Depth/Breadth First Search

Hello all,
I am trying to figure out how to do a depth and breadth first search on a graph of Cities, as defined below. I think that I understand the searches, but I am having an extremely difficult time figuring out how to implement them.
If anyone has any tips, suggestions, or hints (for some reason I think I'm probably just overlooking something simple), I would greatly appreciate them.
Thanks for any help!
* to represent an individual city
public class City {
    String name;
    double latitude;
    double longitude;
    ArrayList<Edge> neighbors;
    //the constructor
    public City(String name, double latitude, double longitude){
        this.name = name;
        this.latitude = latitude;
        this.longitude = longitude;
        this.neighbors = new ArrayList<Edge>();
    //to check if this city is equal to that given city
    public boolean same(City c){
        return this.name.equals(c.name) && this.latitude == c.latitude &&
            this.longitude == c.longitude;
* to represent an edge between two cities
public class Edge {
    City city1;
    City city2;
    double dist;
    boolean visited;
    public Edge(City city1, City city2){
        this.city1 = city1;
        this.city2 = city2;
        this.dist = this.distTo();
        this.visited = false;
     * to find the distance between the two cities in an edge
    public double distTo(){
        return Math.sqrt(((this.city1.latitude - this.city2.latitude) *
                (this.city1.latitude - this.city2.latitude)) +
                ((this.city1.longitude - this.city2.longitude) *
                (this.city1.longitude - this.city2.longitude)));
* to represent a path between two cities as a list
* of edges
public class Graph {
    ArrayList<Edge> alist;
    public Graph(ArrayList<Edge> alist){
        this.alist = alist;
}

http://en.wikipedia.org/wiki/Breadth-first_search (includes algorithm)

Similar Messages

When to go for Breadth first search over depth first search

hi,
under which scenarios breadth first search could be used and under which scenarios depth first search could be used?
what is the difference between these two searches?
Regards,
Ajay.

No real clear-cut rule for when to use one over the other. It depends on the nature of your search and where you would prefer to find results.
The difference is that in breadth-first you first search all immidiate neighbours before searching their neighbours (sort of like preorder traversal). Whereas in depth-first you search some random (or otherwise selected) neighbour and then a neighbour of that node until you can't go deeper (i.e., you've searched a neighbour that has no other neighbours). This probably isn't a very clear explanation, perhaps you'll find this more helpful: http://www.ics.uci.edu/~eppstein/161/960215.html
If you would prefer to find results closer to the origin node, breadth first would be better. If you would prefer to find results further away use depth first.

Depth First Search, Breadth First Search

BANNER
Oracle Database 10g Enterprise Edition Release 10.2.0.5.0 - Prod
PL/SQL Release 10.2.0.5.0 - Production
CORE     10.2.0.5.0     Production
TNS for Linux: Version 10.2.0.5.0 - Production
NLSRTL Version 10.2.0.5.0 - Production
I
have this table that form a tree where record with column value 'more_left' 0 is located more left than 1 (for example, record with id 2 has a parent id 1 and more left than record with id 7 that also has parent id 1):
with t as(
select 2 id,1 parent_id,0 most_left from dual
union all
select 7 id,1 parent_id,1 most_left from dual
union all
select 8 id,1 parent_id,2 most_left from dual
union all
select 3 id,2 parent_id,0 most_left from dual
union all
select 6 id,2 parent_id,1 most_left from dual
union all
select 9 id,8 parent_id,0 most_left from dual
union all
select 12 id,8 parent_id,1 most_left from dual
union all
select 4 id,3 parent_id,0 most_left from dual
union all
select 5 id,3 parent_id,1 most_left from dual
union all
select 10 id,9 parent_id,0 most_left from dual
union all
select 11 id,9 parent_id,1 most_left from dual
select * from t;The problem is to show all the ids, using Breadth First Search and Depth First Search. Tx, in advance.
Edited by: Red Penyon on Apr 12, 2012 3:39 AM

Hi,
I fail to understand how comes there is no row for ID=1 ?
The topmost parent (the root) should be in the table also.
For 11g, this would work (as long as the root is in the table) :Scott@my11g SQL>l
1 WITH t AS
2       (SELECT 1 ID, 0 parent_id, 0 most_left
3          FROM DUAL
4        UNION ALL
5        SELECT 2 ID, 1 parent_id, 0 most_left
6          FROM DUAL
7        UNION ALL
8        SELECT 7 ID, 1 parent_id, 1 most_left
9          FROM DUAL
10        UNION ALL
11        SELECT 8 ID, 1 parent_id, 2 most_left
12          FROM DUAL
13        UNION ALL
14        SELECT 3 ID, 2 parent_id, 0 most_left
15          FROM DUAL
16        UNION ALL
17        SELECT 6 ID, 2 parent_id, 1 most_left
18          FROM DUAL
19        UNION ALL
20        SELECT 9 ID, 8 parent_id, 0 most_left
21          FROM DUAL
22        UNION ALL
23        SELECT 12 ID, 8 parent_id, 1 most_left
24          FROM DUAL
25        UNION ALL
26        SELECT 4 ID, 3 parent_id, 0 most_left
27          FROM DUAL
28        UNION ALL
29        SELECT 5 ID, 3 parent_id, 1 most_left
30          FROM DUAL
31        UNION ALL
32        SELECT 10 ID, 9 parent_id, 0 most_left
33          FROM DUAL
34        UNION ALL
35        SELECT 11 ID, 9 parent_id, 1 most_left
36          FROM DUAL
37 )
38 select
39 rt
40 ,listagg(id,'-') within group (order by lvl,id) BFS
41 ,listagg(id,'-') within group (order by pth,lvl) DFS
42 from (
43 SELECT id, connect_by_root(id) rt, sys_connect_by_path(most_left,'-') pth, level lvl, most_left
44        FROM t
45 CONNECT by nocycle prior ID = parent_id
46 START WITH parent_id = 0
47 )
48* group by rt
Scott@my11g SQL>/
        RT BFS                            DFS
         1 1-2-7-8-3-6-9-12-4-5-10-11     1-2-3-4-5-6-7-8-9-10-11-12But as long as you're 10g, that is no real help.
I'll try to think of a way to get that with 10g.
10g solution for DFS :Scott@my10g SQL>l
1 WITH t AS
2 (
3      SELECT 1 ID, 0 parent_id, 0 most_left FROM DUAL UNION ALL
4      SELECT 2 ID, 1 parent_id, 0 most_left FROM DUAL UNION ALL
5      SELECT 7 ID, 1 parent_id, 1 most_left FROM DUAL UNION ALL
6      SELECT 8 ID, 1 parent_id, 2 most_left FROM DUAL UNION ALL
7      SELECT 3 ID, 2 parent_id, 0 most_left FROM DUAL UNION ALL
8      SELECT 6 ID, 2 parent_id, 1 most_left FROM DUAL UNION ALL
9      SELECT 9 ID, 8 parent_id, 0 most_left FROM DUAL UNION ALL
10      SELECT 12 ID, 8 parent_id, 1 most_left FROM DUAL UNION ALL
11      SELECT 4 ID, 3 parent_id, 0 most_left FROM DUAL UNION ALL
12      SELECT 5 ID, 3 parent_id, 1 most_left FROM DUAL UNION ALL
13      SELECT 10 ID, 9 parent_id, 0 most_left FROM DUAL UNION ALL
14      SELECT 11 ID, 9 parent_id, 1 most_left FROM DUAL
15 )
16 select max(pth) DFS
17 from (
18 select sys_connect_by_path(id,'-') pth
19 from (
20 select id,pth,lvl,most_left, row_number() over (order by pth, lvl) rn
21 from (
22 select id, sys_connect_by_path(most_left,'-') pth, level lvl, most_left
23 from t
24 CONNECT by nocycle prior ID = parent_id
25 START WITH parent_id = 0
26 )
27 order by pth, lvl
28 )
29 connect by prior rn= rn-1
30 start with rn=1
31* )
Scott@my10g SQL>/
DFS
-1-2-3-4-5-6-7-8-9-10-11-12------
10g solution for BFS :Scott@my10g SQL>l
1 WITH t AS
2 (
3      SELECT 1 ID, 0 parent_id, 0 most_left FROM DUAL UNION ALL
4      SELECT 2 ID, 1 parent_id, 0 most_left FROM DUAL UNION ALL
5      SELECT 7 ID, 1 parent_id, 1 most_left FROM DUAL UNION ALL
6      SELECT 8 ID, 1 parent_id, 2 most_left FROM DUAL UNION ALL
7      SELECT 3 ID, 2 parent_id, 0 most_left FROM DUAL UNION ALL
8      SELECT 6 ID, 2 parent_id, 1 most_left FROM DUAL UNION ALL
9      SELECT 9 ID, 8 parent_id, 0 most_left FROM DUAL UNION ALL
10      SELECT 12 ID, 8 parent_id, 1 most_left FROM DUAL UNION ALL
11      SELECT 4 ID, 3 parent_id, 0 most_left FROM DUAL UNION ALL
12      SELECT 5 ID, 3 parent_id, 1 most_left FROM DUAL UNION ALL
13      SELECT 10 ID, 9 parent_id, 0 most_left FROM DUAL UNION ALL
14      SELECT 11 ID, 9 parent_id, 1 most_left FROM DUAL
15 )
16 select max(pth) BFS
17 from (
18 select sys_connect_by_path(id,'-') pth
19 from (
20 select id,pth,lvl,most_left, row_number() over (order by lvl, pth) rn
21 from (
22 select id, sys_connect_by_path(most_left,'-') pth, level lvl, most_left
23 from t
24 CONNECT by nocycle prior ID = parent_id
25 START WITH parent_id = 0
26 )
27 order by lvl, pth
28 )
29 connect by prior rn= rn-1
30 start with rn=1
31* )
Scott@my10g SQL>/
BFS
-1-2-7-8-3-6-9-12-4-5-10-11There might certainly have better ways...

How to extend breadth first Search for Binary Tree to any kind of Tree??

Dear Friends,
I am thinking a problem, How to extend breadth first Search for Binary Tree to any kind of Tree?? ie each node has more than 2 leaves such as 1, 2,3,4 or any,
I have following code to successfully apply for breadth first Search in Binary Tree as follows,
package a.border;
import java.util.ArrayList;
import java.util.LinkedList;
public class Tree
    int root;
    Tree left;
    Tree right;
    static ArrayList<Integer> list = new ArrayList<Integer>();
    static ArrayList<Tree> treeList = new ArrayList<Tree>();
    private static LinkedList<Tree> queue = new LinkedList<Tree>();
     * @param root root value
     * @param left left node
     * @param right right node
    public Tree(int root, Tree left, Tree right)
        this.root = root;
        this.left = left;
        this.right = right;
    /** Creates a new instance of Tree
     * You really should know what this does...
     * @param root
    public Tree(int root)
        this.root = root;
        this.left = null;
        this.right = null;
     * Simply runs a basic left then right traversal.
    public void basicTraversal()
        //Check if we can go left
        if (left != null)
            left.basicTraversal();
        //Add the root
        list.add(root);
        //Check if we can go right
        if (right != null)
            right.basicTraversal();
    public ArrayList<Integer> getBreadthTraversal(ArrayList<Integer> list)
        //Add the root to the arraylist, we know it is always the first entry.
        list.add(root);
        //Basically we add the first set of nodes into the queue for
        //traversing.
        //Query if left exists
        if (left != null)
            //Then add the node into the tree for traversing later
            queue.add(left);
        //Same for right
        if (right != null)
            queue.add(right);
        //Then we call the traverse method to do the rest of the work
        return traverse(list);
    private ArrayList<Integer> traverse(ArrayList<Integer> list)
        //Keep traversing until we run out of people
        while (!queue.isEmpty())
            Tree p = queue.remove();
            //Check if it has any subnodes
            if (p.left != null)
                //Add the subnode to the back of the queue
                queue.add(p.left);
            //Same for left
            if (p.right != null)
                //Same here, no queue jumping!
                queue.add(p.right);
            //Append to the ArrayList
            list.add(p.root);
        //And return
        return list;
     * Makes a tree and runs some operations
     * @param args
    public static void main(String[] args)
         *                             4
         *          t =           2       6
         *                      1   3    5   7
        Tree leaf6 = new Tree(1);
        Tree leaf7 = new Tree(3);
        Tree leaf8 = new Tree(5);
        Tree leaf9 = new Tree(7);
        Tree t4 = new Tree(2, leaf6, leaf7);
        Tree t5 = new Tree(6, leaf8, leaf9);
        Tree t = new Tree(4, t4, t5);
        t.basicTraversal();
        System.out.println("Here is basicTraversal ="+list.toString());
        list.clear();
        t.getBreadthTraversal(list);
        System.out.println("getBreadthTraversal= " +list.toString());
        list.clear();
    }Can Guru help how to update to any kind of tree??
here this code is for the tree like:
         *                             4
         *          t =           2       6
         *                      1   3    5   7
         */But i hope the new code can handle tree like:
         *                             4
         *                           /   | \
         *                          /     |   \
         *          t =            2     8   6
         *                        / | \    |    /| \
         *                      1 11 3 9   5 10 7
         */Thanks

sunnymanman wrote:
Dear Friends,
I am thinking a problem, How to extend breadth first Search for Binary Tree to any kind of Tree?? ...The answer is interfaces.
What do all trees have in common? And what do all nodes in trees have in common?
At least these things:
interface Tree<T> {
    Node<T> getRoot();
interface Node<T> {
    T getData();
    List<Node<T>> getChildren();
}Now write concrete classes implementing these interfaces. Let's start with a binary tree (nodes should have comparable items) and an n-tree:
class BinaryTree<T extends Comparable<T>> implements Tree<T> {
    protected BTNode<T> root;
    public Node<T> getRoot() {
        return root;
class BTNode<T> implements Node<T> {
    private T data;
    private Node<T> left, right;
    public List<Node<T>> getChildren() {
        List<Node<T>> children = new ArrayList<Node<T>>();
        children.add(left);
        children.add(right);
        return children;
    public T getData() {
        return data;
class NTree<T> implements Tree<T> {
    private NTNode<T> root;
    public Node<T> getRoot() {
        return root;
class NTNode<T> implements Node<T> {
    private T data;
    private List<Node<T>> children;
    public List<Node<T>> getChildren() {
        return children;
    public T getData() {
        return data;
}Now with these classes, you can wite a more generic traversal class. Of course, every traversal class (breath first, depth first) will also have something in common: they return a "path" of nodes (if the 'goal' node/data is found). So, you can write an interface like this:
interface Traverser<T> {
    List<Node<T>> traverse(T goal, Tree<T> tree);
}And finally write an implementation for it:
class BreathFirst<T> implements Traverser<T> {
    public List<Node<T>> traverse(T goal, Tree<T> tree) {
        Node<T> start = tree.getRoot();
        List<Node<T>> children = start.getChildren();
        // your algorithm here
        return null; // return your traversal
}... which can be used to traverse any tree! Here's a small demo of how to use it:
public class Test {
    public static void main(String[] args) {
        Tree<Integer> binTree = new BinaryTree<Integer>();
        // populate your binTree
        Tree<Integer> nTree = new NTree<Integer>();
        // populate your nTree
        Traverser<Integer> bfTraverser = new BreathFirst<Integer>();
        // Look for integer 6 in binTree
        System.out.println("bTree bfTraversal -> "+bfTraverser.traverse(6, binTree));
        // Look for integer 6 in nTree
        System.out.println("bTree bfTraversal -> "+bfTraverser.traverse(6, nTree));
}Good luck!

Breadth First Search (More of a Logic Question)

Hey guys, I'm having a logic block with a Breadth First Search using a queue.
Basically the way Breadth First Search works is that it expands all of the nodes one level at a time until it finds the result. I quickly whipped out a flash to demonstrate what I believe is the Breadth First Search.
http://www.nxsupport.com/dimava/bfs.swf
I wrote the code to do the algorithm, However if I just output the queue it shows all of the excess "potential" nodes and not just the direct route. For example, with the flash file linked above, it would show ABECGFC, instead of just ABCD.
I realise that I could trace back from the end to see which of the nodes D is connected to, then which of the nodes C is connected to, etc. But that wouldn't work out if there was more than one path with the same distance leading to the same destination.
Thanks,
Dimava

It's been a long time since college so I may be suggesting a poor way of doing this.
But suppose you have a queue data type. You can use it in two ways: as a representation of a path from your starting node to your ending node, and as a place to hold paths while you're performing a breadth-first search. (Or you could skip the latter and just use recursion.)
You wouldn't keep a list of all possible paths. Rather you'd be storing paths that correspond to nodes currently being examined in your breadth-first search.
So you'd create a queue (representing a path) holding your start node. Then you'd put that queue into the queue that represents your traversal state.
Then while the traversal queue is not empty, you take out a path (a queue), look at its last element (a node), then create new paths that consist of the current path but each terminating with one of the children of the current node. (Well not really children since it's a graph and not a tree, but you know what I mean.) Repeat until the target node is found.

Depth first search and breadth first search

can anyone help for this program. can anyone give me a program for this without using an applet.
thanks!!!

can anyone help for this program. can anyone give me a
program for this without using an applet.
thanks!!!What are you talking about? What program? Depth-first means you recursively search from root to the first leaf, then go up one level, go down as far as you can, repeat. Breadth-first means you search every node on each level before going on to the next level.

Is SQR #include a depth first search or breadth first search alrgorithm

Just wondering if SQR compiler is a dfs or bfs algorithm? I am guessing a dfs since includes can show up at the beginning or end of the sqr or sqc definition but position is relevant. So a few questions that branch off of that one.
I have successfully included file filea.sqc and fileb.sqc in myfile.sqr, then included file subfile.sqc in both filea.sqc and fileb.sqc. There seemed to be no error message, so I am wondering how the file include works. I would think from a compiler point of view, if we included the same procedure or function twice, we should do one of three things - ignore the second instance (do nothing), override the first instance, or throw an error.
So, anyone know how the compiler stack is built?

The documentation doesn't help much, but if you want to prevent an include from being loaded more than once, you should be able to use a combination of compiler directives, i.e. #IFNDEF and #DEFINE.
In your example, if you code subfile.sqc like
#IFNDEF SUBFILE_INCLUDED
#DEFINE SUBFILE_INCLUDED Y
begin-procedure subfile_proc
#ENDIFthen you should only get one copy of the code between the #IFNDEF-#ENDIF statements.
You could probably use something similar to test how SQR handles multiple includes of the same file. Try something like
#IFDEF SUBFILE_INCLUDED
Display 'Subfile included twice'
#ENDIF
#IFNDEF SUBFILE_INCLUDED
#DEFINE SUBFILE_INCLUDED Y
#ENDIFand see if the message is displayed.
Regards,
Bob

Need help finalizing breadth-first search function...

I have a map with bidirectional links and I am to find the shortest path so the first path found will be good for me. I tried follow the algorithm for BFS and it shows below. However, I am not sure when/where/what I should return per loop. I would like to get a list of Nodes (map represented by numeric node numbers). I dont think returning a list is right for this recursion so i put down int for now... I think the steps are right but just not sure how to get the result.... Please help! thanks!
public int(?) getShortestPath() {
          int currentNode = bfsQueue.remove();//already has Root from start
          if (currentNode == destNode) {
               shortestPathFound = true;
          } else {
               ArrayList<Integer> temp = getChildrenLinks(currentNode);
               for (int x=0; x< temp.size(); x++) {
                    bfsQueue.add(temp.get(x));
          if (bfsQueue.size() == 0) {
               return null;
          } else if (shortestPathFound == false){
               getShortestPath();
          return null;
     }

You have almost all variable defined outside your method. You "should" declare them inside. Also, there is no need to make a recursive call (which makes it easier to return your path).
Here's a debugged version of a BFS:
Say you have the following edges:
1 <--> 2
1 <--> 3
2 <--> 5
2 <--> 6
3 <--> 4
4 <--> 5
5 <--> 6
5 <--> 7
7 <--> 8
7 <--> 9
8 <--> 9
8 <--> 10A picture of this would look like:
1 --- 2 --- 6
|      \___ |     9
|          \|    /|
3 --- 4 --- 5   / |
            | / |
            | /   |
            7 --- 8 --- 10And you want to find the shortest path from 1 to 8.
Create a stack that holds all "layers" (a list of numbers) that you are discovering. So you start with an empty STACK and push the start node (in a list) on it:
STACK {
{1} // layer 1
}Pop the last "layer" from your STACK and collect all it's unvisited neighbours. All these unvisited neighbours are the next layer which should be pushed on your STACK. Now your STACK looks like:
STACK {
{1}   // layer 1
{2,3} // layer 2
}Again pop the last layer from your stack (this is now {2,3}) and collect all it's unvisited neighbours and push these on your STACK. Now your STACK looks like:
STACK {
{1}     // layer 1
{2,3}   // layer 2
{4,5,6} // layer 3
}Keep pushing new "layers" on your STACK until you stumble upon your destination node (8 in this example). So, when you get to your destination node, your STACK looks like:
STACK {
{1}     // layer 1
{2,3}   // layer 2
{4,5,6} // layer 3
{7}     // layer 4
{8,9}   // layer 5
}Now that you have gotten to your destination, it's time to find the actual path by back-tracking through your "layers" where you must "jump" from "layer" N to a "layer" N-1 (you can only "jump" to a lower "layer"):
The back-tracking will look like:
STACK {
{1}     // layer 1
   |
{2,3}   // layer 2
{4,5,6} // layer 3
{7}     // layer 4
   |
{8,9}   // layer 5
- 8 (in layer 5) is connected to 7 (in layer 4)
- 7 (in layer 4) is connected to 5 (in layer 3)
- 5 (in layer 3) is connected to 2 (in layer 2)
- 2 (in layer 2) is connected to 1 (in layer 1)So, your shortest path is: [1,2,5,7,8]
HTH

Breadth First Traversal of a Tree...

Hi,
I want to make a program to print every node of a tree in a Breadth First Search (BFS)/Traversal. Since, BFS searches a tree in levels (i.e. 1st the root, then it's Children, then Grand Children etc from left to right), this is where I'm stuck. Here is my TreeNode class:
class TreeNode {
Object element;
TreeNode left;
TreeNode right;
public TreeNode(Object o) {
    element = o;
}Each node has a reference to its Children nodes etc. Here is how my tree looks like. Mine is similar to this: http://www.codeproject.com/KB/vb/SimpleBTree.aspx
All the lectures I have read in the net are talking about Queue, but my problem is reading the tree in BFS Traversal order. It's almost 4 days I'm trying to solve this probelm, but can't come up with something. Any help is greatly appreciated.
Here is my Binary Tree class:
public class BinaryTree {
     private TreeNode root;
     private int size = 0;
     /** Create a default binary tree */
     public BinaryTree() {
     /** Create a binary tree from an array of objects */
    public BinaryTree(Object[] objects) {
      for (int i = 0; i < objects.length; i++)
        insert(objects);
/** Insert element o into the binary tree
* Return true if the element is inserted successfully */
public boolean insert(Object o) {
if (root == null)
root = new TreeNode(o); // Create a new root
else {
// Locate the parent node
TreeNode parent = null;
TreeNode current = root;
while (current != null)
if (((Comparable)o).compareTo(current.element) < 0) {
parent = current;
current = current.left;
else if (((Comparable)o).compareTo(current.element) > 0) {
parent = current;
current = current.right;
else
return false; // Duplicate node not inserted
// Create the new node and attach it to the parent node
if (((Comparable)o).compareTo(parent.element) < 0)
parent.left = new TreeNode(o);
else
parent.right = new TreeNode(o);
size++;
return true; // Element inserted
/** Inorder traversal */
public void inorder() {
inorder(root);
/** Inorder traversal from a subtree */
private void inorder(TreeNode root) {
if (root == null) return;
inorder(root.left);
System.out.print(root.element + " ");
inorder(root.right);
/** Postorder traversal */
public void postorder() {
postorder(root);
/** Postorder traversal from a subtree */
private void postorder(TreeNode root) {
if (root == null) return;
postorder(root.left);
postorder(root.right);
System.out.print(root.element + " ");
/** Preorder traversal */
public void preorder() {
preorder(root);
/** Preorder traversal from a subtree */
private void preorder(TreeNode root) {
if (root == null) return;
System.out.print(root.element + " ");
preorder(root.left);
preorder(root.right);
/** Inner class tree node */
private static class TreeNode {
Object element;
TreeNode left;
TreeNode right;
public TreeNode(Object o) {
element = o;
/** Get the number of nodes in the tree */
public int getSize() {
return size;
/** Search element o in this binary tree */
public boolean search(Object o){
TreeNode temp = root;
boolean found = false;
if(temp == null)
found = false;
else if(((Comparable)(temp.element)).compareTo(o) == 0 && temp != null)
found = true;
else if(((Comparable)temp.element).compareTo(o) > 0){
root = temp.left;
found = search(o);
else{
root = temp.right;
found = search(o);
return found;
/** Display the nodes in breadth-first traversal */
public void breadthFirstTraversal(){
TreeNode temp = root;
// TreeNode leftChild = temp.left;
// TreeNode rightChild = temp.right;
MyQueue s = new MyQueue();
s.enqueue(root);
// while (s.getSize() == 0){
System.out.println(s);
// private void breadthFirstTraversal(TreeNode node){

ChangBroot wrote:
All the lectures I have read in the net are talking about Queue, but my problem is reading the tree in BFS Traversal order. It's almost 4 days I'm trying to solve this probelm, but can't come up with something. Any help is greatly appreciated. One simple strategy is to make an ordinary recursive traversal of the tree. Along in the traversal you keep an array of linked lists. When you visit a node you just add it to the list at the entry in the array corresponding to the depth of this node.
After the traversal you have an array with linked lists. Each list holds the nodes found at a certain tree depth. So each list holds all nodes corresponding to a specific tree "breadth", namely all nodes found at the same tree depth.

Depth First Search w/o Recursion

I am trying to use Depth First Search for 2-D maze, and I was just wondering if there was a straight forward and easy way of doing DFS of a maze without using recursion. Or is recursion the really only way you do DFS without invoking a lot of messy variables to hold temporary information.

itereation instead of recursion...can you be a little more discriptive.
For recursion I can understand how it works because when you pop out of one recursive cycle you got to the next command. So basically
move{
if you can
go left
move()
go right
move()
go up
move()
go down
move()
else
pop out of recursion
hopefully this pseudocode makes some sense. But absically lets say with my maze I go right 3 times then can't go right anymore...well when that happens it will try to go right, because after calling move() under "go left" the next step would just be to go left. But I don't see how this can be done iteratively. If somebody can explain it pseudocode that would be cool too.

Depth first search tree

Hi,
I have stored some states in a vector (States are numbers). In my code i have printed out the parent nodes, the 2 elements after the parent nodes are there child.
The element placed at position 0 in the vector is the root. I do not know how to implement a depth first search tree to address unreachable node. In this case, parent node 5 is unreachable, from other nodes. But i do not know how to do this. I have spent ages reading tutorials/ book, but i cant seem to convert this knowledge into java. My code is very short and therefore easy to understand
import java.awt.*;
import java.util.*;
public class Vec{
public static void main(String argv[]){
    Vec v = new Vec();
    v.remove();
    }//End of main
public void remove(){
        Vector mv = new Vector();
        //Note how a vector can store objects
        //of different types
        mv.addElement(1); // root and parent node of 2 elements below
        mv.addElement(2); // child of above
        mv.addElement(1); // child of above
        mv.addElement(2); // parent of 2 elements below
        mv.addElement(4);
        mv.addElement(2);
        mv.addElement(3); // parent of 2 elements below
        mv.addElement(1);
        mv.addElement(3);
        mv.addElement(4); // parent of 2 elements below
        mv.addElement(3);
        mv.addElement(4);
        mv.addElement(5); // parent of 2 elements below
        mv.addElement(2);
        mv.addElement(4);
// below identifys the parent nodes for you, but doesnt store them as parent nodes
                for(int i=0; i< mv.size(); i++){
        if (i % 3 == 0)
                  System.out.println(mv.elementAt(i));
   }//End of amethod


Ah ok, it's a graph and not a tree. In that case don't use the tree-code I posted, but model your graph as an adjacency matrix (AM) [1]. The AM for the graph you posted in reply #5 would look like this:
Node | 1 2 3 4 5
-----+--------------
   1 | 1 1 0 0 0
     |
   2 | 0 1 0 1 0
     |
   3 | 1 0 1 0 0
     |
   4 | 0 0 1 1 0
     |
   5 | 0 1 1 0 0You can implement this by using a simple 2D array of integers like this:class Graph {
    private int[][] adjacencyMatrix;
    public Graph(int numNodes) {
        adjacencyMatrix = new int[numNodes][numNodes];
    public void addEdge(int from, int to) {
        adjacencyMatrix[from-1][to-1] = 1;
    public boolean isReachable(Integer start, Integer goal) {
        // your algorithm here
        return false;
    public String toString() {
        StringBuilder strb = new StringBuilder();
        for(int i = 0; i < adjacencyMatrix.length; i++) {
            int[] row = adjacencyMatrix;
strb.append((i+1)+" | ");
for(int j = 0; j < row.length; j++) {
strb.append(adjacencyMatrix[i][j]+" ");
strb.append('\n');
return strb.toString();
public static void main(String[] args) {
Graph graph = new Graph(5);
graph.addEdge(1, 1);
graph.addEdge(1, 2);
graph.addEdge(2, 2);
graph.addEdge(2, 4);
graph.addEdge(3, 1);
graph.addEdge(3, 3);
graph.addEdge(4, 3);
graph.addEdge(4, 4);
graph.addEdge(5, 2);
graph.addEdge(5, 3);
System.out.println("Adjacency Matrix:\n"+graph);
System.out.println("Is node 5 reachable from node 1? "+
graph.isReachable(new Integer(1), new Integer(5))); // should be false
System.out.println("Is node 1 reachable from node 5? "+
graph.isReachable(new Integer(5), new Integer(1))); // should be true: a path exists from 5 -> 3 -> 1 and 5 -> 2 -> 4 -> 3 -> 1
Good luck.
[1]
http://en.wikipedia.org/wiki/Adjacency_matrix
http://mathworld.wolfram.com/AdjacencyMatrix.html

8-Puzzle Depth First Search .. Need Help

Hi, I have the following code for doing a depth first search on the 8-Puzzle problem
The puzzle is stored as an object that contains an array such that the puzzle
1--2--3
4--5--6
7--8--[] is stored as (1,2,3,4,5,6,7,8,0)
The class for the puzzle is:
class Puzzle implements {
     Puzzle parent;
     int[] pzle;
     int blank;
     Puzzle(int p1,int p2,int p3,int p4,int p5, int p6, int p7, int p8, int p9){
          pzle = new int[9];
          pzle[0] = p1; if(p1==0){ blank=0;}
          pzle[1] = p2; if(p2==0){ blank=1;}
          pzle[2] = p3; if(p3==0){ blank=2;}
          pzle[3] = p4; if(p4==0){ blank=3;}
          pzle[4] = p5; if(p5==0){ blank=4;}
          pzle[5] = p6; if(p6==0){ blank=5;}
          pzle[6] = p7; if(p7==0){ blank=6;}
          pzle[7] = p8; if(p8==0){ blank=7;}
          pzle[8] = p9; if(p9==0){ blank=8;}
     public Puzzle() {
     public boolean equals(Puzzle p){
          if(p == null) return false;
          for(int i =0;i<9;i++){
               if(p.pzle!=pzle[i]){
                    return false;
          return true;
     public String toString(){
          return pzle[0] + "\t" + pzle[1] + "\t" + pzle [2] + "\n" +
                    pzle[3] + "\t" + pzle[4] + "\t" + pzle [5] + "\n" +
                    pzle[6] + "\t" + pzle[7] + "\t" + pzle [8] + "\n";
     public String printSolution(){
          String ret ="";
          Puzzle st = parent;
          while(st!=null){
               ret = st + "\n=======================\n" + ret;
               st = st.parent;
          ret=ret+this;
          return ret;
     public ArrayList<Puzzle> successors(){
          ArrayList<Puzzle> succ = new ArrayList<Puzzle>();
          int b = blank;
          if (((b-1) % 3) != 2 && b!=0) {
          Puzzle np1 = new Puzzle();
          np1.parent = this;
          np1.pzle = new int[9];
          for(int i =0;i<9;i++){
               np1.pzle[i] = pzle[i];
          np1.pzle[b] = np1.pzle[b-1]; np1.pzle[b-1] = 0;
          np1.blank = b-1;
          if(!np1.equals(this.parent)){
          succ.add(np1);
          if (((b+1) % 3) != 0) {
          Puzzle np2 = new Puzzle();
          np2.parent = this;
          np2.pzle = new int[9];
          for(int i =0;i<9;i++){
               np2.pzle[i] = pzle[i];
          np2.pzle[b] = np2.pzle[b+1]; np2.pzle[b+1] = 0;
          np2.blank = b+1;
          if(!np2.equals(this.parent)){
               succ.add(np2);
          if (b-3 >= 0) {
          Puzzle np3 = new Puzzle();
          np3.parent = this;
          np3.pzle = new int[9];
          for(int i =0;i<9;i++){
               np3.pzle[i] = pzle[i];
          np3.pzle[b] = np3.pzle[b-3]; np3.pzle[b-3] = 0;
          np3.blank = b-3;
          if(!np3.equals(this.parent)){
               succ.add(np3);
          if (b+3 < 9) {
          Puzzle np4 = new Puzzle();
          np4.parent = this;
          np4.pzle = new int[9];
          for(int i =0;i<9;i++){
               np4.pzle[i] = pzle[i];
          np4.pzle[b] = np4.pzle[b+3]; np4.pzle[b+3] = 0;
          np4.blank = b+3;
          if(!np4.equals(this.parent)){
               succ.add(np4);
          return succ;
     The code for the DFS is      public static boolean DFS(Puzzle p, Puzzle goal, ArrayList<Puzzle> closed){
     if(p.equals(goal)){
          sol=p;
          return true;
     for(Puzzle pz : closed){
               if(pz.equals(p)){
                    return false;
     closed.add(p);
                    //Generate all possible puzzles that can be attained
                    ArrayList<Puzzle> succ = p.successors();
                    for(Puzzle puz : succ){
                         if(DFS(puz,goal,closed)){ return true;}
          return false;
}The problem is that when i run this on say the following start and goal:          Puzzle startP = new Puzzle(5,4,0,6,1,8,7,3,2);
          Puzzle goalP =new Puzzle(1,2,3,8,0,4,7,6,5);
          ArrayList<Puzzle> closed = new ArrayList<Puzzle>();
          startP.parent=null;
          boolean t = DFS(startP,goalP,closed);5-4-0
6-1-8
7-3-2
start and goal
1-2-3
8-0-4
7-6-5
it first takes foreever which is expected but it There should be 9! possible states correct? If I print out the size of the closed array (already seen states) it gets up to about 180000 and then returns with no solution found. But this is well under the 9! possible states...
Its hard to know whats wrong and if it is even wrong.. but could someone please look at my code and see if something is off?
I dont think the successor generator is wrong as I tested the blank in every space and it got the correct successors.
Any ideas?

Puzzle(int p1,int p2,int p3,int p4,int p5, int p6, int p7, int p8, int p9){
          pzle = new int[9];
          pzle[0] = p1; if(p1==0){ blank=0;}
          pzle[1] = p2; if(p2==0){ blank=1;}
          pzle[2] = p3; if(p3==0){ blank=2;}
          pzle[3] = p4; if(p4==0){ blank=3;}
          pzle[4] = p5; if(p5==0){ blank=4;}
          pzle[5] = p6; if(p6==0){ blank=5;}
          pzle[6] = p7; if(p7==0){ blank=6;}
          pzle[7] = p8; if(p8==0){ blank=7;}
          pzle[8] = p9; if(p9==0){ blank=8;}
     }vs
Puzzle(int...puzzle) {
assert puzzle.length == 9;
this.puzzle = Arrays.copyOf(puzzle, 9);
for (int blank = 0; blank < 9; ++blank) if (puzzle[blank] == 0) break;
// check values are 0...8
Arrays.sort(puzzle); // sorts the original
for (int i = 0; i < 9; ++i) assert puzzle[i] == i;
Does anyone know if there are any simple applets or programs that do 8 Puzzle on any given start and finish state? Just to see the expected results.Not off hand, but you can always generate your starting state by moving backwards from the finish state.
Also, I dont know..but are there some instances of an 8 Puzzle that cannot be solved at all?? IIRC, you can't transform 123 to 132 leaving the remainder unchanged, though I haven't checked.

Depth First Search help

I have an assignment to do on a graph and have to apply depth first search so it ouputs the vertices in a order. I understand the algorithm and can do it on paper but i cannot do it in code so if anyone can help me on this. Here are the classes i am working with
public class MatrixGraph extends AbstractGraph {
    private double[][] matrix;
    public MatrixGraph(int nV, boolean direct, boolean weight ){
        super(nV, direct, weight);
        matrix = new double[nV][nV];
        // if a weighted graph set all values to Double.POSITIVE_INFINITY
        // otherwise set all values to 0
        //complete this code
    public boolean isEdge(int source, int dest) {
        throw new UnsupportedOperationException("Not supported yet.");
    public void insert(Edge edge) {
        throw new UnsupportedOperationException("Not supported yet.");
    public void remove(Edge edge) {
        throw new UnsupportedOperationException("Not supported yet.");
    public void depthFirstTraversal(int start){
        //Output the vertices in depth first order
public abstract class AbstractGraph implements Graph {
    private int numV;
    private boolean directed;
    private boolean weighted;
    public AbstractGraph(int nV, boolean direct, boolean weight){
        numV = nV;
        directed = direct;
        weighted = weight;
    public int getNumV() {
        return numV;
    public boolean isDirected() {
        return directed;
    public boolean isWeighted() {
        return weighted;
public interface Graph {
    //returns the number of vertices
    int getNumV();
    // determine if this is a directed graph
    boolean isDirected();
    // determine if this is a weighted graph
    boolean isWeighted();
    // determine if an edge exists between source and destination
    boolean isEdge(int source, int dest);
    void insert(Edge edge);
    void remove(Edge edge);
}

JavaLearner2009 wrote:
public void depthFirstTraversal(int start){
//Output the vertices in depth first order
}I need help on this part i don't know how to write it.You mentioned you understand the algorithm and can do it on paper so what have you tried so far?
If you are completely stuck on implementing the traversal in Java, provide pseudo code.
Mel

Depth First Search (LIFO) - stacked based

Hi Everyone,
Herewith attached my workable Breadth fist search (FIFO) code - queue based. Now, I would like to change to LIFO - stacked based.
Anyone here can give some guideline on it?
Many thanks.
Attachments:
Version 1.zip ‏64 KB

I've changed it to queue FIFO. but in the end, it won't show a complete path to reach the destination point. However, queue stack LIFO can show a complete path from starting point till the goal point.
Does that means queue FIFO cannot be used in finding a complete path out? Is there any possible way for this method to find its way out?
Herewith attached queue FIFO.
Attachments:
Queue FIFO.zip ‏64 KB

Bfs - A breadth-first version of find

I'm writing a breadth-first version of find. It's available on the AUR(4), and the code is on GitHub.
Currently it doesn't support any of find's options, but I plan to add support for the common ones. It does support a "-nohidden" flag that filters out dotfiles, something surprisingly difficult to do with find. It also supports colorization, respecting LS_COLORS.
The reason I'm writing it is mainly to integrate it with fzf, a terminal fuzzy finder. fzf uses find by default, which means it will often fully explore a very deep directory tree before reaching the nearby file I'm looking for. bfs ensures that shallower files always show up before deeper ones, which usually means it finds the file I want sooner. The colorization is also nice:
$ bfs -color -nohidden | fzf --ansi
Last edited by tavianator (2015-06-20 04:42:39)

ChangBroot wrote:
All the lectures I have read in the net are talking about Queue, but my problem is reading the tree in BFS Traversal order. It's almost 4 days I'm trying to solve this probelm, but can't come up with something. Any help is greatly appreciated. One simple strategy is to make an ordinary recursive traversal of the tree. Along in the traversal you keep an array of linked lists. When you visit a node you just add it to the list at the entry in the array corresponding to the depth of this node.
After the traversal you have an array with linked lists. Each list holds the nodes found at a certain tree depth. So each list holds all nodes corresponding to a specific tree "breadth", namely all nodes found at the same tree depth.

Depth/Breadth First Search

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