Develop Reference

Tim Sort vs LSD Radix Sort for Suffix Arrays?

I'm confused on why tim sort which makes my suffix array implement O(nlognlogn) faster than LSD radix sort which should make the implementation O(nlogn). Perhaps is it that radix sort simply works better with larger string lengths?
here is my radix sort implementation in java and I don't know if I'm just not making it as efficient as it could be:
class Suffix {
int index;
int[] rank = new int[2];
public Suffix(int i, int r1, int r2){
index = i;
rank[0] = r1;
rank[1] = r2;
}
}
public class Main {
public static int getMax(Suffix[] suffixArray, int index){
int max = Integer.MIN_VALUE;
for(Suffix i : suffixArray){
if(i.rank[index] > max)max = i.rank[index];
}
return max;
}
public static void countingSort(Suffix[] suffixes, int exp, int index){
int[] count = new int[10];
Suffix[] sorted = new Suffix[suffixes.length];
for(Suffix i : suffixes)count[(i.rank[index]/exp) % 10]++;
for(int i = 1; i < 10; i++)count[i] += count[i - 1];
for(int i = suffixes.length - 1; i >= 0; i--){
Suffix current = suffixes[i];
int position = count[(current.rank[index]/exp) % 10] - 1;
sorted[position] = current;
count[(current.rank[index]/exp) % 10]--;
}
for(int i = 0; i < suffixes.length; i++)suffixes[i] = sorted[i];
}
public static void radixSort(Suffix[] suffixes, int index){
int max = getMax(suffixes, index);
for(int exp = 1; max/exp > 0; exp *= 10)countingSort(suffixes, exp, index);
}
public static void sort(Suffix[] suffixes){
int positiveCounter = 0;
for(Suffix i : suffixes){
if(i.rank[1] >= 0)positiveCounter++;
}
Suffix[] positives = new Suffix[positiveCounter];
int positivesIndex = 0;
int tempIndex = 0;
for(Suffix i : suffixes){
if( i.rank[1] >= 0){
positives[positivesIndex] = i;
positivesIndex++;
}
else {
suffixes[tempIndex] = i;
tempIndex++;
}
}
radixSort(positives, 1);
for(int i = 0; i < positives.length; i++){
suffixes[tempIndex] = positives[i];
tempIndex++;
}
radixSort(suffixes, 0);
}

I am facing timeout for my solution for a HackerRank

I am solving a problem on hackerrank
Hacker Rank ICPC Team Problem
I have created the following code as solution for problem.
import java.math.BigInteger;
import java.util.Scanner;
public class ACMICPCTeam {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc=new Scanner(System.in);
int n=sc.nextInt(),m=sc.nextInt(),count=0,maxCount=0,teams=0;
sc.nextLine();
String subjectArray[]=new String[n];
for(int i=0;i<n;i++){
subjectArray[i]=sc.nextLine();
}
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
String temp=""+(new BigInteger(subjectArray[i]).add(new BigInteger(subjectArray[j])));
//System.out.println(temp);
count=temp.replace("0","").length();
if(count>maxCount)
{
maxCount=count;
teams=1;
}
else if(count==maxCount)
{
teams++;
}
}
}
System.out.println(maxCount);
System.out.println(teams);
sc.close();
}
}
So what I am trying to do is I am adding the two teams subjects and I am counting non-zeros of resultant string. The highest count is number of subjects and the occurrence of highest counts are teams which know max number of subject. Even after spending a lot time I am not able to any better solution than this one still I am facing time out as it is not efficient.
I have gone through forum of the question but it was of no help.

Don't use string logic for this.
Parse the string into a BitSet, before entering your loops, i.e. as you read them.
Then use methods or(BitSet set), and cardinality().
I just completed challenge doing that. No timeouts.

Your solution is not optimal you should try something better.
You can utilize BigInteger method or BitSet class to make it easy.
For forming a team you have to use bitwise OR
Here are solutions--
// 1st approach
static int[] acmTeam(String[] topic) {
int n = topic.length;
BigInteger[] bi = new BigInteger[n];
for (int i = 0; i < n; i++)
bi[i] = new BigInteger(topic[i], 2);
int maxTopic = 0;
int teamCount = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
BigInteger iuj = bi[i].or(bi[j]);
int bitCount = iuj.bitCount();
if (bitCount > maxTopic) {
maxTopic = bitCount;
teamCount = 1;
} else if (bitCount == maxTopic) {
teamCount++;
}
}
}
int result[] = { maxTopic, teamCount };
return result;
}
// 2nd approach--using java BitSet class
static int[] acmTeamUsingBitSet(String[] topic) {
int teamCount = 0, maxTopic = 0;
int size = topic.length;
BitSet[] bitset = new BitSet[size];
for (int i = 0; i < size; i++) {
BigInteger b1 = new BigInteger(topic[i], 2);
bitset[i] = BitSet.valueOf(b1.toByteArray());
}
for (int i = 0; i < size - 1; i++) {
BitSet bitset1 = bitset[i];
for (int j = i + 1; j < size; j++) {
BitSet bitset2 = bitset[j];
BitSet tmpset = new BitSet();
tmpset.or(bitset1);
tmpset.or(bitset2);
if (tmpset.cardinality() > maxTopic) {
maxTopic = tmpset.cardinality();
teamCount = 1;
} else if (maxTopic == tmpset.cardinality()) {
teamCount++;
}
}
}
int result[] = { maxTopic, teamCount };
return result;
}
You can refer this link for a detailed video explanation.

I got good result using Java 8.
static int[] acmTeam(String[] topic) {
List<List<Integer>> res = IntStream.range(0, topic.length)
.mapToObj(s -> IntStream.range(0, topic[s].length()).boxed()
.collect(Collectors.groupingBy(i -> topic[s].charAt(i))))
.map(m -> m.get('1'))
.collect(toList());
long maxTopic = 0;
int teamCount = 0;
for (int i = 0; i < res.size(); i++) {
for (int j = i + 1; j < res.size(); j++) {
long topics = Stream.concat(res.get(i).stream(), res.get(j).stream()).distinct().count();
if (topics > maxTopic) {
maxTopic = topics;
teamCount = 1;
} else if (topics == maxTopic) {
teamCount++;
}
}
}
return new int[]{(int) maxTopic, teamCount};
}

Program with threads for matrix multiplication

I'm trying to create a Java program with threads for matrix multiplication. This is the source code:
import java.util.Random;
public class MatrixTest {
//Creating the matrix
static int[][] mat = new int[3][3];
static int[][] mat2 = new int[3][3];
static int[][] result = new int[3][3];
public static void main(String[] args) {
//Creating the object of random class
Random rand = new Random();
//Filling first matrix with random values
for (int i = 0; i < mat.length; i++) {
for (int j = 0; j < mat[i].length; j++) {
mat[i][j] = rand.nextInt(10);
}
}
//Filling second matrix with random values
for (int i = 0; i < mat2.length; i++) {
for (int j = 0; j < mat2[i].length; j++) {
mat2[i][j] = rand.nextInt(10);
}
}
try {
//Object of multiply Class
Multiply multiply = new Multiply(3, 3);
//Threads
MatrixMultiplier thread1 = new MatrixMultiplier(multiply);
MatrixMultiplier thread2 = new MatrixMultiplier(multiply);
MatrixMultiplier thread3 = new MatrixMultiplier(multiply);
//Implementing threads
Thread th1 = new Thread(thread1);
Thread th2 = new Thread(thread2);
Thread th3 = new Thread(thread3);
//Starting threads
th1.start();
th2.start();
th3.start();
th1.join();
th2.join();
th3.join();
} catch (Exception e) {
e.printStackTrace();
}
//Printing the result
System.out.println("\n\nResult:");
for (int i = 0; i < result.length; i++) {
for (int j = 0; j < result[i].length; j++) {
System.out.print(result[i][j] + " ");
}
System.out.println();
}
}//End main
}//End Class
//Multiply Class
class Multiply extends MatrixTest {
private int i;
private int j;
private int chance;
public Multiply(int i, int j) {
this.i = i;
this.j = j;
chance = 0;
}
//Matrix Multiplication Function
public synchronized void multiplyMatrix() {
int sum = 0;
int a = 0;
for (a = 0; a < i; a++) {
sum = 0;
for (int b = 0; b < j; b++) {
sum = sum + mat[chance][b] * mat2[b][a];
}
result[chance][a] = sum;
}
if (chance >= i)
return;
chance++;
}
}//End multiply class
//Thread Class
class MatrixMultiplier implements Runnable {
private final Multiply mul;
public MatrixMultiplier(Multiply mul) {
this.mul = mul;
}
#Override
public void run() {
mul.multiplyMatrix();
}
}
I just tried on Eclipse and it works, but now I want to create another version of that program in which, I use one thread for each cell that I'll have on the result matrix. For example I've got two 3x3 matrices. So the result matrix will be 3x3. Then, I want to use 9 threads to calculate each one of the 9 cells of the result matrix.
Can anyone help me?

You can create n Threads as follows (Note: numberOfThreads is the number of threads that you want to create. This will be the number of cells):
List<Thread> threads = new ArrayList<>(numberOfThreads);
for (int x = 0; x < numberOfThreads; x++) {
Thread t = new Thread(new MatrixMultiplier(multiply));
t.start();
threads.add(t);
}
for (Thread t : threads) {
t.join();
}

Please use the new Executor framework to create Threads, instead of manually doing the plumbing.
ExecutorService executor = Executors.newFixedThreadPool(numberOfThreadsInPool);
for (int i = 0; i < numberOfThreads; i++) {
Runnable worker = new Thread(new MatrixMultiplier(multiply));;
executor.execute(worker);
}
executor.shutdown();
while (!executor.isTerminated()) {
}

With this code i think that i resolve my problem. I don't use synchronized in the methods but i think that is not necessary in that case.
import java.util.Scanner;
class MatrixProduct extends Thread {
private int[][] A;
private int[][] B;
private int[][] C;
private int rig, col;
private int dim;
public MatrixProduct(int[][] A, int[][] B, int[][] C, int rig, int col, int dim_com) {
this.A = A;
this.B = B;
this.C = C;
this.rig = rig;
this.col = col;
this.dim = dim_com;
}
public void run() {
for (int i = 0; i < dim; i++) {
C[rig][col] += A[rig][i] * B[i][col];
}
System.out.println("Thread " + rig + "," + col + " complete.");
}
}
public class MatrixMultiplication {
public static void main(String[] args) {
Scanner In = new Scanner(System.in);
System.out.print("Row of Matrix A: ");
int rA = In.nextInt();
System.out.print("Column of Matrix A: ");
int cA = In.nextInt();
System.out.print("Row of Matrix B: ");
int rB = In.nextInt();
System.out.print("Column of Matrix B: ");
int cB = In.nextInt();
System.out.println();
if (cA != rB) {
System.out.println("We can't do the matrix product!");
System.exit(-1);
}
System.out.println("The matrix result from product will be " + rA + " x " + cB);
System.out.println();
int[][] A = new int[rA][cA];
int[][] B = new int[rB][cB];
int[][] C = new int[rA][cB];
MatrixProduct[][] thrd = new MatrixProduct[rA][cB];
System.out.println("Insert A:");
System.out.println();
for (int i = 0; i < rA; i++) {
for (int j = 0; j < cA; j++) {
System.out.print(i + "," + j + " = ");
A[i][j] = In.nextInt();
}
}
System.out.println();
System.out.println("Insert B:");
System.out.println();
for (int i = 0; i < rB; i++) {
for (int j = 0; j < cB; j++) {
System.out.print(i + "," + j + " = ");
B[i][j] = In.nextInt();
}
}
System.out.println();
for (int i = 0; i < rA; i++) {
for (int j = 0; j < cB; j++) {
thrd[i][j] = new MatrixProduct(A, B, C, i, j, cA);
thrd[i][j].start();
}
}
for (int i = 0; i < rA; i++) {
for (int j = 0; j < cB; j++) {
try {
thrd[i][j].join();
} catch (InterruptedException e) {
}
}
}
System.out.println();
System.out.println("Result");
System.out.println();
for (int i = 0; i < rA; i++) {
for (int j = 0; j < cB; j++) {
System.out.print(C[i][j] + " ");
}
System.out.println();
}
}
}

Consider Matrix.java and Main.java as follows.
public class Matrix extends Thread {
private static int[][] a;
private static int[][] b;
private static int[][] c;
/* You might need other variables as well */
private int i;
private int j;
private int z1;
private int s;
private int k;
public Matrix(int[][] A, final int[][] B, final int[][] C, int i, int j, int z1) { // need to change this, might
// need some information
a = A;
b = B;
c = C;
this.i = i;
this.j = j;
this.z1 = z1; // a[0].length
}
public void run() {
synchronized (c) {
// 3. How to allocate work for each thread (recall it is the run function which
// all the threads execute)
// Here this code implements the allocated work for perticular thread
// Each element of the resulting matrix will generate by a perticular thread
for (s = 0, k = 0; k < z1; k++)
s += a[i][k] * b[k][j];
c[i][j] = s;
}
}
public static int[][] returnC() {
return c;
}
public static int[][] multiply(final int[][] a, final int[][] b) {
/*
* check if multipication can be done, if not return null allocate required
* memory return a * b
*/
final int x = a.length;
final int y = b[0].length;
final int z1 = a[0].length;
final int z2 = b.length;
if (z1 != z2) {
System.out.println("Cannnot multiply");
return null;
}
final int[][] c = new int[x][y];
int i, j;
// 1. How to use threads to parallelize the operation?
// Every element in the resulting matrix will be determined by a different
// thread
// 2. How may threads to use?
// x * y threads are used to generate the result.
for (i = 0; i < x; i++)
for (j = 0; j < y; j++) {
try {
Matrix temp_thread = new Matrix(a, b, c, i, j, z1);
temp_thread.start();
// 4. How to synchronize?
// synchronized() is used with join() to guarantee that the perticular thread
// will be accessed first
temp_thread.join();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
return Matrix.returnC();
}
}
You can use Main.java to give 2 matrices that need to be multiplied.
class Main {
public static int[][] a = {
{1, 1, 1},
{1, 1, 1},
{1, 1, 1}};
public static int[][] b = {
{1},
{1},
{1}};
public static void print_matrix(int[][] a) {
for (int i = 0; i < a.length; i++) {
for (int j = 0; j < a[i].length; j++)
System.out.print(a[i][j] + " ");
System.out.println();
}
}
public static void main(String[] args) {
int[][] x = Matrix.multiply(a, b);
print_matrix(x); // see if the multipication is correct
}
}

In simple terms, what you all need to do is,
1) Create n (no of cells in resultant matrix) threads. Assign their roles. (Ex: Consider M X N, where M and N are matrices. 'thread1' is responsible for the multiplication of M's row_1 elements with N's column_1 elements and storing the result. This is the value for the resultant matrix's cell_1.)
2) Start each thread's process. (by start() method)
3) Wait until all the threads finish their processes and store the resultant value of each cell. Because those processes should be finished before displaying the resultant matrix. (You can do this by join() methods, and other possibilities too)
4) Now, you can display the resultant matrix.
Note:
1) Since, in this example, the shared resources (M and N) are only used to read only purpose, you don't need to use 'synchronized' methods to access them.
2) You can see, in this program, there are a group of threads running and all of them needs to achieve a specific status by their own, before continuing the next step of the whole program. This multi-threaded programming model is known as a Barrier.

Tried below code in eclipse as per thread for each cell. It works fine, you can check it.
class ResMatrix {
static int[][] arrres = new int[2][2];
}
class Matrix {
int[][] arr = new int[2][2];
void setV(int v) {
//int tmp = v;
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
arr[i][j] = v;
v = v + 1;
}
}
}
int[][] getV() {
return arr;
}
}
class Mul extends Thread {
public int row;
public int col;
Matrix m;
Matrix m1;
Mul(int row, int col, Matrix m, Matrix m1) {
this.row = row;
this.col = col;
this.m = m;
this.m1 = m1;
}
public void run() {
//System.out.println("Started Thread: " + Thread.currentThread().getName());
int tmp = 0;
for (int i = 0; i < 2; i++) {
tmp = tmp + this.m.getV()[row][i] * this.m1.getV()[i][col];
}
ResMatrix.arrres[row][col] = tmp;
System.out.println("Started Thread END: " + Thread.currentThread().getName());
}
}
public class Test {
//static int[][] arrres =new int[2][2];
public static void main(String[] args) throws InterruptedException {
Matrix mm = new Matrix();
mm.setV(1);
Matrix mm1 = new Matrix();
mm1.setV(2);
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
Mul mul = new Mul(i, j, mm, mm1);
mul.start();
// mul.join();
}
}
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
System.out.println("VALUE: " + ResMatrix.arrres[i][j]);
}
}
}
}

In my solution I assigned to each worker a number of rows numRowForThread equals to: (number of rows of matA) / (number of threads).
public class MatMulConcur {
private final static int NUM_OF_THREAD = 1;
private static Mat matC;
public static Mat matmul(Mat matA, Mat matB) {
matC = new Mat(matA.getNRows(), matB.getNColumns());
return mul(matA, matB);
}
private static Mat mul(Mat matA, Mat matB) {
int numRowForThread;
int numRowA = matA.getNRows();
int startRow = 0;
Worker[] myWorker = new Worker[NUM_OF_THREAD];
for (int j = 0; j < NUM_OF_THREAD; j++) {
if (j < NUM_OF_THREAD - 1) {
numRowForThread = (numRowA / NUM_OF_THREAD);
} else {
numRowForThread = (numRowA / NUM_OF_THREAD) + (numRowA % NUM_OF_THREAD);
}
myWorker[j] = new Worker(startRow, startRow + numRowForThread, matA, matB);
myWorker[j].start();
startRow += numRowForThread;
}
for (Worker worker : myWorker) {
try {
worker.join();
} catch (InterruptedException e) {
}
}
return matC;
}
private static class Worker extends Thread {
private int startRow, stopRow;
private Mat matA, matB;
public Worker(int startRow, int stopRow, Mat matA, Mat matB) {
super();
this.startRow = startRow;
this.stopRow = stopRow;
this.matA = matA;
this.matB = matB;
}
#Override
public void run() {
for (int i = startRow; i < stopRow; i++) {
for (int j = 0; j < matB.getNColumns(); j++) {
double sum = 0;
for (int k = 0; k < matA.getNColumns(); k++) {
sum += matA.get(i, k) * matB.get(k, j);
}
matC.set(i, j, sum);
}
}
}
}
}
where for the class Mat, I used this implementation:
public class Mat {
private double[][] mat;
public Mat(int n, int m) {
mat = new double[n][m];
}
public void set(int i, int j, double v) {
mat[i][j] = v;
}
public double get(int i, int j) {
return mat[i][j];
}
public int getNRows() {
return mat.length;
}
public int getNColumns() {
return mat[0].length;
}
}

Java: method in static initialization block is slower than in main methd

For some reason, my method "bishops" runs much faster when called from the main method than from the static initialization block. Is this normal, or a bug?
public class Magic
{
public static void main(String[] args)
{
bishops();
}
public static void bishops()
{
//PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("bishops.txt")));
BISHOP_SHIFTS = new int[64];
BISHOP_COMBOS = new long[64][];
for (int square = 0; square < 64; square++) {System.out.println("bbb " + square);
int NUMBER = bitCount(BISHOP_ATTACKS[square]);
BISHOP_SHIFTS[square] = 64 - NUMBER;
long x = BISHOP_ATTACKS[square];
long[] MAPS = new long[NUMBER];
for (int n = 0; n < NUMBER; n++) {
int i = bitScan(x);
MAPS[n] = (1L << i);
x -= MAPS[n];
}
int C = 1 << NUMBER;
BISHOP_COMBOS[square] = new long[C];
for (int i = 0; i < C; i++) {
BISHOP_COMBOS[square][i] = 0;
int j = i;
for (int n = 0; n < NUMBER; n++) {
if ((j & 1) == 1)
BISHOP_COMBOS[square][i] |= MAPS[n];
j >>>= 1;
}
//out.println("SQUARE " + square);
//out.println(toBitboardString(BISHOP_COMBOS[square][i]));
//out.println();
}
}
//out.close();
bishopMagics();
}
public static void bishopMagics()
{
BISHOP_MAGICS = new long[64];
Random r = new Random();
for (int square = 0; square < 64; square++) {System.out.println("asdffff " + square);
int i;
int LENGTH = BISHOP_COMBOS[square].length;
long magic;
do {
magic = r.nextLong() & r.nextLong() & r.nextLong();
//final int COUNT = bitCount(BISHOP_MASKS[square]);
boolean[] used = new boolean[LENGTH];
for (int j = 0; j < used.length; j++)
used[j] = false;
for (i = 0; i < LENGTH; i++) {
int index = (int) ((BISHOP_COMBOS[square][i] * magic) >>> BISHOP_SHIFTS[square]);
if (used[index])
break;
else
used[index] = true;
}
} while (i < LENGTH);
BISHOP_MAGICS[square] = magic;
System.out.println(magic);
}
//bishopTable();
}
/*
* Lots of stuff omitted
*/
static
{
//bishops();
}
}

It will run much faster the second time than the first as the JVM warms up (loads class es and compiles code). The static block is always called first.
Try running it twice from the main() or the static block and see how long it takes each time
BTW: I would take out any logging to the console as this can slow down the code dramatically.

Optimizing N queens puzzle

I'm trying to solve the problem of positioning N queens on NxN board without row, column and diagonal conflicts. I use an algorithm with minimizing the conflicts. Firstly, on each column randomly a queen is positioned. After that, of all conflict queens randomly one is chosen and for her column are calculated the conflicts of each possible position. Then, the queen moves to the best position with min number of conflicts. It works, but it runs extremely slow. My goal is to make it run fast for 10000 queens. Would you, please, suggest me some improvements or maybe notice some mistakes in my logic?
Here is my code:
public class Queen {
int column;
int row;
int d1;
int d2;
public Queen(int column, int row, int d1, int d2) {
super();
this.column = column;
this.row = row;
this.d1 = d1;
this.d2 = d2;
}
#Override
public String toString() {
return "Queen [column=" + column + ", row=" + row + ", d1=" + d1
+ ", d2=" + d2 + "]";
}
#Override
public boolean equals(Object obj) {
return ((Queen)obj).column == this.column && ((Queen)obj).row == this.row;
}
}
And:
import java.util.HashSet;
import java.util.Random;
public class SolveQueens {
public static boolean printBoard = false;
public static int N = 100;
public static int maxSteps = 2000000;
public static int[] queens = new int[N];
public static Random random = new Random();
public static HashSet<Queen> q = new HashSet<Queen>();
public static HashSet rowConfl[] = new HashSet[N];
public static HashSet d1Confl[] = new HashSet[2*N - 1];
public static HashSet d2Confl[] = new HashSet[2*N - 1];
public static void init () {
int r;
rowConfl = new HashSet[N];
d1Confl = new HashSet[2*N - 1];
d2Confl = new HashSet[2*N - 1];
for (int i = 0; i < N; i++) {
r = random.nextInt(N);
queens[i] = r;
Queen k = new Queen(i, r, i + r, N - 1 + i - r);
q.add(k);
if (rowConfl[k.row] == null) {
rowConfl[k.row] = new HashSet<Queen>();
}
if (d1Confl[k.d1] == null) {
d1Confl[k.d1] = new HashSet<Queen>();
}
if (d2Confl[k.d2] == null) {
d2Confl[k.d2] = new HashSet<Queen>();
}
((HashSet<Queen>)rowConfl[k.row]).add(k);
((HashSet<Queen>)d1Confl[k.d1]).add(k);
((HashSet<Queen>)d2Confl[k.d2]).add(k);
}
}
public static void print () {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
System.out.print(queens[i] == j ? "♕ " : "◻◻◻ ");
}
System.out.println();
}
System.out.println();
}
public static boolean checkItLinear() {
Queen r = choseConflictQueen();
if (r == null) {
return true;
}
Queen newQ = findNewBestPosition(r);
q.remove(r);
q.add(newQ);
rowConfl[r.row].remove(r);
d1Confl[r.d1].remove(r);
d2Confl[r.d2].remove(r);
if (rowConfl[newQ.row] == null) {
rowConfl[newQ.row] = new HashSet<Queen>();
}
if (d1Confl[newQ.d1] == null) {
d1Confl[newQ.d1] = new HashSet<Queen>();
}
if (d2Confl[newQ.d2] == null) {
d2Confl[newQ.d2] = new HashSet<Queen>();
}
((HashSet<Queen>)rowConfl[newQ.row]).add(newQ);
((HashSet<Queen>)d1Confl[newQ.d1]).add(newQ);
((HashSet<Queen>)d2Confl[newQ.d2]).add(newQ);
queens[r.column] = newQ.row;
return false;
}
public static Queen choseConflictQueen () {
HashSet<Queen> conflictSet = new HashSet<Queen>();
boolean hasConflicts = false;
for (int i = 0; i < 2*N - 1; i++) {
if (i < N && rowConfl[i] != null) {
hasConflicts = hasConflicts || rowConfl[i].size() > 1;
conflictSet.addAll(rowConfl[i]);
}
if (d1Confl[i] != null) {
hasConflicts = hasConflicts || d1Confl[i].size() > 1;
conflictSet.addAll(d1Confl[i]);
}
if (d2Confl[i] != null) {
hasConflicts = hasConflicts || d2Confl[i].size() > 1;
conflictSet.addAll(d2Confl[i]);
}
}
if (hasConflicts) {
int c = random.nextInt(conflictSet.size());
return (Queen) conflictSet.toArray()[c];
}
return null;
}
public static Queen findNewBestPosition(Queen old) {
int[] row = new int[N];
int min = Integer.MAX_VALUE;
int minInd = old.row;
for (int i = 0; i < N; i++) {
if (rowConfl[i] != null) {
row[i] = rowConfl[i].size();
}
if (d1Confl[old.column + i] != null) {
row[i] += d1Confl[old.column + i].size();
}
if (d2Confl[N - 1 + old.column - i] != null) {
row[i] += d2Confl[N - 1 + old.column - i].size();
}
if (i == old.row) {
row[i] = row[i] - 3;
}
if (row[i] <= min && i != minInd) {
min = row[i];
minInd = i;
}
}
return new Queen(old.column, minInd, old.column + minInd, N - 1 + old.column - minInd);
}
public static void main(String[] args) {
long startTime = System.currentTimeMillis();
init();
int steps = 0;
while(!checkItLinear()) {
if (++steps > maxSteps) {
init();
steps = 0;
}
}
long endTime = System.currentTimeMillis();
System.out.println("Done for " + (endTime - startTime) + "ms\n");
if(printBoard){
print();
}
}
}
Edit:
Here is my a-little-bit-optimized solution with removing some unused objects and putting the queens on diagonal positions when initializing.
import java.util.Random;
import java.util.Vector;
public class SolveQueens {
public static boolean PRINT_BOARD = true;
public static int N = 10;
public static int MAX_STEPS = 5000;
public static int[] queens = new int[N];
public static Random random = new Random();
public static int[] rowConfl = new int[N];
public static int[] d1Confl = new int[2*N - 1];
public static int[] d2Confl = new int[2*N - 1];
public static Vector<Integer> conflicts = new Vector<Integer>();
public static void init () {
random = new Random();
for (int i = 0; i < N; i++) {
queens[i] = i;
}
}
public static int getD1Pos (int col, int row) {
return col + row;
}
public static int getD2Pos (int col, int row) {
return N - 1 + col - row;
}
public static void print () {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
System.out.print(queens[i] == j ? "Q " : "* ");
}
System.out.println();
}
System.out.println();
}
public static boolean hasConflicts() {
generateConflicts();
if (conflicts.isEmpty()) {
return false;
}
int r = random.nextInt(conflicts.size());
int conflQueenCol = conflicts.get(r);
int currentRow = queens[conflQueenCol];
int bestRow = currentRow;
int minConfl = getConflicts(conflQueenCol, queens[conflQueenCol]) - 3;
int tempConflCount;
for (int i = 0; i < N ; i++) {
tempConflCount = getConflicts(conflQueenCol, i);
if (i != currentRow && tempConflCount <= minConfl) {
minConfl = tempConflCount;
bestRow = i;
}
}
queens[conflQueenCol] = bestRow;
return true;
}
public static void generateConflicts () {
conflicts = new Vector<Integer>();
rowConfl = new int[N];
d1Confl = new int[2*N - 1];
d2Confl = new int[2*N - 1];
for (int i = 0; i < N; i++) {
int r = queens[i];
rowConfl[r]++;
d1Confl[getD1Pos(i, r)]++;
d2Confl[getD2Pos(i, r)]++;
}
for (int i = 0; i < N; i++) {
int conflictsCount = getConflicts(i, queens[i]) - 3;
if (conflictsCount > 0) {
conflicts.add(i);
}
}
}
public static int getConflicts(int col, int row) {
return rowConfl[row] + d1Confl[getD1Pos(col, row)] + d2Confl[getD2Pos(col, row)];
}
public static void main(String[] args) {
long startTime = System.currentTimeMillis();
init();
int steps = 0;
while(hasConflicts()) {
if (++steps > MAX_STEPS) {
init();
steps = 0;
}
}
long endTime = System.currentTimeMillis();
System.out.println("Done for " + (endTime - startTime) + "ms\n");
if(PRINT_BOARD){
print();
}
}
}

Comments would have been helpful :)
Rather than recreating your conflict set and your "worst conflict" queen everything, could you create it once, and then just update the changed rows/columns?
EDIT 0:
I tried playing around with your code a bit. Since the code is randomized, it's hard to find out if a change is good or not, since you might start with a good initial state or a crappy one. I tried making 10 runs with 10 queens, and got wildly different answers, but results are below.
I psuedo-profiled to see which statements were being executed the most, and it turns out the inner loop statements in chooseConflictQueen are executed the most. I tried inserting a break to pull the first conflict queen if found, but it didn't seem to help much.
Grouping only runs that took more than a second:
I realize I only have 10 runs, which is not really enough to be statistically valid, but hey.
So adding breaks didn't seem to help. I think a constructive solution will likely be faster, but randomness will again make it harder to check.

Your approach is good : Local search algorithm with minimum-conflicts constraint. I would suggest try improving your initial state. Instead of randomly placing all queens, 1 per column, try to place them so that you minimize the number of conflicts. An example would be to try placing you next queen based on the position of the previous one ... or maybe position of previous two ... Then you local search will have less problematic columns to deal with.

If you randomly select, you could be selecting the same state as a previous state. Theoretically, you might never find a solution even if there is one.
I think you woud be better to iterate normally through the states.
Also, are you sure boards other than 8x8 are solvable?
By inspection, 2x2 is not, 3x3 is not, 4x4 is not.