Create endless grid spiral in java

Question

I want to create endless spiral in java like this:

Endless spiral

I simply want to send integer from 1-∞ number and get it's spiral map. Example:

getXZForMap(0); ----> (0;0)

getXZForMap(5); ----> (-1;0)

Came up with this code now, but do not know how to go further:

public int[] getXZForMap(int i){

    int[] k = new int[2];

    if(i > 0){

        double s = i/8;
        int stage = (int) Math.ceil(s);
        int corner_size = ((stage*8)+4)/4;



    }
    else{
        k[0] = 0;
        k[1] = 0;
    }

    return k;
}

You forgot to paste the code wherein you made a serious and concerted effort to solve this problem yourself. — Kon, Jul 26 '17 at 00:00
Using sin(_θ_) and cos(_θ_) (along with radius _r_) works fine if you're trying to draw an _actual_ spiral, but it looks like you're trying to draw a grid "in spiral order". If you try to round your polar co-ordinates to Cartesian integers, you'll end up with all kinds of awful edge cases. You're much better off thinking in terms of integer-valued (_x_, _y_) ordered pairs. — Kevin J. Chase, Jul 26 '17 at 00:48
Possible duplicate of [Algorithm for iterating over an outward spiral on a discrete 2D grid from the origin](https://stackoverflow.com/questions/3706219/algorithm-for-iterating-over-an-outward-spiral-on-a-discrete-2d-grid-from-the-or) — Kevin J. Chase, Jul 28 '17 at 20:12

score 2 · Answer 1 · answered Jul 26 '17 at 00:38

2

Think algorithmically. Identify the pattern.

The pattern is that you walk the circumference, one layer at a time.

In round 0, you are at the center point (0,0).
In round 1, you walk the 8 cells around it.
In round 2, you walk the 16 cells around that.
In round 3, you walk the 24 cells around that.

Now see the pattern:

In each round, the walk on each side is 2 more steps, totally 8 more steps per round.
Each round starts 1 cell to the right of the upper-left corner.
Walking on each side starts 1 cell from the corner.

So, now that you see the pattern, you can do it like this:

Calculate which round you are in.
For that round, calculate which side you are on.
Then calculate how many steps you've taken on that side.
Finally, calculate the coordinate from that.

answered Jul 26 '17 at 00:38

Andreas

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My friend came up with this code: `public int[] getXZForMap(int index){ int x=0, y=0, dist=0, direction=1; for(int i = 0;true; i++){ if((distance+1)*direction > index){ x+=(((++dist-index)*direction)); break; }else{ x +=(++dist * direction); if((distance+1)*direction > index){ y+=(((++dist-index)*direction)); break; }else{ y +=(++dist * direction); direction *=-1; } } } int[] k = new int[2]; k[0]=x; k[1]=y; return k; }` – TerZer Jul 26 '17 at 00:47
2

Don't show code in a comment. If you already have code: 1) Why didn't you show it in the question. 2) *What* is your question, i.e. what answer(s) are you looking for? – Andreas Jul 26 '17 at 00:50
1

@TerZer Cool. What did _you_ come up with? Post _that_ in your question, then you will get answers you'll like. The more effort you put into writing your question, the more we put into answers. – Jul 26 '17 at 00:51
@Andreas this code, does not work how I want. I simply want to send integer from 1-∞ number and get it's spiral map. Example: getXZForMap(0); ----> (0;0) getXZForMap(5); ----> (-1;0) Will add this to question – TerZer Jul 26 '17 at 11:33
1

*"I simply want"* Then simply write the code to do that, though I wouldn't use the word "simply" for your requirement. With this answer, I gave you the outline of one way to solve the problem. I'm not going to give you the code, because this is an assignment/challenge for *you* to solve, not us. Your friend came up with another way to do it. It's up to you to decide how to proceed from here. – Andreas Jul 26 '17 at 13:54

score 0 · Accepted Answer · answered Jul 26 '17 at 17:15

import java.util.Scanner;

class spiral 
{
    private static String getXZForMap(int np)
    {         
        // (dx, dy) is a vector - direction in which we move right now
        int dx = 0;
        int dy = 1;
        // length of current segment
        int segment_length = 1;

        // current position (x, y) and how much of current segment we passed
        int x = 0;
        int y = 0;
        int segment_passed = 0;
        if (np == 0){
            return ("(" + y + ";" + x + ")");
        }
        for (int n = 0; n < np; ++n) {
            // make a step, add 'direction' vector (dx, dy) to current position (x, y)
            x += dx;
            y += dy;
            ++segment_passed;

            if (segment_passed == segment_length) {
                // done with current segment
                segment_passed = 0;

                // 'rotate' directions
                int buffer = dy;
                dy = -dx;
                dx = buffer;

                // increase segment length if necessary
                if (dx == 0) {
                    ++segment_length;
                }
            }
        }
        return("(" + y + ";" + x + ")");
    }

    public static void main(String[] args){

        Scanner sc = new Scanner(System.in);
        int NUMBER_OF_POINTS = Integer.valueOf(args[0]);   // or delete this line  
        String spiral_map = getXZForMap(NUMBER_OF_POINTS); // and put your int here
        System.out.println(spiral_map);
    } 
}

Adapted from this answer: Algorithm for iterating over an outward spiral on a discrete 2D grid from the origin

Demo:

$ java spiral 0
(0;0)
$ java spiral 5
(-1;0)

Create endless grid spiral in java

2 Answers2