Is it expected that the following test should fail?
The test compares results of a 2D and a 3D AffineTransformation. Both are constructed to have unit scaling and zero offsets in the y and z direction, but to have non-zero and non-unity scaling and offset in the x direction. All other off-diagonal elements are zero. It is my belief that these transformations are identical in the x and y directions, and hence should produce identical results.
Furthermore I have found that the test passes if I use this Kernel:
using K = CGAL::Exact_predicates_exact_constructions_kernel;
Is it to be expected that the test passes if I use this Kernel? Should the test fail with either kernel or pass with either kernel?
TEST(TransformerTest, testCGALAffine) {
using K = CGAL::Exact_predicates_inexact_constructions_kernel;
using Float = typename K::FT;
using Transformation_2 = K::Aff_transformation_2;
using Transformation_3 = K::Aff_transformation_3;
using Point_2 = typename K::Point_2;
using Point_3 = typename K::Point_3;
double lowerCorner(17.005142946538115);
double upperCorner(91.940521484752139);
int resolution = 48;
double tmpScaleX((upperCorner - lowerCorner) / resolution);
Float scaleX(tmpScaleX);
Float zero(0);
Float unit(1);
// create a 2D voxel to world transform
Transformation_2 transformV2W_2(scaleX, zero, Float(lowerCorner),
zero, unit, zero,
unit);
// create it's inverse: a 2D world to voxel transform
auto transformW2V_2 = transformV2W_2.inverse();
// create a 3D voxel to world transform
Transformation_3 transformV2W_3(scaleX, zero, zero, Float(lowerCorner),
zero, unit, zero, zero,
zero, zero, unit, zero,
unit);
// create it's inverse: a 3D world to voxel transform
auto transformW2V_3 = transformV2W_3.inverse();
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 2; ++j) {
EXPECT_EQ(transformV2W_2.cartesian(i, j), transformV2W_3.cartesian(i, j)) << i << ", " << j;
EXPECT_EQ(transformW2V_2.cartesian(i, j), transformW2V_3.cartesian(i, j)) << i << ", " << j;
}
}
std::mt19937_64 rng(0);
std::uniform_real_distribution<double> randReal(0, resolution);
// compare the results of 2D and 3D transformations of random locations
for (int i = 0; i < static_cast<int>(1e4); ++i) {
Float x(randReal(rng));
Float y(randReal(rng));
auto world_2 = transformV2W_2(Point_2(x, y));
auto world_3 = transformV2W_3(Point_3(x, y, 0));
EXPECT_EQ(world_2.x(), world_3.x()) << world_2 << ", " << world_3;
auto voxel_2 = transformW2V_2(world_2);
auto voxel_3 = transformW2V_3(world_3);
EXPECT_EQ(voxel_2.x(), voxel_3.x()) << voxel_2 << ", " << voxel_3;
}
}
Related
Find the nth int with 10 set bits
n is an int in the range 0<= n <= 30 045 014
The 0th int = 1023, the 1st = 1535 and so on
snob() same number of bits,
returns the lowest integer bigger than n with the same number of set bits as n
int snob(int n) {
int a=n&-n, b=a+n;
return b|(n^b)/a>>2;
}
calling snob n times will work
int nth(int n){
int o =1023;
for(int i=0;i<n;i++)o=snob(o);
return o;
}
example
https://ideone.com/ikGNo7
Is there some way to find it faster?
I found one pattern but not sure if it's useful.
using factorial you can find the "indexes" where all 10 set bits are consecutive
1023 << x = the (x+10)! / (x! * 10!) - 1 th integer
1023<<1 is the 10th
1023<<2 is the 65th
1023<<3 the 285th
...
Btw I'm not a student and this is not homework.
EDIT:
Found an alternative to snob()
https://graphics.stanford.edu/~seander/bithacks.html#NextBitPermutation
int lnbp(int v){
int t = (v | (v - 1)) + 1;
return t | ((((t & -t) / (v & -v)) >> 1) - 1);
}
I have built an implementation that should satisfy your needs.
/** A lookup table to see how many combinations preceeded this one */
private static int[][] LOOKUP_TABLE_COMBINATION_POS;
/** The number of possible combinations with i bits */
private static int[] NBR_COMBINATIONS;
static {
LOOKUP_TABLE_COMBINATION_POS = new int[Integer.SIZE][Integer.SIZE];
for (int bit = 0; bit < Integer.SIZE; bit++) {
// Ignore less significant bits, compute how many combinations have to be
// visited to set this bit, i.e.
// (bit = 4, pos = 5), before came 0b1XXX and 0b1XXXX, that's C(3, 3) + C(4, 3)
int nbrBefore = 0;
// The nth-bit can be only encountered after pos n
for (int pos = bit; pos < Integer.SIZE; pos++) {
LOOKUP_TABLE_COMBINATION_POS[bit][pos] = nbrBefore;
nbrBefore += nChooseK(pos, bit);
}
}
NBR_COMBINATIONS = new int[Integer.SIZE + 1];
for (int bits = 0; bits < NBR_COMBINATIONS.length; bits++) {
NBR_COMBINATIONS[bits] = nChooseK(Integer.SIZE, bits);
assert NBR_COMBINATIONS[bits] > 0; // Important for modulo check. Otherwise we must use unsigned arithmetic
}
}
private static int nChooseK(int n, int k) {
assert k >= 0 && k <= n;
if (k > n / 2) {
k = n - k;
}
long nCk = 1; // (N choose 0)
for (int i = 0; i < k; i++) {
// (N choose K+1) = (N choose K) * (n-k) / (k+1);
nCk *= (n - i);
nCk /= (i + 1);
}
return (int) nCk;
}
public static int nextCombination(int w, int n) {
// TODO: maybe for small n just advance naively
// Get the position of the current pattern w
int nbrBits = 0;
int position = 0;
while (w != 0) {
final int currentBit = Integer.lowestOneBit(w); // w & -w;
final int bitPos = Integer.numberOfTrailingZeros(currentBit);
position += LOOKUP_TABLE_COMBINATION_POS[nbrBits][bitPos];
// toggle off bit
w ^= currentBit;
nbrBits++;
}
position += n;
// Wrapping, optional
position %= NBR_COMBINATIONS[nbrBits];
// And reverse lookup
int v = 0;
int m = Integer.SIZE - 1;
while (nbrBits-- > 0) {
final int[] bitPositions = LOOKUP_TABLE_COMBINATION_POS[nbrBits];
// Search for largest bitPos such that position >= bitPositions[bitPos]
while (Integer.compareUnsigned(position, bitPositions[m]) < 0)
m--;
position -= bitPositions[m];
v ^= (0b1 << m--);
}
return v;
}
Now for some explanation. LOOKUP_TABLE_COMBINATION_POS[bit][pos] is the core of the algorithm that makes it as fast as it is. The table is designed so that a bit pattern with k bits at positions p_0 < p_1 < ... < p_{k - 1} has a position of `\sum_{i = 0}^{k - 1}{ LOOKUP_TABLE_COMBINATION_POS[i][p_i] }.
The intuition is that we try to move back the bits one by one until we reach the pattern where are all bits are at the lowest possible positions. Moving the i-th bit from position to k + 1 to k moves back by C(k-1, i-1) positions, provided that all lower bits are at the right-most position (no moving bits into or through each other) since we skip over all possible combinations with the i-1 bits in k-1 slots.
We can thus "decode" a bit pattern to a position, keeping track of the bits encountered. We then advance by n positions (rolling over in case we enumerated all possible positions for k bits) and encode this position again.
To encode a pattern, we reverse the process. For this, we move bits from their starting position forward, as long as the position is smaller than what we're aiming for. We could, instead of a linear search through LOOKUP_TABLE_COMBINATION_POS, employ a binary search for our target index m but it's hardly needed, the size of an int is not big. Nevertheless, we reuse our variant that a smaller bit must also come at a less significant position so that our algorithm is effectively O(n) where n = Integer.SIZE.
I remain with the following assertions to show the resulting algorithm:
nextCombination(0b1111111111, 1) == 0b10111111111;
nextCombination(0b1111111111, 10) == 0b11111111110;
nextCombination(0x00FF , 4) == 0x01EF;
nextCombination(0x7FFFFFFF , 4) == 0xF7FFFFFF;
nextCombination(0x03FF , 10) == 0x07FE;
// Correct wrapping
nextCombination(0b1 , 32) == 0b1;
nextCombination(0x7FFFFFFF , 32) == 0x7FFFFFFF;
nextCombination(0xFFFFFFEF , 5) == 0x7FFFFFFF;
Let us consider the numbers with k=10 bits set.
The trick is to determine the rank of the most significant one, for a given n.
There is a single number of length k: C(k, k)=1. There are k+1 = C(k+1, k) numbers of length k + 1. ... There are C(m, k) numbers of length m.
For k=10, the limit n are 1 + 10 + 55 + 220 + 715 + 2002 + 5005 + 11440 + ...
For a given n, you easily find the corresponding m. Then the problem is reduced to finding the n - C(m, k)-th number with k - 1 bits set. And so on recursively.
With precomputed tables, this can be very fast. 30045015 takes 30 lookups, so that I guess that the worst case is 29 x 30 / 2 = 435 lookups.
(This is based on linear lookups, to favor small values. By means of dichotomic search, you reduce this to less than 29 x lg(30) = 145 lookups at worse.)
Update:
My previous estimates were pessimistic. Indeed, as we are looking for k bits, there are only 10 determinations of m. In the linear case, at worse 245 lookups, in the dichotomic case, less than 50.
(I don't exclude off-by-one errors in the estimates, but clearly this method is very efficient and requires no snob.)
I need the method to check sum CRC8.
I found this code, but it's not working:
- (int)crc8Checksum:(NSString*)dataFrame{
char j;
int crc8 = 0;
int x = 0;
for (int i = 0; i < [dataFrame length]; i++){
x = [dataFrame characterAtIndex:i];
for (int k = 0; k < 8; k++){
j = 1 & (x ^ crc8);
crc8 = floor0(crc8 / 2) & 0xFF;
x = floor0(x / 2) & 0xFF;
if (j != 0 ){
crc8 = crc8 ^ 0x8C;
}
}
}
return crc8;
}
Help me please!
What do you mean "it's not working"? There are 14 different CRC-8 definitions in this catalog, and probably many more out there in the wild. Do you have some CRC values you are comparing to? Is there documentation on what CRC you actually need? What are your test messages and corresponding expected CRCs?
You can't just pick some random CRC-8 code and expect it to do what you need.
That particular code computes a CRC-8/MAXIM in the linked catalog. However it is truly awful code. With unnecessary divides and floors. Here is a better, simpler, faster inner loop:
crc8 ^= x;
for (int k = 0; k < 8; k++)
crc8 = crc8 & 1 ? (crc8 >> 1) ^ 0x8c : crc8 >> 1;
You can get it faster still with tables and algorithms that compute the CRC a byte at a time or a machine word at a time.
The x in the code has its own problems, since an NSString can be a string of unicode characters, so characterAtIndex may not return a byte, and length may not return the number of bytes. You need a way to get the message as a series of bytes.
I'm trying to compute batch 1D FFTs using cufftPlanMany. The data set comes from a 3D field, stored in a 1D array, where I want to compute 1D FFTs in the x and y direction. The data is stored as shown in the figure below; continuous in x then y then z.
Doing batch FFTs in the x-direction is (I believe) straighforward; with input stride=1, distance=nx and batch=ny * nz, it computes the FFTs over elements {0,1,2,3}, {4,5,6,7}, ..., {28,29,30,31}. However, I can't think of a way to achieve the same for the FFTs in the y-direction. A batch for each xy plane is again straightforward (input stride=nx, dist=1, batch=nx results in FFTs over {0,4,8,12}, {1,5,9,13}, etc.). But with batch=nx * nz, going from {3,7,11,15} to {16,20,24,28}, the distance is larger than 1. Can this somehow be done with cufftPlanMany?
I think that the short answer to your question (possibility of using a single cufftPlanMany to perform 1D FFTs of the columns of a 3D matrix) is NO.
Indeed, transformations performed according to cufftPlanMany, that you call like
cufftPlanMany(&handle, rank, n,
inembed, istride, idist,
onembed, ostride, odist, CUFFT_C2C, batch);
must obey the Advanced Data Layout. In particular, 1D FFTs are worked out according to the following layout
input[b * idist + x * istride]
where b addresses the b-th signal and istride is the distance between two consecutive items in the same signal. If the 3D matrix has dimensions M * N * Q and if you want to perform 1D transforms along the columns, then the distance between two consecutive elements will be M, while the distance between two consecutive signals will be 1. Furthermore, the number of batched executions must be set equal to M. With those parameters, you are able to cover only one slice of the 3D matrix. Indeed, if you try increasing M, then the cuFFT will start trying to compute new column-wise FFTs starting from the second row. The only solution to this problem is an iterative call to cufftExecC2C to cover all the Q slices.
For the record, the following code provides a fully worked example on how performing 1D FFTs of the columns of a 3D matrix.
#include <thrust/device_vector.h>
#include <cufft.h>
/********************/
/* CUDA ERROR CHECK */
/********************/
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, const char *file, int line, bool abort=true)
{
if (code != cudaSuccess)
{
fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) exit(code);
}
}
int main() {
const int M = 3;
const int N = 4;
const int Q = 2;
thrust::host_vector<float2> h_matrix(M * N * Q);
for (int k=0; k<Q; k++)
for (int j=0; j<N; j++)
for (int i=0; i<M; i++) {
float2 temp;
temp.x = (float)(j + k * M);
//temp.x = 1.f;
temp.y = 0.f;
h_matrix[k*M*N+j*M+i] = temp;
printf("%i %i %i %f %f\n", i, j, k, temp.x, temp.y);
}
printf("\n");
thrust::device_vector<float2> d_matrix(h_matrix);
thrust::device_vector<float2> d_matrix_out(M * N * Q);
// --- Advanced data layout
// input[b * idist + x * istride]
// output[b * odist + x * ostride]
// b = signal number
// x = element of the b-th signal
cufftHandle handle;
int rank = 1; // --- 1D FFTs
int n[] = { N }; // --- Size of the Fourier transform
int istride = M, ostride = M; // --- Distance between two successive input/output elements
int idist = 1, odist = 1; // --- Distance between batches
int inembed[] = { 0 }; // --- Input size with pitch (ignored for 1D transforms)
int onembed[] = { 0 }; // --- Output size with pitch (ignored for 1D transforms)
int batch = M; // --- Number of batched executions
cufftPlanMany(&handle, rank, n,
inembed, istride, idist,
onembed, ostride, odist, CUFFT_C2C, batch);
for (int k=0; k<Q; k++)
cufftExecC2C(handle, (cufftComplex*)(thrust::raw_pointer_cast(d_matrix.data()) + k * M * N), (cufftComplex*)(thrust::raw_pointer_cast(d_matrix_out.data()) + k * M * N), CUFFT_FORWARD);
cufftDestroy(handle);
for (int k=0; k<Q; k++)
for (int j=0; j<N; j++)
for (int i=0; i<M; i++) {
float2 temp = d_matrix_out[k*M*N+j*M+i];
printf("%i %i %i %f %f\n", i, j, k, temp.x, temp.y);
}
}
The situation is different for the case when you want to perform 1D transforms of the rows. In that case, the distance between two consecutive elements is 1, while the distance between two consecutive signals is M. This allows you to set a number of N * Q transformations and then invoking cufftExecC2C only one time. For the record, the code below provides a full example of 1D transformations of the rows of a 3D matrix.
#include <thrust/device_vector.h>
#include <cufft.h>
/********************/
/* CUDA ERROR CHECK */
/********************/
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, const char *file, int line, bool abort=true)
{
if (code != cudaSuccess)
{
fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) exit(code);
}
}
int main() {
const int M = 3;
const int N = 4;
const int Q = 2;
thrust::host_vector<float2> h_matrix(M * N * Q);
for (int k=0; k<Q; k++)
for (int j=0; j<N; j++)
for (int i=0; i<M; i++) {
float2 temp;
temp.x = (float)(j + k * M);
//temp.x = 1.f;
temp.y = 0.f;
h_matrix[k*M*N+j*M+i] = temp;
printf("%i %i %i %f %f\n", i, j, k, temp.x, temp.y);
}
printf("\n");
thrust::device_vector<float2> d_matrix(h_matrix);
thrust::device_vector<float2> d_matrix_out(M * N * Q);
// --- Advanced data layout
// input[b * idist + x * istride]
// output[b * odist + x * ostride]
// b = signal number
// x = element of the b-th signal
cufftHandle handle;
int rank = 1; // --- 1D FFTs
int n[] = { M }; // --- Size of the Fourier transform
int istride = 1, ostride = 1; // --- Distance between two successive input/output elements
int idist = M, odist = M; // --- Distance between batches
int inembed[] = { 0 }; // --- Input size with pitch (ignored for 1D transforms)
int onembed[] = { 0 }; // --- Output size with pitch (ignored for 1D transforms)
int batch = N * Q; // --- Number of batched executions
cufftPlanMany(&handle, rank, n,
inembed, istride, idist,
onembed, ostride, odist, CUFFT_C2C, batch);
cufftExecC2C(handle, (cufftComplex*)(thrust::raw_pointer_cast(d_matrix.data())), (cufftComplex*)(thrust::raw_pointer_cast(d_matrix_out.data())), CUFFT_FORWARD);
cufftDestroy(handle);
for (int k=0; k<Q; k++)
for (int j=0; j<N; j++)
for (int i=0; i<M; i++) {
float2 temp = d_matrix_out[k*M*N+j*M+i];
printf("%i %i %i %f %f\n", i, j, k, temp.x, temp.y);
}
}
I guess, idist=nx*nz could also jump a whole plane and batch=nz would then cover one yx plane. The decision should be made according to whether nx or nz is larger.
This code works for multiply 2 matrix which it send matrix a and matrix b and them pointer to matrixMultiply method.
But I have trouble at the line "matrix12[i][j] += matrix1[i][k] * matrix2[k];"
double **matrixMultiply(double *matrix1,double *matrix2,int row1,int col1,int col2){
double **matrix12 = malloc(sizeof(double*)*row1);
for (int i=0; i<row1; i++){
matrix12[i] = malloc(sizeof(double*)col2);
for (int j=0; j<col2; j++){
matrix12[i][j] = 0.0;
for (int k=0; k<col1; k++){
matrix12[i][j] += matrix1[i][k] * matrix2[k]; //invalid operands to binary expression
}
}
}
return matrix12;
}
double *kmult = *matrixMultiply(a, b, 4, 4, 4,);
Ps.This code declare in ViewController.m
Are you sure Matrix1 is a 2-dimension matrix? It's declared at a 1-dimension matrix: double *matrix1 instead of "double **matrix1"
That's probably why the line matrix12[i][j] += matrix1[i][k] * matrix2[k]; doesn't work.
I want to calculate the product A^T*A ( A is 2000x1000 Matrix). Also i only want to solve the upper triangular Matrix. In the inner loop i have to solve the dot product of two vectors.
Now, here is the problem. Using cblas ddot() is not faster than calculating the dot product with a loop. How is this possible? (using Intel Core (TM)i7 CPU M620 #2,67GHz, 1,92GB RAM)
The problem is caused essentially by matrix size, not by ddot. Your matrices are so large that they do not fit in the cache memory. The solution is to rearrange the three nested loops such that as much as possible can be done with a line in cache, so reducing cache refreshes. A model implementation follows for both the ddot and an daxpy approach. On my computer the time consumption was about 15:1.
In other words: never, never, never program a matrix multiplication along the "row times column" scheme that we learned in school.
/*
Matrix product of A^T * A by two methods.
1) "Row times column" as we learned in school.
2) With rearranged loops such that need for cash refreshes is reduced
(this can be improved even more).
Compile: gcc -o aT_a aT_a.c -lgslcblas -lblas -lm
*/
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <cblas.h>
#define ROWS 2000
#define COLS 1000
static double a[ROWS][COLS];
static double c[COLS][COLS];
static void dot() {
int i, j;
double *ai, *bj;
ai = a[0];
for (i=0; i<COLS; i++) {
bj = a[0];
for (j=0; j<COLS; j++) {
c[i][j] = cblas_ddot(ROWS,ai,COLS,bj,COLS);
bj += 1;
}
ai += 1;
}
}
static void axpy() {
int i, j;
double *ci, *bj, aij;
for (i=0; i<COLS; i++) {
ci = c[i];
for (j=0; j<COLS; j++) ci[j] = 0.;
for (j=0; j<ROWS; j++) {
aij = a[j][i];
bj = a[j];
cblas_daxpy(COLS,aij,bj,1,ci,1);
}
}
}
int main(int argc, char** argv) {
clock_t t0, t1;
int i, j;
for (i=0; i<ROWS; ++i)
for (j=0; j<COLS; ++j)
a[i][j] = i+j;
t0 = clock();
dot();
t0 = clock();
printf("Time for DOT : %f sec.\n",(double)t0/CLOCKS_PER_SEC);
axpy();
t1 = clock();
printf("Time for AXPY: %f sec.\n",(double)(t1-t0)/CLOCKS_PER_SEC);
return 0;
}
The CBLAS dot product is effectively just a computation in slightly unrolled loop. The netlib Fortran is just this:
DO I = MP1,N,5
DTEMP = DTEMP + DX(I)*DY(I) + DX(I+1)*DY(I+1) +
$ DX(I+2)*DY(I+2) + DX(I+3)*DY(I+3) + DX(I+4)*DY(I+4)
END DO
ie. just a loop unrolled to a stride of 5.
If you must use a ddot style dot product for your operation, you might get a performance boost by re-writing your loop to use SSE2 intrinsics:
#include <emmintrin.h>
double ddotsse2(const double *x, const double *y, const int n)
{
double result[2];
int n2 = 2 * (n/2);
__m128d dtemp;
if ( (n % 2) == 0) {
dtemp = _mm_setzero_pd();
} else {
dtemp = _mm_set_sd(x[n] * y[n]);
}
for(int i=0; i<n2; i+=2) {
__m128d x1 = _mm_loadr_pd(x+i);
__m128d y1 = _mm_loadr_pd(y+i);
__m128d xy = _mm_mul_pd(x1, y1);
dtemp = _mm_add_pd(dtemp, xy);
}
_mm_store_pd(&result[0],dtemp);
return result[0] + result[1];
}
(not tested, never been compiled, buyer beware).
This may or may be faster than the standard BLAS implementation. You may also want to investigate whether further loop unrolling could improve performance.
If you're not using SSE2 intrinsics or using a data type that may not boost performance with them, you can try to transpose the matrix for an easy improvement in performance for larger matrix multiplications with cblas_?dot. Performing the matrix multiplication in blocks also helps.
void matMulDotProduct(int n, float *A, float* B, int a_size, int b_size, int a_row, int a_col, int b_row, int b_col, float *C) {
int i, j, k;
MKL_INT incx, incy;
incx = 1;
incy = b_size;
//copy out multiplying matrix from larger matrix
float *temp = (float*) malloc(n * n * sizeof(float));
for (i = 0; i < n; ++i) {
cblas_scopy(n, &B[(b_row * b_size) + b_col + i], incy, &temp[i * n], 1);
}
//transpose
mkl_simatcopy('R', 'T', n, n, 1.0, temp, 1, 1);
for (i = 0; i < n; i+= BLOCK_SIZE) {
for (j = 0; j < n; j++) {
for (k = 0; k < BLOCK_SIZE; ++k) {
C[((i + k) * n) + j] = cblas_sdot(n, &A[(a_row + i + k) * a_size + a_col], incx, &temp[n * j], 1);
}
}
}
free(temp);
}
On my machine, this code is about 1 order of magnitude faster than the the 3 loop code (but also 1 order of magnitude slower than cblas_?gemm call) for single precision floats and 2K by 2K matrices. (I'm using Intel MKL).