How to make dynamic nested for loop in java?

Question

This is for a Brute-Force Algorithm I'm trying. Trying every combination.

The code produces every combination of moves of length 10 from 4 moves(up, down, left, right). So that's why the nested for loops are 10 deep. But what if there are less or more moves required like 9 or 11 or 12. How can I make the nested for loop dynamic. I'm having trouble with recursion.

int available = new int[] {TILT_MOVE.UP, TILT_MOVE.DOWN, TILT_MOVE.LEFT, TILT_MOVE.RIGHT}.length;

    int n = 10;
    int total = 1;

    for (int i = 0; i < n; i++) {
        total *= available;
    }

    int combinations[][] = new int[total][n];
    int count = 0;

    //My head got burnt up with the implementation of maze tilting logic so this is the best I can do for now.
    for (int a = 0; a < available; a++) {
        for (int b = 0; b < available; b++) {
            for (int c = 0; c < available; c++) {
                for (int d = 0; d < available; d++) {
                    for (int e = 0; e < available; e++) {
                        for (int f = 0; f < available; f++) {
                            for (int g = 0; g < available; g++) {
                                for (int h = 0; h < available; h++) {
                                    for (int i = 0; i < available; i++) {
                                        for (int j = 0; j < available; j++) {   
                                            int[] moves = new int[n];
                                            moves[0] = a;
                                            moves[1] = b;
                                            moves[2] = c;
                                            moves[3] = d;
                                            moves[4] = e;
                                            moves[5] = f;
                                            moves[6] = g;
                                            moves[7] = h;
                                            moves[8] = i;
                                            moves[9] = j;

                                            combinations[count++] = moves;

                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }

"But what if there are less or more moves required like 9 or 11 or 12." - Then you use recursion. --- "I'm having trouble with recursion." - In the beginning, everyone has. Shying away from it won't make it better. Practice it! You will understand it eventually =) — Turing85
– Turing85, Commented Jun 15, 2018 at 15:35
@Carcigenicate Very likely? It would take a lot more than 12 levels... — shmosel
– shmosel, Commented Jun 15, 2018 at 15:53
@shmosel Ya, I'm not not why I wrote very likely. If you can guarantee a max of 12 levels, it would be very unlikely. — Carcigenicate
– Carcigenicate, Commented Jun 15, 2018 at 15:58
what are the values of TILT_MOVE.UP, TILT_MOVE.DOWN, TILT_MOVE.LEFT, TILT_MOVE.RIGHT — Youcef LAIDANI
– Youcef LAIDANI, Commented Jun 15, 2018 at 16:32
In general if you are having 4 or more nested loops you probably need to rethink your logic — Bayko
– Bayko, Commented Jun 15, 2018 at 19:24

Michael · Accepted Answer · 2018-06-15 16:45:45Z

4

I believe this should work:

for (int i = 0; i < total; i++)
{
    int[] moves = new int[n];
    for (int move = 0; move < n; ++move)
    {
        moves[move] = (i / (int) Math.pow(available, move)) % available;
    }
    combinations[i] = moves;
}

edited Jun 15, 2018 at 16:45

answered Jun 15, 2018 at 15:41

Michael

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2 Comments

user4408936 Over a year ago

It doesn't work. These is an example move from my code [0 0 3 2 3 3 2 2 1 1 ]. This is from your code [0 9 49 349 1349 61349 61349 61349 61349 61349 ]. The size 10 array only needs values from 0 - 3 which corresponds to TILT_UP, TILT_DOWN, TILT_RIGHT, TILT_LEFT.

Michael Over a year ago

@user4408936 Sorry, yes it wasn't correct before. I've updated my answer.

Brent Sandstrom · Accepted Answer · 2018-06-15 19:25:44Z

1

You're trying to find a path through a maze?

If so, stop thinking of the maze as a human maze. To a computer a maze is a graph, where each node is possible location.

The solution to the maze is a sequence of moves that leads the balls to the hole(s). This can be thought of as a tree, where each node represents {tilt-direction, ballA-position, ballB-position}.

In this more complicated version of the maze you'll need to account for the ability to tilt when one ball is already stuck up against a barrier. Only when both balls have a wall to their left do you no longer have a possible left move.

You'll also need to make sure that you don't keep recurring over the same moves. If ballA and B have both been on the same space during the same turn don't continue trying this move.

edited Jun 15, 2018 at 19:25

answered Jun 15, 2018 at 15:46

Brent Sandstrom

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5 Comments

k_ssb Over a year ago

Rather, a maze is a graph. In particular, you'll have to remember which nodes are already visited (to avoid infinite recursion), which is more complicated than simple tree traversal.

Turing85 Over a year ago

@pkpnd you mean Dijkstra's algorithm? =)

k_ssb Over a year ago

@Turing85 I was thinking BFS as a step up from a tree traversal algorithm.

user4408936 Over a year ago

@Turing85 yes a maze. But it's quite different. There are two balls and two holes and also a few 1x1 blocks. I've already solve it using the code above. Now just I'm learning how to do that code in a more dynamic way because it seems most of the sample problem given by the company I'm applying for uses code like that.

Brent Sandstrom Over a year ago

@pkpnd You're right I worded this wrong. The maze is a graph, but the solution can be displayed as a tree

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