Develop Reference

How to find most profitable Path in 2-Dimensional Array

I'm trying to implement a game where the viable moves are down-left and down-right.
The parameter for the function is for the size of the array, so if you pass 4 it will be a 4 by 4 array.
The starting position is the top row from any column. Every element in the array is a number in the range 1-100, taken from a file. I need to find the resulting value for the most profitable route from any starting column.
My current implementation will compare the right position and left position and move to whichever is higher. The problem is, for example, if the left position is lower in value than the right, but the left position will provide more profit in the long run since it can access higher value elements, my algorithm fails.
Here is a demo:
84 (53) 40 62
*42* 14 [41] 57
76 *47* 80 [95]
If we start at number 53. The numbers enclosed in * are the moves that my algorithm will take, but the numbers enclosed in [] are the moves my algorithm should take.
This is my code:
import java.util.ArrayList;
import java.util.Scanner;
public class bestPathGame{
private int[][] grid;
private int n;
public bestPathGame(int num){
Scanner input = new Scanner(System.in);
n = num;
grid = new int[n][n];
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++){
grid[i][j] = input.nextInt();
}
}
}
public static void main(String[] args){
bestPathGame obj = new bestPathGame(Integer.parseInt(args[0]));
obj.bestPath();
}
private boolean moveLeftBetter(int r,int c){
if(c <= 0){
return false;
} else if (c >= n -1 ){
return true;
}
return grid[r][c-1] > grid[r][c+1];
}
public void bestPath(){
ArrayList<Integer> allOptions = new ArrayList<>();
for(int k = 0; k < n; k++){
int row = 0;
int col = k;
int collection = grid[row][col];
while(row < n - 1){
row += 1;
if(moveLeftBetter(row,col)){
col-=1;
} else{
col+=1;
}
collection += grid[row][col];
}
allOptions.add(collection);
}
System.out.println(allOptions.stream().reduce((a,b)->Integer.max(a,b)).get());
}
}

Greedy algorithm vs Dynamic programming
There's an issue with the logic of your solution.
Basically, what you are implemented is a called a greedy algorithm. At each step of iteration, you are picking a result that optimal locally, assuming that this choice will lead to the optimal global result. I.e. your code is based on the assumption that by choosing a local maximum between the two columns, you will get the correct global maximum.
As a consequence, your code in the bestPath() method almost at each iteration will discard a branch of paths based on only one next value. This approach might lead to incorrect results, especially with large matrixes.
Greedy algorithms are rarely able to give an accurate output, usually their result is somewhat close but not precise. As an upper-hand, they run fast, typically in O(n) time.
For this problem, you need to use a dynamic programming (DP).
In short, DP is an enhanced brute-force approach which cashes the results and reuses them instead of recalculating the same values multiple times. And as well, as a regular brute-force DP algorithms are always checking all possible combinations.
There are two major approaches in dynamic programming: tabulation and memoization (take a look at this post for more information).
Tabulation
While implementing a tabulation first you need to create an array which then need to be prepopulated (completely or partially). Tabulation is also called the bottom-up approach because calculation start from the elementary edge cases. Every possible outcome is being computed based on the previously obtained values while iterating over this array. The final result will usually be stored in the last cell (in this case in the last row).
To implement the tabulation, we need to create the matrix of the same size, and copy all the values from the given matrix into it. Then row by row every cell will be populated with the maximum possible profit that could be obtained by reaching this cell from the first row.
I.e. every iteration will produce a solution for a 2D-array, that continuously increases by one row at each step. It'll start from the array that consists of only one first row (no changes are needed), then to get the profit for every cell in the second row it's values has to be combined with the best values from the first row (that will be a valid solution for 2D-array of size 2 * n), and so on. That way, solution gradually develops, and the last row will contain the maximum results for every cell.
That how the code will look like:
public static int getMaxProfitTabulation(int[][] matrix) {
int[][] tab = new int[matrix.length][matrix.length];
for (int row = 0; row < tab.length; row++) { // populating the tab to preserve the matrix intact
tab[row] = Arrays.copyOf(matrix[row], matrix[row].length);
}
for (int row = 1; row < tab.length; row++) {
for (int col = 0; col < tab[row].length; col++) {
if (col == 0) { // index on the left is invalid
tab[row][col] += tab[row - 1][col + 1];
} else if (col == matrix[row].length - 1) { // index on the right is invalid
tab[row][col] += tab[row - 1][col - 1];
} else {
tab[row][col] += Math.max(tab[row - 1][col - 1], tab[row - 1][col + 1]); // max between left and right
}
}
}
return getMax(tab);
}
Helper method responsible for extracting the maximum value from the last row (if you want to utilize streams for that, use IntStream.of(tab[tab.length - 1]).max().orElse(-1);).
public static int getMax(int[][] tab) {
int result = -1;
for (int col = 0; col < tab[tab.length - 1].length; col++) {
result = Math.max(tab[tab.length - 1][col], result);
}
return result;
}
Memoization
The second option is to use Memoization, also called the top-down approach.
As I said, DP is an improved brute-force algorithm and memoization is based on the recursive solution that generates all possible outcomes, that is enhanced by adding a HashMap that stores all previously calculated results for every cell (i.e. previously encountered unique combination of row and column).
Recursion starts with the first row and the base-case of recursion (condition that terminates the recursion and is represented by a simple edge-case for which output is known in advance) for this task is when the recursive call hits the last row row == matrix.length - 1.
Otherwise, HashMap will be checked whether it already contains a result. And if it not the case all possible combination will be evaluated and the best result will be placed into the HashMap in order to be reused, and only the then the method returns.
Note that tabulation is usually preferred over memoization, because recursion has significant limitations, especially in Java. But recursive solutions are sometimes easier to came up with, so it's completely OK to use it when you need to test the idea or to prove that an iterative solution is working correctly.
The implementation will look like that.
public static int getMaxProfitMemoization(int[][] matrix) {
int result = 0;
for (int i = 0; i < matrix[0].length; i++) {
result = Math.max(result, maxProfitHelper(matrix, 0, i, new HashMap<>()));
}
return result;
}
public static int maxProfitHelper(int[][] matrix, int row, int col,
Map<String, Integer> memo) {
if (row == matrix.length - 1) { // base case
return matrix[row][col];
}
String key = getKey(row, col);
if (memo.containsKey(key)) { // if cell was already encountered result will be reused
return memo.get(key);
}
int result = matrix[row][col]; // otherwise result needs to be calculated
if (col == matrix[row].length - 1) { // index on the right is invalid
result += maxProfitHelper(matrix, row + 1, col - 1, memo);
} else if (col == 0) { // index on the left is invalid
result += maxProfitHelper(matrix, row + 1, col + 1, memo);
} else {
result += Math.max(maxProfitHelper(matrix, row + 1, col - 1, memo),
maxProfitHelper(matrix, row + 1, col + 1, memo));
}
memo.put(key, result); // placing result in the map
return memo.get(key);
}
public static String getKey(int row, int col) {
return row + " " + col;
}
Method main() and a matrix-generator used for testing purposes.
public static void main(String[] args) {
int[][] matrix = generateMatrix(100, new Random());
System.out.println("Tabulation: " + getMaxProfitTabulation(matrix));
System.out.println("Memoization: " + getMaxProfitMemoization(matrix));
}
public static int[][] generateMatrix(int size, Random random) {
int[][] result = new int[size][size];
for (int row = 0; row < result.length; row++) {
for (int col = 0; col < result[row].length; col++) {
result[row][col] = random.nextInt(1, 101);
}
}
return result;
}

How do you print a grid of asterisks using multidimensional arrays?

I am new to programming and I have an exercise that's killing me. How can you print a grid (5-by-6) which consists of asterisks alone? [Later on, these asterisks will have to be replaced by letters which are read in with StdIn.readInt() and a switch statement, but for now I at least need to understand how to print the grid]. I would appreciate any help so much!
More specifically, the grid should look like this:
//THIS ISN'T THE CODE; JUST AN ILLUSTRATION OF WHAT SHOULD BE PRINTED
0 1 2 3 4 5
0 * * * * * *
1 * * * * * *
2 * * * * * *
3 * * * * * *
4 * * * * * *
//I AM SUPPOSED TO START WITH SOMETHING LIKE THIS:
public class Grid {
static int X = 6;
static int Y = 7;
public static void main(String[]args) {
int [][] grid = new int [X][Y];

This could have been done in many ways, but this is my way of doing it:
When you want to print a grid, you have to use 2 nested for loops.
Let's see what happens when you use 2 nested for loops:
for(int i = 0; i < 6; i++){
for(int j = 0; j < 7; j++){
}
}
We start with the first loop:
for i = 0, we will enter the second loop and iterate from 0 to 6.
for i = 1, we will enter the second loop and iterate from 0 to 6.
...
for i = 5, we will enter the second loop and iterate from 0 to 6.
What you should notice is that j will iterate and take values from 0 to 6 with each value of i.
Going back to your question, and comparing it by what i just showed, you should notice that for each line, you are printing 7 values (of a column).
Let's assume i is the number of lines, and j is the index of each value in that line (column).
public static void printGrid() {
for (int i = 0; i < 6; i++) {
System.out.println();
for (int j = 0; j < 7; j++) {
System.out.print("*");
}
}
}
This code prints on each line (i), 7 asterixes (j).
And each time i is incrementing, we are going back to the next line System.out.println(). That's why we put it inside the for loop with i.
In your situation, we have to tweak this code a little bit to be able to print the numbers on the sides, and that space at the top left corner.
The explanation is in the comments in my code.
public class Question_55386466{
static int X = 6;
static int Y = 7;
public static void printGrid() {
System.out.print(" "); // Printing the space on the top left corner
for (int i = 0; i < X; i++) {
if (i > 0) { // Printing the numbers column on the left, taking i>0 to start from the second line (i == 1)
System.out.println(); // Going to the next line after printing the whole line
System.out.print(i - 1);//Printing the numbers of the column. Taking i-1 because we start the count for i == 1 not 0
}
for (int j = 0; j < Y; j++) {
if (i == 0)
System.out.print(j + " ");//Print the first line numbers.
else
System.out.print(" * "); //if the line isn't the first line(i == 0), print the asterixes.
}
}
}
You can always edit the values of X and Y and get the desired result.
And later you can give this method your array as a parameter and print each element instead of the asterixes.

How can I set a grid with the value of my 2D array?(Java, Android Studio)

I have a school project, I have to build a Tetris Game.
So I began with the creation of my menu with the different level, when I click on one level i go to my second activity (the game area) and I have also created my custom block.
My problem is a visual issue, indeed I don't know what type of layout I have to use for my surface game (gridlayout, linearlayout, grid etc. ...).
And then how to affect my blocks custom in this surface game, in this layout?
See the result expected.
enter image description here

I'm not sure I understand what you want entirely but ill give it a shot from what i think you mean.
You should use a nested for loop to do it, if your array was int[10,20](not right syntax but i cant be bothered to count the actual size of your array).
you should go:
(pseudocode)
also assume your resolution is 100, 200
For(int i = 1 To 10){
For(int k = 1 To 20){
DrawSquare(i*10, k*10, "block type")
k = k + 1
}
i = i + 1
}
Then it will fill your 100, 200 area with the block type specified. Now if you want to load what block type you want freom the array you can just call the array in the block type.
DrawSquare(i*10, k*10, Array[i,k])
Obviously bear in mind that its all pseudocode to display the logic.
Hope this helps

Building on Valhalla's answer with a Java-specific example, it's still a bit unclear what you're asking, but assuming you want to initialise the grid to begin with you can use this code:
private final int columns = 10;
private final int rows = columns * 2;
private int[][] grid;
private void initialise() {
grid = new int[columns][rows];
for (int i = 0; i < columns; i++) {
for(int j = 0; j < rows; j++) {
grid[i][j] = 0;
}
}
}
And assuming that you have a block that starts at the top, and falls one square with each iteration provided there's nothing underneath, you can try this:
private void blockFall() {
// Start from 1 row above the bottom and parse upwards
// so a block won't drop right to the bottom on a single iteration
for (int i = 0; i < columns; i++) {
for(int j = rows - 2; j >= 0; j--) {
if (grid[i][j] > 0 && grid[i][j+1] == 0) {
grid[i][j+1] = grid[i][j];
grid[i][j] = 0;
}
}
}
}

Dynamically change the number of nested for loops

I don't know if this is a stupid question, but I need to dynamically change the number of for-loops without using recursion.
For example, if n=3, I need 3 nested for-loops.
for(int i=0; i<size; i++){
for(int j=0; j<size-1; j++){
for(int k=0; k<size-2; k++){
//do something
}
}
}
If n=5:
for(int i=0; i<size; i++){
for(int j=0; j<size-1; j++){
for(int k=0; k<size-2; k++){
for(int l=0; l<size-3; l++){
for(int m=0; m<size-4; m++){
//do something
}
}
}
}
}
Is there any way to achieve this without recursion?
Another question: what is the use of Multiple Dispatch in Java? I'm trying to code something in ONE METHOD, and it should run different events in different cases of the parameter. NO IF STATEMENTS / TERNARY OPERATORS / CASES.
NOTE: I can ONLY have one method (part of the problem), and cannot use recursion. Sorry.

Think about how many times you run through this loop. It looks like (size!) / (size - n)!:
int numLoops = 1;
for (int i = 0; i < n; i++) {
numLoops*= (size - i);
}
for (int i = 0; i < numLoops; i++) {
//do something
}

It depends what exactly you're trying to do. Recursion can always be replaced with iteration (see this post for examples using a Stack to store state).
But perhaps the modulo (%) operator could work here? i.e. Have a single loop that increments a variable (i) and then the other variables are calculated using modulo (i % 3 etc). You could use a Map to store the values of the variables indirectly, if there are a varying number of variables.

You have to create array of loop counters and increment it manually.
Quick and dirty example:
public static void nestedFors(int n, int size) {
assert n > size;
assert size > 0;
int[] i = new int[n];
int l = n - 1;
while(l >= 0) {
if(l == n - 1) {
System.out.println(Arrays.toString(i));
}
i[l]++;
if(i[l] == size - l) {
i[l] = 0;
l--;
} else if(l < n - 1) {
l++;
}
}
}
Replace System.out.println(Arrays.toString(i)) with your own code.
You can check it here: http://ideone.com/IKbDUV

It's a bit convoluted, but: here is a way to do it without recursion, in one function and without ifs.
public static void no_ifs_no_recursion(int n){
int[] helper = new int[n-1];
int[] pointers = new int[n]; //helper for printing the results
int totalsize = 1;
for (int loops = 2; loops <= n; loops++){
helper[n - loops] = totalsize;
totalsize*=loops;
}
for (int i=0; i<totalsize; i++){
int carry = i;
for (int loops = 0; loops < n-1; loops++){
pointers[loops] = carry/helper[loops];
carry = carry - (pointers[loops]*helper[loops]);
}
System.out.println(Arrays.toString(pointers));
//or do something else with `i` -> `pointers[0]`, `j` -> `pointers[1]`, `k` -> `pointers[2]` etc..
}
}

I think you need a backtracking algorithm.
But then you would replace your nested loops with recursion.
I don't want to post links here as seems moderators don't like that.
Look at "eight queens puzzle" (you can Google it), you will get my idea.
I know this idea works as I've posed this same question (which you have) to myself on many occasions, and I've applied it several times successfully.
Here is a small example (I changed it as the previous one was a bit complex).
public class Test001 {
public static void main(String[] args) {
loop(0, 5, 10);
}
/**
* max_level - the max count of nesting loops
* size - the size of the collection
*
* 0 - top most level
* level 1 - nested into 0
* level 2 - nested into 1
* ...
* and so on.
*/
private static void loop(int level, int max_level, int size){
if (level > max_level) return;
for (int i=0; i<size-level; i++){
System.out.println("Now at level: " + level + " counter = " + i);
loop(level + 1, max_level, size);
}
}
}
But this still uses recursion.

How to make n nested for loops recursively?

I have a method that must do the following:
for (int a01 = 1; a01 <= 25; a01++) {
for (int a02 = a01 + 1; a02 <= 25; a02++) {
for (int a03 = a02 + 1; a03 <= 25; a03++) {
...
System.out.println(a01 + "," + a02 + "," + ... + "," + a015);
}
}
}
I'd like to specify the number of nested for's (in the case above, I want 15 nested for's).
Is there a way to use recursive programming here?

Yes. This can be performed by recursive programming.
I assume you do not like to WRITE DOWN these nested for's in source code - as in your example, because this is really ugly programming - like the commentors explain.
The following (pseudo Java-like) code illustrates it. I assume a fixed depth for the nesting. Then you actually like to loop over an integer vector of dimension depth.
int[] length = new int[depth];
int[] counters = new int[depth];
The array counters has to be initialised to 0 (Arrays.fill(counters,0)). The array length has to be initialised to the number of iterations for the respective for loop.
I assume that you like to perform a certain operation within the inner loop. I will call this
performOperation(int[] counters);
- it depends on the multi-dimensional counter, i.e. the counters of the outer for's.
Then you can run the nested for loops by calling
nestedLoopOperation(counters, length, 0);
where
void nestedLoopOperation(int[] counters, int[] length, int level) {
if(level == counters.length) performOperation(counters);
else {
for (counters[level] = 0; counters[level] < length[level]; counters[level]++) {
nestedLoopOperation(counters, length, level + 1);
}
}
}
In your case your System.out.println() would be
performOperation(int[] counters) {
String counterAsString = "";
for (int level = 0; level < counters.length; level++) {
counterAsString = counterAsString + counters[level];
if (level < counters.length - 1) counterAsString = counterAsString + ",";
}
System.out.println(counterAsString);
}

I created this program to show all the different possible combination of cards (non repeating). It uses recursive for loops. Maybe it can help you.
//I'm lazy, so yeah, I made this import...
import static java.lang.System.out;
class ListCombinations {
//Array containing the values of the cards
static Symbol[] cardValues = Symbol.values();
//Array to represent the positions of the cards,
//they will hold different card values as the program executes
static Symbol[] positions = new Symbol[cardValues.length];
//A simple counter to show the number of combinations
static int counter = 1;
/*Names of cards to combine, add as many as you want, but be careful, we're
talking about factorials here, so 4 cards = 24 different combinations (4! = 24),
but 8 cards = 40320 combinations and 13 cards = 6.23 billion combinations!*/
enum Symbol {
AofSpades, TwoofSpades, ThreeofSpades, FourofSpades
}
public static void main(String args[]) {
//I send an argument of 0 because that is the first location which
//we want to add value to. Every recursive call will then add +1 to the argument.
combinations(0);
}
static void combinations(int locNumber) {
/* I use a recursive (repeating itself) method, since nesting for loops inside
* each other looks nasty and also requires one to know how many cards we will
* combine. I used 4 cards so we could nest 4 for loops one after another, but
* as I said, that's nasty programming. And if you add more cards, you would
* need to nest more for loops. Using a recursive method looks better, and gives
* you the freedom to combine as many cards as you want without changing code. */
//Recursive for loop nesting to iterate through all possible card combinations
for(int valueNumber = 0; valueNumber < cardValues.length; valueNumber++) {
positions[locNumber] = cardValues[valueNumber];
if (locNumber < (cardValues.length-1)) {
combinations(locNumber + 1);
}
//This if statement grabs and displays card combinations in which no card value
// is repeated in the current "positions" array. Since in a single deck,
// there are no repeated cards. It also appends the combination count at the end.
if (locNumber == (cardValues.length-1) && repeatedCards(positions)) {
for (int i = 0; i < cardValues.length; i++) {
out.print(positions[i]);
out.print(" ");
}
out.printf("%s", counter);
counter++;
out.println();
}
}
}
static boolean repeatedCards(Symbol[] cards) {
/*Method used to check if any cards are repeated in the current "positions" array*/
boolean booleanValue = true;
for(int i = 0; i < cardValues.length; i++) {
for(int j = 0; j < cardValues.length; j++) {
if(i != j && cards[i] == cards[j]) {
booleanValue = false;
}
}
}
return booleanValue;
}
}