5
\$\begingroup\$

I believe this is considered value noise or possibly gradient noise in that I simply interpolate between random values (always returning the same values per coordinate).

I am using this currently to generate a basic heightmap in a 3D game that expands outward as the player moves (being nearly infinite, only restricted by variable limitations).

I contain my game's map within chunks, 32x32 (1024) data points per chunk representing height. I call the function GetNoise2D() with the X and Z values of every data point.

This is currently my largest bottleneck. I could probably just do the 4 corners per chunk and interpolate between them to get reasonable looking terrain, but to put it simply, I'd rather not.

Does anyone see any noticeable performance issues with the algorithm or the concept after initialization? I call on the algorithm using the function GetHeight() which combines two calls to the algorithm.

private float GetHeight(int X, int Z) {
    float fNoise = Noise.GetNoise2D(X * .02f, Z * .02f) * .5f;
    fNoise += Noise.GetNoise2D(X * .04f, Z * .04f) * .5f;

    //Scale noise from 0-1 to 0-20
    return fNoise * 20f;
}


public class clsNoise2D {
    readonly byte[] Permutations = new byte[512];
    readonly float[] Values = new float[256];
    float xLerpAmount, yLerpAmount, v00, v10, v01, v11;
    //pX, pXa, dX, and dY are helper values to reduce operations
    int pX, pXa; int dX, dY;

    public Random random;

    public clsNoise2D(int iSeed) {
        random = new Random(iSeed);

        //Randomize permutations array with values 0-255
        List<byte> listByte = new List<byte>();
        for (int i = 0; i < 256; i++) listByte.Add((byte)i);
        for (int i = 256; i > 0; i--) { Permutations[256 - i] = listByte[random.Next(i)]; listByte.Remove(Permutations[256 - i]); }

        //Take permutations array up to 512 elements to reduce wrapping needs in GetNoise2D call
        for (int i = 256; i < 512; i++) { Permutations[i] = Permutations[i - 256]; }

        //Set values to be between 0 and 1 incrementally from 0/255 through 255/255.
        for (int i = 0; i < 256; i++) { Values[i] = (i / 255f); }
    }

    public float GetNoise2D(float CoordX, float CoordY) {
        //Get floor value of inputs
        dX = (int)Math.Floor(CoordX); dY = (int)Math.Floor(CoordY);

        //Get fractional value of inputs
        xLerpAmount = CoordX - dX; yLerpAmount = CoordY - dY;

        //Wrap floored values to byte values
        dX = dX & 255; dY = dY & 255;

        //Start permutation/value pulling
        pX = Permutations[dX]; pXa = Permutations[dX + 1];

        v00 = Values[Permutations[(dY + pX)]];
        v10 = Values[Permutations[(dY + pXa)]];
        v01 = Values[Permutations[(dY + 1 + pX)]];
        v11 = Values[Permutations[(dY + 1 + pXa)]];

        //Smooth lerp amounts by cosine function
        xLerpAmount = (1f - (float)Math.Cos(xLerpAmount * Math.PI)) * .5f;
        yLerpAmount = (1f - (float)Math.Cos(yLerpAmount * Math.PI)) * .5f;

        //Return 2D interpolation for v00, v01, v10, and v11
        return (v00 * (1 - xLerpAmount) * (1 - yLerpAmount) +
                v10 * xLerpAmount * (1 - yLerpAmount) +
                v01 * (1 - xLerpAmount) * yLerpAmount +
                v11 * xLerpAmount * yLerpAmount);
    }
}

Edit: To clarify the question itself, is there a very noticeable performance mistake currently being made OR is there a completely different way to achieve identical or nearly identical values that SHOULD knock the performance out of the park?

\$\endgroup\$
2
  • \$\begingroup\$ In what way is this a bottleneck? If I take your sample and run it over a sample set of 1024 values (as I think you are doing) then it takes milliseconds for it to complete, so I think the question is how fast do you expect it to perform and how are you sure this is the slowest part of your program? \$\endgroup\$ Nov 14, 2013 at 17:16
  • \$\begingroup\$ This is being ran over 1024 values per chunk. 17x17 chunks. If the player is moving fast (ie. flying) it is possible to be generating up to 10 chunks per frame. Although I don't like it, I am currently interpolating using only the algorithm above in the corners of each chunk (and it works well enough I believe). I was simply hoping to not have to. \$\endgroup\$
    – Mythics
    Nov 14, 2013 at 19:41

2 Answers 2

1
\$\begingroup\$

According to Visual Studio, the most expensive calls are indeed the lines where you call Math.Cos.

Visual Studio performance analysis for <code>GetNoise2D</code>

You could probably shave a couple of cycles by creating a lookup table which would return the result of that whole expression, (1-cos(w * PI))/2, just by indexing an array.

\$\endgroup\$
2
\$\begingroup\$

I'm not sure if Cos and Sin are efficient here, they might be a ferformance killer. Consider making a table of sin/cos values for come range of angles.

\$\endgroup\$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.