(source, destination) and it's type (tree, back, forward, cross)?
Here you go. Code in Java
import java.util.ArrayList;
import java.util.List;
class Node {
public String name;
public List<Node> connections = new ArrayList<>();
boolean visited = false;
Node(String name) {
this.name = name;
}
}
class DFS {
// Main part.
public static void search(Node root) {
if (root == null) {
return;
}
root.visited = true;
for (Node node : root.connections) {
if (!node.visited) {
// Print result.
System.out.println(root.name + "->" + node.name);
search(node);
}
}
}
}
public class App {
public static void main(String[] args) {
Node a = new Node("a");
Node b = new Node("b");
Node c = new Node("c");
Node d = new Node("d");
Node e = new Node("e");
a.connections.add(b);
b.connections.add(a);
b.connections.add(c);
b.connections.add(d);
c.connections.add(b);
c.connections.add(d);
d.connections.add(b);
d.connections.add(c);
d.connections.add(e);
DFS.search(d);
}
}
Nice question.
This is the solution based on the source you posted as comment.
IMPORTANT: There is an error on the start/end table, third row third column should be "end[u] < end[v]" instead of "end[u] > end[v]"
void main(G, s){
for each node v in G{
explored[v]=false
parent[v]=null
start[v]=end[v]=null
}
Global clock = 0
DFS(G, s, s)
}
void DFS(G, s, parent){
explored[s] = true;
parent[s] = parent
start[s]=clock
clock++
for each u=(s,v){
if(explored[v] == false){
DFS(G, v)
}
print (s + "-->" + v +"type: " + getType(s,v))
}
end[s]=clock
clock++
}
String getType(s, v){
if(start[s]<start[v]){
if(end[s]>end[v]) return "Tree edge"
else return "Forward edge"
else{
if(end[s]<end[v]) return "Back edge"
else return "Cross edge"
}
}
Related
In the CTCI solution for checking to see if there is a route between two nodes, a 3 state enum is defined. However, it appears that what is really important is a binary state of (visited = true|false). Is this true? If not, why is it necessary to distinguish between 3 separate states?
public enum State {
Unvisited, Visited, Visiting;
}
public static boolean search(Graph g,Node start,Node end) {
LinkedList<Node> q = new LinkedList<Node>();
for (Node u : g.getNodes()) {
u.state = State.Unvisited;
}
start.state = State.Visiting;
q.add(start);
Node u;
while(!q.isEmpty()) {
u = q.removeFirst();
if (u != null) {
for (Node v : u.getAdjacent()) {
if (v.state == State.Unvisited) {
if (v == end) {
return true;
} else {
v.state = State.Visiting;
q.add(v);
}
}
}
u.state = State.Visited;
}
}
return false;
}
I was wondering if in a way to avoid having to deal with the root as a special case in a Binary Search Tree I could use some sort of sentinel root node?
public void insert(int value) {
if (root == null) {
root = new Node(value);
++size;
} else {
Node node = root;
while (true) {
if (value < node.value) {
if (node.left == null) {
node.left = new Node(value);
++size;
return;
} else {
node = node.left;
}
} else if (value > node.value) {
if (node.right == null) {
node.right = new Node(value);
++size;
return;
} else {
node = node.right;
}
} else return;
}
}
}
For instance, in the insert() operation I have to treat the root node in a special way. In the delete() operation the same will happen, in fact, it will be way worse.
I've thought a bit regarding the issue but I couldn't come with any good solution. Is it because it is simply not possible or am I missing something?
The null node itself is the sentinel, but instead of using null, you can use an instance of a Node with a special flag (or a special subclass), which is effectively the null node. A Nil node makes sense, as that is actually a valid tree: empty!
And by using recursion you can get rid of the extra checks and new Node littered all over (which is what I presume is really bothering you).
Something like this:
class Node {
private Value v;
private boolean is_nil;
private Node left;
private Node right;
public void insert(Value v) {
if (this.is_nil) {
this.left = new Node(); // Nil node
this.right = new Node(); // Nil node
this.v = v;
this.is_nil = false;
return;
}
if (v > this.v) {
this.right.insert(v);
} else {
this.left.insert(v);
}
}
}
class Tree {
private Node root;
public Tree() {
root = new Node(); // Nil Node.
}
public void insert(Value v) {
root.insert(v);
}
}
If you don't want to use recursion, your while(true) is kind of a code smell.
Say we keep it as null, we can perhaps refactor it as.
public void insert(Value v) {
prev = null;
current = this.root;
boolean left_child = false;
while (current != null) {
prev = current;
if (v > current.v) {
current = current.right;
left_child = false;
} else {
current = current.left;
left_child = true;
}
}
current = new Node(v);
if (prev == null) {
this.root = current;
return;
}
if (left_child) {
prev.left = current;
} else {
prev.right = current;
}
}
The root will always be a special case. The root is the entry point to the binary search tree.
Inserting a sentinel root node means that you will have a root node that is built at the same time as the tree. Furthermore, the sentinel as you mean it will just decrease the balance of the tree (the BST will always be at the right/left of its root node).
The only way that pops in my mind to not manage the root node as a special case during insert/delete is to add empty leaf nodes. In this way you never have an empty tree, but instead a tree with an empty node.
During insert() you just replace the empty leaf node with a non-empty node and two new empty leafs (left and right).
During delete(), as a last step (if such operation is implemented as in here) you just empty the node (it becomes an empty leaf) and trim its existing leafs.
Keep in mind that if you implement it this way you will have more space occupied by empty leaf nodes than by nodes with meaningful data. So, this implementation has sense only if space is not an issue.
The code would look something like this:
public class BST {
private Node root;
public BST(){
root = new Node();
}
public void insert(int elem){
root.insert(elem);
}
public void delete(int elem){
root.delete(elem);
}
}
public class Node{
private static final int EMPTY_VALUE = /* your empty value */;
private int element;
private Node parent;
private Node left;
private Node right;
public Node(){
this(EMPTY_VALUE, null, null, null);
}
public Node(int elem, Node p, Node l, Node r){
element = elem;
parent = p;
left = l;
right = r;
}
public void insert(int elem){
Node thisNode = this;
// this cycle goes on until an empty node is found
while(thisNode.element != EMPTY_VALUE){
// follow the correct path for the insertion here
}
// insert new element here
// thisNode is an empty node at this point
thisNode.element = elem;
thisNode.left = new Node();
thisNode.right = new Node();
thisNode.left.parent = thisNode;
thisNode.right.parent = thisNode;
}
public void delete(int elem){
// manage delete here
}
}
I am looking at the code to do this in CC150. One of its method is as follows, it does this by retrieving the tail of the left sub-tree.
public static BiNode convert(BiNode root) {
if (root == null) {
return null;
}
BiNode part1 = convert(root.node1);
BiNode part2 = convert(root.node2);
if (part1 != null) {
concat(getTail(part1), root);
}
if (part2 != null) {
concat(root, part2);
}
return part1 == null ? root : part1;
}
public static BiNode getTail(BiNode node) {
if (node == null) {
return null;
}
while (node.node2 != null) {
node = node.node2;
}
return node;
}
public static void concat(BiNode x, BiNode y) {
x.node2 = y;
y.node1 = x;
}
public class BiNode {
public BiNode node1;
public BiNode node2;
public int data;
public BiNode(int d) {
data = d;
}
}
What I don't understand is the Time Complexity the author gives in the book O(n^2). What I came up with is T(N) = 2*T(N/2) + O(N/2), O(N/2) is consumed by the getting tail reference because it needs to traverse a list length of O(N/2). So by Master Theorem, it should be O(NlogN). Did I do anything wrong? Thank you!
public static BiNode convert(BiNode root) {//worst case BST everything
if (root == null) { // on left branch (node1)
return null;
}
BiNode part1 = convert(root.node1);//Called n times
BiNode part2 = convert(root.node2);//Single call at beginning
if (part1 != null) {
concat(getTail(part1), root);// O(n) every recursive call
} // for worst case so 1 to n
// SEE BELOW
if (part2 != null) {
concat(root, part2);
}
return part1 == null ? root : part1;
}
public static BiNode getTail(BiNode node) {//O(n)
if (node == null) {
return null;
}
while (node.node2 != null) {
node = node.node2;
}
return node;
}
public static void concat(BiNode x, BiNode y) {//O(1)
x.node2 = y;
y.node1 = x;
}
SAMPLE TREE:
4
/
3
/
2
/
1
As you can see, in the worst case scenario (Big-Oh is not average case), the BST would be structured using only the node1 branch(es). Thus, the recursion would have a have to run getTail() with '1 + 2 + ... + N' problem sizes to complete the conversion.
Which is O(n^2)
I'm trying to figure out how to invoke a Boolean method as true or false. Here is what I have thus far:
import java.util.Scanner;
import java.text.DecimalFormat;
public class AssignmentThree {
public static void main(String[]args) {
Scanner scan = new Scanner(System.in);
System.out.print("Enter the length of Edge 1: ");
double edge1 = scan.nextDouble();
System.out.print("Enter the length of Edge 2: ");
double edge2 = scan.nextDouble();
System.out.print("Enter the length of Edge 3: ");
double edge3 = scan.nextDouble();
scan.close();
DecimalFormat Dec = new DecimalFormat("###,###,##0.00");
if(isValid (edge1, edge2, edge3) == true) {
System.out.println("The area of the triangle is " + Dec.format(area(edge1, edge2, edge3)));
}
if(isValid (edge1, edge2, edge3) == false) {
System.out.println("This is not a valid traingle");
}
}
public static boolean isValid(double side1, double side2, double side3) {
//testing to see if the triangle is real
if(side1+side2>side3 || side1+side3>side2 || side2+side3>side1) {
return true;
} else {
return false;
}
}
I'm not sure if the problem is in my method or not but when the program is ran it simply prints the area, and never if its valid or not.
Let E be a given directed edge set. Suppose it is known that the edges in E can form a directed tree T with all the nodes (except the root node) has only 1 in-degree. The problem is how to efficiently traverse the edge set E, in order to find all the paths in T?
For example, Given a directed edge set E={(1,2),(1,5),(5,6),(1,4),(2,3)}. We know that such a set E can generate a directed tree T with only 1 in-degree (except the root node). Is there any fast method to traverse the edge set E, in order to find all the paths as follows:
Path1 = {(1,2),(2,3)}
Path2 = {(1,4)}
Path3 = {(1,5),(5,6)}
By the way, suppose the number of edges in E is |E|, is there complexity bound to find all the paths?
I have not worked on this kind of problems earlier. So just tried out a simple solution. Check this out.
public class PathFinder
{
private static Dictionary<string, Path> pathsDictionary = new Dictionary<string, Path>();
private static List<Path> newPaths = new List<Path>();
public static Dictionary<string, Path> GetBestPaths(List<Edge> edgesInTree)
{
foreach (var e in edgesInTree)
{
SetNewPathsToAdd(e);
UpdatePaths();
}
return pathsDictionary;
}
private static void SetNewPathsToAdd(Edge currentEdge)
{
newPaths.Clear();
newPaths.Add(new Path(new List<Edge> { currentEdge }));
if (!pathsDictionary.ContainsKey(currentEdge.PathKey()))
{
var pathKeys = pathsDictionary.Keys.Where(c => c.Split(",".ToCharArray())[1] == currentEdge.StartPoint.ToString()).ToList();
pathKeys.ForEach(key => { var newPath = new Path(pathsDictionary[key].ConnectedEdges); newPath.ConnectedEdges.Add(currentEdge); newPaths.Add(newPath); });
pathKeys = pathsDictionary.Keys.Where(c => c.Split(",".ToCharArray())[0] == currentEdge.EndPoint.ToString()).ToList();
pathKeys.ForEach(key => { var newPath = new Path(pathsDictionary[key].ConnectedEdges); newPath.ConnectedEdges.Insert(0, currentEdge); newPaths.Add(newPath); });
}
}
private static void UpdatePaths()
{
Path oldPath = null;
foreach (Path newPath in newPaths)
{
if (!pathsDictionary.ContainsKey(newPath.PathKey()))
pathsDictionary.Add(newPath.PathKey(), newPath);
else
{
oldPath = pathsDictionary[newPath.PathKey()];
if (newPath.PathWeights < oldPath.PathWeights)
pathsDictionary[newPath.PathKey()] = newPath;
}
}
}
}
public static class Extensions
{
public static bool IsNullOrEmpty(this IEnumerable<object> collection) { return collection == null || collection.Count() > 0; }
public static string PathKey(this ILine line) { return string.Format("{0},{1}", line.StartPoint, line.EndPoint); }
}
public interface ILine
{
int StartPoint { get; }
int EndPoint { get; }
}
public class Edge :ILine
{
public int StartPoint { get; set; }
public int EndPoint { get; set; }
public Edge(int startPoint, int endPoint)
{
this.EndPoint = endPoint;
this.StartPoint = startPoint;
}
}
public class Path :ILine
{
private List<Edge> connectedEdges = new List<Edge>();
public Path(List<Edge> edges) { this.connectedEdges = edges; }
public int StartPoint { get { return this.IsValid ? this.connectedEdges.First().StartPoint : 0; } }
public int EndPoint { get { return this.IsValid ? this.connectedEdges.Last().EndPoint : 0; } }
public bool IsValid { get { return this.EdgeCount > 0; } }
public int EdgeCount { get { return this.connectedEdges.Count; } }
// For now as no weights logics are defined
public int PathWeights { get { return this.EdgeCount; } }
public List<Edge> ConnectedEdges { get { return this.connectedEdges; } }
}
I think DFS(Depth First Search) should suit your requirements. Have a look at it here - Depth First Search - Wikipedia. You can tailor it to print the paths in the format that you require. As regards the complexity, since every node in your tree has in-degree one , the number of edges for your tree is bounded as - |E| = O(|V|). Since DFS operates with a complexity of O(|V|+|E|), your overall complexity comes out to be O(|V|).
I did this question as a part of a my assignment. The gentleman above has correctly pointed out to use pathID. You must visit each edge atleast once hence the complexity bound is O(V+E) but for tree E=O(V) therefore the complexity is O(v). I will give you a glimpse since the details are bit involved -
you will label each path with a unique ID and the path are alloted IDs in the incremental values such as 0,1,2.... A pathID of a path is the sum of weights of the edges on the path. So using DFS allocate weights to the path. You may begin by using 0 for edges until you encounter your first path and then you keep adding 1 and so on. You will also have to argue the correctness and properly allocate the weights. DFS will do the trick.