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

Similar Messages

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.

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 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, 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...

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 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

DEPTH FIRST SEARCH

hi,
could u help me i want to read tree(node and edges betwwen node) i want to use DFS but it's unclear how to DFS to read data stored in file
do I need to store tree information in file or create it in the same jave
could u help me please

Melanie_Green wrote:
ali99099 wrote:
hi,
could u help me i want to read tree(node and edges betwwen node) i want to use DFS but it's unclear how to DFS to read data stored in file
do I need to store tree information in file or create it in the same jave
could u help me pleaseI don't think I can explain DFS better then the 1 quadrillion sites out there, however I could sum up the algorithm in less then a sentence. Left most child before Right child then Parent, rinse repeat.
You also mention data is stored in a file, as we have NFC what type of data is stored in the file or how the data is stored I can not give you an answer.
p.s. Sentences start with a capital letter.
Melp.s. "then" is an adverb, you want the conjunction "than"

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)

Using depth first traversal to add a new node to a tree with labels

Hello,
I'm currently trying to work my way through Java and need some advice on using and traversing trees. I've written a basic JTree program, which allows the user to add and delete nodes. Each new node is labelled in a sequential order and not dependent upon where they are added to the tree.
Basically, what is the best way to add and delete these new nodes with labels that reflect their position in the tree in a depth-first traversal?
ie: the new node's label will correctly reflect its position in the tree and the other labels will change to reflect this addition of a new node.
I've searched Google and can't seem to find any appropriate examples for this case.
My current code is as follows,
import java.awt.*;
import java.awt.event.*;
import javax.swing.*;
import javax.swing.event.*;
import javax.swing.tree.*;
public class BasicTreeAddDelete extends JFrame implements ActionListener
    private JTree tree;
    private DefaultTreeModel treeModel;
    private JButton addButton;
    private JButton deleteButton;
    private int newNodeSuffix = 1;
    public BasicTreeAddDelete()
        setTitle("Basic Tree with Add and Delete Buttons");
        DefaultMutableTreeNode rootNode = new DefaultMutableTreeNode("Root");
        treeModel = new DefaultTreeModel(rootNode);
        tree = new JTree(treeModel);
        JScrollPane scrollPane = new JScrollPane(tree);
        getContentPane().add(scrollPane, BorderLayout.CENTER);
        JPanel panel = new JPanel();
        addButton = new JButton("Add Node");
        addButton.addActionListener(this);
        panel.add(addButton);
        getContentPane().add(panel, BorderLayout.SOUTH);
        deleteButton = new JButton("Delete Node");
        deleteButton.addActionListener(this);
        panel.add(deleteButton);
        getContentPane().add(panel, BorderLayout.SOUTH);
        setDefaultCloseOperation(EXIT_ON_CLOSE);
        setSize(400, 300);
        setVisible(true);
    public void actionPerformed(ActionEvent event)
        DefaultMutableTreeNode selectedNode = (DefaultMutableTreeNode)tree.getLastSelectedPathComponent();
        if(event.getSource().equals(addButton))
            if (selectedNode != null)
                // add the new node as a child of a selected node at the end
                DefaultMutableTreeNode newNode = new DefaultMutableTreeNode("New Node" + newNodeSuffix++);
                  treeModel.insertNodeInto(newNode, selectedNode, selectedNode.getChildCount());
                  //make the node visible by scrolling to it
                TreeNode[] totalNodes = treeModel.getPathToRoot(newNode);
                TreePath path = new TreePath(totalNodes);
                tree.scrollPathToVisible(path);
        else if(event.getSource().equals(deleteButton))
            //remove the selected node, except the parent node
            removeSelectedNode();
    public void removeSelectedNode()
        DefaultMutableTreeNode selectedNode = (DefaultMutableTreeNode)tree.getLastSelectedPathComponent();
        if (selectedNode != null)
            //get the parent of the selected node
            MutableTreeNode parent = (MutableTreeNode)(selectedNode.getParent());
            // if the parent is not null
            if (parent != null)
                //remove the node from the parent
                treeModel.removeNodeFromParent(selectedNode);
    public static void main(String[] arg)
        BasicTreeAddDelete basicTree = new BasicTreeAddDelete();
}      Thank you for any help.

> Has anybody got any advice, help or know of any
examples for this sort of problem.
Thank you.
Check this site: http://www.apl.jhu.edu/~hall/java/Swing-Tutorial/Swing-Tutorial-JTree.html

What is the non-recursive stack based equivalent of this function?

what is the non-recursive stack based equivalent of this function?
     static void _try (int n)
          int i; if (n==4) print(); else for (i=0; i<4; i++) if (is_free(i,n)) {
               x[n] = i;
               _try(n+1);
     }

It goes infinite no output. Thanks though.
public class CopyOfDamen { // x[i] = x coordinate of queen in row i.
     static int N = 4; static Stack stack = new Stack(); static int [] x = new int[8]; public static void main(String [] par) { _try(); }
     // prints field
     static void print ()
          int i,j;
          System.out.print ("+----------------+\n");
          for (i=0; i<8; i++) {
               System.out.print ("|");
               for (j=0; j<8; j++)
                    if (j==x) System.out.print ("<>"); else System.out.print (" ");
               System.out.print ("|\n");
          System.out.print ("+----------------+\n\n");
     // tests, whether (ix, iy) is beaten by queens 0...(iy-1)
     static boolean is_free (int ix, int iy)
          int i;
          for (i=0; i<iy; i++)
               if ((x[i]==ix) || (Math.abs(x[i]-ix)==Math.abs(i-iy))) return false;
          return true;
     // tries to place queen n on row n
     static void _try () {
          int i = 0, n = 0;
call:
          for(;;) { // forever
               if (n == N) {
                    print();
               } else {
                    for (;i < N; i++) {
                         if (is_free(i,n)) {
                              x[n] = i;
                              System.out.print(x[n] + " ");
                              n++;
                              stack.push(i);
                              i = 0;
                              continue call; // call _try (but first save state and initiate new state)
                    } System.out.println();
               // _try returns (check termination criterion and restore state)
               n--;
               if (n < 0) break; // terminate
               i = stack.pop();
} class Stack {
     int StackSize = 32, top = 0; int [] stack = new int[StackSize]; public Stack() {} void push(int x) { if (top < StackSize) stack[top++] = x; } int pop() { if (top >= 1) return stack[--top]; return -1; }

Implementation of depth-first algorithm

Hi, I am trying to implement depth-first algorithm.(directed graph, no loops). Here what I have so far.
//form a one-element stack with the root node.
Stack st=new st();
boolean goalflg=false;
st.push(?)//do I need to clone the value from the tree, if yes, how can I do this? Should it be a Spanning Tree or not?
//repeat until the first path in the stack terminates at the goal node or the stack is empty
while(st.size()!=null || goalflg == true){
/*-remove the first math from the Stack;create new paths by extending the first path to all neighbors of the terminal node * /
int temp=st.pop();
nodeRight=tree.goRight();// I aren't sure nodeLeft=tree.goLeft();
//reject all new paths with loops (how can I do this?)
//if the goal is found-success, else -failure
if (temp==goalNode)
{goalflg=true;}
else{        //add the new paths to the front of stack
st.push(nodeRight);
st.push(nodeLeft);
Please,please help!
Sincerely,
Byryndyk
P.S. I check out google, but it has mostly the theory about this method. What I am interested is how to make it alive.

This algorithm for DFS has no constraints on edges as it works purely by inspecting adjacency lists; if a directed edge, e, goes from vertex u to vertex v, then v is adjacent to u, but u is not adjacent to v i.e. you can not travel this edge from v to u.
     public VertexModelInterface[] depthFirstSearch(VertexModelInterface from)
          visited.add(from);
          VertexModelInterface[] adjacencyList = (VertexModelInterface[]) from.getAdjacencyList().toArray(new VertexModelInterface[0]);
          for (int index = 0; index < adjacencyList.length; index++)
               if (!visited.contains(adjacencyList[index]))
                    depthFirstSearch(adjacencyList[index]);
               } // end if
          } // end for
          return (VertexModelInterface[]) visited.toArray(new VertexModelInterface[0]);
     } // end depthFirstSearch
visited is a list containing vertices. Each vertex has an adjacency list containing references to adjacent vertices.
Hope this helps anyone whos looking for a neat recursive (implicit stack) implementation. I also have implementations for BFS, Cut-Edge-Detection and Fleury's algorithm.
Kind regards,
Darren Bishop.
[email protected]

Create URL,Telephone number as paramater Account Search and result based on parameter of telephone number ?

Hi Experts,
create URL,Telephone number as paramater Account Search and result based on parameter of telephone number ?..
Previous my thread was locked...Now i want to say thanks to KALYANI L and Richa Dameja,..Now This code i have followed now its working fine..Thanks Great help to Kalyani L..and Richa Dameja.. i have implemented the DO_INIT_CONTEXT.....Now getting result...
As suggest Kalyani L i have implemented DO_INTI_CONTEXT method,Now its working fine....
Thanks For support.
Thanks
kalpana
Message was edited by: Andrei Vishnevsky
Disussion is locked.
Reason: Re: create URL,Telephone number as paramater Account Search and result based on parameter of telephone number ?

Hello Kalpana,
I've already locked your previous discussion and thought that I gave pretty clear warning.
First of all "do my job" posts are not welcomed on SCN.
Second point: you're incorrect in choosing SCN space with such questions. Here is a little which is related to IC in your task.
Third point is that if somebody has an answer to your exact question then he will give you it if he wants. There is no reason to post-post-post messages asking for help or hurry. Your "urgent requirement" is not the reason either.
Fourth one is: according to The SCN Rules of Engagement you need to do the search before posting. Almost all of your questions regarding this topic has an answer already.
Locking the discussion again. If you continue to post such questions I will need to report this situation to SAP CRM space editors and global moderators.

Depth First Search (LIFO) - stacked based

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