Random positions on hex grid around one vector

I`m not so good at math as I would like to be, so I need a little help . What I need is to randomly generate positions (Vector2) around one position (also Vector2) in certain distance - four hexes around - as it is shown on this image:
https://dl.dropboxusercontent.com/u/52159221/HexGridPaintedPrzykladNaForum.jpg

For now I create a random array of Vector2 in non normalized hex grid, but in that case in corners objects lands far to far that I want. I appriciate any help

 

If it didn't have to be a hex shape, I'd say just do a circle and then convert it to a tile... but to do hex shape I think I'd create a random number from 0 to how many tiles are within that area, then just convert that to a row/column number.

 

It looks to me like you can just walk four cells in every direction systematically starting 0 left and 4 down, then 1 left and 3 down, then 2 left and 2 down etc. Grab all those cells and you should have your shape.

 

*imaginaryhuman Yes, for now it is the only way known to me to do it, but I hoped it could be a little more universal - this method assumes 4 hex distance from center always, plus writing it will be somewhat long in codeing.

*secondbreakfast, do You mean from 4,4, to 4,0 / from 4,4 to 8,0 etc. ? I have to admit - I don`t understand exacly what You mean

 

I'll try to explain better. There will be multiple loops to go through. There are a few ways to systematically do it, but here is one way. We will cover the diamond of tiles to the left of the center, then the diamond to the right of the center, then the triangle on top, then the triangle on bottom.

For the diamond shapes:
Start at 4,4 and then we'll move left and up-left with a max of 4 collecting all the tiles along the way

outer loop starts here
Move left 1 times that leaves you at 4,3
inner loop for moving up-left starts here
Move up-left again that you at 3,3
Move up-left again that you at 2,3
Move up-left again that you at 1,3
We're at the max of 4 now so we're done with the inner loop
Continue at the position where we left before we started going up-left
inner loop for moving down-left starts here
...do the same, but going downward. 5,2 then 6,1 then 7,0
Continue at the position where we left before we started going down-left
Move left 1 time that leaves you at 4,2
inner loop for moving up-left starts here
Move up-left again that you at 3,2
Move up-left again that you at 2,2
Continue at the position where we left before we started going up-left
Move left 1 time that leaves you at 4,1
Move up-left again that you at 3,1
Continue at the position where we left before we started going up-left
Move left 1 time that leaves you at 4,0

The right side diamond would be the opposite.

For the triangle shapes you want to move up-left starting at 4,4 then go up-right on each iteration similar to the diamond

Here's some pseudocode for the left side diamond:
//go left
for (a=1; a <= 4; a++) {
//go up-left
for (b=0; b <= 4-a; b++) {
get tile left "a" spaces and up-left "b" spaces
}
//go down-left
for (b=1; b <= 4-a-1; b++) {
get tile left "a" spaces and down-left "b" spaces
}
}

Here's some pseudocode for the upper side triangle:
//go up-left
for (a=0; a <= 4; a++) {
//go up-right
for (b=0; b <= 4-a; b++) {
get tile up-left "a" spaces and up-right "b" spaces
}
}


//repeat those for the right diamond and bottom triangle and you have then walked over every tile

 

I think an easy way would be to take a square that fits your hexagon, generate a vector2 in that square and then just regenerate if you fall in 1 of the 2 corners that don't fall in the hexagon.

In the case of the image you posted your square would run from ( 0, 0 ) to ( 8, 8 ).

So just normalize that and what you have is ( -diameter, -diameter ) to ( diameter, diameter ).

So what you can do to see if your vector2 falls in one of the corners. You can add the coordinates and the result should be no smaller then -diameter and no bigger then +diameter.

And that's it. Should be a tiny piece of code.

 

Thank You very much Seconbreakefast I`ll try to build a code on this, as it is understable enought for me and easy to implement althought little time consumeing But it should work perfectly!

SpacemanBoots, there is a bit of problem with this - it is same as imaginaryhuman sad - to generate array of all hexes and from it (because You have to do it anyway to know witch ones aren`t on square) and than get random value.

 

I don't know what this means... You don't have to generate an array.

Like I said, you only have to generate a number that's in between ( -diameter, -diameter ) and ( diameter, diameter ).
So that's something like:

Code (CSharp):

1. new Vector2( Random.Range( -diameter, diameter+1 ), Random.Range( -diameter, diameter+1 ) );

There is no generating of arrays involved.

 

Yes, but how will You know witch Vector2 are out of area? I mean without array

 

So in the hex is, coord.x + coord.y <= diameter AND >= -diameter.

 

Can you retrieve the center of each hex given it's x,y coordinates?
If so I'd just write a small function that returns all the neighbouring cells, something pseudo-like this:

Code (Pseudo):

1. findNeighbours(x,y)
2. {
3.   return new int[12] {
4.             x-1, y,
5.             x-1, y+1,
6.             x, y+1,
7.             x+1, y,
8.             x+1, y-1,
9.             x, y-1};
10. }
11.  

Where each n entry is the X of a cell and each n+1 is the Y index.
Then iteratively call the function for how many "steps" you're allowed to travel (from your example it seems like it's 4) and append results to some main array.
The final array (cleaned of duplicates) should encompass the area you're looking for, so you can just pick a random number in the range [0:array.Length/2].

Makes sense?

 

I finally did it - I`ve created method witch calculates all positions around point, if anyone need it, here it is:

Code (CSharp):

1.     public static List<Vector2> hexAroundPoint2d(Vector2 point,int distanceFromCenter)
2.     {
3.         List<Vector2> tablicaVec = new List<Vector2>();
4.         for (int i = 0;i<=distanceFromCenter;i++)
5.         {
6.             for (int j = 1;j<= i;j++)
7.             {
8.                 tablicaVec.Add(new Vector2(point.x - i + j,point.y - j));
9.             }
10.         }
11.  
12.         for (int i = 1;i<=distanceFromCenter;i++)
13.         {
14.             for (int j = 1;j<= i;j++)
15.             {
16.                 tablicaVec.Add(new Vector2(point.x + j,point.y - i));
17.             }
18.         }
19.  
20.         for (int i = 1;i<=distanceFromCenter;i++)
21.         {
22.             for (int j = 1;j<= i;j++)
23.             {
24.                 tablicaVec.Add(new Vector2(point.x + i,point.y - i + j));
25.             }
26.         }
27.  
28.         for (int i = 1;i<=distanceFromCenter;i++)
29.         {
30.             for (int j = 1;j<= i;j++)
31.             {
32.                 tablicaVec.Add(new Vector2(point.x + i - j,point.y + j));
33.             }
34.         }
35.  
36.         for (int i = 1;i<=distanceFromCenter;i++)
37.         {
38.             for (int j = 1;j<= i;j++)
39.             {
40.                 tablicaVec.Add(new Vector2(point.x - j,point.y + i));
41.             }
42.         }
43.  
44.         for (int i = 1;i<=distanceFromCenter;i++)
45.         {
46.             for (int j = 1;j<= i;j++)
47.             {
48.                 tablicaVec.Add(new Vector2(point.x - i,point.y + i - j));
49.             }
50.         }
51.  
52.         return tablicaVec;
53.     }

 

It is very interesting solution! It works definetly, but it takes some of cpu time if numbers are higher. But it made me thinking, and finally I found right solution, so thank You

 

Check this on how to post code: http://forum.unity3d.com/threads/using-code-tags-properly.143875/

 

Right, thank You, changes are done