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Find the max sum path from bottom left side to top right cell

I have to find the maximum sum path in a 2d Array(n,m) given which has a value from 0 to 99999. 0 means wall. We have t start from the left bottom side of the array and must reach the right top cell(0,m-1). You can go up/down/right and can visit each cell once. Below is the code without any blocks .My problem is that i cant move from left bottom to right top cell . I also created left array(lest side of the main array) so that i can start from the best value possible .Sorry am not good programmer :).
Code
public static int maxvalue(int [][]field,int[] left)
{
for(int i=field.length-1;i>0 && left[i]!=-1;i--)
{
System.out.println( "Startpos "+i+" 0");
int distance =max(i,0,field,0,field.length-1);
if(distance>maxvalue)
maxvalue=distance;
}
return maxvalue;
}
public static int max(int r, int c,int [][]field ,int destR, int destC)
{
if(r>destR|| c>destC)
return 0;
if(r==0 && c==field[0].length)
return field[r][c];
int sum1=max(r-1,c,field,destR,destC); // up
System.out.println(sum1);
int sum2= max(r+1,c,field,destR,destC); //down
int sum3= max(r,c+1,field,destR,destC); //right
return field[r][c]+Math.max(sum1, Math.max(sum2, sum3));
}
Sample
Input
0 1 2 3
2 0 2 4
3 3 0 3
4 2 1 2
Output
25
How to do solve this question? if all the path is blocked then print No Solution.

Have you first tried to solve it by yourself?
It looks like a bit of work but it is not impossible.
What I would use is 3 int variables : xPosition, yPosition and Sum;
Go on and test the values of xPosition+1, yPosition-1 in priority and then the rest (because you want to reach xPosition == array.length - 1 && yPosition == 0.) and if you find a 0, test the other possibilities and exclude the ones you already passed by.
Each time you find a good path, add the value of the cell to your sum.
Reset it to 0 once you're blocked.

For every element in the array, you have to find the maximum of the adjacent elements and also check the boundary conditions. I hope this code will help you.
public class StackOverFlow {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int m = in.nextInt();
Integer [][] array = new Integer[n][m];
boolean [][] visited = new boolean[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
array[i][j] = in.nextInt();
}
}
int i = n-1, j =0;
visited[i][j] = true;
int sum = array[i][j];
while(true)
{
int max = -1;
int maxi = 0, maxj = 0;
if(i-1 >= 0 && i-1<= n-1 && j>=0 && j<= m-1 && array[i-1][j] != null && array[i-1][j]>max && !visited[i-1][j])
{
max = array[i-1][j];
maxi = i-1;
maxj=j;
}
if(i+1 >= 0 && i+1<= n-1 && j>=0 && j<= m-1 &&array[i+1][j] != null && array[i+1][j]>max && !visited[i+1][j])
{
max = array[i+1][j];
maxi = i+1;
maxj=j;
}
if(i >= 0 && i<= n-1 && j-1>=0 && j-1<= m-1 && array[i][j-1] != null && array[i][j-1]>max && !visited[i][j-1])
{
max = array[i][j-1];
maxi = i;
maxj=j-1;
}
if(i >= 0 && i<= n-1 && j+1>=0 && j+1<= m-1 && array[i][j+1] != null && array[i][j+1]>max && !visited[i][j+1])
{
max = array[i][j+1];
maxi = i;
maxj=j+1;
}
i = maxi;
j = maxj;
visited[i][j] = true;
sum += max;
if(i == 0 && j == m-1)
break;
}
System.out.println(sum);
}
}

How to print out an X using nested loops

I have searched through to find a simple solution to this problem.
I have a method called
printCross(int size,char display)
It accepts a size and prints an X with the char variable it receives of height and width of size.
The calling method printShape(int maxSize, char display) accepts the maximum size of the shape and goes in a loop, sending multiples of 2 to the printCross method until it gets to the maximum.
Here is my code but it is not giving me the desired outcome.
public static void drawShape(char display, int maxSize)
{
int currentSize = 2; //start at 2 and increase in multiples of 2 till maxSize
while(currentSize<=maxSize)
{
printCross(currentSize,display);
currentSize = currentSize + 2;//increment by multiples of 2
}
}
public static void printCross(int size, char display)
{
for (int row = 0; row<size; row++)
{
for (int col=0; col<size; col++)
{
if (row == col)
System.out.print(display);
if (row == 1 && col == 5)
System.out.print(display);
if (row == 2 && col == 4)
System.out.print(display);
if ( row == 4 && col == 2)
System.out.print(display);
if (row == 5 && col == 1)
System.out.print(display);
else
System.out.print(" ");
}
System.out.println();
}
}
Is it because I hardcoded the figures into the loop? I did a lot of math but unfortunately it's only this way that I have been slightly close to achieving my desired output.
If the printCross() method received a size of 5 for instance, the output should be like this:
x x
x x
x
x x
x x
Please I have spent weeks on this and seem to be going nowhere. Thanks

The first thing you have to do is to find relationships between indices. Let's say you have the square matrix of length size (size = 5 in the example):
0 1 2 3 4
0 x x
1 x x
2 x
3 x x
4 x x
What you can notice is that in the diagonal from (0,0) to (4,4), indices are the same (in the code this means row == col).
Also, you can notice that in the diagonal from (0,4) to (4,0) indices always sum up to 4, which is size - 1 (in the code this is row + col == size - 1).
So in the code, you will loop through rows and then through columns (nested loop). On each iteration you have to check if the conditions mentioned above are met. The logical OR (||) operator is used to avoid using two if statements.
Code:
public static void printCross(int size, char display)
{
for (int row = 0; row < size; row++) {
for (int col = 0; col < size; col++) {
if (row == col || row + col == size - 1) {
System.out.print(display);
} else {
System.out.print(" ");
}
}
System.out.println();
}
}
Output: (size = 5, display = 'x')
x x
x x
x
x x
x x

Instead of giving a direct answer, I will give you some hints.
First, you are right to use nested for loops.
However as you noticed, you determine when to print 'x' for the case of 5.
Check that 'x' is printed if and only if row = col or row + col = size - 1

for your printCross method, try this:
public static void printCross(int size, char display) {
if( size <= 0 ) {
return;
}
for( int row = 0; row < size; row++ ) {
for( int col = 0; col < size; col++ ) {
if( col == row || col == size - row - 1) {
System.out.print(display);
}
else {
System.out.print(" ");
}
}
System.out.println();
}
}
ah, I got beaten to it xD

Here's a short, ugly solution which doesn't use any whitespace strings or nested looping.
public static void printCross(int size, char display) {
for (int i = 1, j = size; i <= size && j > 0; i++, j--) {
System.out.printf(
i < j ? "%" + i + "s" + "%" + (j - i) + "s%n"
: i > j ? "%" + j + "s" + "%" + (i - j) + "s%n"
: "%" + i + "s%n", //intersection
display, display
);
}
}

Lte's try this simple code to print cross pattern.
class CrossPattern {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
System.out.println("enter the number of rows=column");
int n = s.nextInt();
int i, j;
s.close();
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
if (j == i) {
System.out.print("*");
} else if (j == n - (i - 1)) {
System.out.print("*");
} else {
System.out.print(" ");
}
}
System.out.println();
}
}
}

java:generalized 8 queen to work for any initial state using depth first search [duplicate]

i am try to implement 8 queen using depth search for any initial state it work fine for empty board(no queen on the board) ,but i need it to work for initial state if there is a solution,if there is no solution for this initial state it will print there is no solution
Here is my code:
public class depth {
public static void main(String[] args) {
//we create a board
int[][] board = new int[8][8];
board [0][0]=1;
board [1][1]=1;
board [2][2]=1;
board [3][3]=1;
board [4][4]=1;
board [5][5]=1;
board [6][6]=1;
board [7][7]=1;
eightQueen(8, board, 0, 0, false);
System.out.println("the solution as pair");
for(int i=0;i<board.length;i++){
for(int j=0;j<board.length;j++)
if(board[i][j]!=0)
System.out.println(" ("+i+" ,"+j +")");
}
System.out.println("the number of node stored in memory "+count1);
}
public static int count1=0;
public static void eightQueen(int N, int[][] board, int i, int j, boolean found) {
long startTime = System.nanoTime();//time start
if (!found) {
if (IsValid(board, i, j)) {//check if the position is valid
board[i][j] = 1;
System.out.println("[Queen added at (" + i + "," + j + ")");
count1++;
PrintBoard(board);
if (i == N - 1) {//check if its the last queen
found = true;
PrintBoard(board);
double endTime = System.nanoTime();//end the method time
double duration = (endTime - startTime)*Math.pow(10.0, -9.0);
System.out.print("total Time"+"= "+duration+"\n");
}
//call the next step
eightQueen(N, board, i + 1, 0, found);
} else {
//if the position is not valid & if reach the last row we backtracking
while (j >= N - 1) {
int[] a = Backmethod(board, i, j);
i = a[0];
j = a[1];
System.out.println("back at (" + i + "," + j + ")");
PrintBoard(board);
}
//we do the next call
eightQueen(N, board, i, j + 1, false);
}
}
}
public static int[] Backmethod(int[][] board, int i, int j) {
int[] a = new int[2];
for (int x = i; x >= 0; x--) {
for (int y = j; y >= 0; y--) {
//search for the last queen
if (board[x][y] != 0) {
//deletes the last queen and returns the position
board[x][y] = 0;
a[0] = x;
a[1] = y;
return a;
}
}
}
return a;
}
public static boolean IsValid(int[][] board, int i, int j) {
int x;
//check the queens in column
for (x = 0; x < board.length; x++) {
if (board[i][x] != 0) {
return false;
}
}
//check the queens in row
for (x = 0; x < board.length; x++) {
if (board[x][j] != 0) {
return false;
}
}
//check the queens in the diagonals
if (!SafeDiag(board, i, j)) {
return false;
}
return true;
}
public static boolean SafeDiag(int[][] board, int i, int j) {
int xx = i;
int yy = j;
while (yy >= 0 && xx >= 0 && xx < board.length && yy < board.length) {
if (board[xx][yy] != 0) {
return false;
}
yy++;
xx++;
}
xx = i;
yy = j;
while (yy >= 0 && xx >= 0 && xx < board.length && yy < board.length) {
if (board[xx][yy] != 0) {
return false;
}
yy--;
xx--;
}
xx = i;
yy = j;
while (yy >= 0 && xx >= 0 && xx < board.length && yy < board.length) {
if (board[xx][yy] != 0) {
return false;
}
yy--;
xx++;
}
xx = i;
yy = j;
while (yy >= 0 && xx >= 0 && xx < board.length && yy < board.length) {
if (board[xx][yy] != 0) {
return false;
}
yy++;
xx--;
}
return true;
}
public static void PrintBoard(int[][] board) {
System.out.print(" ");
for (int j = 0; j < board.length; j++) {
System.out.print(j);
}
System.out.print("\n");
for (int i = 0; i < board.length; i++) {
System.out.print(i);
for (int j = 0; j < board.length; j++) {
if (board[i][j] == 0) {
System.out.print(" ");
} else {
System.out.print("Q");
}
}
System.out.print("\n");
}
}
}
for example for this initial state it give me the following error:
Exception in thread "main" java.lang.StackOverflowError
i am stuck, i think the error is infinite call for the method how to solve this problem.
any idea will be helpful,thanks in advance.
note:the broad is two dimensional array,when i put (1) it means there queen at this point.
note2:
we i put the initial state as the following it work:
board [0][0]=1;
board [1][1]=1;
board [2][2]=1;
board [3][3]=1;
board [4][4]=1;
board [5][5]=1;
board [6][6]=1;
board [7][1]=1;

[EDIT: Added conditional output tip.]
To add to #StephenC's answer:
This is a heck of a complicated piece of code, especially if you're not experienced in programming Java.
I executed your code, and it outputs this over and over and over and over (and over)
back at (0,0)
01234567
0
1 Q
2 Q
3 Q
4 Q
5 Q
6 Q
7 Q
back at (0,0)
And then crashes with this
Exception in thread "main" java.lang.StackOverflowError
at java.nio.Buffer.<init>(Unknown Source)
...
at java.io.PrintStream.print(Unknown Source)
at java.io.PrintStream.println(Unknown Source)
at Depth.eightQueen(Depth.java:56)
at Depth.eightQueen(Depth.java:60)
at Depth.eightQueen(Depth.java:60)
at Depth.eightQueen(Depth.java:60)
at Depth.eightQueen(Depth.java:60)
...
My first instinct is always to add some System.out.println(...)s to figure out where stuff is going wrong, but that won't work here.
The only two options I see are to
Get familiar with a debugger and use it to step through and analyze why it's never stopping the loop
Break it down man! How can you hope to deal with a massive problem like this without breaking it into digestible chunks???
Not to mention that the concept of 8-queens is complicated to begin with.
One further thought:
System.out.println()s are not useful as currently implemented, because there's infinite output. A debugger is the better solution here, but another option is to somehow limit your output. For example, create a counter at the top
private static final int iITERATIONS = 0;
and instead of
System.out.println("[ANUMBERFORTRACING]: ... USEFUL INFORMATION ...")
use
conditionalSDO((iITERATIONS < 5), "[ANUMBERFORTRACING]: ... USEFUL INFORMATION");
Here is the function:
private static final void conditionalSDO(boolean b_condition, String s_message) {
if(b_condition) {
System.out.println(s_message);
}
}
Another alternative is to not limit the output, but to write it to a file.
I hope this information helps you.
(Note: I edited the OP's code to be compilable.)

You asked for ideas on how to solve it (as distinct from solutions!) so, here's a couple of hints:
Hint #1:
If you get a StackOverflowError in a recursive program it can mean one of two things:
your problem is too "deep", OR
you've got a bug in your code that is causing it to recurse infinitely.
In this case, the depth of the problem is small (8), so this must be a recursion bug.
Hint #2:
If you examine the stack trace, you will see the method names and line numbers for each of the calls in the stack. This ... and some thought ... should help you figure out the pattern of recursion in your code (as implemented!).
Hint #3:
Use a debugger Luke ...
Hint #4:
If you want other people to read your code, pay more attention to style. Your indentation is messed up in the most important method, and you have committed the (IMO) unforgivable sin of ignoring the Java style rules for identifiers. A method name MUST start with a lowercase letter, and a class name MUST start with an uppercase letter.
(I stopped reading your code very quickly ... on principle.)

Try to alter your method IsValid in the lines where for (x = 0; x < board.length - 1; x++).
public static boolean IsValid(int[][] board, int i, int j) {
int x;
//check the queens in column
for (x = 0; x < board.length - 1; x++) {
if (board[i][x] != 0) {
return false;
}
}
//check the queens in row
for (x = 0; x < board.length - 1; x++) {
if (board[x][j] != 0) {
return false;
}
}
//check the queens in the diagonals
if (!SafeDiag(board, i, j)) {
return false;
}
return true;
}

Creating a random colour grid with all adjacent colours different

So I have the following problem set to me: Write a program that takes an integer command-line argument N, and uses two nested for loops to print an N-by-N board that alternates between 6 colours randomly separated by spaces. The colours are denoted by letters (like 'r' for RED, 'b' for BLUE). You are not allowed to have two of the same colour next to eachother.
So, I know I probably need arrays to get around this problem. I tried several methods that all came up wrong. The following is one of my recent attempts, but I am unsure as how to now go through the grid and correct it. What the code does is make every row randomized with no colour left or right the same, but the columns are not fixed.
Note that I am a first year CS student with no programming history. I am guessing the solution to this problem isnt too complex, however, I cant see a simple solution...
int N = StdIn.readInt();
int array1[] = new int[N];
for (int column = 0; column < N; column++) {
int x = 0;
for (int row = 0; row < N; row++) {
int c = (int) (Math.random() * 6 + 1);
while (x == c) {
c = (int) (Math.random() * 6 + 1);
array1[row] = c;
}
if (c == 1) {
System.out.print("R ");
}
if (c == 2) {
System.out.print("O ");
}
if (c == 3) {
System.out.print("Y ");
}
if (c == 4) {
System.out.print("G ");
}
if (c == 5) {
System.out.print("B ");
}
if (c == 6) {
System.out.print("I ");
}
x = c;
}
System.out.println();
}
}

this was my solution for the problem. Quite convoluted though, but the logic behind it is straightforward. Each time you assign a new colour to your 2D array, you need only check the value of the array to the top and to the left of the position where you want to assign a new colour. You can only do this after you have assigned colours to the first row of the array however so you need to create separate conditions for the first row.
public class ColourGrid {
public static void main(String[] args) {
int N = Integer.parseInt(args[0]);
char[][] clrGrid = new char[N][N];
char colours[] = {'r','b','y','w','o','g'} ;
for (int counter = 0 ; counter < N; counter++) {
for (int counter2 = 0 ; counter2 < N; counter2++) {
if (counter == 0 && counter2 == 0) {
clrGrid[counter][counter2] = colours[(int)(Math.random()* 5 + 1)] ;
}
else if (counter != 0 && counter2 == 0) {
clrGrid[counter][counter2] = colours[(int)(Math.random()* 5 + 1)] ;
while (clrGrid[counter][counter2] == clrGrid[(counter)-1][counter2]) {
clrGrid[counter][counter2] = colours[(int)(Math.random()* 5 + 1)] ;
}
}
else if (counter == 0 && counter2 != 0) {
clrGrid[counter][counter2] = colours[(int)(Math.random()* 5 + 1)] ;
while (clrGrid[counter][counter2] == clrGrid[(counter)][counter2-1]) {
clrGrid[counter][counter2] = colours[(int)(Math.random()* 5 + 1)] ;
}
}
else if (counter != 0 && counter2 != 0) {
clrGrid[counter][counter2] = colours[(int)(Math.random()* 5 + 1)] ;
while (clrGrid[counter][counter2] == clrGrid[(counter)-1][counter2] || clrGrid[counter][counter2] == clrGrid[counter][(counter2)-1]) {
clrGrid[counter][counter2] = colours[(int)(Math.random()* 5 + 1)] ;
}
}
else {
clrGrid[counter][counter2] = colours[(int)(Math.random()* 5 + 1)] ;
}
}
}
for (int counter = 0 ; counter < N; counter++) {
System.out.println("");
for (int counter2 = 0 ; counter2 < N; counter2++) {
System.out.print(clrGrid[counter][counter2] + " ");
}
}
}
}

My assignment is to make a house using for loops. The house should look like this

*
*****
*********
*********
**** ***
**** ***
so far i only have this
for (int i=1; i<10; i += 4)
{
for (int j=0; j<i; j++)
{
System.out.print("*");
}
System.out.println("");
}
}
}

The simplest decision will be:
for (int y = 0; y < 6; y++) {
int shift = y < 2 ? 4 / (y + 1) : 0;
for (int x = 0; x < 9 - shift; x++) System.out.print(x >= shift && (y < 4 || (x < 4 || x > 5)) ? "*" : " ");
System.out.println();
}

I think Andre's answer is the most concise one, but if you want to have configurable home building you can use next one(try to change HEIGHT/WIDTH to see effect):
public class House {
public static void main(String[] args) {
final int HEIGHT = 6;
final int WIDTH = 9;
for (int i = 0; i < HEIGHT * 2; i += 2) {
for (int j = 0; j < WIDTH; j++) {// check for roof
if ((i + (i % 2) + (WIDTH) / 2) < j // right slope
|| (i + (i % 2) + j) < (WIDTH) / 2)// left slope
{
System.out.print(" ");
} else {
if ((i / 2 >= HEIGHT * 2 / 3) && (j >= WIDTH / 2) && j < WIDTH / 2 + HEIGHT / 3) {// check for door
System.out.print(" ");
} else {// solid then
System.out.print("*");
}
}
}
System.out.println();
}
}
}
EDIT - answer to comment:
Try to run next two example and compare output:
public static void main(String[] args) {
final int SIZE = 9;
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
System.out.print(i < j ? "+" : "-");
}
System.out.println();
}
}
and
public static void main(String[] args) {
final int SIZE = 9;
for (int i = 0; i < SIZE; i++) {
for (int j = 0; j < SIZE; j++) {
System.out.print(i < SIZE - j - 1 ? "+" : "-");
}
System.out.println();
}
}
First one will give you right slope and second left one. It all come from geometric properties of points. In first case all points will have bigger value on on x axis than on y axis. In second both x and y in sum won't exceed SIZE.
You can try to modify boolean expression inside of if() statement and see what happens, but I'd encourage you to get piece of paper and try to play with paper and pen and see what properties certain points have. Let me know if you need more explanation.

You could use a two-dimensional array like this:
char matrice [][]= {{' ',' ',' ',' ' '*', ' ',' ',' ',' '},
{' ',' ','*','*', '*', '*','*',' ',' '}};
(And so on). You basically draw your house using your array indexes.
Now you can parse each line using System.out.print() when you have to print a character, and System.out.println("") between each row.
It would look like this:
for(char[] line : house){
for(char d : line){
System.out.print(d);
}
System.out.println("");
}
You should take a look at the for-each statement documentation if you're not familiar with it.