below is the code to find if a Hamiltonian path exist in a graph using backtracking. And as per the code below time complexity comes out to be O(V^2)
, where V
is total number to vertices. But Hamiltonian problem is NP-Complete. which as per my understanding, a problem that cannot be solved in polynomial time n^k
, where n
is input and k
is some constant. I have tested the code below and is working fine. So did I calculate time complexity wrong ?
public boolean check() {
Stack<Node> nodeStack = new Stack<>();
nodeStack.add(root);
root.setIsOnStack();
while (!nodeStack.isEmpty()) {
Node currentNode = nodeStack.peek();
for (Entry<Node, Boolean> entry : currentNode.getNeighbourList().entrySet()) {
Node currentNeighbourer = entry.getKey();
if (!currentNeighbourer.isOnStack()) {
if (!entry.getValue()) {
nodeStack.push(currentNeighbourer);
currentNeighbourer.setIsOnStack();
break;
}
} else if (currentNeighbourer == root && nodeStack.size() == noOfVertices) {
return true;
}
}
if (currentNode == nodeStack.peek()) {
for (Entry<Node, Boolean> entry : currentNode.getNeighbourList().entrySet()) {
currentNode.setNodeIsNotVisited(entry.getKey());
}
nodeStack.pop();
currentNode.setIsNotOnStack();
Node nodeOnTop = nodeStack.peek();
nodeOnTop.setNodeIsVisited(currentNode);
}
}
return false;
}
Node class:
public class Node {
private final char label;
private Map<Node, Boolean> neighbourList;
private boolean isOnStack;
public Node(char label) {
this.label = label;
this.isOnStack = false;
neighbourList = new LinkedHashMap<>();
}
public char getLabel() {
return label;
}
public void addNeighbour(Node node) {
neighbourList.put(node, false);
}
public boolean isOnStack() {
return isOnStack;
}
public void setIsOnStack() {
isOnStack = true;
}
public void setIsNotOnStack() {
isOnStack = false;
}
public Map<Node, Boolean> getNeighbourList() {
return neighbourList;
}
public void setNodeIsVisited(Node node) {
neighbourList.replace(node, true);
}
public void setNodeIsNotVisited(Node node) {
neighbourList.replace(node, false);
}
public boolean isNodeVisited(Node node) {
return neighbourList.get(node);
}
}