# hash space partitioning explanation needed

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Hi, sorry for stupid question

After lots of time of reading blogs and books about spatial hash partitioning on bradphase collision Im still cant understand some things. Maybe Im not too smart for that or its my english language knowledge.. So could someone to help finaly get point.

does hash partitioning is some kind similar to uniform grid, becouse it still has uniform cells, but  every cell has allmost unique hash value and stored not as map[w][h] but in list[hash]. So it could be used on endless world and empty cells were not stored.

Yes?

Edited by serumas

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I think there are many different ways to skin this cat. If you're really interested, Christer Ericson's book "Real-Time Collision Detection" has a great chapter on grids.

For a normal grid, you map 2d world coords (x,y) to 2d grid coords (u,v) via something like: u = floor(x/cellsize), v = floor(y/cellsize).

For a hashed grid, your 2d grid is only virtual, and it maps onto a 1d list of cells.

The way I approach this is to map the virtual (u,v) grid coordinates to a cell index i via something like: i = (u + (v * stride)) % list_length (where stride is the number of columns in the virtual grid). I don't know if this makes sense, but it works for me. AFAICR Ericson provides some other hash functions.

If your input coordinates may be negative, you can either just use abs() -- which is simple but ends up mapping points around the origin to the same cell, which may or may not be a big problem -- or you can handle them like you would a 2d rotation, i.e computing the naive (u,v) coordinates but then adding some multiple of the virtual grid dimensions to shift things into the proper range, so that u is in [0,num_columns-1] and v is in [0, num_rows-1].

Note that since multiple cells in the virtual grid can map to the same physical cell, you can't assume two objects in the same cell are necessarily close to each other, so you should do e.g an AABB test to make sure they're near each other, before a narrow-phase test.

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So my statement is correct. Thanks I just wanted to make shure about that...