I'm trying to write a program that'll find the MST of a given undirected weighted graph with Kruskal's and Prim's algorithms. I've successfully implemented Kruskal's algorithm in the program, but I'm having trouble with Prim's. To be more precise, I can't figure out how to actually build the Prim function so that it'll iterate through all the vertices in the graph. I'm getting some IndexOutOfBoundsException errors during program execution. I'm not sure how much information is needed for others to get the idea of what I have done so far, but hopefully there won't be too much useless information.
This is what I have so far:
I have a Graph, Edge and a Vertex class.
Vertex class mostly just an information storage that contains the name (number) of the vertex.
Edge class can create a new Edge that has gets parameters (Vertex start, Vertex end, int edgeWeight). The class has methods to return the usual info like start vertex, end vertex and the weight.
Graph class reads data from a text file and adds new Edges to an ArrayList. The text file also tells us how many vertecis the graph has, and that gets stored too.
In the Graph class, I have a Prim() -method that's supposed to calculate the MST:
public ArrayList<Edge> Prim(Graph G) {
ArrayList<Edge> edges = G.graph; // Copies the ArrayList with all edges in it.
ArrayList<Edge> MST = new ArrayList<Edge>();
Random rnd = new Random();
Vertex startingVertex = edges.get(rnd.nextInt(G.returnVertexCount())).returnStartingVertex(); // This is just to randomize the starting vertex.
// This is supposed to be the main loop to find the MST, but this is probably horribly wrong..
while (MST.size() < returnVertexCount()) {
Edge e = findClosestNeighbour(startingVertex);
MST.add(e);
visited.add(e.returnStartingVertex());
visited.add(e.returnEndingVertex());
edges.remove(e);
}
return MST;
}
The method findClosesNeighbour() looks like this:
public Edge findClosestNeighbour(Vertex v) {
ArrayList<Edge> neighbours = new ArrayList<Edge>();
ArrayList<Edge> edges = graph;
for (int i = 0; i < edges.size() -1; ++i) {
if (edges.get(i).endPoint() == s.returnVertexID() && !visited(edges.get(i).returnEndingVertex())) {
neighbours.add(edges.get(i));
}
}
return neighbours.get(0); // This is the minimum weight edge in the list.
}
ArrayList<Vertex> visited and ArrayList<Edges> graph get constructed when creating a new graph.
Visited() -method is simply a boolean check to see if ArrayList visited contains the Vertex we're thinking about moving to. I tested the findClosestNeighbour() independantly and it seemed to be working but if someone finds something wrong with it then that feedback is welcome also.
Mainly though as I mentioned my problem is with actually building the main loop in the Prim() -method, and if there's any additional info needed I'm happy to provide it.
Thank you.
Edit: To clarify what my train of thought with the Prim() method is. What I want to do is first randomize the starting point in the graph. After that, I will find the closest neighbor to that starting point. Then we'll add the edge connecting those two points to the MST, and also add the vertices to the visited list for checking later, so that we won't form any loops in the graph.
Here's the error that gets thrown:
Exception in thread "main" java.lang.IndexOutOfBoundsException: Index: 0, Size: 0
at java.util.ArrayList.rangeCheck(Unknown Source)
at java.util.ArrayList.get(Unknown Source)
at Graph.findClosestNeighbour(graph.java:203)
at Graph.Prim(graph.java:179)
at MST.main(MST.java:49)
Line 203: return neighbour.get(0); in findClosestNeighbour()
Line 179: Edge e = findClosestNeighbour(startingVertex); in Prim()
Vertex startingVertex = edges.get(rnd.nextInt(G.returnVertexCount())).returnStartingVertex();
This uses the vertex count to index an edge list, mixing up vertices and edges.
// This is supposed to be the main loop to find the MST, but this is probably horribly wrong..
while (MST.size() < returnVertexCount()) {
Edge e = findClosestNeighbour(startingVertex);
MST.add(e);
visited.add(e.returnStartingVertex());
visited.add(e.returnEndingVertex());
edges.remove(e);
}
This shouldn't be passing the same startingVertex to findClosestNeighbour each time.
public Edge findClosestNeighbour(Vertex v) {
ArrayList<Edge> neighbours = new ArrayList<Edge>();
ArrayList<Edge> edges = graph;
for (int i = 0; i < edges.size() -1; ++i) {
if (edges.get(i).endPoint() == s.returnVertexID() && !visited(edges.get(i).returnEndingVertex())) {
neighbours.add(edges.get(i));
}
}
return neighbours.get(0); // This is the minimum weight edge in the list.
}
What is s here? This doesn't look like it's taking the edge weights into account. It's skipping the last edge, and it's only checking the ending vertex, when the edges are non-directional.
// Simple weighted graph representation
// Uses an Adjacency Linked Lists, suitable for sparse graphs /*undirected
9
A
B
C
D
E
F
G
H
I
A B 1
B C 2
C E 7
E G 1
G H 8
F H 3
F D 4
D E 5
I F 9
I A 3
A D 1
This is the graph i used saved as graph.txt
*/
import java.io.*;
import java.util.Scanner;
class Heap
{
private int[] h; // heap array
private int[] hPos; // hPos[h[k]] == k
private int[] dist; // dist[v] = priority of v
private int MAX;
private int N; // heap size
// The heap constructor gets passed from the Graph:
// 1. maximum heap size
// 2. reference to the dist[] array
// 3. reference to the hPos[] array
public Heap(int maxSize, int[] _dist, int[] _hPos)
{
N = 0;
MAX = maxSize;
h = new int[maxSize + 1];
dist = _dist;
hPos = _hPos;
}
public boolean isEmpty()
{
return N == 0;
}
public void siftUp( int k)
{
int v = h[k];
h[0] = 0;
dist[0] = Integer.MIN_VALUE;
//vertex using dist moved up heap
while(dist[v] < dist[h[k/2]]){
h[k] = h[k/2]; //parent vertex is assigned pos of child vertex
hPos[h[k]] = k;//hpos modified for siftup
k = k/2;// index of child assigned last parent to continue siftup
}
h[k] = v;//resting pos of vertex assigned to heap
hPos[v] = k;//index of resting pos of vertex updated in hpos
//display hpos array
/* System.out.println("\nThe following is the hpos array after siftup: \n");
for(int i = 0; i < MAX; i ++){
System.out.println("%d", hPos[i]);
}
System.out.println("\n Following is heap array after siftup: \n");
for (int i = 0; i < MAX; i ++ ){
System.out.println("%d" , h[i]);
}*/
}
//removing the vertex at top of heap
//passed the index of the smallest value in heap
//siftdown resizes and resorts heap
public void siftDown( int k)
{
int v, j;
v = h[k];
while(k <= N/2){
j = 2 * k;
if(j < N && dist[h[j]] > dist[h[j + 1]]) ++j; //if node is > left increment j child
if(dist[v] <= dist[h[j]]) break;//if sizeof parent vertex is less than child stop.
h[k] = h[j];//if parent is greater than child then child assigned parent pos
hPos[h[k]] = k;//update new pos of last child
k = j;//assign vertex new pos
}
h[k] = v;//assign rest place of vertex to heap
hPos[v] = k;//update pos of the vertex in hpos array
}
public void insert( int x)
{
h[++N] = x;//assign new vertex to end of heap
siftUp( N);//pass index at end of heap to siftup
}
public int remove()
{
int v = h[1];
hPos[v] = 0; // v is no longer in heap
h[N+1] = 0; // put null node into empty spot
h[1] = h[N--];//last node of heap moved to top
siftDown(1);//pass index at top to siftdown
return v;//return vertex at top of heap
}
}
class Graph {
class Node {
public int vert;
public int wgt;
public Node next;
}
// V = number of vertices
// E = number of edges
// adj[] is the adjacency lists array
private int V, E;
private Node[] adj;
private Node z;
private int[] mst;
// used for traversing graph
private int[] visited;
private int id;
// default constructor
public Graph(String graphFile) throws IOException
{
int u, v;
int e, wgt;
Node t;
FileReader fr = new FileReader(graphFile);
BufferedReader reader = new BufferedReader(fr);
String splits = " +"; // multiple whitespace as delimiter
String line = reader.readLine();
String[] parts = line.split(splits);
System.out.println("Parts[] = " + parts[0] + " " + parts[1]);
V = Integer.parseInt(parts[0]);
E = Integer.parseInt(parts[1]);
// create sentinel node
z = new Node();
z.next = z;
// create adjacency lists, initialised to sentinel node z
adj = new Node[V+1];
for(v = 1; v <= V; ++v)
adj[v] = z;
// read the edges
System.out.println("Reading edges from text file");
for(e = 1; e <= E; ++e)
{
line = reader.readLine();
parts = line.split(splits);
u = Integer.parseInt(parts[0]);
v = Integer.parseInt(parts[1]);
wgt = Integer.parseInt(parts[2]);
System.out.println("Edge " + toChar(u) + "--(" + wgt + ")--" + toChar(v));
// write code to put edge into adjacency matrix
t = new Node(); t.vert = v; t.wgt = wgt; t.next = adj[u]; adj[u] = t;
t = new Node(); t.vert = u; t.wgt = wgt; t.next = adj[v]; adj[v] = t;
}
}
// convert vertex into char for pretty printing
private char toChar(int u)
{
return (char)(u + 64);
}
// method to display the graph representation
public void display() {
int v;
Node n;
for(v=1; v<=V; ++v){
System.out.print("\nadj[" + toChar(v) + "] ->" );
for(n = adj[v]; n != z; n = n.next)
System.out.print(" |" + toChar(n.vert) + " | " + n.wgt + "| ->");
}
System.out.println("");
}
//use the breath first approach to add verts from the adj list to heap
//uses 3 arrays where array = # of verts in graph
//parent array to keep track of parent verts
// a dist matrix to keep track of dist between it and parent
//hpos array to track pos of vert in the heap
public void MST_Prim(int s)
{
int v, u;
int wgt, wgt_sum = 0;
int[] dist, parent, hPos;
Node t;
//declare 3 arrays
dist = new int[V + 1];
parent = new int[V + 1];
hPos = new int[V +1];
//initialise arrays
for(v = 0; v <= V; ++v){
dist[v] = Integer.MAX_VALUE;
parent[v] = 0;
hPos[v] = 0;
}
dist[s] = 0;
//d.dequeue is pq.remove
Heap pq = new Heap(V, dist, hPos);
pq.insert(s);
while (! pq.isEmpty())
{
// most of alg here
v = pq.remove();
wgt_sum += dist[v];//add the dist/wgt of vert removed to mean spanning tree
//System.out.println("\nAdding to MST edge {0} -- ({1}) -- {2}", toChar(parent[v]), dist[v], toChar[v]);
dist[v] = -dist[v];//mark it as done by making it negative
for(t = adj[v]; t != z; t = t.next){
u = t.vert;
wgt = t.wgt;
if(wgt < dist[u]){ //weight less than current value
dist[u] = wgt;
parent[u] = v;
if(hPos[u] == 0)// not in heap insert
pq.insert(u);
else
pq.siftUp(hPos[u]);//if already in heap siftup the modified heap node
}
}
}
System.out.print("\n\nWeight of MST = " + wgt_sum + "\n");
//display hPos array
/*System.out.println("\nhPos array after siftUp: \n");
for(int i = 0; i < V; i ++){
System.out.println("%d", hPos[i]);
}*/
mst = parent;
}
public void showMST()
{
System.out.print("\n\nMinimum Spanning tree parent array is:\n");
for(int v = 1; v <= V; ++v)
System.out.println(toChar(v) + " -> " + toChar(mst[v]));
System.out.println("");
}
}
public class PrimLists {
public static void main(String[] args) throws IOException
{
int s = 2;
String fname = "graph.txt";
Graph g = new Graph(fname);
g.display();
}
}
Related
I need to get something on this format:
C(N)=2C(N/2)+N
C(N)=O(N〖log〗_2 N)
But I don't understand how the values work because my method has a loop that normaly wont pass through all vertexs and a loop inside of it that normally goes all the way but the number of iterations varies a lot.
...
private static boolean bfs(int[] pathArray, int[] dist) {
//Will make sure nodes are checked in the right order
LinkedList<Integer> queue = new LinkedList<>();
//Starts the array dist[] that has the distance of the vertex to the 1st Critical point; and the array path[] that has the previous vertex;
Arrays.fill(dist, -1);
Arrays.fill(pathArray, -1);
//Make CriticalPoint1 the starting point of the bfs
dist[criticalPoint1] = 0;
queue.add(criticalPoint1);
int adj;
int k;
//Goes to all the adjacent vertexes and sees if any of them are the destination, if they are not adds them to the queue to check their adjacent vertexes
while (!queue.isEmpty()) {
// k is the current vertex
k = queue.remove();
for (Graph.Edge<Integer> allAdjIterator : graph.getVertex(k).getAdjacencies()) {
// adj is the current adjacent vertex
adj = allAdjIterator.getIdAdj();
if (dist[adj]==-1 && !allPassedEdges.contains(graph.getEdge(k,adj))) {
dist[adj] = dist[k] + 1;
pathArray[adj] = k;
queue.add(adj);
if (adj == criticalPoint2) { return true; }
}
}
}
//Critical points aren't connected
return false;
}
...
I've got this code to find a minimum spanning tree for given graph. I need to revert this algorithm to get an hamiltonian cycle instead of hamiltonian path. Can u help me to revert this code to get an hamiltonian cycle? I need it to Travelling Salesman Problem. I tried to remove and change this fragment of code responsible to discarding next_edge if edge causes cycle and add additional edge from vertices 1-2, but when i removed it, a program returns arrayIndexOutOfBoundsException, and when I changed if(x != y) to if(x==y) or delete this conditional, output graph were refilled by 0.
I'm very beginner in programming and I can't resolve this problem by myself.
It is the code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Graph {
// A class to represent a graph edge
class Edge implements Comparable<Edge> {
int src, dest, weight;
// Comparator function used for sorting edges
// based on their weight
public int compareTo(Edge compareEdge) {
return this.weight - compareEdge.weight;
}
};
// A class to represent a subset for union-find
class subset {
int parent, rank;
};
int V, E; // V-> no. of vertices & E->no.of edges
Edge edge[]; // collection of all edges
// Creates a graph with V vertices and E edges
Graph(int v, int e) {
V = v;
E = e;
edge = new Edge[E];
for (int i = 0; i < e; ++i) {
edge[i] = new Edge();
}
}
// A utility function to find set of an element i
// (uses path compression technique)
int find(subset subsets[], int i) {
// find root and make root as parent of i (path compression)
if (subsets[i].parent != i) {
subsets[i].parent = find(subsets, subsets[i].parent);
}
return subsets[i].parent;
}
// A function that does union of two sets of x and y
// (uses union by rank)
void Union(subset subsets[], int x, int y) {
int xroot = find(subsets, x);
int yroot = find(subsets, y);
// Attach smaller rank tree under root of high rank tree
// (Union by Rank)
if (subsets[xroot].rank < subsets[yroot].rank) {
subsets[xroot].parent = yroot;
} else if (subsets[xroot].rank > subsets[yroot].rank) {
subsets[yroot].parent = xroot;
}
// If ranks are same, then make one as root and increment
// its rank by one
else {
subsets[yroot].parent = xroot;
subsets[xroot].rank++;
}
}
// The main function to construct MST using Kruskal's algorithm
void KruskalMST() {
Edge result[] = new Edge[V]; // Tnis will store the resultant MST
int e = 0; // An index variable, used for result[]
int i = 0; // An index variable, used for sorted edges
for (i = 0; i < V; ++i) {
result[i] = new Edge();
}
// Step 1: Sort all the edges in non-decreasing order of their
// weight. If we are not allowed to change the given graph, we
// can create a copy of array of edges
Arrays.sort(edge);
// Allocate memory for creating V ssubsets
subset subsets[] = new subset[V];
for (i = 0; i < V; ++i) {
subsets[i] = new subset();
}
// Create V subsets with single elements
for (int v = 0; v < V; ++v) {
subsets[v].parent = v;
subsets[v].rank = 0;
}
i = 0; // Index used to pick next edge
// Number of edges to be taken is equal to V-1
while (e < V - 1) {
// Step 2: Pick the smallest edge. And increment
// the index for next iteration
Edge next_edge = new Edge();
next_edge = edge[i++];
int x = find(subsets, next_edge.src);
int y = find(subsets, next_edge.dest);
// If including this edge does't cause cycle,
// include it in result and increment the index
// of result for next edge
if (x != y) {
result[e++] = next_edge;
Union(subsets, x, y);
}
// Else discard the next_edge
}
// print the contents of result[] to display
// the built MST
System.out.println("Following are the edges in " +
"the constructed MST");
for (i = 0; i < e; ++i)
System.out.println(result[i].src + " -- " +
result[i].dest + " == " + result[i].weight);
}
// Driver Program
public static void main (String[] args) {
/* Let us create following weighted graph
10
0--------1
| \ |
6| 5\ |15
| \ |
2--------3
4 */
int V = 4; // Number of vertices in graph
int E = 5; // Number of edges in graph
Graph graph = new Graph(V, E);
// add edge 0-1
graph.edge[0].src = 0;
graph.edge[0].dest = 1;
graph.edge[0].weight = 10;
// add edge 0-2
graph.edge[1].src = 0;
graph.edge[1].dest = 2;
graph.edge[1].weight = 6;
// add edge 0-3
graph.edge[2].src = 0;
graph.edge[2].dest = 3;
graph.edge[2].weight = 5;
// add edge 1-3
graph.edge[3].src = 1;
graph.edge[3].dest = 3;
graph.edge[3].weight = 15;
// add edge 2-3
graph.edge[4].src = 2;
graph.edge[4].dest = 3;
graph.edge[4].weight = 4;
graph.KruskalMST();
}
}
public class Problem3 {
public static void main (String [] args){
Scanner sc= new Scanner (System.in);
System.out.println("Enter no. of Islands");
int n= sc.nextInt();
Graph g = new Graph (n);
System.out.println("Enter no. of one-way bridges");
int m= sc.nextInt();
System.out.println("Enter no. of island you want to be intially on");
int r= sc.nextInt();
try{ for (int i=0; i<m;i++){
System.out.println("This one-way bridge connects between");
int u = sc.nextInt();
int v = sc.nextInt();
if(u == v || u>n || v>n){ throw new Bounds("");}
else{ g.addEgde(u-1, v-1);}
}
g.topoSort();}
catch(IndexOutOfBoundsException e){
System.out.println("Please enter a valid input!");
}
catch(Bounds e){
System.out.println("Please enter a valid input!");
}
}
public static class Bounds extends Exception{
public Bounds (String message){
super(message);
}}
static class Graph {
int V;
LinkedList<Integer>[] adjList;
Graph(int V) {
this.V = V;
adjList = new LinkedList[V];
for (int i = 0; i < V; i++) {
adjList[i] = new LinkedList<>();
}
}
public void addEgde(int u, int v) {
adjList[u].addFirst(v);
}
public void topoSort() {
boolean[] visited = new boolean[V];
stack stack = new stack();
for (int i = 0; i < V; i++) {
if (!visited[i]) {
topoSortRE(i, visited, stack);
}
}
System.out.println("Topological Sort: ");
int size = stack.size();
for (int i = 0; i <size ; i++) {
System.out.print(stack.pop()+ 1 + " ");
}
}
public void topoSortRE(int s, boolean[] visited, stack stack) {
visited[s] = true;
for (int i = 0; i < adjList[s].size(); i++) {
int vertex = adjList[s].get(i);
if (!visited[vertex])
topoSortRE(vertex, visited, stack);
}
stack.push(s);
}}}
The following code is an attempt to solve the following problem:
There are many islands that are connected by one-way bridges, that is, if a bridge connects
islands a and b, then you can only use the bridge to go from a to b but you cannot travel back
by using the same. If you are on island a, then you select (uniformly and randomly) one of
the islands that are directly reachable from a through the one-way bridge and move to that
island. You are stuck on an island if you cannot move any further. It is guaranteed that if
there is a directed path from one island to the second there is no path that leads from the
second back to the first. In other words the formed graph is a Directed Acyclic Graph.
Find the island that you are most likely to get stuck on; that is the island that you can
possibly reach with the maximum number of paths from all other islands.
Input format:
First line: Three integers n (the number of islands), m (the number of one-way bridges), and r
(the index of the island you are initially on)
Next m lines: Two integers ui and vi representing a one-way bridge from island ui to vi.
Output format:
Print the index of the island that you are most likely to get stuck on. If there are multiple
islands, then print them in the increasing order of indices (space separated values in a single
line).
Sample input
5, 7, 1
(1, 2)
(1, 3)
(1, 4)
(1, 5)
(2, 4)
(2, 5)
(3, 4)
Sample output
4
I wrote the code to topologically sort the graph but I am having issues with how to make the input r the intial island and also how to make the output be the most probable island to be stuck on. I know that the island I'm most likely to be stuck on is the island that has the most indegrees and no outdegrees but don't know how to implement that.
For each node make a value (real number) representing probability that you will reach it from your starting island. At first, set this value for initial node to 1 and 0 for other nodes.
During the topological sort, when you're in node v, add its probability value divided by number of neighbors to each neighbor's value (in other words, since you know that the probability of getting to v is v.value, then the probability of reaching its neighbor should be increased by v.value * ppb of choosing this neighbor, that is 1 / #_of_neighbors). In this way, whenever you're in some node during topological sort, its value will be equal to the total probability of reaching it.
Your answer is an ending island (node with outdegree 0) with largest value.
Your topological sort seems wrong, you're doing something that looks like DFS. In topological sort you want to visit each vertex after visiting all vertices with an edge ending in it.
About implementation, I changed your DFS into topological sort and added those probabilities I've talked about. I left the part about choosing best ending vertex to you, I don't think doing all the work for someone is educational in any way. Also, I do not guarantee that my changes below do not contain any spelling mistakes, etc. I did my best, but I have not run this code.
static class Graph {
int V;
LinkedList<Integer>[] adjList;
Graph(int V) {
this.V = V;
adjList = new LinkedList[V];
probability = new double[V];
for (int i = 0; i < V; i++) {
adjList[i] = new LinkedList<>();
}
}
public void addEgde(int u, int v) {
adjList[u].addFirst(v);
}
public void topoSort(int start) {
double[] probability;
probability[start] = 1;
int[] indegree = new int[V];
stack stack = new stack();
for (int i = 0; i < V; i++) {
probability[i] = 0;
for (int j = 0; j < adjList[i].size(); ++j) {
indegree[adjList[i][j]] += 1;
}
}
probability[start] = 1;
for(int i = 0; i < V; ++i)
{
if(indegree[i] == 0)
stack.push(i);
}
while(stack.size())
{
int v = stack.pop();
for (int i = 0; i < adjList[v].size(); ++i)
{
indegree[adjList[v][i]] -= 1;
probability[adjList[v][i]] += probability[v] / (double)(adjList[v].size());
if(indegree[adjList[v][i]] == 0)
stack.push(adjList[v][i]);
}
}
//probability array now contains all probabilities to visit each node
}
}
}
In Prim's algorithm, it is recommended to maintain the invariant in the following way :
When a vertice v is added to the MST:
For each edge (v,w) in the unexplored tree:
1. Delete w from the min heap.
2. Recompute the key[w] (i.e. it's value from the unexplored tree
to the explored one).
3. Add the value back to the heap.
So, basically this involves deletion from the heap (and heapify which takes O(logn)) and then reinserting (again O(logn))
Instead, if I use the following approach:
For each edge (v,w) in the unexplored tree:
1. Get the position of the node in the heap(array) using HashMap -> O(1)
2. Update the value in place.
3. Bubble up or bubble down accordingly. -> O(logn)
Which gives better constants than the previous one.
The controversial part is the 3rd part where Im supposed to bubble up or down.
My implementation is as follows :
public int heapifyAt(int index){
// Bubble up
if(heap[index].edgeCost < heap[(int)Math.floor(index/2)].edgeCost){
while(heap[index].edgeCost < heap[(int)Math.floor(index/2)].edgeCost){
swap(index, (int)Math.floor(index/2));
index = (int)Math.floor(index/2);
}
}else{
// Bubble down
while(index*2 + 2 < size && (heap[index].edgeCost > heap[index*2 + 1].edgeCost|| heap[index].edgeCost > heap[index*2 + 2].edgeCost)){
if(heap[index*2 + 1].edgeCost < heap[index*2 + 2].edgeCost){
//swap with left child
swap(index, index*2 + 1);
index = index*2 + 1;
}else{
//swap with right child
swap(index, index*2 + 2);
index = index*2 + 2;
}
}
}
return index;
}
And I'am plucking from the heap this way :
public AdjNode pluck(){
AdjNode min = heap[0];
int minNodeNumber = heap[0].nodeNumber;
AdjNode toRet = new AdjNode(min.nodeNumber, min.edgeCost);
heap[0].edgeCost = INF; // set this to infinity, so it'll be at the bottom
// of the heap.
heapifyat(0);
visited.add(minNodeNumber);
updatevertices(minNodeNumber); // Update the adjacent vertices
return toRet;
}
And updating the plucked vertices this way :
public void updatevertices(int pluckedNode){
for(AdjNode adjacentNode : g.list[pluckedNode]){
if(!visited.contains(adjacentNode.nodeNumber)){ // Skip the nodes that are already visited
int positionInHeap = map.get(adjacentNode.nodeNumber); // Retrive the position from HashMap
if(adjacentNode.edgeCost < heap[positionInHeap].edgeCost){
heap[positionInHeap].edgeCost = adjacentNode.edgeCost; // Update if the cost is better
heapifyAt(positionInHeap); // Now this will go bottom or up, depending on the value
}
}
}
}
But when I execute it on large graph, the code fails, There are small values in the bottom of heap and large values at the top. But the heapifyAt() API seems to work fine. So I am unable to figure out is my approach wrong or my code?
Moreover, if I replace the heapifyAt() API by siftDown(), i.e. construct the heap, it works fine, but it doesnt make sense calling siftDown() that takes O(n) time for every updates which can be processed in logarithmic time.
In short : Is it possible to update the values in Heap both way, or the algorithm is wrong, since that's why it is recommended to first remove the element from Heap and reinsert it.
EDIT : Complete code:
public class Graph1{
public static final int INF = 9999999;
public static final int NEGINF = -9999999;
static class AdjNode{
int nodeNumber;
int edgeCost;
AdjNode next;
AdjNode(int nodeNumber, int edgeCost){
this.nodeNumber = nodeNumber;
this.edgeCost = edgeCost;
}
}
static class AdjList implements Iterable<AdjNode>{
AdjNode head;
AdjList(){
}
public void add(int to, int cost){
if(head==null){
head = new AdjNode(to, cost);
}else{
AdjNode temp = head;
while(temp.next!=null){
temp = temp.next;
}
temp.next = new AdjNode(to, cost);
}
}
public Iterator<AdjNode> iterator(){
return new Iterator<AdjNode>(){
AdjNode temp = head;
public boolean hasNext(){
if(head==null){
return false;
}
return temp != null;
}
public AdjNode next(){
AdjNode ttemp = temp;
temp = temp.next;
return ttemp;
}
public void remove(){
throw new UnsupportedOperationException();
}
};
}
public void printList(){
AdjNode temp = head;
if(head==null){
System.out.println("List Empty");
return;
}
while(temp.next!=null){
System.out.print(temp.nodeNumber + "|" + temp.edgeCost + "-> ");
temp = temp.next;
}
System.out.println(temp.nodeNumber + "|" + temp.edgeCost);
}
}
static class Heap{
int size;
AdjNode[] heap;
Graph g;
int pluckSize;
Set<Integer> visited = new HashSet<Integer>();
HashMap<Integer, Integer> map = new HashMap<>();
Heap(){
}
Heap(Graph g){
this.g = g;
this.size = g.numberOfVertices;
this.pluckSize = size - 1;
heap = new AdjNode[size];
copyElements();
constructHeap();
}
public void copyElements(){
AdjList first = g.list[0];
int k = 0;
heap[k++] = new AdjNode(0, NEGINF); //First entry
for(AdjNode nodes : first){
heap[nodes.nodeNumber] = nodes;
}
for(int i=0; i<size; i++){
if(heap[i]==null){
heap[i] = new AdjNode(i, INF);
}
}
}
public void printHashMap(){
System.out.println("Priniting HashMap");
for(int i=0; i<size; i++){
System.out.println(i + " Pos in heap :" + map.get(i));
}
line();
}
public void line(){
System.out.println("*******************************************");
}
public void printHeap(){
System.out.println("Printing Heap");
for(int i=0; i<size; i++){
System.out.println(heap[i].nodeNumber + " | " + heap[i].edgeCost);
}
line();
}
public void initializeMap(){
for(int i=0; i<size; i++){
map.put(heap[i].nodeNumber, i);
}
}
public void swap(int one, int two){
AdjNode first = heap[one];
AdjNode second = heap[two];
map.put(first.nodeNumber, two);
map.put(second.nodeNumber, one);
AdjNode temp = heap[one];
heap[one] = heap[two];
heap[two] = temp;
}
public void constructHeap(){
for(int i=size-1; i>=0; i--){
int temp = i;
while(heap[temp].edgeCost < heap[(int)Math.floor(temp/2)].edgeCost){
swap(temp, (int)Math.floor(temp/2));
temp = (int)Math.floor(temp/2);
}
}
initializeMap();
}
public void updatevertices(int pluckedNode){
for(AdjNode adjacentNode : g.list[pluckedNode]){
if(!visited.contains(adjacentNode.nodeNumber)){
int positionInHeap = map.get(adjacentNode.nodeNumber);
if(adjacentNode.edgeCost < heap[positionInHeap].edgeCost){
// //System.out.println(adjacentNode.nodeNumber + " not visited, Updating vertice " + heap[positionInHeap].nodeNumber + " from " + heap[positionInHeap].edgeCost + " to " + adjacentNode.edgeCost);
// heap[positionInHeap].edgeCost = INF;
// //heap[positionInHeap].edgeCost = adjacentNode.edgeCost;
// int heapifiedIndex = heapifyAt(positionInHeap); // This code follows my logic
// heap[heapifiedIndex].edgeCost = adjacentNode.edgeCost; // (which doesnt work)
// //heapifyAt(size - 1);
heap[positionInHeap].edgeCost = adjacentNode.edgeCost;
//heapifyAt(positionInHeap);
constructHeap(); // When replaced by SiftDown,
} // works as charm
}
}
}
public void printSet(){
Iterator<Integer> it = visited.iterator();
System.out.print("Printing set : [");
while(it.hasNext()){
System.out.print((int)it.next() + ", ");
}
System.out.println("]");
}
public AdjNode pluck(){
AdjNode min = heap[0];
int minNodeNumber = heap[0].nodeNumber;
AdjNode toRet = new AdjNode(min.nodeNumber, min.edgeCost);
heap[0].edgeCost = INF;
constructHeap();
visited.add(minNodeNumber);
updatevertices(minNodeNumber);
return toRet;
}
public int heapifyAt(int index){
if(heap[index].edgeCost < heap[(int)Math.floor(index/2)].edgeCost){
while(heap[index].edgeCost < heap[(int)Math.floor(index/2)].edgeCost){
swap(index, (int)Math.floor(index/2));
index = (int)Math.floor(index/2);
}
}else{
if(index*2 + 2 < size){
while(index*2 + 2 < size && (heap[index].edgeCost > heap[index*2 + 1].edgeCost|| heap[index].edgeCost > heap[index*2 + 2].edgeCost)){
if(heap[index*2 + 1].edgeCost < heap[index*2 + 2].edgeCost){
//swap with left child
swap(index, index*2 + 1);
index = index*2 + 1;
}else{
//swap with right child
swap(index, index*2 + 2);
index = index*2 + 2;
}
}
}
}
return index;
}
}
static class Graph{
int numberOfVertices;
AdjList[] list;
Graph(int numberOfVertices){
list = new AdjList[numberOfVertices];
for(int i=0; i<numberOfVertices; i++){
list[i] = new AdjList();
}
this.numberOfVertices = numberOfVertices;
}
public void addEdge(int from, int to, int cost){
this.list[from].add(to, cost);
this.list[to].add(from, cost);
}
public void printGraph(){
System.out.println("Printing Graph");
for(int i=0; i<numberOfVertices; i++){
System.out.print(i + " = ");
list[i].printList();
}
}
}
public static void prims(Graph graph, Heap heap){
int totalMin = INF;
int tempSize = graph.numberOfVertices;
while(tempSize>0){
AdjNode min = heap.pluck();
totalMin += min.edgeCost;
System.out.println("Added cost : " + min.edgeCost);
tempSize--;
}
System.out.println("Total min : " + totalMin);
}
public static void main(String[] args) throws Throwable {
Scanner in = new Scanner(new File("/home/mayur/Downloads/PrimsInput.txt"));
Graph graph = new Graph(in.nextInt());
in.nextInt();
while(in.hasNext()){
graph.addEdge(in.nextInt() - 1, in.nextInt() - 1, in.nextInt());
}
Heap heap = new Heap(graph);
prims(graph, heap);
}
}
With a proper implementation of heap, you should be able to bubble up and down. Heap preserves a group of elements using an order that applies to both directions and bubbling up and down are essentially the same, apart from the direction in which you move.
As to your implementation, I believe you are correct but one, seemingly minor issue: indexing.
If you look around for array implementations of heap, you will notice that in most cases the root is located at index 1, instead of 0. The reason for that being, in a 1-indexed array you preserve the following relation between parent p and children c1 and c2.
heap[i] = p
heap[2 * i] = c1
heap[2 * i + 1] = c2
It is trivial to draw an array on a piece of paper and see that this relation holds, if you have the root at heap[1]. The children of root, at index 1, are located at indices 2 and 3. Children of the node at index 2 are at indices 4 & 5, while children of the node at index 3 are at indices 6 & 7, and so on.
This relation helps you get to the children or the parent of any node at i, without having to keep track of where they are. (i.e. parent is at floor(i/2) and children are at 2i and 2i+1)
What you seem to have tried is a 0-indexed implementation of heap. Consequently you had to use a slightly different relation given below for parent p and children c1 and c2.
heap[i] = p
heap[2 * i + 1] = c1
heap[2 * i + 2] = c2
This seems to be ok when accessing the children. For example, the children of root, at index 0, are located at indices 1 and 2. Children of the node at index 1 are at indices 3 & 4, while children of the node at index 2 are at indices 5 & 6, and so on. However there is a pickle when accessing the parent of a node. If you consider node 3 and take floor(3/2), you do get index 1, which is the parent of 1. However, if you take the node at index 4, floor(4/2) gives you index 2, which is not the parent of the node at index 4.
Obviously, this adaptation of the index relationship between a parent and its children does not work for both children. Unlike the 1-indexed heap implementation, you can not treat both children the same while accessing their parents. Therefore, the problem lies specifically in your bubbling up part, without necessarily being related to bubbling up operation. As a matter of fact, though I haven't tested your code, the bubbling up portion of heapifyAt function seems to be correct.(i.e. except the indexing, of course)
Now, you may keep using a 0-indexed heap and adapt your code so that whenever you are looking for a node's parent, you implicitly check whether it is the right (i.e. not as in correct but as in the opposite of left) child of that parent and use floor((i-1)/2) if it is. Checking whether a node is the right child is trivial: just look if it is even or not. (i.e. as you index right children with 2i + 2, they will always be even)
However I recommend you take a different approach and instead use a 1-indexed array implementation of heap. The elegance of the array implementation of heap is that you can treat each node the same and you don't have to do anything different based on its index or location, with the root of the heap perhaps being the only possible exception to this.
I have an assignment where I have to compute the Chromatic Polynomials of a graph. According to the “Fundamental Reduction Theorem”:
A recursive way of computing the chromatic polynomial is based on edge contraction: for a pair of adjacent vertices u and v the graph Guv is obtained by merging the two vertices and removing the edge between them. Then the chromatic polynomial satisfies the recurrence relation:
P(G, x) = P(G − uv, x) − P(Guv, x)
where u and v are adjacent vertices and G − uv is the graph with the edge uv removed.
Likewise, Chromatic Polynomials can be found when by:
A recursive way of computing the chromatic polynomial is based on edge contraction: for a pair of non-adjacent vertices u and v the graph Guv is obtained by merging the two vertices and adding the edge between them. Then the chromatic polynomial satisfies the recurrence relation
P(G, x) = P(G + uv, x) + P(Guv, x)
where u and v are adjacent vertices and G + uv is the graph with the edge uv added.
It was determined for this assignment that when we want to make null graphs based on the previous formula was when the edges of the graph is <= (the number of vertices) * ((number of vertices-1)/4). This is the “break point” for determining if it is more efficient to go toward null graphs or complete graphs.
Additionally, I have a class for creating Polynomials. So I don’t have to post two classes I’ll the Polynomial class here summarize here:
Constructors:
Polynomial() – creates a constant polynomial P(x) = 1.
Polynomial(int a)- creates a linear polynomial of the form P(x) = x + a.
Polynomial(int a, int b) – creates a polynomial of the form a * x^b.
int degree() – returns the degree of the polynomial.
Polynomial plus(Polynomial b) – returns the sum of this polynomial and b, i.e, c = this + b
Polynomial minus(Polynomial b) – returns the difference of this polynomial and , i.e, c = this–b
Polynomial times(Polynomial b) – returns the product of this polynomial and b, return (this*b)
Polynomial compose(Polynomial b)- returns the composite of this polynomial and b. return this(b(x)) – compute using Horner’s method.
boolean equals(Polynomial b)- returns true whenever this polynomial and b are identical.
int evaluate(int x)- evaluate this polynomial at x. return this(x).
Polynomial differentiate()- return the derivative of the polynomial.
String toString()- return the textual representation of this polynomial.
The following is the Graph class. It uses a Bag data structure array that is implemented using a linked list. So for example, all vertices that are adjacent to vertex 0 will be in the Bag located at Bag[0]. I need help doing the recursive part where it computes a Chromatic Polynomial. I have some comments of questions I have in the chromaticPolynomial method as well as mergeVertices method. I am not sure if what I am thinking in these two methods is correct and would appreciate anyone’s help on it. Thanks in advance.
import java.util.Arrays;
public class Graph {
private final int V;
private int E;
private Bag<Integer>[] adj;
/**
* Create an empty graph with V vertices.
*/
public Graph(int V) {
if (V < 0) throw new RuntimeException("Number of vertices must be nonnegative");
this.V = V;
this.E = 0;
adj = (Bag<Integer>[]) new Bag[V];
for (int v = 0; v < V; v++) {
adj[v] = new Bag<Integer>();
}
}
/**
* Create a random graph with V vertices and E edges.
* Expected running time is proportional to V + E.
*/
public Graph(int V, int E) {
this(V);
if (E < 0) throw new RuntimeException("Number of edges must be nonnegative");
for (int i = 0; i < E; i++) {
int v = (int) (Math.random() * V);
int w = (int) (Math.random() * V);
addEdge(v, w);
}
}
/**
* Create a digraph from input stream.
*/
public Graph(In in) {
this(in.readInt());
int E = in.readInt();
for (int i = 0; i < E; i++) {
int v = in.readInt();
int w = in.readInt();
addEdge(v, w);
}
}
/**
* Copy constructor.
*/
public Graph(Graph G) {
this(G.V());
this.E = G.E();
for (int v = 0; v < G.V(); v++) {
// reverse so that adjacency list is in same order as original
Stack<Integer> reverse = new Stack<Integer>();
for (int w : G.adj[v]) {
reverse.push(w);
}
for (int w : reverse) {
adj[v].add(w);
}
}
}
/**
* Return the number of vertices in the graph.
*/
public int V() { return V; }
/**
* Return the number of edges in the graph.
*/
public int E() { return E; }
/**
* Add the edge v-w to graph.
*/
public void addEdge(int v, int w) {
E++;
adj[v].add(w);
adj[w].add(v);
}
/**
* Remove edge from v-w to graph.
*/
public void removeEdge(int v, int w){
E--;
adj[v].remove(w);
adj[w].remove(v);
}
public Polynomial chromaticPolynomial(Graph G){
// null graph is reached, return a polynomial that is G1.minus(G2)
if(G.V() == 0) return new Polynomial(); // how do I create a Chromatic Polynomial when this condition is reached?
// complete graph is reached, return a polynomial that is G1.plus(G2). If the number of vertices in a complete graph is n, then the number of edges is equal to n(n-1)/2.
if(G.E() == ((G.V() - (G.V() - 1))/2)) return new Polynomial(); // how do I create a Chromatic Polynomial when this condtion is reached?
// head toward a null graph.
if(G.E() <= G.V()*(G.V()-1/4)){
/**
for every vertex (u) and every edge (v) from that vertex, create a graph that has an edge removed (G1)
and a graph with merged vertices (G2), then recurse until a null graph is reached.
*/
for(int u = 0; u < G.V(); u++){
for(int v : adj[u]){
Graph G1 = new Graph(G);
Graph G2 = new Graph(G);
G1.removeEdge(u,v);
G2.mergeVertices(u,v);
chromaticPolynomial(G1);
chromaticPolynomial(G2);
// reutrn a polynomial here?
}
}
}
// head towards a complete graph.
else{
/**
for every vertex (u) and every edge (v) from that vertex, create a graph that has an edge added (G1) and
a graph with merged verticies (G2), then recures unitl all these graphs are a complete graph.
*/
for(int u = 0; u < G.V(); u++){
for(int v : adj[u]){
Graph G1 = new Graph(G);
Graph G2 = new Graph(G);
G1.addEdge(u,v);
G2.mergeVertices(u,v);
chromaticPolynomial(G1);
chromaticPolynomial(G2);
// return a polynomial here?
}
}
}
// return nothing. It didn't work.
return null;
}
public void mergeVertices(int vert1, int vert2){
// edge has already been removed between these two.
// just have to renumber the vertices so one is "removed"
/**
does this look correct?
*/
for(int u = 0; u < this.V(); u++){
for(int w: adj[u]){
if(w > vert1 || w > vert2) w--;
if(u > vert1 || u > vert2) u--;
}
}
}
/**
* Return the list of neighbors of vertex v as in Iterable.
*/
public Iterable<Integer> adj(int v) {
return adj[v];
}
/**
* Return a string representation of the graph.
*/
public String toString() {
StringBuilder s = new StringBuilder();
String NEWLINE = System.getProperty("line.separator");
s.append(V + " vertices, " + E + " edges " + NEWLINE);
for (int v = 0; v < V; v++) {
s.append(v + ": ");
for (int w : adj[v]) {
s.append(w + " ");
}
s.append(NEWLINE);
}
return s.toString();
}