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Depth-First Search puzzle

Not sure how to implement depth first search algorithmn into my code. Here is an example of breadth first search algorithm solving a 8 puzzle:
public direction[] Solve(nPuzzle puzzle) {
//This method uses the fringe as a queue.
//Therefore, nodes are searched in order of cost, with the lowest cost
// unexplored node searched next.
//-----------------------------------------
//put the start state in the Fringe to get explored.
addToFrontier(puzzle.StartState);
ArrayList<PuzzleState> newStates = new ArrayList<PuzzleState>();
while(Frontier.size() > 0)
{
//get the next item off the fringe
PuzzleState thisState = popFrontier();
//is it the goal item?
if(thisState.equals(puzzle.GoalState))
{
//We have found a solution! return it!
return thisState.GetPathToState();
}
else
{
//This isn't the goal, just explore the node
newStates = thisState.explore();
for(int i = 0; i < newStates.size(); i++)
{
//add this state to the fringe, addToFringe() will take care of duplicates
addToFrontier(newStates.get(i));
}
}
}
//No solution found and we've run out of nodes to search
//return null.
return null;
}
There is more code here
Would really appreciate any help

a. The main difference in implementation is: in DFS the nodes to be explored are stored in a stck, while in BFS they are stored in a queue.
b. For the puzzle you are trying to solve I think BFS is more appropriate because you are searching for the shortest path.
For more help post mcve so we do not have to do a lot of guesswork.

Longest path in directed acyclic graph in jgrapht library

I want to find the longest path in directed (acyclic) graph. Let's say that I know starting node - sink. The path should begin from this point.
I was thinking that I can set weights of edges to -1. There is many methods of finding all shortest path but you have to pass ending point. Is it possible to get the shortest path (no matter the end node)?
DirectedAcyclicGraph graph = new DirectedAcyclicGraph<Integer, DefaultEdge>(DefaultEdge.class);
graph.addVertex(1);
graph.addVertex(2);
graph.addVertex(3);
graph.addVertex(4);
graph.addVertex(5);
graph.addVertex(6);
graph.addVertex(7);
graph.addVertex(8);
try {
graph.addDagEdge(1, 2);
graph.addDagEdge(2, 3);
graph.addDagEdge(3, 4);
graph.addDagEdge(5, 6);
graph.addDagEdge(2, 7);
graph.addDagEdge(7, 8);
} catch(Exception e) {
}
//????????????????
Let's say that I'd like to find longest path for node nr 1 (sink). So this algoritm shoud give me 1-2-3-4-5-6.

I was looking for an answer to a similar question to calculate parallel build groupings in Jenkins from a DAG of Git repos. To solve it, I applied the algorithm described here and here. The code below is written in Groovy, so you'll have to convert to Java. The result is a Map of vertices to their respective max depths. From that you can get the single largest value. If instead you want to know the max depth from a specific vertex in the graph, you can first prune the graph into a subgraph rooted by your desired source vertex and then run the method below on the subgraph.
def calcDepths(g) {
Map<String, Integer> vertexToDepthMap = new HashMap<>()
Iterator<String> iterator = new TopologicalOrderIterator<String, DefaultEdge>(g)
iterator.each { v ->
Set<String> predecessors = Graphs.predecessorListOf(g, v).toSet()
Integer maxPredecessorDepth = -1
predecessors.each { predecessor ->
maxPredecessorDepth = Math.max(maxPredecessorDepth, vertexToDepthMap.get(predecessor))
}
vertexToDepthMap.put(v, maxPredecessorDepth + 1)
}
return vertexToDepthMap
}

You can use AllDirectedPaths algorithm to resolve all paths, sort the results by path length in reverse order and get the first:
AllDirectedPaths<String, DefaultEdge> paths = new AllDirectedPaths<String, DefaultEdge>(graph);
GraphPath<String, DefaultEdge> longestPath = paths.getAllPaths(source, target, true, null)
.stream()
.sorted((GraphPath<String, DefaultEdge> path1, GraphPath<String, DefaultEdge> path2)-> new Integer(path2.getLength()).compareTo(path1.getLength()))
.findFirst().get();
System.out.println(longestPath.getLength() + " " + longestPath);

Connecting nodes in a matrix for a graph

Hi I'm working on this little project which requires me to build a matrix in Java which resembles a chess board. I'm supposed to get the Knight to get from a point to another(In the way Knight moves). So I need to find the shortest way to get there in the end.
My problem is, I can't get to connect the edges to get to that point. I can find out if the vertex is a valid move but I can't seem to find a way to create nodes to get to that point. For Example,
0 XXXXX
1 XXXOX
2 XXXXX
3 XXKXX
4 XXXXX
5 XXXXX
I need to create nodes that connect K to O to find out shortest distance later.
PS. I'll be okay with just hints of how to get there or just some tips. Don't really need the exact code. Thank you very much!
I know it's a bad representation of matrix up there but spare me the critique please

A classic Breadth-First-Search is probably the simplest approach:
class Location {
int x;
int y;
List<Location> adjacent() {
// TODO return list of locations reachable in a single step
}
}
List<Location> findShortestPath(Location start, Location destination) {
Location[][] previous = new Location[8][8];
Deque<Location> queue = new ArrayDeque<>();
queue.add(start);
do {
Location loc = queue.poll();
for (Location n : loc.neighbors()) {
if (previous[n.x][n.y] == null) {
previous[n.x][n.y] = loc;
queue.add(n);
if (n.x == destination.x && n.y == destination.y) {
// we've found a way, let's reconstruct the list of steps
List<Location> path = new ArrayList<>();
for (Location l = n; l != start; l = previous[l.x][l.y]) {
path.add(l);
}
path.reverse();
return path;
}
}
}
} while (!queue.isEmpty());
return null; // no path exists
}
This code enumerates all paths from the start location. Therefore, if there is a path to destination, it will find it. In addition, because paths are enumerated in order or ascending length, the first such path will be a shortest one.

The chess board can be implemented by a 2d array. Each cell in the matrix can be considered to be a node (or vertex) in the graph. Edge is composed of two nodes (in this case two cells) one being the from or source [ lets call it Nod A] and other being the to or neighbor or destination node [ Lets call it node B].
Edge exits if there is a possibility of moving from node A to node B.
You can use Dijkstra's algorithm.
http://krishnalearnings.blogspot.in/2015/07/implementation-in-java-for-dijkstras.html
For Node with the Knight's position you can see the possibilities of the cells where Knight can move to and add in the Min Heap. The weight of each edge is constant. You just need to update the cost of the Node.

Efficient algorithm to find all the paths from A to Z?

With a set of random inputs like this (20k lines):
A B
U Z
B A
A C
Z A
K Z
A Q
D A
U K
P U
U P
B Y
Y R
Y U
C R
R Q
A D
Q Z
Find all the paths from A to Z.
A - B - Y - R - Q - Z
A - B - Y - U - Z
A - C - R - Q - Z
A - Q - Z
A - B - Y - U - K - Z
A location cannot appear more than once in the path, hence A - B - Y - U - P - U - Z is not valid.
Locations are named AAA to ZZZ (presented here as A - Z for simplicity) and the input is random in such a way that there may or may not be a location ABC, all locations may be XXX (unlikely), or there may not be a possible path at all locations are "isolated".
Initially I'd thought that this is a variation of the unweighted shortest path problem, but I find it rather different and I'm not sure how does the algorithm there apply here.
My current solution goes like this:
Pre-process the list such that we have a hashmap which points a location (left), to a list of locations (right)
Create a hashmap to keep track of "visited locations". Create a list to store "found paths".
Store X (starting-location) to the "visited locations" hashmap.
Search for X in the first hashmap, (Location A will give us (B, C, Q) in O(1) time).
For-each found location (B, C, Q), check if it is the final destination (Z). If so store it in the "found paths" list. Else if it doesn't already exist in "visited locations" hashmap, Recurl to step 3 now with that location as "X". (actual code below)
With this current solution, it takes forever to map all (not shortest) possible routes from "BKI" to "SIN" for this provided data.
I was wondering if there's a more effective (time-wise) way of doing it. Does anyone know of a better algorithm to find all the paths from an arbitrary position A to an arbitrary position Z ?
Actual Code for current solution:
import java.util.*;
import java.io.*;
public class Test {
private static HashMap<String, List<String>> left_map_rights;
public static void main(String args[]) throws Exception {
left_map_rights = new HashMap<>();
BufferedReader r = new BufferedReader(new FileReader("routes.text"));
String line;
HashMap<String, Void> lines = new HashMap<>();
while ((line = r.readLine()) != null) {
if (lines.containsKey(line)) { // ensure no duplicate lines
continue;
}
lines.put(line, null);
int space_location = line.indexOf(' ');
String left = line.substring(0, space_location);
String right = line.substring(space_location + 1);
if(left.equals(right)){ // rejects entries whereby left = right
continue;
}
List<String> rights = left_map_rights.get(left);
if (rights == null) {
rights = new ArrayList<String>();
left_map_rights.put(left, rights);
}
rights.add(right);
}
r.close();
System.out.println("start");
List<List<String>> routes = GetAllRoutes("BKI", "SIN");
System.out.println("end");
for (List<String> route : routes) {
System.out.println(route);
}
}
public static List<List<String>> GetAllRoutes(String start, String end) {
List<List<String>> routes = new ArrayList<>();
List<String> rights = left_map_rights.get(start);
if (rights != null) {
for (String right : rights) {
List<String> route = new ArrayList<>();
route.add(start);
route.add(right);
Chain(routes, route, right, end);
}
}
return routes;
}
public static void Chain(List<List<String>> routes, List<String> route, String right_most_currently, String end) {
if (right_most_currently.equals(end)) {
routes.add(route);
return;
}
List<String> rights = left_map_rights.get(right_most_currently);
if (rights != null) {
for (String right : rights) {
if (!route.contains(right)) {
List<String> new_route = new ArrayList<String>(route);
new_route.add(right);
Chain(routes, new_route, right, end);
}
}
}
}
}

As I understand your question, Dijkstras algorithm cannot be applied as is, since shortest path problem per definition finds a single path in a set of all possible paths. Your task is to find all paths per-se.
Many optimizations on Dijkstras algorithm involve cutting off search trees with higher costs. You won't be able to cut off those parts in your search, as you need all findings.
And I assume you mean all paths excluding circles.
Algorithm:
Pump network into a 2dim array 26x26 of boolean/integer. fromTo[i,j].
Set a 1/true for an existing link.
Starting from the first node trace all following nodes (search links for 1/true).
Keep visited nodes in a some structure (array/list). Since maximal
depth seems to be 26, this should be possible via recursion.
And as #soulcheck has written below, you may think about cutting of paths you have aleady visted. You may keep a list of paths towards the destination in each element of the array. Adjust the breaking condition accordingly.
Break when
visiting the end node (store the result)
when visiting a node that has been visted before (circle)
visiting a node for which you have already found all paths to the destination and merge your current path with all the existing ones from that node.
Performance wise I'd vote against using hashmaps and lists and prefer static structures.
Hmm, while re-reading the question, I realized that the name of the nodes cannot be limited to A-Z. You are writing something about 20k lines, with 26 letters, a fully connected A-Z network would require far less links. Maybe you skip recursion and static structures :)
Ok, with valid names from AAA to ZZZ an array would become far too large. So you better create a dynamic structure for the network as well. Counter question: regarding performance, what is the best data structure for a less popuplate array as my algorithm would require? I' vote for an 2 dim ArrayList. Anyone?

What you're proposing is a scheme for DFS, only with backtracking.It's correct, unless you want to permit cyclic paths (you didn't specify if you do).
There are two gotchas, though.
You have to keep an eye on nodes you already visited on current path (to eliminate cycles)
You have to know how to select next node when backtracking, so that you don't descend on the same subtree in the graph when you already visited it on the current path.
The pseudocode is more or less as follows:
getPaths(A, current_path) :
if (A is destination node): return [current_path]
for B = next-not-visited-neighbor(A) :
if (not B already on current path)
result = result + getPaths(B, current_path + B)
return result
list_of_paths = getPaths(A, [A])
which is almost what you said.
Be careful though, as finding all paths in complete graph is pretty time and memory consuming.
edit
For clarification, the algorithm has Ω(n!) time complexity in worst case, as it has to list all paths from one vertex to another in complete graph of size n, and there are at least (n-2)! paths of form <A, permutations of all nodes except A and Z, Z>. No way to make it better if only listing the result would take as much.

Your data is essentially an adjacency list which allows you to construct a tree rooted at the node corresponding to A. In order to obtain all the paths between A & Z, you can run any tree traversal algorithm.
Of course, when you're building the tree you have to ensure that you don't introduce cycles.

I would proceed recursively where I would build a list of all possible paths between all pairs of nodes.
I would start by building, for all pairs (X, Y), the list L_2(X, Y) which is the list of paths of length 2 that go from X to Y; that's trivial to build since that's the input list you are given.
Then I would build the lists L_3(X, Y), recursively, using the known lists L_2(X, Z) and L_2(Z, Y), looping over Z. For example, for (C, Q), you have to try all Z in L_2(C, Z) and L_2(Z, Q) and in this case Z can only be R and you get L_3(C, Q) = {C -> R -> Q}. For other pairs, you might have an empty L_3(X, Y), or there could be many paths of length 3 from X to Y.
However you have to be careful here when building the paths here since some of them must be rejected because they have cycles. If a path has twice the same node, it is rejected.
Then you build L_4(X, Y) for all pairs by combining all paths L_2(X, Z) and L_3(Z, Y) while looping over all possible values for Z. You still remove paths with cycles.
And so on... until you get to L_17576(X, Y).
One worry with this method is that you might run out of memory to store those lists. Note however that after having computed the L_4's, you can get rid of the L_3's, etc. Of course you don't want to delete L_3(A, Z) since those paths are valid paths from A to Z.
Implementation detail: you could put L_3(X, Y) in a 17576 x 17576 array, where the element at (X, Y) is is some structure that stores all paths between (X, Y). However if most elements are empty (no paths), you could use instead a HashMap<Pair, Set<Path>>, where Pair is just some object that stores (X, Y). It's not clear to me if most elements of L_3(X, Y) are empty, and if so, if it is also the case for L_4334(X, Y).
Thanks to #Lie Ryan for pointing out this identical question on mathoverflow. My solution is basically the one by MRA; Huang claims it's not valid, but by removing the paths with duplicate nodes, I think my solution is fine.
I guess my solution needs less computations than the brute force approach, however it requires more memory. So much so that I'm not even sure it is possible on a computer with a reasonable amount of memory.

different for loops java

I'm having some difficulties with the following problem:
I'm making a little game where you're at a specific spot and each spot has each some possible directions.
The available directions are N(ord),E(ast),S,W . I use the function getPosDirections to get the possible directions of that spot. The function returns the directions into an ArrayList<String> e.g. for spot J3: [E,W]
Now the game goes like this: 2 dice will be rolled so you get a number between 2 and 12, this number represents the number of steps you can make.
What I want is an ArrayList of all the possible routes
clarification of all the possible routes:
When I'm at the current position I check what the possibilities are from there. Let's say that's go East and go West. So we get 2 new positions and from there on we need to check for the next possibilities again for both positions (until we took x directions)
(x equals the number thrown by the dice).
e.g.: I throw 3 and I'm currently at spot J3:
[[E,N,E],[E,N,S],[E,S,E],[E,S,S],[W,N,E],[W,N,S],[W,S,E],[W,S,S]]
How would obtain the last mentioned Array(list)?

First, you might wish to think about your approach some more. In the worst case (a 12 is rolled, and all 4 directions are possible at every location), there will be 4^12 ~= 160 million routes. Is it really necessary to iterate over them all? And is it necessary to fill about 1 GB of memory to store that list?
Next, it is probably a good idea to represent directions in a type-safe manner, for instance using an enum.
That being said, recursion is your friend:
private void iteratePaths(Location currentLoc, List<Direction> currentPath, List<List<Direction>> allPaths, int pathLength) {
if (currentPath.size() >= pathLength) {
allPaths.add(new ArrayList<Direction>(currentPath));
return;
}
for (Direction d : currentLoc.getPosDirections()) {
currentPath.add(d);
Location newLoc = currentLoc.walk(d);
iteratePaths(newLoc, currentPath, allPaths, pathLength);
currentPath.remove(currentPath.size() - 1);
}
}
public void List<List<Direction>> getAllPaths(Location loc, int length) {
List<List<Direction>> allPaths = new ArrayList<List<Direction>>();
List<Direction> currentPath = new ArrayList<Direction>();
iteratePaths(loc, currentPath, allPaths, length);
return allPaths;
}

You can assume that your field of spots is a complete graph. Then you need to implement BFS or DFS with saving pathes.
You can implement all logic in any of these algorithms (like getting a list of possible directions from a certain node).