Count neighboring cells - java

I don't understand what is wrong with this loop, it normaly would check the surrounding cells of (x,y).
When calling this function, there is only 4 passages in the loop and not 8
This is probably a stupid mistake but I cant see it !
private int countAliveNeighbors(int x, int y) {
int count = 0;
for (int i = Math.max(0, x - 1); i < Math.min(grid.getWidth()-1, x + 1); i++) {
for (int j = Math.max(0, y - 1); j < Math.min(grid.getHeight()-1, y + 1); j++) {
if (!(i == x && j == y)) {
if (grid.isAlive(i, j))
count++;
}
}
}
return count;
}

Your upper limit in your for statements should by checking for <= not just less-than. By testing for less-than you stop before getting to x+1 or y+1. It should look like this:
for (int i = Math.max(0, x - 1); i <= Math.min(grid.getWidth()-1, x + 1); i++) {
for (int j = Math.max(0, y - 1); j <= Math.min(grid.getHeight()-1, y + 1); j++) {

Related

Checking for out of bounds in a 2D array

I'm trying to check the neighboring values of each element in a 2D array but am getting an IndexOutOfBoundsException when I reach the sides of the array or a corner. For example if my array is:
|2|4|2|7|8|
|8|1|0|5|6|
|0|3|1|5|2|
|1|9|7|2|0|
I know that all the neighbors of 8 are 7,5 and 6, but my if statements don't check the bounds properly. The code I have for this is:
int numOfRows = imageArray.length;
int numOfColumns = imageArray[0].length;
for(int i = 0; i < numOfRows; i++)
for(int j = 0; j < numOfColumns; j++)
if((j+1) < numOfColumns-1)
if((i+1) < numOfRows-1)
if((j-1) > 0 )
if((i-1) > 0 )
if((i+1) < numOfColumns-1 && (j+1) < numOfRows-1)
if((i-1) >= 0 && (j-1) >= 0)
if((i+1) < numOfColumns-1 && (j-1) >= 0)
if((i-1) >= 0 && (j+1) < numOfRows-1)
I've been working on this for a while and have gone through many different techniques to solve this. Any help would be great. Thanks.
If you're trying to get all the neighbor cells and do something with them, for example add them, then you need to do some sort of bounds checking, for example something modified from this could work:
for (int i = 0; i < numOfRows; i++) {
for (int j = 0; j < numOfCols; j++) {
// check all bounds out of range:
int iMin = Math.max(0, i - 1);
int iMax = Math.min(numOfRows - 1, i + 1);
int jMin = Math.max(0, j - 1);
int jMax = Math.min(numOfCols - 1, j + 1);
// loop through the above numbers safely
for (int innerI = iMin; innerI <= iMax; innerI++) {
for (int innerJ = jMin; innerJ <= jMax; innerJ++) {
if (i != innerI && j != innerJ) {
// do what needs to be done
}
}
}
}
}
Caveat: code has not been compiled nor tested and is mainly to show you the idea of what can be done rather than a copy-paste solution

My recursive function for finding the number of paths in a grid runs infinitely

I'm working on a couple of Project Euler problems and want to test my solution. My recursive function never ends even with reachable base cases.
in a 20x20 grid I am using x and y coordinates to navigate up and left to find the number of paths from (19,19) to (0,0). My base case is to return 1 when we reach (0,0). Otherwise I add the current count to the recursive call.
Function:
private static int numPaths(int x, int y, int pathsFound)
{
if(x == 0 && y == 0)
return 1;
else
{
if(x > 0)
{
pathsFound += numPaths(x - 1, y, pathsFound);
}
if(y > 0)
{
pathsFound += numPaths(x, y - 1, pathsFound);
}
}
return pathsFound;
}
Main:
int x = 19;
int y = 19;
System.out.println("Answer: " + numPaths(x, y, 0));
Is there a flaw in my recursive logic, or is just taking a very long time to compute? If you know the solution to this Euler problem, please do not post it.
https://projecteuler.net/problem=15
So if anyone is interested, I looked into memoization and came up with an elegant solution without recursion.
Function:
private static BigInteger numberPaths(ArrayList<ArrayList<BigInteger>> grid)
{
for(int i = 0; i <= 20; ++i)
{
for(int j = 0; j <= 20; ++j)
{
int x = j;
int y = i;
if(x - 1 < 0 || y - 1 < 0)
{
grid.get(x).set(y, BigInteger.ONE);
}
else
{
BigInteger topVal = grid.get(x - 1).get(y);
BigInteger leftVal = grid.get(x).get(y - 1);
grid.get(x).set(y, topVal.add(leftVal));
}
}
}
return grid.get(20).get(20); //the solution
}
Main:
ArrayList<ArrayList<BigInteger>> grid = new ArrayList<>();
for(int i = 0; i <= 20; ++i)
{
ArrayList<BigInteger> column = new ArrayList<>();
for(int j = 0; j <= 20; ++j)
{
column.add(BigInteger.valueOf(0));
}
grid.add(column);
}
System.out.println("Answer: " + numberPaths(grid));

Traversing a 2D array matrix diagonally from bottom left to upper right

I have a 3x4 matrix represented by a 2D array:
. 0 1 2 3
0 a c f i
1 b e h k
2 d g j l
and my approach to traverse the diagonal slice was to treat each slice as a sum, like this:
a = (0+0) = 0
b,c = (0+1),(1+0) = 1
d,e,f = (0+2),(1+1),(2+0) = 2
g,h,i = (1+2),(2+1),(3+0) = 3
j, k = (2+2),(3+1) = 4
l = (3+2) = 5
However, my code right now prints it in the opposite way that I want it to, which is from upper right to bottom left.
Current Output is:
acbfedihgkjl
Desired Output is:
abcdefghijkl
for (int sum = 0; sum <= numRows + numColumns - 2; sum++) {
for (int i = 0; i < numRows; i++) {
int j = sum - i;
if ((i >= 0 && i < numRows) && (j >= 0 && j < numColumns)) {
System.out.print(array[i][j]);
}
}
}
Can somebody point me in the right direction on how to fix my code to get the output that I want?
While it isn't very pretty, I think this will do it:
int i = 0;
int j = 0;
while (true) {
System.out.println("" + array[i][j]);
--i;
++j;
if (i < 0) {
if (j == numCols)
break;
i = Math.min(j, numRows - 1);
j = Math.max(j - numCols + 2, 0);
} else if (j >= numCols) {
if (i == numRows - 2)
break;
i = numRows - 1;
j = Math.max(j + 2 - numCols + i, 0);
}
}
int i = 0;
int j = 0;
int n = 0;
int x = 3;
int y = 4;
int newSize = Math.max(x,y) * Math.max(x,y);
while(n < newSize){
if(i <= x && j <= y)
System.out.println(array[i][j]);
n++;
if(i == 0) {
i = n:
j = 0;
} else {
--i;
++j;
}
}

Searching a word in a given string array

You are given a 2D array as a string and a word via keyboard. The word
can be in any way (all 8 neighbors to be considered) but you can’t use
same character twice while matching. Return word's first and last
character's index as (x,y). If match is not found return -1.
That's the question. I'm having trouble with searching. I tried that:
int x=0,y=0;
for(int f=0; f<WordinArray.length; f++){
for(int i=0; i<matrix.length; i++){
for(int j=0; j<matrix[0].length; j++){
if(matrix[i][j].equals(WordinArray[f])){
x=i; y=j;
System.out.print("("+x+","+y+")");
}
}
}
}
But, That code is not working as it is supposed to. How else I can write this searching code?
Referring to Sixie's code
Assuming this is a valid input/output to your program?
Size:
4x4
Matrix:
a b c d
e f g h
i j k l
m n o p
Word: afkp
(0,0)(3,3)
I edited your code, so that it should work for input on this form (it is case sensitive at the moment, but can easily be changed by setting .toLowerCase()
Scanner k = new Scanner(System.in);
System.out.println("Size: ");
String s = k.nextLine();
s.toUpperCase();
int Xindex = s.indexOf('x');
int x = Integer.parseInt(s.substring(0, Xindex));
int y = Integer.parseInt(s.substring(Xindex + 1));
System.out.println("Matrix:");
char[][] matrix = new char[x][y];
for (int i = 0; i < x; i++) {
for (int p = 0; p < y; p++) {
matrix[i][p] = k.next().charAt(0);
}
}
System.out.print("Word: ");
String word = k.next();
int xStart = -1, yStart = -1;
int xEnd = -1, yEnd = -1;
// looping through the matrix
for (int i = 0; i < x; i++) {
for (int j = 0; j < y; j++) {
// when a match is found at the first character of the word
if (matrix[i][j] == word.charAt(0)) {
int tempxStart = i;
int tempyStart = j;
// calculating all the 8 normals in the x and y direction
// (the 8 different directions from each cell)
for (int normalX = -1; normalX <= 1; normalX++) {
for (int normalY = -1; normalY <= 1; normalY++) {
// go in the given direction for the whole length of
// the word
for (int wordPosition = 0; wordPosition < word
.length(); wordPosition++) {
// calculate the new (x,y)-position in the
// matrix
int xPosition = i + normalX * wordPosition;
int yPosition = j + normalY * wordPosition;
// if the (x,y)-pos is inside the matrix and the
// (x,y)-vector normal is not (0,0) since we
// dont want to check the same cell over again
if (xPosition >= 0 && xPosition < x
&& yPosition >= 0 && yPosition < y
&& (normalX != 0 || normalY != 0)) {
// if the character in the word is not equal
// to the (x,y)-cell break out of the loop
if (matrix[xPosition][yPosition] != word
.charAt(wordPosition))
break;
// if the last character in the word is
// equivalent to the (x,y)-cell we have
// found a full word-match.
else if (matrix[xPosition][yPosition] == word
.charAt(wordPosition)
&& wordPosition == word.length() - 1) {
xStart = tempxStart;
yStart = tempyStart;
xEnd = xPosition;
yEnd = yPosition;
}
} else
break;
}
}
}
}
}
}
System.out.println("(" + xStart + "," + yStart + ")(" + xEnd + ","
+ yEnd + ")");
k.close();
I think you need to plan your algorithm a bit more carefully before you start writing code. If I were doing it, my algorithm might look something like this.
(1) Iterate through the array, looking for the first character of the word.
(2) Each time I find the first character, check out all 8 neighbours, to see if any is the second character.
(3) Each time I find the second character as a neighbour of the first, iterate along the characters in the array, moving in the correct direction, and checking each character against the word.
(4) If I have matched the entire word, then print out the place where I found the match and stop.
(5) If I have reached the edge of the grid, or found a character that doesn't match, then continue with the next iteration of loop (2).
Once you have your algorithm nailed down, think about how to convert each step to code.
If I understood your question right. This is a quick answer I made now.
int H = matrix.length;
int W = matrix[0].length;
int xStart = -1, yStart = -1;
int xEnd = -1, yEnd = -1;
String word = "WordLookingFor".toLowerCase();
for (int i = 0; i < H; i++) {
for (int j = 0; j < W; j++) {
if (matrix[i][j] == word.charAt(0)) {
int tempxStart = i;
int tempyStart = j;
for (int x = -1; x <= 1; x++) {
for (int y = -1; y <= 1; y++) {
for (int k = 0; k < word.length(); k++) {
int xx = i+x*k;
int yy = j+y*k;
if(xx >= 0 && xx < H && yy >= 0 && yy < W && (x != 0 || y != 0)) {
if(matrix[xx][yy] != word.charAt(k))
break;
else if (matrix[xx][yy] == word.charAt(k) && k == word.length()-1) {
xStart = tempxStart;
yStart = tempyStart;
xEnd = xx;
yEnd = yy;
}
} else
break;
}
}
}
}
}
}
A little trick I used for checking all the 8 neighbors is to use two for-loops to create all the directions to go in:
for (int x = -1; x <= 1; x++) {
for (int y = -1; y <= 1; y++) {
if(x !=0 || y != 0)
System.out.println(x + ", " + y);
}
}
This creates
-1, -1
-1, 0
-1, 1
0, -1
0, 1
1, -1
1, 0
1, 1
Notice: All but 0,0 (you don't want to revisit the same cell).
The rest of the code is simply traversing though the matrix of characters, and though the whole length of the word you are looking for until you find (or maybe you don't find) a full match.
This time the problem is that how could I print word's first and last
letter's indexes. I tried various ways like printing after each word
was searched. But, all of them didn't work. I am about to blow up.
int[] values = new int[2];
for(int i=0; i<matrix.length; i++){
for(int j=0; j<matrix[0].length; j++){
if(Character.toString(word.charAt(0)).equals(matrix[i][j]) == true || Character.toString(ReversedWord.charAt(0)).equals(matrix[i][j]) == true ){
System.out.print("("+ i + "," +j+")");
//First letter is found.Continue.
for(int p=1; p<word.length(); p++){
try{
for (int S = -1; S <= 1; S++) {
for (int SS = -1; SS <= 1; SS++) {
if(S !=0 || SS != 0)
if(matrix[i+S][j+SS].equals(Character.toString(word.charAt(p))) && blocksAvailable[i+S][j+SS] == true ||
matrix[i+S][j+SS].equals(Character.toString(ReversedWord.charAt(p))) && blocksAvailable[i+S][j+SS] == true) {
values[0] = i+S;
values[1] = j+SS;
blocksAvailable[i+S][j+SS] = false;
}
}
}
}catch (ArrayIndexOutOfBoundsException e) {}

Looping Algorithm

How do I make this:
*******
-*****-
--***--
---*---
--***--
-*****-
*******
The following is my code that I have written to try to accomplish the above, but it is not working as expected:
public static void stars(/*int jmlBaris*/) {
for ( int i = 7; i >= 1; i-=2) {
for (int j = 1; j <= i; j++) {
System.out.print("*");
}
System.out.println("");
}
for (int i = 1; i <= 7; i+=2) {
for (int j = 1; j <= i; j++){
System.out.print("*");
}
System.out.println("");
}
}
public static void main(String[] args) {
stars();
}
}
This is how I might write it.
// three loops
public static void stars(int size) {
for (int y = 0; y < size; y++) {
for (int i = 0; i < y && i < size - y - 1; i++)
System.out.print(' ');
for (int i = Math.min(y, size - y - 1); i < Math.max(y + 1, size - y); i++)
System.out.print('*');
System.out.println();
}
}
or
// two loops
public static void stars(int size) {
for (int y = 0; y < size; y++) {
for (int x = 0; x < size; x++)
System.out.print(
(x >= y && x < size - y) ||
(x >= size - y - 1 && x <= y) ? '*' : ' ');
System.out.println();
}
}
or
// one loop
public static void stars(int size) {
for (int i = 0; i < size * size; i++) {
int y = i / size, x = i % size;
System.out.print(
(x >= y && x < size - y) ||
(x >= size - y - 1 && x <= y) ? '*' : ' ');
if (x == size - 1)
System.out.println();
}
}
Note: Whether this uses one, two or three loops, the time complexity is O(N^2). A simple way to determine this is the number of stars produced is O(N^2) no matter how it is done.
I would do something like this with substrings.
String a = "*******"; //7 stars
String blank = " "; //7 spaces
int j = 7;
for (int i = 0; i < 7; i++) {
if (i > j){
System.out.print(blank.substring(0,i));
System.out.println(a.substring(i,j));
}
else{
System.out.print(blank.substring(0,j));
System.out.println(a.substring(j,i));
}
j--;
}
System.out.println(a);
**Previous edit wouldn't have worked. Changes made.
This works.
Try something like this code I compiled on IDEOne (it seems to work, though):
http://ideone.com/9xZ1YB
class Main
{
public static void main(String[] args)
{
stars();
}
static void stars()
{
final int MAX_WIDTH = 7;
for (int i = 0; i < 7; ++i)
{
int width;
if (i < 3) width = MAX_WIDTH - i * 2;
else if (i > 3) width = (i - 3) * 2 + 1;
else width = 1;
// Before spaces
for (int j = 0; j < (MAX_WIDTH - width) / 2; ++j)
{
System.out.print(" ");
}
// Stars
for (int j = 0; j < width; ++j)
{
System.out.print("*");
}
// After spaces
for (int j = 0; j < (MAX_WIDTH - width) / 2; ++j)
{
System.out.print(" ");
}
System.out.println();
}
}
}
For a beginner in algorithms I would recommend you to break down the structure in sub-parts and then try to solve the pattern.
For this specific pattern it could be broken down into several triangles. Each triangle is then solved by different for loops as shown in the image below.
public static void printPattern(int num) {
// this loop generates first 4 lines
for (int i = 0; i < num / 2 + 1; i++) {
// draws the red triangle of '-'
for (int j = 0; j < i; j++) {
System.out.print("-");
}
// draws the green triangle of '*'
for (int j = i; j < num / 2 + 1; j++) {
System.out.print("*");
}
// draws the blue triangle of '*'
for (int j = i + 1; j < num / 2 + 1; j++) {
System.out.print("*");
}
// draws the orange triangle of '-'
for (int j = 0; j < i; j++) {
System.out.print("-");
}
System.out.println();
}
/* this loop generates last 3 lines */
for (int i = 0; i < num / 2; i++) {
// draws the green triangle of '-'
for (int j = i + 1; j < num / 2; j++) {
System.out.print("-");
}
// draws the red triangle of '*'
for (int j = 0; j < i + 2; j++) {
System.out.print("*");
}
// draws the orange triangle of '*'
for (int j = 0; j < i + 1; j++) {
System.out.print("*");
}
// draws the blue triangle of '-'
for (int j = i + 1; j < num / 2; j++) {
System.out.print("-");
}
System.out.println();
}
}
Using similar technique you could generate any pattern.
If I understood you right, your problem is to print indent in lines 2-7.
Imagine same problem with asterisk symbol replaced by 'x' and whitespace replaced by '-'. Then you need to draw
xxxxxxx
-xxxxx-
--xxx--
---x---
--xxx--
-xxxxx-
xxxxxxx
That means you should output 0, 1, 2 space(s) before asterisks in first, second, thrid strings respectively. I let details for you to figure them out.
public static void stars(/*int jmlBaris*/){
String starstr = "*";
String blank = "_";
int spaceBlank;;
for(int i=7; i>=1;i-=2){
spaceBlank = (7-i)*.5;
String starrep = StringUtils.repeat(starstr, i);
String blankrep = StrinUtils.repeat(blank, spacesBlank);
system.out.println(blankrep + starrep + blankrep);
}
for(int j=3 j<=7; j+=2){
spaceBlank = (7-j)*.5;
starrep = StringUtils.repeat(starstr, j);
String blankrep = StrinUtils.repeat(blank, spacesBlank);
system.out.println(blankrep + starrep + blankrep);
}
}
public static void main(String[] args){
stars();
}
You have little missing to put space on your code. I don't care about right space, who can see that? But left space is very important!!
Try this:
public static void stars(/*int jmlBaris*/) {
for ( int i = 7; i >= 1; i-=2) {
for (int k = 0; k < ((7-i) / 2); k++){ /* Missing Here */
System.out.print(" "); /* Missing Here */
} /* Missing Here */
for (int j = 1; j <= i; j++) {
System.out.print("*");
}
System.out.println("");
}
for (int i = 1; i <= 7; i+=2) {
for (int k = 0; k < ((7-i) / 2); k++){ /* Missing Here */
System.out.print(" "); /* Missing Here */
} /* Missing Here */
for (int j = 1; j <= i; j++){
System.out.print("*");
}
System.out.println("");
}
}
int N = 7;
for (int y=0; y<N; y++)
{
for (int x=0; x<N; x++)
System.out.print( (y-x)*(N-y-x-1)<=0 ? '*' : '-');
System.out.println();
}
or, more symmetrically,
int n = 3;
for (int y=-n; y<=n; y++)
{
for (int x=-n; x<=n; x++)
System.out.print( y*y>=x*x ? '*' : '-');
System.out.println();
}

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