changing each letter to every other possible lowercase letter and checking if the new word exists in the dictionary.

Parameters:

• const string &word: The current word to transform.
• const unordered_set<string> &dict: The set containing all valid words (dictionary).

Returns: vector<string> containing all valid one-letter transformed words.

Function: findWordLadder

Purpose: Uses BFS to find the shortest transformation sequence from the starting word to the ending word.

Parameters:

• const string &start: The starting word.
• const string &end: The target word.
• const vector<string> &wordList: The list of valid words (dictionary).

Returns: vector<string> representing the shortest transformation sequence. Returns an empty vector if no sequence exists.

Main Function

Purpose: Sets up the problem by defining the starting and ending words, the word list (dictionary), and invokes the findWordLadder function to obtain the transformation sequence. It then outputs the result.

Flow:

1. Defines the starting word start and the target word end.
2. Initializes the word list wordList containing all valid intermediate words.
3. Calls findWordLadder to find the shortest transformation sequence.
4. Checks if a valid sequence is found and prints the sequence; otherwise, indicates that no sequence exists.

How the Program Works

1. The program starts by defining the starting word, ending word, and the dictionary of valid words.
2. It then initializes a queue for BFS and an unordered map to keep track of each word’s parent, which is essential for reconstructing the path once the target word is found.
3. Using BFS, the program explores all possible one-letter transformations of the current word. For each valid transformation found in the dictionary, it is added to the queue for further exploration.
4. If the target word is encountered during the BFS traversal, the program reconstructs the path from the starting word to the target word by backtracking using the parent map.
5. If the BFS completes without finding the target word, the program concludes that no transformation sequence exists.

Sample Output

Shortest transformation sequence:
hit hot dot dog cog 
    

How to Run the Program

1. Ensure you have a C++ compiler installed (e.g., g++).
2. Copy the provided code into a file named word_ladder.cpp.
3. Open the terminal and navigate to the directory containing the file.
4. Compile the program using the command:

   g++ -o word_ladder word_ladder.cpp

5. Run the compiled program using:

   ./word_ladder

Optimizing for Larger Dictionaries

While BFS efficiently finds the shortest path in many cases, larger dictionaries can lead to increased memory usage and longer processing times. To optimize performance for larger word lists, consider the following enhancements:

• Bidirectional BFS: Perform BFS simultaneously from both the start and end words. This approach reduces the search space significantly.
• Preprocessing: Create a pattern-based adjacency list where words differing by one letter share common patterns. This allows for faster lookup of neighbors.
• Heuristic Improvements: Implement heuristic-based methods like A* search by estimating the distance to the target word.

Conclusion

The Word Ladder problem serves as an excellent example of how graph traversal algorithms like BFS can be applied to solve real-world problems. By representing words as nodes and valid transformations as edges, BFS efficiently explores all possible paths to find the shortest transformation sequence. Understanding and implementing such algorithms not only enhances problem-solving skills but also has practical implications in various computational fields.