We have introduced Branch and Bound and discussed the 0/1 Knapsack problem in the below posts.
- Branch and Bound | Set 1 (Introduction with 0/1 Knapsack)
- Branch and Bound | Set 2 (Implementation of 0/1 Knapsack)
In this puzzle solution of the 8 puzzle problem is discussed.
Given a 3×3 board with 8 tiles (every tile has one number from 1 to 8) and one empty space. The objective is to place the numbers on tiles to match the final configuration using the empty space. We can slide four adjacent (left, right, above, and below) tiles into the empty space.
For example,
1. DFS (Brute-Force)
We can perform a depth-first search on state-space (Set of all configurations of a given problem i.e. all states that can be reached from the initial state) tree.
In this solution, successive moves can take us away from the goal rather than bringing us closer. The search of state-space tree follows the leftmost path from the root regardless of the initial state. An answer node may never be found in this approach.
2. BFS (Brute-Force)
We can perform a Breadth-first search on the state space tree. This always finds a goal state nearest to the root. But no matter what the initial state is, the algorithm attempts the same sequence of moves like DFS.
3. Branch and Bound
The search for an answer node can often be speeded by using an “intelligent” ranking function, also called an approximate cost function to avoid searching in sub-trees that do not contain an answer node. It is similar to the backtracking technique but uses a BFS-like search.
There are basically three types of nodes involved in Branch and Bound
1. Live node is a node that has been generated but whose children have not yet been generated.
2. E-node is a live node whose children are currently being explored. In other words, an E-node is a node currently being expanded.
3. Dead node is a generated node that is not to be expanded or explored any further. All children of a dead node have already been expanded.
Cost function:
Each node X in the search tree is associated with a cost. The cost function is useful for determining the next E-node. The next E-node is the one with the least cost. The cost function is defined as
C(X) = g(X) + h(X) where
g(X) = cost of reaching the current node
from the root
h(X) = cost of reaching an answer node from X.
The ideal Cost function for an 8-puzzle Algorithm :
We assume that moving one tile in any direction will have a 1 unit cost. Keeping that in mind, we define a cost function for the 8-puzzle algorithm as below:
c(x) = f(x) + h(x) where
f(x) is the length of the path from root to x
(the number of moves so far) and
h(x) is the number of non-blank tiles not in
their goal position (the number of mis-
-placed tiles). There are at least h(x)
moves to transform state x to a goal state
An algorithm is available for getting an approximation of h(x) which is an unknown value.
Complete Algorithm:
/* Algorithm LCSearch uses c(x) to find an answer node
* LCSearch uses Least() and Add() to maintain the list
of live nodes
* Least() finds a live node with least c(x), deletes
it from the list and returns it
* Add(x) adds x to the list of live nodes
* Implement list of live nodes as a min-heap */
struct list_node
{
list_node *next;
// Helps in tracing path when answer is found
list_node *parent;
float cost;
}
algorithm LCSearch(list_node *t)
{
// Search t for an answer node
// Input: Root node of tree t
// Output: Path from answer node to root
if (*t is an answer node)
{
print(*t);
return;
}
E = t; // E-node
Initialize the list of live nodes to be empty;
while (true)
{
for each child x of E
{
if x is an answer node
{
print the path from x to t;
return;
}
Add (x); // Add x to list of live nodes;
x->parent = E; // Pointer for path to root
}
if there are no more live nodes
{
print ("No answer node");
return;
}
// Find a live node with least estimated cost
E = Least();
// The found node is deleted from the list of
// live nodes
}
}
The below diagram shows the path followed by the above algorithm to reach the final configuration from the given initial configuration of the 8-Puzzle. Note that only nodes having the least value of cost function are expanded.
// Program to print path from root node to destination node // for N*N -1 puzzle algorithm using Branch and Bound // The solution assumes that instance of puzzle is solvable #include <bits/stdc++.h> using namespace std;
#define N 3 // state space tree nodes struct Node
{ // stores the parent node of the current node
// helps in tracing path when the answer is found
Node* parent;
// stores matrix
int mat[N][N];
// stores blank tile coordinates
int x, y;
// stores the number of misplaced tiles
int cost;
// stores the number of moves so far
int level;
}; // Function to print N x N matrix int printMatrix( int mat[N][N])
{ for ( int i = 0; i < N; i++)
{
for ( int j = 0; j < N; j++)
printf ( "%d " , mat[i][j]);
printf ( "\n" );
}
} // Function to allocate a new node Node* newNode( int mat[N][N], int x, int y, int newX,
int newY, int level, Node* parent)
{ Node* node = new Node;
// set pointer for path to root
node->parent = parent;
// copy data from parent node to current node
memcpy (node->mat, mat, sizeof node->mat);
// move tile by 1 position
swap(node->mat[x][y], node->mat[newX][newY]);
// set number of misplaced tiles
node->cost = INT_MAX;
// set number of moves so far
node->level = level;
// update new blank tile coordinates
node->x = newX;
node->y = newY;
return node;
} // bottom, left, top, right int row[] = { 1, 0, -1, 0 };
int col[] = { 0, -1, 0, 1 };
// Function to calculate the number of misplaced tiles // ie. number of non-blank tiles not in their goal position int calculateCost( int initial[N][N], int final[N][N])
{ int count = 0;
for ( int i = 0; i < N; i++)
for ( int j = 0; j < N; j++)
if (initial[i][j] && initial[i][j] != final[i][j])
count++;
return count;
} // Function to check if (x, y) is a valid matrix coordinate int isSafe( int x, int y)
{ return (x >= 0 && x < N && y >= 0 && y < N);
} // print path from root node to destination node void printPath(Node* root)
{ if (root == NULL)
return ;
printPath(root->parent);
printMatrix(root->mat);
printf ( "\n" );
} // Comparison object to be used to order the heap struct comp
{ bool operator()( const Node* lhs, const Node* rhs) const
{
return (lhs->cost + lhs->level) > (rhs->cost + rhs->level);
}
}; // Function to solve N*N - 1 puzzle algorithm using // Branch and Bound. x and y are blank tile coordinates // in initial state void solve( int initial[N][N], int x, int y,
int final[N][N])
{ // Create a priority queue to store live nodes of
// search tree;
priority_queue<Node*, std::vector<Node*>, comp> pq;
// create a root node and calculate its cost
Node* root = newNode(initial, x, y, x, y, 0, NULL);
root->cost = calculateCost(initial, final);
// Add root to list of live nodes;
pq.push(root);
// Finds a live node with least cost,
// add its childrens to list of live nodes and
// finally deletes it from the list.
while (!pq.empty())
{
// Find a live node with least estimated cost
Node* min = pq.top();
// The found node is deleted from the list of
// live nodes
pq.pop();
// if min is an answer node
if (min->cost == 0)
{
// print the path from root to destination;
printPath(min);
return ;
}
// do for each child of min
// max 4 children for a node
for ( int i = 0; i < 4; i++)
{
if (isSafe(min->x + row[i], min->y + col[i]))
{
// create a child node and calculate
// its cost
Node* child = newNode(min->mat, min->x,
min->y, min->x + row[i],
min->y + col[i],
min->level + 1, min);
child->cost = calculateCost(child->mat, final);
// Add child to list of live nodes
pq.push(child);
}
}
}
} // Driver code int main()
{ // Initial configuration
// Value 0 is used for empty space
int initial[N][N] =
{
{1, 2, 3},
{5, 6, 0},
{7, 8, 4}
};
// Solvable Final configuration
// Value 0 is used for empty space
int final[N][N] =
{
{1, 2, 3},
{5, 8, 6},
{0, 7, 4}
};
// Blank tile coordinates in initial
// configuration
int x = 1, y = 2;
solve(initial, x, y, final);
return 0;
} |
// Java Program to print path from root node to destination node // for N*N -1 puzzle algorithm using Branch and Bound // The solution assumes that instance of puzzle is solvable import java.io.*;
import java.util.*;
class GFG
{ public static int N = 3 ;
public static class Node
{
// stores the parent node of the current node
// helps in tracing path when the answer is found
Node parent;
int mat[][] = new int [N][N]; // stores matrix
int x, y; // stores blank tile coordinates
int cost; // stores the number of misplaced tiles
int level; // stores the number of moves so far
}
// Function to print N x N matrix
public static void printMatrix( int mat[][]){
for ( int i = 0 ; i < N; i++){
for ( int j = 0 ; j < N; j++){
System.out.print(mat[i][j]+ " " );
}
System.out.println( "" );
}
}
// Function to allocate a new node
public static Node newNode( int mat[][], int x, int y,
int newX, int newY, int level,
Node parent){
Node node = new Node();
node.parent = parent; // set pointer for path to root
// copy data from parent node to current node
node.mat = new int [N][N];
for ( int i = 0 ; i < N; i++){
for ( int j = 0 ; j < N; j++){
node.mat[i][j] = mat[i][j];
}
}
// move tile by 1 position
int temp = node.mat[x][y];
node.mat[x][y] = node.mat[newX][newY];
node.mat[newX][newY]=temp;
node.cost = Integer.MAX_VALUE; // set number of misplaced tiles
node.level = level; // set number of moves so far
// update new blank tile coordinates
node.x = newX;
node.y = newY;
return node;
}
// bottom, left, top, right
public static int row[] = { 1 , 0 , - 1 , 0 };
public static int col[] = { 0 , - 1 , 0 , 1 };
// Function to calculate the number of misplaced tiles
// ie. number of non-blank tiles not in their goal position
public static int calculateCost( int initialMat[][], int finalMat[][])
{
int count = 0 ;
for ( int i = 0 ; i < N; i++)
for ( int j = 0 ; j < N; j++)
if (initialMat[i][j]!= 0 && initialMat[i][j] != finalMat[i][j])
count++;
return count;
}
// Function to check if (x, y) is a valid matrix coordinate
public static int isSafe( int x, int y)
{
return (x >= 0 && x < N && y >= 0 && y < N)? 1 : 0 ;
}
// print path from root node to destination node
public static void printPath(Node root){
if (root == null ){
return ;
}
printPath(root.parent);
printMatrix(root.mat);
System.out.println( "" );
}
// Comparison object to be used to order the heap
public static class comp implements Comparator<Node>{
@Override
public int compare(Node lhs, Node rhs){
return (lhs.cost + lhs.level) > (rhs.cost+rhs.level)? 1 :- 1 ;
}
}
// Function to solve N*N - 1 puzzle algorithm using
// Branch and Bound. x and y are blank tile coordinates
// in initial state
public static void solve( int initialMat[][], int x,
int y, int finalMat[][])
{
// Create a priority queue to store live nodes of search tree
PriorityQueue<Node> pq = new PriorityQueue<>( new comp());
// create a root node and calculate its cost
Node root = newNode(initialMat, x, y, x, y, 0 , null );
root.cost = calculateCost(initialMat,finalMat);
// Add root to list of live nodes;
pq.add(root);
// Finds a live node with least cost,
// add its childrens to list of live nodes and
// finally deletes it from the list.
while (!pq.isEmpty())
{
Node min = pq.peek(); // Find a live node with least estimated cost
pq.poll(); // The found node is deleted from the list of live nodes
// if min is an answer node
if (min.cost == 0 ){
printPath(min); // print the path from root to destination;
return ;
}
// do for each child of min
// max 4 children for a node
for ( int i = 0 ; i < 4 ; i++)
{
if (isSafe(min.x + row[i], min.y + col[i])> 0 )
{
// create a child node and calculate
// its cost
Node child = newNode(min.mat, min.x, min.y, min.x + row[i],min.y + col[i], min.level + 1 , min);
child.cost = calculateCost(child.mat, finalMat);
// Add child to list of live nodes
pq.add(child);
}
}
}
}
//Driver Code
public static void main (String[] args)
{
// Initial configuration
// Value 0 is used for empty space
int initialMat[][] =
{
{ 1 , 2 , 3 },
{ 5 , 6 , 0 },
{ 7 , 8 , 4 }
};
// Solvable Final configuration
// Value 0 is used for empty space
int finalMat[][] =
{
{ 1 , 2 , 3 },
{ 5 , 8 , 6 },
{ 0 , 7 , 4 }
};
// Blank tile coordinates in initial
// configuration
int x = 1 , y = 2 ;
solve(initialMat, x, y, finalMat);
}
} // This code is contributed by shruti456rawal |
# Python3 program to print the path from root # node to destination node for N*N-1 puzzle # algorithm using Branch and Bound # The solution assumes that instance of # puzzle is solvable # Importing copy for deepcopy function import copy
# Importing the heap functions from python # library for Priority Queue from heapq import heappush, heappop
# This variable can be changed to change # the program from 8 puzzle(n=3) to 15 # puzzle(n=4) to 24 puzzle(n=5)... n = 3
# bottom, left, top, right row = [ 1 , 0 , - 1 , 0 ]
col = [ 0 , - 1 , 0 , 1 ]
# A class for Priority Queue class priorityQueue:
# Constructor to initialize a
# Priority Queue
def __init__( self ):
self .heap = []
# Inserts a new key 'k'
def push( self , k):
heappush( self .heap, k)
# Method to remove minimum element
# from Priority Queue
def pop( self ):
return heappop( self .heap)
# Method to know if the Queue is empty
def empty( self ):
if not self .heap:
return True
else :
return False
# Node structure class node:
def __init__( self , parent, mat, empty_tile_pos,
cost, level):
# Stores the parent node of the
# current node helps in tracing
# path when the answer is found
self .parent = parent
# Stores the matrix
self .mat = mat
# Stores the position at which the
# empty space tile exists in the matrix
self .empty_tile_pos = empty_tile_pos
# Stores the number of misplaced tiles
self .cost = cost
# Stores the number of moves so far
self .level = level
# This method is defined so that the
# priority queue is formed based on
# the cost variable of the objects
def __lt__( self , nxt):
return self .cost < nxt.cost
# Function to calculate the number of # misplaced tiles ie. number of non-blank # tiles not in their goal position def calculateCost(mat, final) - > int :
count = 0
for i in range (n):
for j in range (n):
if ((mat[i][j]) and
(mat[i][j] ! = final[i][j])):
count + = 1
return count
def newNode(mat, empty_tile_pos, new_empty_tile_pos,
level, parent, final) - > node:
# Copy data from parent matrix to current matrix
new_mat = copy.deepcopy(mat)
# Move tile by 1 position
x1 = empty_tile_pos[ 0 ]
y1 = empty_tile_pos[ 1 ]
x2 = new_empty_tile_pos[ 0 ]
y2 = new_empty_tile_pos[ 1 ]
new_mat[x1][y1], new_mat[x2][y2] = new_mat[x2][y2], new_mat[x1][y1]
# Set number of misplaced tiles
cost = calculateCost(new_mat, final)
new_node = node(parent, new_mat, new_empty_tile_pos,
cost, level)
return new_node
# Function to print the N x N matrix def printMatrix(mat):
for i in range (n):
for j in range (n):
print ( "%d " % (mat[i][j]), end = " " )
print ()
# Function to check if (x, y) is a valid # matrix coordinate def isSafe(x, y):
return x > = 0 and x < n and y > = 0 and y < n
# Print path from root node to destination node def printPath(root):
if root = = None :
return
printPath(root.parent)
printMatrix(root.mat)
print ()
# Function to solve N*N - 1 puzzle algorithm # using Branch and Bound. empty_tile_pos is # the blank tile position in the initial state. def solve(initial, empty_tile_pos, final):
# Create a priority queue to store live
# nodes of search tree
pq = priorityQueue()
# Create the root node
cost = calculateCost(initial, final)
root = node( None , initial,
empty_tile_pos, cost, 0 )
# Add root to list of live nodes
pq.push(root)
# Finds a live node with least cost,
# add its children to list of live
# nodes and finally deletes it from
# the list.
while not pq.empty():
# Find a live node with least estimated
# cost and delete it from the list of
# live nodes
minimum = pq.pop()
# If minimum is the answer node
if minimum.cost = = 0 :
# Print the path from root to
# destination;
printPath(minimum)
return
# Generate all possible children
for i in range ( 4 ):
new_tile_pos = [
minimum.empty_tile_pos[ 0 ] + row[i],
minimum.empty_tile_pos[ 1 ] + col[i], ]
if isSafe(new_tile_pos[ 0 ], new_tile_pos[ 1 ]):
# Create a child node
child = newNode(minimum.mat,
minimum.empty_tile_pos,
new_tile_pos,
minimum.level + 1 ,
minimum, final,)
# Add child to list of live nodes
pq.push(child)
# Driver Code # Initial configuration # Value 0 is used for empty space initial = [ [ 1 , 2 , 3 ],
[ 5 , 6 , 0 ],
[ 7 , 8 , 4 ] ]
# Solvable Final configuration # Value 0 is used for empty space final = [ [ 1 , 2 , 3 ],
[ 5 , 8 , 6 ],
[ 0 , 7 , 4 ] ]
# Blank tile coordinates in # initial configuration empty_tile_pos = [ 1 , 2 ]
# Function call to solve the puzzle solve(initial, empty_tile_pos, final) # This code is contributed by Kevin Joshi |
using System;
using System.Collections.Generic;
class GFG
{ public static int N = 3;
public class Node
{
// Stores the parent node of the current node
// Helps in tracing the path when the answer is found
public Node parent;
public int [,] mat = new int [N, N]; // Stores the matrix
public int x, y; // Stores blank tile coordinates
public int cost; // Stores the number of misplaced tiles
public int level; // Stores the number of moves so far
}
// Function to print N x N matrix
public static void PrintMatrix( int [,] mat)
{
for ( int i = 0; i < N; i++)
{
for ( int j = 0; j < N; j++)
{
Console.Write(mat[i, j] + " " );
}
Console.WriteLine();
}
}
// Function to allocate a new node
public static Node NewNode( int [,] mat, int x, int y, int newX, int newY, int level, Node parent)
{
Node node = new Node();
node.parent = parent; // Set pointer for the path to root
// Copy data from the parent node to the current node
node.mat = new int [N, N];
for ( int i = 0; i < N; i++)
{
for ( int j = 0; j < N; j++)
{
node.mat[i, j] = mat[i, j];
}
}
// Move the tile by 1 position
int temp = node.mat[x, y];
node.mat[x, y] = node.mat[newX, newY];
node.mat[newX, newY] = temp;
node.cost = int .MaxValue; // Set the number of misplaced tiles
node.level = level; // Set the number of moves so far
// Update new blank tile coordinates
node.x = newX;
node.y = newY;
return node;
}
// Bottom, left, top, right
public static int [] row = { 1, 0, -1, 0 };
public static int [] col = { 0, -1, 0, 1 };
// Function to calculate the number of misplaced tiles
// i.e., the number of non-blank tiles not in their goal position
public static int CalculateCost( int [,] initialMat, int [,] finalMat)
{
int count = 0;
for ( int i = 0; i < N; i++)
{
for ( int j = 0; j < N; j++)
{
if (initialMat[i, j] != 0 && initialMat[i, j] != finalMat[i, j])
{
count++;
}
}
}
return count;
}
// Function to check if (x, y) is a valid matrix coordinate
public static int IsSafe( int x, int y)
{
return (x >= 0 && x < N && y >= 0 && y < N) ? 1 : 0;
}
// Print the path from the root node to the destination node
public static void PrintPath(Node root)
{
if (root == null )
{
return ;
}
PrintPath(root.parent);
PrintMatrix(root.mat);
Console.WriteLine();
}
// Comparison object to be used to order the heap
public class Comp : IComparer<Node>
{
public int Compare(Node lhs, Node rhs)
{
return (lhs.cost + lhs.level) > (rhs.cost + rhs.level) ? 1 : -1;
}
}
// Function to solve N*N - 1 puzzle algorithm using
// Branch and Bound. x and y are blank tile coordinates
// in the initial state
public static void Solve( int [,] initialMat, int x, int y, int [,] finalMat)
{
// Create a priority queue to store live nodes of the search tree
var pq = new SortedSet<Node>( new Comp());
// Create a root node and calculate its cost
Node root = NewNode(initialMat, x, y, x, y, 0, null );
root.cost = CalculateCost(initialMat, finalMat);
// Add the root to the list of live nodes
pq.Add(root);
// Find a live node with the least cost,
// add its children to the list of live nodes, and
// finally remove it from the list
while (pq.Count > 0)
{
Node min = pq.Min; // Find a live node with the least estimated cost
pq.Remove(min); // The found node is removed from the list of live nodes
// If min is an answer node
if (min.cost == 0)
{
PrintPath(min); // Print the path from the root to the destination
return ;
}
// Do for each child of min
// Max 4 children for a node
for ( int i = 0; i < 4; i++)
{
if (IsSafe(min.x + row[i], min.y + col[i]) > 0)
{
// Create a child node and calculate its cost
Node child = NewNode(min.mat, min.x, min.y, min.x + row[i], min.y + col[i], min.level + 1, min);
child.cost = CalculateCost(child.mat, finalMat);
// Add the child to the list of live nodes
pq.Add(child);
}
}
}
}
// Driver Code
public static void Main( string [] args)
{
// Initial configuration
// Value 0 is used for empty space
int [,] initialMat =
{
{1, 2, 3},
{5, 6, 0},
{7, 8, 4}
};
// Solvable Final configuration
// Value 0 is used for empty space
int [,] finalMat =
{
{1, 2, 3},
{5, 8, 6},
{0, 7, 4}
};
// Blank tile coordinates in the initial configuration
int x = 1, y = 2;
Solve(initialMat, x, y, finalMat);
}
} |
// Program to print path from root node to destination node // for N*N - 1 puzzle algorithm using Branch and Bound // The solution assumes that the instance of the puzzle is solvable const N = 3; // State space tree nodes class Node { constructor(mat, x, y, level, parent) {
// Stores the parent node of the current node
// Helps in tracing the path when the answer is found
this .parent = parent;
// Stores matrix
this .mat = mat.map(row => [...row]);
// Stores blank tile coordinates
this .x = x;
this .y = y;
// Stores the number of misplaced tiles
this .cost = Infinity;
// Stores the number of moves so far
this .level = level;
}
} // Function to print N x N matrix function printMatrix(mat) {
for (let i = 0; i < N; i++) {
console.log(mat[i].join( ' ' ));
}
console.log( '\n' );
} // Function to allocate a new node function newNode(mat, x, y, newX, newY, level, parent) {
const node = new Node(mat, x, y, level, parent);
// Move tile by 1 position
[node.mat[x][y], node.mat[newX][newY]] = [node.mat[newX][newY], node.mat[x][y]];
// Update new blank tile coordinates
node.x = newX;
node.y = newY;
return node;
} // Bottom, left, top, right const row = [1, 0, -1, 0]; const col = [0, -1, 0, 1]; // Function to calculate the number of misplaced tiles // i.e., number of non-blank tiles not in their goal position function calculateCost(initial, final) {
let count = 0;
for (let i = 0; i < N; i++)
for (let j = 0; j < N; j++)
if (initial[i][j] && initial[i][j] !== final[i][j])
count++;
return count;
} // Function to check if (x, y) is a valid matrix coordinate function isSafe(x, y) {
return x >= 0 && x < N && y >= 0 && y < N;
} // Print path from root node to destination node function printPath(root) {
if (!root) return ;
printPath(root.parent);
printMatrix(root.mat);
} // Comparison object to be used to order the heap class comp { static compare(lhs, rhs) {
return (lhs.cost + lhs.level) > (rhs.cost + rhs.level);
}
} // Function to solve N*N - 1 puzzle algorithm using // Branch and Bound. x and y are blank tile coordinates // in the initial state function solve(initial, x, y, final) {
// Create an array to store live nodes of the search tree
const pq = [];
// Create a root node and calculate its cost
const root = newNode(initial, x, y, x, y, 0, null );
root.cost = calculateCost(initial, final);
// Add root to the array of live nodes
pq.push(root);
// Find a live node with the least cost,
// add its children to the array of live nodes, and
// finally delete it from the array
while (pq.length > 0) {
// Find a live node with the least estimated cost
pq.sort(comp.compare);
const min = pq.shift();
// If min is an answer node
if (min.cost === 0) {
// Print the path from root to destination
printPath(min);
return ;
}
// Do for each child of min
// Max 4 children for a node
for (let i = 0; i < 4; i++) {
if (isSafe(min.x + row[i], min.y + col[i])) {
// Create a child node and calculate its cost
const child = newNode(min.mat, min.x,
min.y, min.x + row[i],
min.y + col[i],
min.level + 1, min);
child.cost = calculateCost(child.mat, final);
// Add child to the array of live nodes
pq.push(child);
}
}
}
} // Driver code // Initial configuration // Value 0 is used for empty space const initial = [ [1, 2, 3],
[5, 6, 0],
[7, 8, 4]
]; // Solvable Final configuration // Value 0 is used for empty space const final = [ [1, 2, 3],
[5, 8, 6],
[0, 7, 4]
]; // Blank tile coordinates in the initial configuration const startX = 1, startY = 2; solve(initial, startX, startY, final); // This code is contributed by shivamgupta310570 |
Output :
1 2 3
5 6 0
7 8 4
1 2 3
5 0 6
7 8 4
1 2 3
5 8 6
7 0 4
1 2 3
5 8 6
0 7 4
The time complexity of this algorithm is O(N^2 * N!) where N is the number of tiles in the puzzle, and the space complexity is O(N^2).
Sources:
www.cs.umsl.edu/~sanjiv/classes/cs5130/lectures/bb.pdf
https://www.seas.gwu.edu/~bell/csci212/Branch_and_Bound.pdf