I am trying to find the time complexity of the recursive function below. I've tried to draw the tree, but it is confusing because in the if condition the function is called once, and otherwise twice.
To give some context, the function is called on nodes of a tree. The task is to calculate the max rating of each node. The rule is that if you add some node to the rating you can't add it's children to the node, but if you don't add it than you can which children to add or don't.
Here is the function:
static int solve(Node node, boolean take) {
int result;
if(take) {
result = node.rating;
for(Node child : node.children) {
result += solve(child, false);
}
return result;
}
result = 0;
for(Node child : node.children) {
result += Math.max(solve(child, true), solve(child, false));
}
return result;
}
I know how to bind a property, but how can I bind the call of a fuction?
For example: I have a ObjectProperty which points to a file. Now, I want to bind the path to its folder? If the value of ObjectProperty is C:\\user\Desktop\text.txt, the bonding should point to C:\\user\Desktop.
I thought I can call getParentFile() within the binding.
There a many ways to map an ObjectProperty, take a look at the class Bindings.
(All examples assume that you have a ObjectProperty<File> file)
Bindings.createObjectBinding(Callable<T> func, Observable... dependencies)
ObjectBinding<File> parent = Bindings.createObjectBinding(() -> {
File f = file.getValue();
return f == null ? null : f.getParentFile();
}, file);
Bindings.select(ObservableValue<?> root, String... steps)
ObjectBinding<File> parent = Bindings.select(file, "parentFile");
This will print a warning on the error-stream when file is null.
You can also create your own mapping method (which is similar to createObjectBinding):
public static <T,R> ObjectBinding<R> map(ObjectProperty<T> property, Function<T,R> function) {
return new ObjectBinding<R>() {
{
bind(property);
}
#Override
protected R computeValue() {
return function.apply(property.getValue());
}
#Override
public void dispose() {
unbind(property);
}
};
}
And use it
ObjectBinding<File> parent = map(file, f -> f == null ? null : f.getParentFile());
Is it possible to sum up the nodes of a tree using Java 8 streams, if possible in a one liner ?
Here is a node class
public class Node
{
private int nodeNum;
ArrayList<Node> children = new ArrayList<>();
public Node(int num)
{
this.nodeNum = num;
}
public int getNodeNum()
{
return nodeNum;
}
public boolean addNode(Node node)
{
return children.add(node);
}
public ArrayList<Node> getNodes()
{
return this.children;
}
}
Normal way to solve this is using a recursion and sum up the node , like the code below.
int getNodeSum(Node node)
{
int total = 0;
if(node.children.isEmpty())
return node.getNodeNum();
else
{
for(Node tempNode:node.children)
{
total+= getNodeSum(tempNode);
}
return total+node.getNodeNum();
}
}
We can use streams to sum up the immediate child nodes but I'm not getting how to move deep and do it recursively using Streams.
This code only solves the problem to a single level. Any ideas?
total = list.stream().filter(Node -> node.children.isEmpty()).map(Node:: getNodeNum).reduce(node.getNodeNum(), (a,b) -> a+b);
One solution to your problem would be to use recursion along with Stream.flatMap.
First, you'd need to add the following helper method to your Node class:
public Stream<Node> allChildren() {
return Stream.concat(
Stream.of(this),
this.children.stream().flatMap(Node::allChildren)); // recursion here
}
This returns a Stream<Node> whose elements are this node and all its descendant nodes.
Then, you could rewrite your getNodeSum method as follows:
int getNodeSum(Node node) {
return node.allChildren()
.mapToInt(Node::getNodeNum)
.sum();
}
This uses the above defined Node.allChildren method along with the Stream.mapToInt and IntStream.sum methods to calculate the total sum.
Alternatively, you could have a Function<Node, Stream<Node>> descendants attribute in your Node class that performs the recursion in place:
private Function<Node, Stream<Node>> descendants =
node -> Stream.concat(
Stream.of(node),
node.children.stream()
.flatMap(this.descendants)); // recursion here: function invoked again
This is a recursive lambda expression, since the function you are defining is at both sides of the = sign. This kind of lambda expressions are allowed only as attributes of a class, i.e. you cannot assign a recursive lambda expression to a local variable.
With that recursive function in place, you could rewrite the allChildren method as follows:
public Stream<Node> allChildren() {
return descendants.apply(this);
}
Finally, the code for your getNodeSum method would be identical to the previous version:
int getNodeSum(Node node) {
return node.allChildren()
.mapToInt(Node::getNodeNum)
.sum();
}
Note: while this approach might result attractive for some people, it might have some drawbacks, i.e. now every instance of the Node class has the descendants attribute, despite not being needed at all. You could circumvect this i.e. by having a Tree class with this recursive function as an attribute, and Node being an inner class (with the descendants attribute removed).
You need to add recusive method for Node class, which wil be join child streams
public Stream<Node> recursiveConcat() {
return Stream.concat(
Stream.of(this),
children.stream().flatMap(Node::recursiveConcat));
}
Then do -
root.recusiveConcat().mapToInt(Node::getNodeNum).sum()
whole code
public class Node {
private int nodeNum;
ArrayList<Node> children = new ArrayList<>();
public Node(int num) {
this.nodeNum = num;
}
public int getNodeNum() {
return nodeNum;
}
public boolean addNode(Node node) {
return children.add(node);
}
public ArrayList<Node> getNodes() {
return this.children;
}
public Stream<Node> recursiveConcat() {
return Stream.concat(
Stream.of(this),
children.stream().flatMap(Node::recursiveConcat));
}
}
Node root = new Node(1);
Node node1 = new Node(2);
Node node2 = new Node(3);
Node node3 = new Node(4);
node2.addNode(node3);
node1.addNode(node2);
root.addNode(node1);
System.out.println(root.recursiveConcat().mapToInt(Node::getNodeNum).sum());
Given a directed graph with
A root node
Some leaves nodes
Multiple nodes can be connected to the same node
Cycles can exist
We need to print all the paths from the root node to all the leaves nodes. This is the closest question I got to this problem
Find all paths between two graph nodes
If you actually care about ordering your paths from shortest path to longest path then it would be far better to use a modified A* or Dijkstra Algorithm. With a slight modification the algorithm will return as many of the possible paths as you want in order of shortest path first. So if what you really want are all possible paths ordered from shortest to longest then this is the way to go. The code I suggested above would be much slower than it needs to be if you care about ordering from shortest to longest, not to mention would take up more space then you'd want in order to store every possible path at once.
If you want an A* based implementation capable of returning all paths ordered from the shortest to the longest, the following will accomplish that. It has several advantages. First off it is efficient at sorting from shortest to longest. Also it computes each additional path only when needed, so if you stop early because you dont need every single path you save some processing time. It also reuses data for subsequent paths each time it calculates the next path so it is more efficient. Finally if you find some desired path you can abort early saving some computation time. Overall this should be the most efficient algorithm if you care about sorting by path length.
import java.util.*;
public class AstarSearch {
private final Map<Integer, Set<Neighbor>> adjacency;
private final int destination;
private final NavigableSet<Step> pending = new TreeSet<>();
public AstarSearch(Map<Integer, Set<Neighbor>> adjacency, int source, int destination) {
this.adjacency = adjacency;
this.destination = destination;
this.pending.add(new Step(source, null, 0));
}
public List<Integer> nextShortestPath() {
Step current = this.pending.pollFirst();
while( current != null) {
if( current.getId() == this.destination )
return current.generatePath();
for (Neighbor neighbor : this.adjacency.get(current.id)) {
if(!current.seen(neighbor.getId())) {
final Step nextStep = new Step(neighbor.getId(), current, current.cost + neighbor.cost + predictCost(neighbor.id, this.destination));
this.pending.add(nextStep);
}
}
current = this.pending.pollFirst();
}
return null;
}
protected int predictCost(int source, int destination) {
return 0; //Behaves identical to Dijkstra's algorithm, override to make it A*
}
private static class Step implements Comparable<Step> {
final int id;
final Step parent;
final int cost;
public Step(int id, Step parent, int cost) {
this.id = id;
this.parent = parent;
this.cost = cost;
}
public int getId() {
return id;
}
public Step getParent() {
return parent;
}
public int getCost() {
return cost;
}
public boolean seen(int node) {
if(this.id == node)
return true;
else if(parent == null)
return false;
else
return this.parent.seen(node);
}
public List<Integer> generatePath() {
final List<Integer> path;
if(this.parent != null)
path = this.parent.generatePath();
else
path = new ArrayList<>();
path.add(this.id);
return path;
}
#Override
public int compareTo(Step step) {
if(step == null)
return 1;
if( this.cost != step.cost)
return Integer.compare(this.cost, step.cost);
if( this.id != step.id )
return Integer.compare(this.id, step.id);
if( this.parent != null )
this.parent.compareTo(step.parent);
if(step.parent == null)
return 0;
return -1;
}
#Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Step step = (Step) o;
return id == step.id &&
cost == step.cost &&
Objects.equals(parent, step.parent);
}
#Override
public int hashCode() {
return Objects.hash(id, parent, cost);
}
}
/*******************************************************
* Everything below here just sets up your adjacency *
* It will just be helpful for you to be able to test *
* It isnt part of the actual A* search algorithm *
********************************************************/
private static class Neighbor {
final int id;
final int cost;
public Neighbor(int id, int cost) {
this.id = id;
this.cost = cost;
}
public int getId() {
return id;
}
public int getCost() {
return cost;
}
}
public static void main(String[] args) {
final Map<Integer, Set<Neighbor>> adjacency = createAdjacency();
final AstarSearch search = new AstarSearch(adjacency, 1, 4);
System.out.println("printing all paths from shortest to longest...");
List<Integer> path = search.nextShortestPath();
while(path != null) {
System.out.println(path);
path = search.nextShortestPath();
}
}
private static Map<Integer, Set<Neighbor>> createAdjacency() {
final Map<Integer, Set<Neighbor>> adjacency = new HashMap<>();
//This sets up the adjacencies. In this case all adjacencies have a cost of 1, but they dont need to.
addAdjacency(adjacency, 1,2,1,5,1); //{1 | 2,5}
addAdjacency(adjacency, 2,1,1,3,1,4,1,5,1); //{2 | 1,3,4,5}
addAdjacency(adjacency, 3,2,1,5,1); //{3 | 2,5}
addAdjacency(adjacency, 4,2,1); //{4 | 2}
addAdjacency(adjacency, 5,1,1,2,1,3,1); //{5 | 1,2,3}
return Collections.unmodifiableMap(adjacency);
}
private static void addAdjacency(Map<Integer, Set<Neighbor>> adjacency, int source, Integer... dests) {
if( dests.length % 2 != 0)
throw new IllegalArgumentException("dests must have an equal number of arguments, each pair is the id and cost for that traversal");
final Set<Neighbor> destinations = new HashSet<>();
for(int i = 0; i < dests.length; i+=2)
destinations.add(new Neighbor(dests[i], dests[i+1]));
adjacency.put(source, Collections.unmodifiableSet(destinations));
}
}
The output from the above code is the following:
[1, 2, 4]
[1, 5, 2, 4]
[1, 5, 3, 2, 4]
Notice that each time you call nextShortestPath() it generates the next shortest path for you on demand. It only calculates the extra steps needed and doesnt traverse any old paths twice. Moreover if you decide you dont need all the paths and end execution early you've saved yourself considerable computation time. You only compute up to the number of paths you need and no more.
Finally it should be noted that the A* and Dijkstra algorithms do have some minor limitations, though I dont think it would effect you. Namely it will not work right on a graph that has negative weights.
Here is a link to JDoodle where you can run the code yourself in the browser and see it working. You can also change around the graph to show it works on other graphs as well: http://jdoodle.com/a/ukx
I wrote countLeaf method in my binary tree class to count every leaves from root.
However, it gave me stack overflow error, but I couldn't figure what I did wrong.
this is the countLeaf class from my binaryTree
public int countLeaf(Node node){
if(root == null){return 0;} // this part work when I create null Tree
else if(root.left == null && root.right == null){
return 1; //this work when I create tree without left and right
}
else {
System.out.print(root.data); // check infinite loop
return countLeaf(root.left) + countLeaf(root.right);
}
}
And this is my main
public static void main (String[] args){
BinaryTree a = new BinaryTree("A?",
new BinaryTree("B?",
new BinaryTree("D"),
new BinaryTree("E")),
new BinaryTree("C?",
new BinaryTree("E"),
new BinaryTree("F")));
System.out.print(a);
int n = a.countLeaf(a.root);
}
and when I run it, it gave me
A?A?A?A?A?A?A?A?A?A?A?A?A? ... and stackoverflow error
why it keep repeating original root instead of follow left or right??
Replace public int countLeaf(Node node) with public int countLeaf(Node root).
I believe it will help.
Anyways, variable node is never used.