1. Introduction to the Problem
The pathfinding problem is about finding a route between specific nodes (vertices) in a given graph.
This problem is one of the fundamental and important problems for evaluating coding tests and algorithm problem-solving abilities.
In this course, we will explain how to solve such problems using Kotlin.
2. Problem Definition
Determine whether a path exists between the start node and the goal node in the given graph,
and if a path exists, print the corresponding path.
Input Format
- The first line: the number of vertices N (1 <= N <= 1000)
- The second line: the number of edges M (0 <= M <= 10000)
- From the third line to M lines: edge information (vertex A, vertex B)
- The last line: start node and goal node (start, goal)
Output Format
- If a path exists, print that path; if not, print “No path exists.”
3. Problem-Solving Strategy
To solve the pathfinding problem, the following algorithms can be used:
- Breadth-First Search (BFS)
- Depth-First Search (DFS)
Here, we will look at how to search for the shortest path using BFS. Since BFS explores in levels,
it is advantageous for finding the shortest path.
4. Algorithm Implementation (Kotlin Example)
import java.util.*
fun bfs(graph: Array>, start: Int, goal: Int): List? {
val queue: Queue = LinkedList()
val visited = BooleanArray(graph.size)
val parent = IntArray(graph.size) { -1 }
queue.add(start)
visited[start] = true
while (queue.isNotEmpty()) {
val current = queue.poll()
if (current == goal) {
return constructPath(parent, start, goal)
}
for (neighbor in graph[current]) {
if (!visited[neighbor]) {
visited[neighbor] = true
parent[neighbor] = current
queue.add(neighbor)
}
}
}
return null
}
fun constructPath(parent: IntArray, start: Int, goal: Int): List {
val path = mutableListOf()
var at = goal
while (at != -1) {
path.add(at)
at = parent[at]
}
path.reverse()
return path
}
fun main() {
val scanner = Scanner(System.`in`)
val N = scanner.nextInt() // Number of vertices
val M = scanner.nextInt() // Number of edges
val graph = Array(N) { mutableListOf() }
for (i in 0 until M) {
val u = scanner.nextInt()
val v = scanner.nextInt()
graph[u].add(v)
graph[v].add(u) // Undirected graph
}
val start = scanner.nextInt()
val goal = scanner.nextInt()
val path = bfs(graph, start, goal)
if (path != null) {
println("Path: ${path.joinToString(" -> ")}")
} else {
println("No path exists.")
}
}
5. Code Explanation
The above code is an implementation of pathfinding based on BFS.
- bfs(graph, start, goal): Searches for a path from the start node to the goal node in the given graph.
It uses a queue to proceed with the search and records visited nodes in the visited array.
When it reaches the goal node, it returns the path based on parent information. - constructPath(parent, start, goal): Reconstructs the path from the goal node to the start node.
It traces the path using the parent information of each node and returns the array.
6. Time Complexity Analysis
The time complexity of the BFS algorithm is O(V + E), where V is the number of vertices and E is the number of edges.
In this problem, there can be a maximum of 1000 vertices and 10000 edges, making it possible to solve efficiently.
7. Conclusion
In this course, we explored how to solve the pathfinding problem using Kotlin.
We presented the most efficient way to search for paths in graphs using the BFS algorithm,
and verified the process of solving the problem with actual code.
Additionally, understanding the implementation of DFS and the differences from BFS can provide further insights.
I hope this course will be of great help in preparing for coding tests and understanding algorithms. Thank you!