## poniedziałek, 15 września 2014

### Finding Two Max Values in an Array in Shader

One of the problems I was faced was to, given four floats, find the two highest values and their indices in the array. This had to be done in a shader program so ideally no loops, arrays indexing or things like that. Fortunately my sort-of-a-boss, KriS, quickly found a solution that I'm presenting here.

Here comes the code:
void GetTwoMaxValues( in float4 values, out float2 weights, out float2 indices )
{
float v1 = max( max( max( values.x, values.y ), values.z ), values.w );
float i1 = 0;
i1 = CndMask( values.y == v1, 1, i1 );
i1 = CndMask( values.z == v1, 2, i1 );
i1 = CndMask( values.w == v1, 3, i1 );

//

values.x = CndMask( i1 == 0, 0, values.x );
values.y = CndMask( i1 == 1, 0, values.y );
values.z = CndMask( i1 == 2, 0, values.z );
values.w = CndMask( i1 == 3, 0, values.w );

float v2 = max( max( max( values.x, values.y ), values.z ), values.w );
float i2 = 0;
i2 = CndMask( values.y == v2, 1, i2 );
i2 = CndMask( values.z == v2, 2, i2 );
i2 = CndMask( values.w == v2, 3, i2 );

//

weights.x = v1;
weights.y = v2;
indices.x = i1;
indices.y = i2;
}
The code finds the first maximum value simply by "iterating" over elements and "max'ing" them. That was simple. But we also need to know the index of the value in the array. We assume it's the 0th index and compare the found value with other values from the array using CndMask statement. If the compared value is equal to the value of the element in the array, we update the index.

To find the second highest values we first zeroe the actual highest element found in the previous pass and do the same procedure as in the previous pass to find the second highest element.

The code can actually be understood much quicker by actually reading it rather than reading explanation trying to explain it :). It doesn't do anything magical in algorithmic terms but shows how to solve (efficiently) this simple problem of two max values in a shader program.

## środa, 30 lipca 2014

### Why Are Kids Happy?

Yes. This blog post is way different from what you (like three or four of you?) have accustomed to see here but I just have the need to share some of my life philosophies/thoughts, not just math/game/graphics-related stuff :).

So, the subject of this post goes "why are kids happy?". Today I loosely pondered a bit on that with my flatmates (who by the way are some of my favorites and closest friends :)). For whatever it may sound, it started with Moomins and Smurfs... Think of these books or TV series or whatever form they assumed. You might have forgotten them but you remember them. Those hippo-shaped humanoids and blue gnoms. What do you think of them? Do you think how high their creators must have been at the time they made their concepts arts? What stuff did these guys inhale to make a story of humanoidal hippos with occasional humans appearing in form of an always-wandering Snufkin or girl/lady(?) called Little My? Not to mention the blue Smurf creatures who were almost always males, with one Smurfette lady who probably had to satisfy all of the other male Smurfs sexual needs. Are any of these or something similar on your mind right now? If yes, that is not a big deal. That's just what adults do. They analyze a whole lot. And then more and more thinking of how the worlds created in Smurfs or Moomins could exist in "real-life". And that's the problem. Kids don't do that. They don't analyze. They don't wonder on whether a village full of male blue gnoms and a single lady makes sense and what they do when lights go off. They just experience this world as it is. Sure, it's easy to manipulate kids this way because they are so trustful, but try from time to time to observe a child or a kid in a casual situation and see what fun he or she has from experiencing it, indulging in the moment.

Kids just happen to live for the moment more often than adults, who spent more time on living somewhere else, whether it be the past or future. I want a new car, a new house, a good-looking partner, a thriving business... Don't you spend too much time on thinking (and worrying!) on all those things? What was the last time you walked barefoot on wet grass? Or just did something that did not quite fit to your daily routine? So next time you catch yourself on thinking what your life is going to be in a year from now, take a bike and go on a trip outside of city. There is so many beautiful green areas around! See them. Feel them. Experience them. Like this little kid who indeed does this, even though it might happen for him/her on a more subconscious than conscious level :).

## środa, 9 kwietnia 2014

### Tangent Basis in Compact Form

Recently at Flying Wild Hog my task was to change the vertex format of meshes. Both static and skinned meshes used 12 bytes for storing normal, tangent, bitangent and color of the vertex. This looked something like this:
UByte4N normal;
UByte4N tangent;
UByte4N bitangent;
Each tangent basis vector is stored explicitly on 24 bits and the last 8 bits of each vector are intended to store one component of the vertex color (red stored in normal's alpha, green in tangent's alpha and blue in bitangent's alpha).

The UByte4N type internally stores one 32-bit unsigned integer (hence the U prefix) and has some helper functions for packing and unpacking four 8-bit values. Obviously, as tangent basis vector's components are floats in normalized $[-1, 1]$ interval (hence the N postfix meaning normalized), there are also functions which convert it to decimal $[0, 255]$ interval.

This form of the tangent basis with color packed together is actually quite nice. The precision is not bad and all channels are fully utilized. The biggest problem is that in order to get color we must read three vectors. We can do better though.

The first thing is to get rid off the bitangent vector altogether. The whole tangent basis can nicely be packed with just normal, tangent and one value that indicates the handedness of the entire basis. To get the handedness of the basis you can compute the determinant of a matrix that is formed of that basis - it will be either $1$ or $-1$ for an orthonormal basis. This way we save two values (one handedness value as opposed to three defining the bitangent vector) and we can extract the color to a separate vector:
UByte4N normal; // .w unused
UByte4N tangent; // .w stores handedness
UByte4N color;
It looks like now the color has been enriched with the alpha channel. Also, the $w$ component of the normal vector is unused. And hey, do we really need to waste 8 bits for the handedness? One value would suffice. Fortunately, GPUs are equipped with RGB10A2 format, where 10 bits are devoted to RGB channels and 2 bits for the alpha channel. This way we can store the tangent vector with greater precision. So, our new tangent basis and color part of the vertex structure is this:
UX10Y10Z10W2N normal; // .w unused
UX10Y10Z10W2N tangent; // .w stores handedness
UByte4N color;
Now this one looks scary at first glance. The U prefix again means that the stored values are unsigned - decimal $[0, 1023]$ interval for the $x$, $y$, and $z$ components and $[0, 3]$ for the $w$ component. Also, the N postfix means the that input float values to the structure must be normalized, that is in $[-1, 1]$ interval.

Finally, some piece of working code:
struct UX10Y10Z10W2N
{
UInt32 m_value;

void Set( const float* xyz, int w )
{
m_value = 0;

for ( int i = 0; i < 3; i++ )
{
float f = ( xyz[ i ] + 1.0f ) * 511.5f;

if ( f < 0.0f )
f = 0.0f;
else if ( f > 1023.0f )
f = 1023.0f;

m_value |= ( ( int )f ) << 10*i;
}

m_value |= w << 30;
}

void Get( float& x, float& y, float& z ) const
{
x = ( float )( ( m_value >> 0  ) & 1023 );
y = ( float )( ( m_value >> 10 ) & 1023 );
z = ( float )( ( m_value >> 20 ) & 1023 );

x = ( x - 511.0f ) / 512.0f;
y = ( y - 511.0f ) / 512.0f;
z = ( z - 511.0f ) / 512.0f;
}

void Get( float& x, float& y, float& z, int& w ) const
{
Get( x, y, z );
w = ( ( m_value >> 30 ) & 3 );
}
};
I'm leaving you without much explanation as it should be quite easy to read. One thing worth noting is that the $w$ component is not passed or read as float but rather as int. That is because I determined it does not make much sense dealing with float here as we have barely 2 bits. I pass here $0$ or $3$ (which will be converted to $1$ in the shader) depending on the handedness of the tangent basis.

## sobota, 22 lutego 2014

### Edge Detection from Depths and Normals

Edge detection is an algorithm that comes handy from time to time in every graphics programmer's life. There are various approaches to this problem and various tasks that we need edge detection for. For instance, we might use edge detection to perform anti-aliasing (like in here http://http.developer.nvidia.com/GPUGems2/gpugems2_chapter09.html) or to draw some stylish outlines. So how can we detect the edges? Some algortihms compare color differences between the pixel we are currently processing and its neighbours (like antialiasing algorithms FXAA and MLAA). Others, like the one I linked a few words before, use the rendered scene's geometry information, like the depth-buffer and normals, to do so. In this post I present a variant of this algorithm.

The algorithm I use needs three pixels. The one we're processing and some two "different" neighbours. You can take the one to the left along with the one to the top for instance. Now, we read normal vectors (either in view or world space) at those pixels and depths. We also recover view or world space positions from the sampled depths (I showed how to do this here http://maxestws.blogspot.com/2013/11/recovering-camera-position-from-depth.html).

So, to sum up at this point, we have three (I assume view space) normals ($N$, $N_{left}$ and $N_{top}$) and positions ($P$, $P_{left}$ and $P_{top}$). Here comes the shader code:
bool IsEdge(
float3 centerNormal,
float3 leftNormal,
float3 topNormal,
float3 centerPosition,
float3 leftPosition,
float3 topPosition,
float normalsDotThreshold,
float distanceThreshold)
{
// normals dot criterium

float centerLeftNormalsDot = dot(centerNormal, leftNormal);
float centerTopNormalsDot = dot(centerNormal, topNormal);

if (centerLeftNormalsDot < normalsDotThreshold ||
centerTopNormalsDot < normalsDotThreshold)
{
return true;
}

// distances difference criterium

float3 v1 = leftPosition - centerPosition;
float3 v2 = topPosition - centerPosition;

if (abs(dot(centerNormal, v1)) > distanceThreshold ||
abs(dot(centerNormal, v2)) > distanceThreshold)
{
return true;
}

//

return false;
}
Good starting value for normalsDotThreshold is $0.98$ and for distanceThreshold it's $0.01$.

The code uses two criterions to determine whether we are on an edge. The first checks dots of the normals. It's purpose is to detect edges that appear on the continous surface but with varying normals, where, for instance, two perpendicular planes meet (think of a side of a box standing on a floor).

The second criterium checks for distances. Imagine two planes parallel to each other but at different heights. When viewed in such a way that an edge of the top plane appears in front of the bottom plane we clearly have, well, an edge to detect. The normals dot product won't work because normals of all pixels here are the same (the planes are parallel). So in this case we need to track vectors from the center pixel to the neighbouring pixels. If they are too far away then we have an edge. Note here that we actually dot those vectors with the center pixel's normal. This is to avoid false positive cases for pixels who are on the same plane but are far away from each other. In that case, obviously, we don't care they are far away. They lie on the same plane so there is no edge.

I know that a picture is worth more than a thousand words. That's why I preferred to described the code with words instead of putting images here. Wait... what? That's right. Your's best bet is to just take this simple piece of code and try it in practice to see exactly how it works. I highly recommend doing the following modifications and see what happens:
1. Leave just the normals dot criterium.
2. Leave just the distances difference criterium.
3. Normalize vectors $v_1$ and $v_2$ in the distances difference criterium.

Have fun :).