Bilinear interpolation in C/C++ and CUDA - c++

Question

I want to emulate the behavior of CUDA bilinear interpolation on CPU, but I found that the return value of tex2D seems not fit to the bilinear formula.
I guess that casting the interpolation coefficients from float to 9-bit fixed point format with 8 bits of fractional value[1] results in different values.
According to the conversion fomula [2, line 106], the result of the conversion will be the same as the input float when the coeffient is 1/2^n, with n=0,1,..., 8, but I still (not always) receive weird values.
Below I report an example of weird values. In this case, weird values always happen when id = 2*n+1, could anyone tell me why?
Src Array:
Src[0][0] = 38;
Src[1][0] = 39;
Src[0][1] = 118;
Src[1][1] = 13;
Texture Definition:
static texture<float4, 2, cudaReadModeElementType> texElnt;
texElnt.addressMode[0] = cudaAddressModeClamp;
texElnt.addressMode[1] = cudaAddressModeClamp;
texElnt.filterMode = cudaFilterModeLinear;
texElnt.normalized = false;
Kernel Function:
static __global__ void kernel_texElnt(float* pdata, int w, int h, int c, float stride/*0.03125f*/) {
const int gx = blockIdx.x*blockDim.x + threadIdx.x;
const int gy = blockIdx.y*blockDim.y + threadIdx.y;
const int gw = gridDim.x * blockDim.x;
const int gid = gy*gw + gx;
if (gx >= w || gy >= h) {
return;
}
float2 pnt;
pnt.x = (gx)*(stride)/*1/32*/;
pnt.y = 0.0625f/*1/16*/;
float4 result = tex2D( texElnt, pnt.x + 0.5, pnt.y + 0.5f);
pdata[gid*3 + 0] = pnt.x;
pdata[gid*3 + 1] = pnt.y;
pdata[gid*3 + 2] = result.x;
}
Bilinear Result of CUDA
id pnt.x pnt.y tex2D
0 0.00000 0.0625 43.0000000
1 0.03125 0.0625 42.6171875
2 0.06250 0.0625 42.6484375
3 0.09375 0.0625 42.2656250
4 0.12500 0.0625 42.2968750
5 0.15625 0.0625 41.9140625
6 0.18750 0.0625 41.9453125
7 0.21875 0.0625 41.5625000
8 0.25000 0.0625 41.5937500
9 0.28125 0.0625 41.2109375
0 0.31250 0.0625 41.2421875
10 0.34375 0.0625 40.8593750
11 0.37500 0.0625 40.8906250
12 0.40625 0.0625 40.5078125
13 0.43750 0.0625 40.5390625
14 0.46875 0.0625 40.1562500
15 0.50000 0.0625 40.1875000
16 0.53125 0.0625 39.8046875
17 0.56250 0.0625 39.8359375
18 0.59375 0.0625 39.4531250
19 0.62500 0.0625 39.4843750
20 0.65625 0.0625 39.1015625
21 0.68750 0.0625 39.1328125
22 0.71875 0.0625 38.7500000
23 0.75000 0.0625 38.7812500
24 0.78125 0.0625 38.3984375
25 0.81250 0.0625 38.4296875
26 0.84375 0.0625 38.0468750
27 0.87500 0.0625 38.0781250
28 0.90625 0.0625 37.6953125
29 0.93750 0.0625 37.7265625
30 0.96875 0.0625 37.3437500
31 1.00000 0.0625 37.3750000
CPU Result:
// convert coefficient ((1-α)*(1-β)), (α*(1-β)), ((1-α)*β), (α*β) to fixed point format
id pnt.x pnt.y tex2D
0 0.00000 0.0625 43.00000000
1 0.03125 0.0625 43.23046875
2 0.06250 0.0625 42.64843750
3 0.09375 0.0625 42.87890625
4 0.12500 0.0625 42.29687500
5 0.15625 0.0625 42.52734375
6 0.18750 0.0625 41.94531250
7 0.21875 0.0625 42.17578125
8 0.25000 0.0625 41.59375000
9 0.28125 0.0625 41.82421875
0 0.31250 0.0625 41.24218750
10 0.34375 0.0625 41.47265625
11 0.37500 0.0625 40.89062500
12 0.40625 0.0625 41.12109375
13 0.43750 0.0625 40.53906250
14 0.46875 0.0625 40.76953125
15 0.50000 0.0625 40.18750000
16 0.53125 0.0625 40.41796875
17 0.56250 0.0625 39.83593750
18 0.59375 0.0625 40.06640625
19 0.62500 0.0625 39.48437500
20 0.65625 0.0625 39.71484375
21 0.68750 0.0625 39.13281250
22 0.71875 0.0625 39.36328125
23 0.75000 0.0625 38.78125000
24 0.78125 0.0625 39.01171875
25 0.81250 0.0625 38.42968750
26 0.84375 0.0625 38.66015625
27 0.87500 0.0625 38.07812500
28 0.90625 0.0625 38.30859375
29 0.93750 0.0625 37.72656250
30 0.96875 0.0625 37.95703125
31 1.00000 0.0625 37.37500000
I leave a simple code on my github [3], after running the program you will got two files in D:\.
Edit 2014/01/20
I run the program with different increments and found the specification of tex2D "when alpha multiplied beta is less than 0.00390625, the return of tex2D does not match the bilinear interpolation formula"

Answer 1

Already satisfactory answers have been provided to this question, so now I just want to give a compendium of hopefully useful information on bilinear interpolation, how can it be implemented in C++ and the different ways it can be done in CUDA.
Maths behind bilinear interpolation
Assume that the original function T(x, y) is sampled at the Cartesian regular grid of points (i, j) with 0 <= i < M1, 0 <= j < M2 and i and j integers. For each value of y, one can first use 0 <= a < 1 to represent an arbitrary point i + a comprised between i and i + 1. Then, a linear interpolation along the y = j axis (which is parallel to the x axis) at that point can be performed obtaining
where r(x,y) is the function interpolating the samples of T(x,y). The same can be done for the line y = j + 1, obtaining
Now, for each i + a, an interpolation along the y axis can be performed on the samples r(i+a,j) and r(i+a,j+1). Accordingly, if one uses 0 <= b < 1 to represent an arbitrary point j + b located between j and j + 1, then a linear interpolation along the x = i + a axis (which is parallel to the y axis) can be worked out, so getting the final result
Note that the relations between i, j, a, b, x and y are the following
C/C++ implementation
Let me stress that this implementation, as well as the following CUDA ones, assume, as done at the beginning, that the samples of T are located on the Cartesian regular grid of points (i, j) with 0 <= i < M1, 0 <= j < M2 and i and j integers (unit spacing). Also, the routine is provided in single precision, complex (float2) arithmetics, but it can be easily cast in other arithmetics of interest.
void bilinear_interpolation_function_CPU(float2 * __restrict__ h_result, float2 * __restrict__ h_data,
float * __restrict__ h_xout, float * __restrict__ h_yout,
const int M1, const int M2, const int N1, const int N2){
float2 result_temp1, result_temp2;
for(int k=0; k<N2; k++){
for(int l=0; l<N1; l++){
const int ind_x = floor(h_xout[k*N1+l]);
const float a = h_xout[k*N1+l]-ind_x;
const int ind_y = floor(h_yout[k*N1+l]);
const float b = h_yout[k*N1+l]-ind_y;
float2 h00, h01, h10, h11;
if (((ind_x) < M1)&&((ind_y) < M2)) h00 = h_data[ind_y*M1+ind_x]; else h00 = make_float2(0.f, 0.f);
if (((ind_x+1) < M1)&&((ind_y) < M2)) h10 = h_data[ind_y*M1+ind_x+1]; else h10 = make_float2(0.f, 0.f);
if (((ind_x) < M1)&&((ind_y+1) < M2)) h01 = h_data[(ind_y+1)*M1+ind_x]; else h01 = make_float2(0.f, 0.f);
if (((ind_x+1) < M1)&&((ind_y+1) < M2)) h11 = h_data[(ind_y+1)*M1+ind_x+1]; else h11 = make_float2(0.f, 0.f);
result_temp1.x = a * h10.x + (-h00.x * a + h00.x);
result_temp1.y = a * h10.y + (-h00.y * a + h00.y);
result_temp2.x = a * h11.x + (-h01.x * a + h01.x);
result_temp2.y = a * h11.y + (-h01.y * a + h01.y);
h_result[k*N1+l].x = b * result_temp2.x + (-result_temp1.x * b + result_temp1.x);
h_result[k*N1+l].y = b * result_temp2.y + (-result_temp1.y * b + result_temp1.y);
}
}
}
The if/else statements within the above code are simply boundary checkings. If the sample falls outside the [0, M1-1] x [0, M2-1], then it is set to 0.
Standard CUDA implementation
This is a "standard" CUDA implementation tracing the above CPU one. No usage of texture memory.
__global__ void bilinear_interpolation_kernel_GPU(float2 * __restrict__ d_result, const float2 * __restrict__ d_data,
const float * __restrict__ d_xout, const float * __restrict__ d_yout,
const int M1, const int M2, const int N1, const int N2)
{
const int l = threadIdx.x + blockDim.x * blockIdx.x;
const int k = threadIdx.y + blockDim.y * blockIdx.y;
if ((l<N1)&&(k<N2)) {
float2 result_temp1, result_temp2;
const int ind_x = floor(d_xout[k*N1+l]);
const float a = d_xout[k*N1+l]-ind_x;
const int ind_y = floor(d_yout[k*N1+l]);
const float b = d_yout[k*N1+l]-ind_y;
float2 d00, d01, d10, d11;
if (((ind_x) < M1)&&((ind_y) < M2)) d00 = d_data[ind_y*M1+ind_x]; else d00 = make_float2(0.f, 0.f);
if (((ind_x+1) < M1)&&((ind_y) < M2)) d10 = d_data[ind_y*M1+ind_x+1]; else d10 = make_float2(0.f, 0.f);
if (((ind_x) < M1)&&((ind_y+1) < M2)) d01 = d_data[(ind_y+1)*M1+ind_x]; else d01 = make_float2(0.f, 0.f);
if (((ind_x+1) < M1)&&((ind_y+1) < M2)) d11 = d_data[(ind_y+1)*M1+ind_x+1]; else d11 = make_float2(0.f, 0.f);
result_temp1.x = a * d10.x + (-d00.x * a + d00.x);
result_temp1.y = a * d10.y + (-d00.y * a + d00.y);
result_temp2.x = a * d11.x + (-d01.x * a + d01.x);
result_temp2.y = a * d11.y + (-d01.y * a + d01.y);
d_result[k*N1+l].x = b * result_temp2.x + (-result_temp1.x * b + result_temp1.x);
d_result[k*N1+l].y = b * result_temp2.y + (-result_temp1.y * b + result_temp1.y);
}
}
CUDA implementation with texture fetch
This is the same implementation as above, but the global memory is now accessed by the texture cache. For example, T[i,j] is accessed as
tex2D(d_texture_fetch_float,ind_x,ind_y);
(where, of course ind_x = i and ind_y = j, and d_texture_fetch_float is assumed to be a global scope variable) instead of
d_data[ind_y*M1+ind_x];
Note that the hard-wired texture filtering capabilities are not exploited here. The routine below has the same precision as the above one and could result somewhat faster than that on old CUDA architectures.
__global__ void bilinear_interpolation_kernel_GPU_texture_fetch(float2 * __restrict__ d_result,
const float * __restrict__ d_xout, const float * __restrict__ d_yout,
const int M1, const int M2, const int N1, const int N2)
{
const int l = threadIdx.x + blockDim.x * blockIdx.x;
const int k = threadIdx.y + blockDim.y * blockIdx.y;
if ((l<N1)&&(k<N2)) {
float2 result_temp1, result_temp2;
const int ind_x = floor(d_xout[k*N1+l]);
const float a = d_xout[k*N1+l]-ind_x;
const int ind_y = floor(d_yout[k*N1+l]);
const float b = d_yout[k*N1+l]-ind_y;
const float2 d00 = tex2D(d_texture_fetch_float,ind_x,ind_y);
const float2 d10 = tex2D(d_texture_fetch_float,ind_x+1,ind_y);
const float2 d11 = tex2D(d_texture_fetch_float,ind_x+1,ind_y+1);
const float2 d01 = tex2D(d_texture_fetch_float,ind_x,ind_y+1);
result_temp1.x = a * d10.x + (-d00.x * a + d00.x);
result_temp1.y = a * d10.y + (-d00.y * a + d00.y);
result_temp2.x = a * d11.x + (-d01.x * a + d01.x);
result_temp2.y = a * d11.y + (-d01.y * a + d01.y);
d_result[k*N1+l].x = b * result_temp2.x + (-result_temp1.x * b + result_temp1.x);
d_result[k*N1+l].y = b * result_temp2.y + (-result_temp1.y * b + result_temp1.y);
}
}
Texture binding can be done according to
void TextureBindingBilinearFetch(const float2 * __restrict__ data, const int M1, const int M2)
{
size_t pitch;
float* data_d;
gpuErrchk(cudaMallocPitch((void**)&data_d,&pitch, M1 * sizeof(float2), M2));
cudaChannelFormatDesc desc = cudaCreateChannelDesc<float2>();
gpuErrchk(cudaBindTexture2D(0,&d_texture_fetch_float,data_d,&desc,M1,M2,pitch));
d_texture_fetch_float.addressMode[0] = cudaAddressModeClamp;
d_texture_fetch_float.addressMode[1] = cudaAddressModeClamp;
gpuErrchk(cudaMemcpy2D(data_d,pitch,data,sizeof(float2)*M1,sizeof(float2)*M1,M2,cudaMemcpyHostToDevice));
}
Note that now we need no if/else boundary checking, because the texture will automatically clamp to zero the samples falling outside the [0, M1-1] x [0, M2-1] sampling region, thanks to the instructions
d_texture_fetch_float.addressMode[0] = cudaAddressModeClamp;
d_texture_fetch_float.addressMode[1] = cudaAddressModeClamp;
CUDA implementation with texture interpolation
This is the last implementation and uses the hard-wired capabilities of texture filtering.
__global__ void bilinear_interpolation_kernel_GPU_texture_interp(float2 * __restrict__ d_result,
const float * __restrict__ d_xout, const float * __restrict__ d_yout,
const int M1, const int M2, const int N1, const int N2)
{
const int l = threadIdx.x + blockDim.x * blockIdx.x;
const int k = threadIdx.y + blockDim.y * blockIdx.y;
if ((l<N1)&&(k<N2)) { d_result[k*N1+l] = tex2D(d_texture_interp_float, d_xout[k*N1+l] + 0.5f, d_yout[k*N1+l] + 0.5f); }
}
Note that the interpolation formula implemented by this feature is the same as derived above, but now
where x_B = x - 0.5 and y_B = y - 0.5. This explains the 0.5 offset in the instruction
tex2D(d_texture_interp_float, d_xout[k*N1+l] + 0.5f, d_yout[k*N1+l] + 0.5f)
In this case, texture binding should be done as follows
void TextureBindingBilinearInterp(const float2 * __restrict__ data, const int M1, const int M2)
{
size_t pitch;
float* data_d;
gpuErrchk(cudaMallocPitch((void**)&data_d,&pitch, M1 * sizeof(float2), M2));
cudaChannelFormatDesc desc = cudaCreateChannelDesc<float2>();
gpuErrchk(cudaBindTexture2D(0,&d_texture_interp_float,data_d,&desc,M1,M2,pitch));
d_texture_interp_float.addressMode[0] = cudaAddressModeClamp;
d_texture_interp_float.addressMode[1] = cudaAddressModeClamp;
d_texture_interp_float.filterMode = cudaFilterModeLinear; // --- Enable linear filtering
d_texture_interp_float.normalized = false; // --- Texture coordinates will NOT be normalized
gpuErrchk(cudaMemcpy2D(data_d,pitch,data,sizeof(float2)*M1,sizeof(float2)*M1,M2,cudaMemcpyHostToDevice));
}
Note that, as already mentioned in the other answers, a and b are stored in 9-bit fixed point format with 8 bits of fractional value, so this approach will be very fast, but less accurate than those above.

Answer 2

The UV interpolants are truncated to 9 bits, not the participating texel values. In Chapter 10 (Texturing) of The CUDA Handbook, this is described in detail (including CPU emulation code) for the 1D case. Code is open source and may be found at https://github.com/ArchaeaSoftware/cudahandbook/blob/master/texturing/tex1d_9bit.cu

Answer 3

Wrong formula of bilinear interpolation makes result of texture fetching weird.
Formula - 1: you can find it in cuda appendix or wiki easily
tex(x,y)=(1−α)(1−β)T[i,j] + α(1−β)T[i+1,j] + (1−α)βT[i,j+1] + αβT[i+1,j+1]
Formula - 2: reduce times of multiply
tex(x,y)=T[i,j] + α(T[i+1,j]-T[i,j]) + β(T[i,j+1]-T[i,j]) + αβ(T[i,j]+T[i+1,j+1] - T[i+1, j]-T[i,j+1])
If you use 9-bit fixed point format to formula 1, you will get mismatch result of texture fetching, but formula 2 works fine.
Conclusion :
If you want to emulate the bilinear interpolation implemented by cuda texture, you should use formula 3. Try it!
Formula - 3:
tex(x,y)=T[i,j] + frac(α)(T[i+1,j]-T[i,j]) + frac(β)(T[i,j+1]-T[i,j]) + frac(αβ)(T[i,j]+T[i+1,j+1] - T[i+1, j]-T[i,j+1])
// frac(x) turns float to 9-bit fixed point format with 8 bits of fraction values.
float frac( float x ) {
float frac, tmp = x - (float)(int)(x);
float frac256 = (float)(int)( tmp*256.0f + 0.5f );
frac = frac256 / 256.0f;
return frac;
}