Develop Reference

DFS and BFS with adjacency matrix counted in Java

Currently trying to build a program that can implement both DFS and BFS in Java by taking in an adjacency matrix from a file and printing out the following info: order that vertices are first encountered, order that vertices become dead ends, number of components, and the tree edges.
Here is my code:
import java.io.File;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.Scanner;
public class ProjectDriver {
public static int count;
public static void main(String[] args) throws FileNotFoundException {
ArrayList<Integer> dfsDeadEnd = new ArrayList<Integer>();
ArrayList<Integer> dfsVertices = new ArrayList<Integer>();
ArrayList<Integer> bfsVertices = new ArrayList<Integer>();
int dfsComponents = 0;
int bfsComponents = 0;
Scanner scanner = new Scanner(new File("sample1.txt"));
int n = 8;
int[][] edges = new int[n][n];
boolean[] visited = new boolean[n];
count = 0;
for(int i=0; i < n; i++) {
for (int j=0; j < n; j++) {
edges[i][j] = scanner.nextInt();
}
}
for(int i = 0; i < n; i++){
visited[i] = false;
}
int[][] BFStreeEdgeGraph = new int[n][n];
int[][] DFStreeEdgeGraph = new int[n][n];
int[][] crossGraph = new int[n][n];
for(int i = 0; i <= n-1; i++){
for(int j = 0; j <= n-1; j++){
DFStreeEdgeGraph[i][j] = 0;
BFStreeEdgeGraph[i][j] = 0;
crossGraph[i][j] = 0;
}
}
for(int i = 0; i <= n-1; i++){
if(visited[i] == false){
dfs(edges,i,visited, dfsVertices, dfsDeadEnd, DFStreeEdgeGraph);
dfsDeadEnd.add(i);
dfsComponents++;
}
}
for(int i = 0; i <= n-1; i++) {
if(visited[i] == false) {
bfs(edges, i, visited, bfsVertices, BFStreeEdgeGraph);
bfsComponents++;
}
}
System.out.println();
System.out.println("DFS: Number of Connected Components: " + dfsComponents);
System.out.print("DFS: Order of First Encountered: ");
for(int i : dfsVertices){
System.out.print((i+1) + " ");
}
System.out.println();
System.out.print("DFS: Order of First Dead-Ends: ");
for(int i : dfsDeadEnd){
System.out.print((i+1) + " ");
}
System.out.println();
System.out.println();
System.out.println("Tree edges:");
displayGraph(DFStreeEdgeGraph, n);
System.out.println();
System.out.println();
System.out.println("BFS: Number of Connected Components: " + bfsComponents);
System.out.print("BFS: Order of First encountered: ");
for(int i : bfsVertices){
System.out.print((i+1) + " ");
}
System.out.println();
System.out.println();
}
public static void dfs(int[][] edges, int vertex, boolean[] visited, ArrayList<Integer> dfsVertices, ArrayList<Integer> dfsDeadEnd, int[][] DFStreeEdgeGraph) {
visited[vertex] = true;
dfsVertices.add(count, vertex);
count = count + 1;
for(int w = 0; w <= edges.length-1; w++) {
if(edges[vertex][w] == 1 && !visited[w]) {
DFStreeEdgeGraph[vertex][w] = 1;
dfs(edges, w, visited, dfsVertices, dfsDeadEnd, DFStreeEdgeGraph);
dfsDeadEnd.add(w);
}
}
}
public static void bfs(int[][] edges, int vertex, boolean[] visited, ArrayList<Integer> bfsVertices, int[][] BFStreeEdgeGraph) {
bfsVertices.add(count, vertex);
count = count + 1;
for(int w = 0; w < edges.length; w++) {
if(edges[vertex][w] != 0 && !visited[w]) {
visited[vertex] = true;
}
}
for(int w = bfsVertices.indexOf(vertex) + 1; w < bfsVertices.size(); w++) {
int value = bfsVertices.get(w);
bfs(edges, value, visited, bfsVertices, BFStreeEdgeGraph);
}
}
public static void displayGraph(int[][] graph, int n) {
for(int i = 0; i <= n-1; ++i){
System.out.print(" ");
for(int j = 0; j <= n-1; ++j){
System.out.print(graph[i][j] + " ");
}
System.out.println();
}
}
}
And here is the output from running my code:
Input graph:
0 1 0 0 1 1 0 0
1 0 0 0 0 1 1 0
0 0 0 1 0 0 1 0
0 0 1 0 0 0 0 1
1 0 0 0 0 1 0 0
1 1 0 0 1 0 0 0
0 1 1 0 0 0 0 1
0 0 0 1 0 0 1 0
DFS: Number of Connected Components: 1
DFS: Order of First Encountered: 1 2 6 5 7 3 4 8
DFS: Order of First Dead-Ends: 5 6 8 4 3 7 2 1
Tree edges:
0 1 0 0 0 0 0 0
0 0 0 0 0 1 1 0
0 0 0 1 0 0 0 0
0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0
0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 0
BFS: Number of Connected Components: 0
BFS: Order of First encountered:
And here are my issues. For the DFS, my order of first encountered should be 1 2 6 7 4 3 5 8, but this is not what I'm getting. Additionally, my order of dead ends should be 8 7 5 4 1 2 6 3, but this is also off. My tree edges managed to come out correct.
For the BFS, I can't get anything to print, and debugging my DFS and BFS methods hasn't given me the answer yet. I'd be very grateful from some help.

Java Multidimensional Array Outprint

I dont understand, why these numbers are printed out. Shouldn't it just out print 3 2 1? Instead, it prints:
3
0
0
0
2
0
0
0
1
Thank you for your help :)
public static void main(String[] args) {
int i, j, n = 3;
int[][] polje = new int[n][n];
polje[0][0] = 3;
polje[1][1] = 2;
polje[2][2] = 1;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
System.out.print(polje[i][j] + " ");
System.out.println();
}
}
}

you have set likewise,
3 0 0
0 2 0
0 0 1
so obviously you get, 3 0 0 0 2 0 0 0 1
if you want to print only 3 2 1 then made this changes only,
if(i == j){
System.out.print(polje[i][j] + " ");
System.out.println();
}

2^N Combinaisons with Integers (Kernel), how to generate them?

(Merry Christmas btw ^^)
Here is my problem (in JAVA) but it's definitely an algorithmic problem and I don't know how to solve it :/
So here it is, with an example (just for information, all my calculs are in Binary, so 1+1 = 0)
let's name variables:
N : the number of elements in kernel.
M : the length of an element in the kernel.
int[][] Kernel:
....
i : 0 1 1 1 0 1 0 1 0 1 1 1 0 1 1 1 0 1 0 1 0 1 1 1 0 (length = M)
i+1 : 1 0 1 0 1 1 0 1 0 1 0 0 0 0 0 1 0 1 0 1 1 0 1 0 1 (length = M)
....
N : ....
My goal with theses things, is to generate all the possible combinaison (so 2^N elements)
and I want to generate them.
By generate I mean exactly this :
Result[0] = 0 0 0 0 0 0 0 0 0 0 0 0 0
Result[1] = Kernel[0]
Result[2] = Kernel[1]
....
Result[i] = Kernel[i-1]
Result[N-1] = Kernel[N-2]
Result[N] = Kernel[0] + Kernel[1]
Result[N+1] = Kernel[0] + Kernel[2]
Result[N+i] = Kernel[0] + Kernel[i]
Result[2N-1] = Kernel[0] + Kernel[N-1]
....
Result[I] = Kernel[0] + Kernel[1] + Kernel[2]
Result[I+1] = Kernel[0] + Kernel[1] + Kernel[i]
Result[I+J] = Kernel[0] + Kernel[1] + Kernel[N-1]
....
Result[2^N+1] = Kernel[0] + Kernel[1] + ... + Kernel[i] + ... + Kernel[N-1]
Here is what I already success to do, but it's not complete and I don't know how to generalize the calcul in order to work with any N...
public static int[][] combinaisons(int[][] kernel) {
/* if the kernel is empty, there is no possible combinaison */
if(kernel.length == 0) return kernel;
/* We allocate the good number of space... */
int[][] result = new int[(int) (Math.pow(2, noyau.length)+1)][];
/* Every element in result has the same length as in kernel's elements. */
for(int i = 0; i < resultat.length; i++) {
result[i] = new int[kernel[0].length];
}
/* The first element of result has to be only 0 0 0 0 0 0 0 */
for(int j = 0; j < kernel[0].length; j++) {
result[0][j] = 0;
}
/* We rewrite the element of kernel because they are a part of the solution... */
for(int i = 0; i < kernel.length; i++) {
for(int j = 0; j < kernel[i].length; j++) {
result[i+1][j] = kernel[i][j];
}
}
/*
I managed to do it when it's the sum of only 2 elements,
but it has to be with 3, 4 ... N-1 :/
*/
for(int i = 0; i < kernel.length; i++) {
for(int j = 0; j < kernel[i].length; j++) {
for(int k = i+1; k < kernel.length; k++) {
result[k*kernel.length+i][j] = (kernel[i][j]+kernel[k][j])%2;
}
}
}
return result;
}
Edit:
About an example, let's give this:
N = 2
M = 4
Kernel:
0 1 1 0
1 0 0 1
In result I want:
0 0 0 0
0 1 1 0
1 0 0 1
1 1 1 1 (the sum of the 2 elements in Kernel)
So this is a simple example (quite particularly values, if you want bigger, just ask :) )
Even if the array at the end seems to be VERY HUGE :) that's exactly what I want to generate (don't care about memory, it will for sure be okay)

I am going to use boolean[][] instead of int[][]. 0 means false, 1 means true.
public static boolean[][] combinations(boolean kernel[][]) {
int n = kernel.length;
int m = kernel[0].length;
int p = 1 << n;
boolean[][] temp = new boolean[p][m];
for (int i = 0; i < p; i++)
for (int j = 0; j < n; j++)
if (((1 << j) & i) != 0)
for (int k = 0; k < m; k++)
temp[i][k] ^= kernel[j][k];
return temp;
}

Print triangle (Half Pyramid) pattern using Java

I have to print a triangle pattern (Half Pyramid) like
1
0 1
1 0 1
0 1 0 1
I tried with this program
class tri{
public static void main(String arg[]){
int i,j,a = 1, b =0, c=0;
for(i=1; i<=4; i++){
for(j=1; j<=i; j++){
System.out.print(a+ " ");
c = a;
a = b;
b = c;
}
System.out.println();
}
}
}
but this prints pattern as shown in image
please if some one could help me editing that code to bring the pattern

The shortest form would be
String str = "";
for (int i = 1; i <= 4; i++) {
str = (i % 2) + " " + str;
System.out.println(str);
}
This will give output as you desired
1
0 1
1 0 1
0 1 0 1

You need to set starting values correctly. Because what you are doing is continuously swapping
Say row two 0 1
last element = 1, (a = 1, b = 0) and on swapping (a = 0, b = 1) for next row first element.
However this is incorrect as it was supposed to start with (a = 1) and not (a = 0) from previous state.
int i,j,a = 1, b =0, c=0;
for (i = 1; i <= 4; i++){
if (i % 2 == 0) {
a = 0;
b = 1;
} else {
a = 1;
b = 0;
}
for(j=1; j<=i; j++) {
System.out.print(a+ " ");
c = a;
a = b;
b = c;
}
System.out.println();
}
You can also switch between 0 and 1 using XOR :
int i, j, a = 1;
for (i = 1; i <= 4; i++){
a = i % 2;
for(j=1; j<=i; j++) {
System.out.print(a+ " ");
a = a ^ 1;
}
System.out.println();
}
However Shorter solution would be :
String str = "";
for (int i = 1; i <= 4; i++) {
str = (i % 2) + " " + str;
System.out.println(str);
}
output :
1
0 1
1 0 1
0 1 0 1

You can use boolean flag for this to check if you are current starting at 1 or 0;
sample:
boolean flag = true;
for(int i=1; i<=4; i++){
for(int j=1; j<=i; j++){
if(flag)
System.out.print("1 ");
else
System.out.print("0 ");
flag = !flag;
}
if((i % 2) == 0)
flag = true;
else
flag = false;
System.out.println();
}
result:
1
0 1
1 0 1
0 1 0 1

int x=1,y=1;
for(int i=1;i<8;i++){
for(int k=0;k<i;k++){
y=(k==0) ? x:y;
System.out.print(y+" ");
y=(y==1) ? 0:1;
}
System.out.println("");
x=(x==1) ? 0:1;
}
output---

write my anwser here, seen #sujithvm solution is more short and efficient.
int sideLength = 4;
for(int i = 0 ; i < sideLength ; i++)
{
for(int j = 0 ; j <= i ; j++)
{
System.out.print((i + j + 1) % 2 + " ");
}
System.out.println();
}

Complete Java program for beginners:
public class PrintPattern15
{
public static void main(String args[])
{
int n = 5;
PrintPattern15 d = new PrintPattern15();
d.printPattern(n);
}
public void printPattern(int noOfRows)
{
for(int i = 1; i <= noOfRows; i++ )
{
printRows(i);
}
}
public void printRows(int startPt)
{
for(int i = startPt ; i >= 1; i--)
{
System.out.print(i%2);
}
System.out.println();
}
}

This Works for me:
public class DrawPattern {
public static void main(String[] args) {
int i, j;
int num = 7;
for (i = 0; i <num; i++) {
for (j = 0; j < i; j++) {
if (isConditionMatch(num, i, j)) {
System.out.print("0");
} else {
System.out.print("1");
}
}
System.out.println();
}
}
private static boolean isConditionMatch(int num, int i, int j) {
return (i%2 != 0 && j%2 !=0 || (i%2 == 0 && j%2==0));
}
}
Output:
1
01
101
0101
10101
010101

Code
public class pra1 {
public static void main(String[] args) {
int space, rows=11, k=0;
// Scanner scan = new Scanner(System.in);
// System.out.print("Enter Number of Rows : ");
// rows = scan.nextInt();
// rowa=6;
for(int i=1; i<=rows; i++)
{
for(space=1; space<=(rows-i); space++)
{
System.out.print(" ");
}
while(k != (2*i-1))
{ // int p=k;
if(k%2==0)
{
System.out.print("1 ");
// System.out.print(" ");
}if(k%2!=0 )
{
System.out.print("0 ");
// System.out.print(" ");
}
else{
System.out.print("");
}
// p--;
k++;
}
k = 0;
System.out.println();
}
}
}
Output
1
1 0 1
1 0 1 0 1
1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

This is how I solved the problem:
int k = 0;
for (i = 1; i <= 4; i++){
k = (i%2 == 0)? 1:0;
for(j=1; j<=i; j++) {
System.out.print(k+ " ");
k = k==0?1:0;
}
System.out.println();
}

Flood fill for 2d int array optimization in Java

I am writing my own implementation of the Flood Fill.
I came up with this code:
public static void fillArea(int x, int y, int original, int fill, int[][] arr) {
Stack<int[]> q = new Stack<int[]>();
int[] w = new int[2]; //to the west
int[] e = new int[2]; //to the east
if (arr[x][y] != original) {
return;
}
q.push(new int[]{x, y});
while (!q.isEmpty()) {
int[] pos = (int[]) q.pop();
int i = pos[0];
int j = pos[1];
if (arr[i][j] == original) {
e[0] = i;
e[1] = j;
w[0] = i;
w[1] = j;
}
while (w[1] > 0 && arr[w[0]][w[1] - 1] == original) { // to the west
w[1] -= 1;
}
while (e[1] < arr[0].length - 1 && arr[e[0]][e[1] + 1] == original) { // to the east
e[1] += 1;
}
for (int a = w[1]; a <= e[1]; a++) { // for every point between west and east
arr[w[0]][a] = fill;
if (w[0] > 0 && arr[w[0] - 1][a] == original) { //check if we can go north
q.push(new int[]{(w[0] - 1), a});
}
if (w[0] < arr.length - 1 && arr[w[0] + 1][a] == original) {//check if we can go south
q.push(new int[]{(w[0] + 1), a});
}
}
}
return;
}
Params are:
start point coords, value we want to change, new value we want to see as a result of filling, original array.
It is implementation of Wiki pseudocode:
Flood-fill (node, target-color, replacement-color):
1. Set Q to the empty queue.
2. If the color of node is not equal to target-color, return.
3. Add node to Q.
4. For each element N of Q:
5. If the color of N is equal to target-color:
6. Set w and e equal to N.
7. Move w to the west until the color of the node to the west of w no longer matches target-color.
8. Move e to the east until the color of the node to the east of e no longer matches target-color.
9. For each node n between w and e:
10. Set the color of n to replacement-color.
11. If the color of the node to the north of n is target-color, add that node to Q.
12. If the color of the node to the south of n is target-color, add that node to Q.
13. Continue looping until Q is exhausted.
14. Return.
I am using Stack instead of Queue, because it seems that Stack is much faster, or I have a problem with my code.
The problem is : it is very slow.
Is there anything I can do about performance? I can trade memory for the performance, but native recursion get me stackoverflow.

Ok, here we go :).
I simulated stack with array, which I hope would be much faster. I did not followed exactly the pseudocode, but it works and I think it should be fast.
public static void main(String[] args) {
int testArray[][] = new int[10][10];
for (int i = 0; i < 10; i++) {
testArray[i][i] = 1;
testArray[9-i][i] = 1;
}
testArray[4][7] = 1;
System.out.println("Array before");
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
System.out.print(testArray[j][i] + " ");
}
System.out.println("");
}
fillArea(6,8,0,7,testArray);
System.out.println("Array after");
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
System.out.print(testArray[j][i] + " ");
}
System.out.println("");
}
}
public static void fillArea(int x, int y, int original, int fill, int[][] arr) {
int maxX = arr.length - 1;
int maxY = arr[0].length - 1;
int[][] stack = new int[(maxX+1)*(maxY+1)][2];
int index = 0;
stack[0][0] = x;
stack[0][1] = y;
arr[x][y] = fill;
while (index >= 0){
x = stack[index][0];
y = stack[index][1];
index--;
if ((x > 0) && (arr[x-1][y] == original)){
arr[x-1][y] = fill;
index++;
stack[index][0] = x-1;
stack[index][1] = y;
}
if ((x < maxX) && (arr[x+1][y] == original)){
arr[x+1][y] = fill;
index++;
stack[index][0] = x+1;
stack[index][1] = y;
}
if ((y > 0) && (arr[x][y-1] == original)){
arr[x][y-1] = fill;
index++;
stack[index][0] = x;
stack[index][1] = y-1;
}
if ((y < maxY) && (arr[x][y+1] == original)){
arr[x][y+1] = fill;
index++;
stack[index][0] = x;
stack[index][1] = y+1;
}
}
}
This code having this output :
Array before
1 0 0 0 0 0 0 0 0 1
0 1 0 0 0 0 0 0 1 0
0 0 1 0 0 0 0 1 0 0
0 0 0 1 0 0 1 0 0 0
0 0 0 0 1 1 0 0 0 0
0 0 0 0 1 1 0 0 0 0
0 0 0 1 0 0 1 0 0 0
0 0 1 0 1 0 0 1 0 0
0 1 0 0 0 0 0 0 1 0
1 0 0 0 0 0 0 0 0 1
Array after
1 0 0 0 0 0 0 0 0 1
0 1 0 0 0 0 0 0 1 0
0 0 1 0 0 0 0 1 0 0
0 0 0 1 0 0 1 0 0 0
0 0 0 0 1 1 0 0 0 0
0 0 0 0 1 1 0 0 0 0
0 0 0 1 7 7 1 0 0 0
0 0 1 7 1 7 7 1 0 0
0 1 7 7 7 7 7 7 1 0
1 7 7 7 7 7 7 7 7 1

This version is even faster
public static void floodFill(int y, int x, byte originalvalue, byte newvalue, byte[][] arr) {
Deque queue = new ArrayDeque();
queue.add(new int[]{y, x});
while (!queue.isEmpty()) {
int[] t = (int[]) queue.poll();
y = t[0];
x = t[1];
if (arr[y][x] == originalvalue) {
arr[y][x] = newvalue;
for (int i = 0; i
< 8; i++) {
if (x + dx[i] < arr[0].length && y + dy[i] < arr.length && x + dx[i] > -1 && y + dy[i] > -1 && arr[y + dy[i]][x + dx[i]] == originalvalue) {
queue.add(new int[]{y + dy[i], x + dx[i]});
}
}
}
}
}