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I'm using a macro to compute the size of a bit-field. Would it be possible to use a template?
Code example:
#include <stdio.h>
/*!
#param reg a bit-fields structure
#param field name of one of the fields in 'reg'
*/
#define GET_FIELD_MAX(reg, field) ({ \
decltype(reg) r; \
r.field = ~0UL; \
r.field; })
struct A {
unsigned a : 3;
unsigned b : 4;
};
int main()
{
A x;
printf("b size = %d", GET_FIELD_MAX(x, b));
}
Output:
b size = 15
I'm building a CUDA kernel to compute the numerical N*N jacobian of a function, using finite differences; in the example I provided, it is the square function (each entry of the vector is squared). The host coded allocates in linear memory, while I'm using a 2-dimensional indexing in the kernel.
My issue is that I haven't found a way to sum on the diagonal of the matrices cudaMalloc'ed. My attempt has been to use the statement threadIdx.x == blockIdx.x as a condition for the diagonal, but instead it evaluates to true only for them both at 0.
Here is the kernel and EDIT: I posted the whole code as an answer, based on the suggestions in the comments (the main() is basically the same, while the kernel is not)
template <typename T>
__global__ void jacobian_kernel (
T * J,
const T t0,
const T tn,
const T h,
const T * u0,
const T * un,
const T * un_old)
{
T cgamma = 2 - sqrtf(2);
const unsigned int t = threadIdx.x;
const unsigned int b = blockIdx.x;
const unsigned int tid = t + b * blockDim.x;
/*__shared__*/ T temp_sx[BLOCK_SIZE][BLOCK_SIZE];
/*__shared__*/ T temp_dx[BLOCK_SIZE][BLOCK_SIZE];
__shared__ T sm_temp_du[BLOCK_SIZE];
T* temp_du = &sm_temp_du[0];
if (tid < N )
{
temp_sx[b][t] = un[t];
temp_dx[b][t] = un[t];
if ( t == b )
{
if ( tn == t0 )
{
temp_du[t] = u0[t]*0.001;
temp_sx[b][t] += temp_du[t]; //(*)
temp_dx[b][t] -= temp_du[t];
temp_sx[b][t] += ( abs( temp_sx[b][t] ) < 10e-6 ? 0.1 : 0 );
temp_dx[b][t] += ( abs( temp_dx[b][t] ) < 10e-6 ? 0.1 : 0 );
temp_sx[b][t] = ( temp_sx[b][t] == 0 ? 0.1 : temp_sx[b][t] );
temp_dx[b][t] = ( temp_dx[b][t] == 0 ? 0.1 : temp_dx[b][t] );
}
else
{
temp_du[t] = MAX( un[t] - un_old[t], 10e-6 );
temp_sx[b][t] += temp_du[t];
temp_dx[b][t] -= temp_du[t];
}
}
__syncthreads();
//J = f(tn, un + du)
d_func(tn, (temp_sx[b]), (temp_sx[b]), 1.f);
d_func(tn, (temp_dx[b]), (temp_dx[b]), 1.f);
__syncthreads();
J[tid] = (temp_sx[b][t] - temp_dx[b][t]) * powf((2 * temp_du[t]), -1);
//J[tid]*= - h*cgamma/2;
//J[tid]+= ( t == b ? 1 : 0);
//J[tid] = temp_J[tid];
}
}
The general procedure for computing the jacobian is
Copy un into every row of temp_sx and temp_dx
Compute du as a 0.01 magnitude from u0
Sum du to the diagonal of temp_sx, subtract du from the diagonal of temp_dx
Compute the square function on each entry of temp_sx and temp_dx
Subtract them and divide every entry by 2*du
This procedure can be summarized with (f(un + du*e_i) - f(un - du*e_i))/2*du.
My problem is to sum du to the diagonal of the matrices of temp_sx and temp_dx like I tried in (*). How can I achieve that?
EDIT: Now calling 1D blocks and threads; in fact, .y axis wasn't used at all in the kernel. I'm calling the kernel with a fixed amount of shared memory
Note that in int main() I'm calling the kernel with
#define REAL sizeof(float)
#define N 32
#define BLOCK_SIZE 16
#define NUM_BLOCKS ((N*N + BLOCK_SIZE - 1)/ BLOCK_SIZE)
...
dim3 dimGrid(NUM_BLOCKS,);
dim3 dimBlock(BLOCK_SIZE);
size_t shm_size = N*N*REAL;
jacobian_kernel <<< dimGrid, dimBlock, size_t shm_size >>> (...);
So that I attempt to deal with block-splitting the function calls. In the kernel to sum on the diagonal I used if(threadIdx.x == blockIdx.x){...}. Why isn't this correct? I'm asking it because while debugging and making the code print the statement, It only evaluates true if they both are 0. Thus du[0] is the only numerical value and the matrix becomes nan. Note that this approach worked with the first code I built, where instead I called the kernel with
jacobian_kernel <<< N, N >>> (...)
So that when threadIdx.x == blockIdx.x the element is on the diagonal. This approach doesn't fit anymore though, since now I need to deal with larger N (possibly larger than 1024, which is the maximum number of threads per block).
What statement should I put there that works even if the matrices are split into blocks and threads?
Let me know if I should share some other info.
Here is how I managed to solve my problem, based on the suggestion in the comments on the answer. The example is compilable, provided you put helper_cuda.h and helper_string.h in the same directory or you add -I directive to the CUDA examples include path, installed along with the CUDA toolkit. The relevant changes are only in the kernel; there's a minor change in the main() though, since I was calling double the resources to execute the kernel, but the .y axis of the grid of thread blocks wasn't even used at all, so it didn't generate any error.
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <math.h>
#include <assert.h>
#include <cuda.h>
#include <cuda_runtime.h>
#include "helper_cuda.h"
#include "helper_string.h"
#include <fstream>
#ifndef MAX
#define MAX(a,b) ((a > b) ? a : b)
#endif
#define REAL sizeof(float)
#define N 128
#define BLOCK_SIZE 128
#define NUM_BLOCKS ((N*N + BLOCK_SIZE - 1)/ BLOCK_SIZE)
template <typename T>
inline void printmatrix( T mat, int rows, int cols);
template <typename T>
__global__ void jacobian_kernel ( const T * A, T * J, const T t0, const T tn, const T h, const T * u0, const T * un, const T * un_old);
template<typename T>
__device__ void d_func(const T t, const T u[], T res[], const T h = 1);
template<typename T>
int main ()
{
float t0 = 0.; //float tn = 0.;
float h = 0.1;
float* u0 = (float*)malloc(REAL*N); for(int i = 0; i < N; ++i){u0[i] = i+1;}
float* un = (float*)malloc(REAL*N); memcpy(un, u0, REAL*N);
float* un_old = (float*)malloc(REAL*N); memcpy(un_old, u0, REAL*N);
float* J = (float*)malloc(REAL*N*N);
float* A = (float*)malloc(REAL*N*N); host_heat_matrix(A);
float *d_u0;
float *d_un;
float *d_un_old;
float *d_J;
float *d_A;
checkCudaErrors(cudaMalloc((void**)&d_u0, REAL*N)); //printf("1: %p\n", d_u0);
checkCudaErrors(cudaMalloc((void**)&d_un, REAL*N)); //printf("2: %p\n", d_un);
checkCudaErrors(cudaMalloc((void**)&d_un_old, REAL*N)); //printf("3: %p\n", d_un_old);
checkCudaErrors(cudaMalloc((void**)&d_J, REAL*N*N)); //printf("4: %p\n", d_J);
checkCudaErrors(cudaMalloc((void**)&d_A, REAL*N*N)); //printf("4: %p\n", d_J);
checkCudaErrors(cudaMemcpy(d_u0, u0, REAL*N, cudaMemcpyHostToDevice)); assert(d_u0 != NULL);
checkCudaErrors(cudaMemcpy(d_un, un, REAL*N, cudaMemcpyHostToDevice)); assert(d_un != NULL);
checkCudaErrors(cudaMemcpy(d_un_old, un_old, REAL*N, cudaMemcpyHostToDevice)); assert(d_un_old != NULL);
checkCudaErrors(cudaMemcpy(d_J, J, REAL*N*N, cudaMemcpyHostToDevice)); assert(d_J != NULL);
checkCudaErrors(cudaMemcpy(d_A, A, REAL*N*N, cudaMemcpyHostToDevice)); assert(d_A != NULL);
dim3 dimGrid(NUM_BLOCKS); std::cout << "NUM_BLOCKS \t" << dimGrid.x << "\n";
dim3 dimBlock(BLOCK_SIZE); std::cout << "BLOCK_SIZE \t" << dimBlock.x << "\n";
size_t shm_size = N*REAL; //std::cout << shm_size << "\n";
//HERE IS A RELEVANT CHANGE OF THE MAIN, SINCE I WAS CALLING
//THE KERNEL WITH A 2D GRID BUT WITHOUT USING THE .y AXIS,
//WHILE NOW THE GRID IS 1D
jacobian_kernel <<< dimGrid, dimBlock, shm_size >>> (d_A, d_J, t0, t0, h, d_u0, d_un, d_un_old);
checkCudaErrors(cudaMemcpy(J, d_J, REAL*N*N, cudaMemcpyDeviceToHost)); //printf("4: %p\n", d_J);
printmatrix( J, N, N);
checkCudaErrors(cudaDeviceReset());
free(u0);
free(un);
free(un_old);
free(J);
}
template <typename T>
__global__ void jacobian_kernel (
const T * A,
T * J,
const T t0,
const T tn,
const T h,
const T * u0,
const T * un,
const T * un_old)
{
T cgamma = 2 - sqrtf(2);
const unsigned int t = threadIdx.x;
const unsigned int b = blockIdx.x;
const unsigned int tid = t + b * blockDim.x;
/*__shared__*/ T temp_sx[BLOCK_SIZE][BLOCK_SIZE];
/*__shared__*/ T temp_dx[BLOCK_SIZE][BLOCK_SIZE];
__shared__ T sm_temp_du;
T* temp_du = &sm_temp_du;
//HERE IS A RELEVANT CHANGE (*)
if ( t < BLOCK_SIZE && b < NUM_BLOCKS )
{
temp_sx[b][t] = un[t]; //printf("temp_sx[%d] = %f\n", t,(temp_sx[b][t]));
temp_dx[b][t] = un[t];
//printf("t = %d, b = %d, t + b * blockDim.x = %d \n",t, b, tid);
//HERE IS A NOTE (**)
if ( t == b )
{
//printf("t = %d, b = %d \n",t, b);
if ( tn == t0 )
{
*temp_du = u0[t]*0.001;
temp_sx[b][t] += *temp_du;
temp_dx[b][t] -= *temp_du;
temp_sx[b][t] += ( abs( temp_sx[b][t] ) < 10e-6 ? 0.1 : 0 );
temp_dx[b][t] += ( abs( temp_dx[b][t] ) < 10e-6 ? 0.1 : 0 );
temp_sx[b][t] = ( temp_sx[b][t] == 0 ? 0.1 : temp_sx[b][t] );
temp_dx[b][t] = ( temp_dx[b][t] == 0 ? 0.1 : temp_dx[b][t] );
}
else
{
*temp_du = MAX( un[t] - un_old[t], 10e-6 );
temp_sx[b][t] += *temp_du;
temp_dx[b][t] -= *temp_du;
}
;
}
//printf("du[%d] %f\n", tid, (*temp_du));
__syncthreads();
//printf("temp_sx[%d][%d] = %f\n", b, t, temp_sx[b][t]);
//printf("temp_dx[%d][%d] = %f\n", b, t, temp_dx[b][t]);
//d_func(tn, (temp_sx[b]), (temp_sx[b]), 1.f);
//d_func(tn, (temp_dx[b]), (temp_dx[b]), 1.f);
matvec_dev( tn, A, (temp_sx[b]), (temp_sx[b]), N, N, 1.f );
matvec_dev( tn, A, (temp_dx[b]), (temp_dx[b]), N, N, 1.f );
__syncthreads();
//printf("temp_sx_later[%d][%d] = %f\n", b, t, (temp_sx[b][t]));
//printf("temp_sx_later[%d][%d] - temp_dx_later[%d][%d] = %f\n", b,t,b,t, (temp_sx[b][t] - temp_dx[b][t]) / 2 * *temp_du);
//if (t == b ) printf( "2du[%d]^-1 = %f\n",t, powf((2 * *temp_du), -1));
J[tid] = (temp_sx[b][t] - temp_dx[b][t]) / (2 * *temp_du);
}
}
template<typename T>
__device__ void d_func(const T t, const T u[], T res[], const T h )
{
__shared__ float temp_u;
temp_u = u[threadIdx.x];
res[threadIdx.x] = h*powf( (temp_u), 2);
}
template <typename T>
inline void printmatrix( T mat, int rows, int cols)
{
std::ofstream matrix_out;
matrix_out.open( "heat_matrix.txt", std::ofstream::out);
for( int i = 0; i < rows; i++)
{
for( int j = 0; j <cols; j++)
{
double next = mat[i + N*j];
matrix_out << ( (next >= 0) ? " " : "") << next << " ";
}
matrix_out << "\n";
}
}
The relevant change is on (*). Before I used if (tid < N) which has two downsides:
First, it is wrong, since it should be tid < N*N, as my data is 2D, while tid is a global index which tracks all the data.
Even if I wrote tid < N*N, since I'm splitting the function calls into blocks, the t < BLOCK_SIZE && b < NUM_BLOCKS seems clearer to me in how the indexing is arranged in the code.
Moreover, the statement t == b in (**) is actually the right one to operate on the diagonal elements of the matrix. The fact that it was evaluated true only on 0 was because of my error right above.
Thanks for the suggestions!
1, It is a pity that memset(void* dst, int value, size_t size) fools a lot of people when they first use this function! 2nd parameter "int value" should be "uchar value" to describe the real operation inside.
Don't misunderstand me, I am asking a memset-like function!
2, I know there are some c++ candy function like std::fill_n(my_array, array_length, constant_value);
even a pure c function in OS X: memset_pattern4(grid, &pattern, sizeof grid);
mentioned in a perfect thread Why is memset() incorrectly initializing int?.
So, is there a similar c function in runtime library of visual studio like memset_pattern4()?
3, for somebody asked why i wouldn't use a for-loop to set integer by integer. here is my answer: memset turns to a better performance when setting big trunk(10K?) of memory at least in x86.
http://www.gamedev.net/topic/472631-performance-of-memset/page-2 gives more discussion, although without a conclusion(I doubt there will be).
4, said function can be used to simplify counting sort by avoiding useless Fibonacci accumulation.
Original:
for (int i = 0; i < SRC_ARRY_SIZE; i++)
counter_arry[src_arry[i]]++;
for (int i = SRC_LOW_BOUND; i < SRC_HI_BOUND; i++)//forward fabnacci??
counter_arry[i+1] += counter_arry[i];
for (int i = 0; i < SRC_ARRY_SIZE; i++)
{
value = src_arry[i];
map = --counter_arry[value];//then counter down!
temp[map] = value;
}
Expected:
for (int i = 0; i < SRC_ARRY_SIZE; i++)
counter_arry[src_arry[i]]++;
for (int i = SRC_LOW_BOUND; i < SRC_HI_BOUND+1; i++)//forward fabnacci??
{
memset_4(cur_ptr,i, counter_arry[i]);
cur_ptr += counter_arry[i];
}
Thanks for your kindly review and reply!
Here's an implementation of memset_pattern4() that you might find useful. It's nothing like Darwin's SSE assembly language version, except that it has the same interface.
#include <string.h>
#include <stdint.h>
/*
A portable implementation of the Darwin memset_patternX() family of functions:
These are analogous to memset(), except that they fill memory with a replicated
pattern either 4, 8, or 16 bytes long. b points to a buffer of size len bytes
which is to be filled. The second parameter points to the pattern. If the
buffer length is not an even multiple of the pattern length, the last instance
of the pattern will be truncated. Neither the buffer nor the pattern pointer
need be aligned.
*/
/*
alignment utility macros stolen from Linux
see https://lkml.org/lkml/2006/11/25/2 for a discussion of why typeof() is used
*/
#if !_MSC_VER
#define __ALIGN_KERNEL(x, a) __ALIGN_KERNEL_MASK(x, (typeof(x))(a) - 1)
#define __ALIGN_KERNEL_MASK(x, mask) (((x) + (mask)) & ~(mask))
#define ALIGN(x, a) __ALIGN_KERNEL((x), (a))
#define __ALIGN_MASK(x, mask) __ALIGN_KERNEL_MASK((x), (mask))
#define PTR_ALIGN(p, a) ((typeof(p))ALIGN((uintptr_t)(p), (a)))
#define IS_ALIGNED(x, a) (((x) & ((typeof(x))(a) - 1)) == 0)
#define IS_PTR_ALIGNED(p, a) (IS_ALIGNED((uintptr_t)(p), (a)))
#else
/* MS friendly versions */
/* taken from the DDK's fltKernel.h header */
#define IS_ALIGNED(_pointer, _alignment) \
((((uintptr_t) (_pointer)) & ((_alignment) - 1)) == 0)
#define ROUND_TO_SIZE(_length, _alignment) \
((((uintptr_t)(_length)) + ((_alignment)-1)) & ~(uintptr_t) ((_alignment) - 1))
#define __ALIGN_KERNEL(x, a) ROUND_TO_SIZE( (x), (a))
#define ALIGN(x, a) __ALIGN_KERNEL((x), (a))
#define PTR_ALIGN(p, a) ALIGN((p), (a))
#define IS_PTR_ALIGNED(p, a) (IS_ALIGNED((uintptr_t)(p), (a)))
#endif
void nx_memset_pattern4(void *b, const void *pattern4, size_t len)
{
enum { pattern_len = 4 };
unsigned char* dst = (unsigned char*) b;
unsigned const char* src = (unsigned const char*) pattern4;
if (IS_PTR_ALIGNED( dst, pattern_len) && IS_PTR_ALIGNED( src, pattern_len)) {
/* handle aligned moves */
uint32_t val = *((uint32_t*)src);
uint32_t* word_dst = (uint32_t*) dst;
size_t word_count = len / pattern_len;
dst += word_count * pattern_len;
len -= word_count * pattern_len;
for (; word_count != 0; --word_count) {
*word_dst++ = val;
}
}
else {
while (pattern_len <= len) {
memcpy(dst, src, pattern_len);
dst += pattern_len;
len -= pattern_len;
}
}
memcpy( dst, src, len);
}
I'm working in C on a PC, trying to leverage as little C++ as possible, working with binary data stored in unsigned char format, although other formats are certainly possible if worthwhile. The goal is subtracting two signed integer values (which can be ints, signed ints, longs, signed longs, signed shorts, etc.) in binary without converting to other data formats. The raw data is just prepackaged as unsigned char, though, with the user basically knowing which of the signed integer formats should be used for reading (i.e. we know how many bytes to read at once). Even though data is stored as an unsigned char array, data are meant to be read signed as two's-complement integers.
One common way we're often taught in school is adding the negative. Negation, in turn, is often taught to be performed as flipping bits and adding 1 (0x1), resulting in two additions (perhaps a bad thing?); or, as other posts point out, flipping bits past the first zero starting from the MSB. I'm wondering if there is a more efficient way, that may not be easily described as a pen-and-paper operation, but works because of the way data is stored in bit format. Here are some prototypes I've written, which may not be the most efficient way, but which summarizes my progress so far based on textbook methodology.
The addends are passed by reference in case I have to manually extend them to balance their length. Any and all feedback will be appreciated! Thanks in advance for considering.
void SubtractByte(unsigned char* & a, unsigned int & aBytes,
unsigned char* & b, unsigned int & bBytes,
unsigned char* & diff, unsigned int & nBytes)
{
NegateByte(b, bBytes);
// a - b == a + (-b)
AddByte(a, aBytes, b, bBytes, diff, nBytes);
// Restore b to its original state so input remains intact
NegateByte(b, bBytes);
}
void AddByte(unsigned char* & a, unsigned int & aBytes,
unsigned char* & b, unsigned int & bBytes,
unsigned char* & sum, unsigned int & nBytes)
{
// Ensure that both of our addends have the same length in memory:
BalanceNumBytes(a, aBytes, b, bBytes, nBytes);
bool aSign = !((a[aBytes-1] >> 7) & 0x1);
bool bSign = !((b[bBytes-1] >> 7) & 0x1);
// Add bit-by-bit to keep track of carry bit:
unsigned int nBits = nBytes * BITS_PER_BYTE;
unsigned char carry = 0x0;
unsigned char result = 0x0;
unsigned char a1, b1;
// init sum
for (unsigned int j = 0; j < nBytes; ++j) {
for (unsigned int i = 0; i < BITS_PER_BYTE; ++i) {
a1 = ((a[j] >> i) & 0x1);
b1 = ((b[j] >> i) & 0x1);
AddBit(&a1, &b1, &carry, &result);
SetBit(sum, j, i, result==0x1);
}
}
// MSB and carry determine if we need to extend:
if (((aSign && bSign) && (carry != 0x0 || result != 0x0)) ||
((!aSign && !bSign) && (result == 0x0))) {
++nBytes;
sum = (unsigned char*)realloc(sum, nBytes);
sum[nBytes-1] = (carry == 0x0 ? 0x0 : 0xFF); //init
}
}
void FlipByte (unsigned char* n, unsigned int nBytes)
{
for (unsigned int i = 0; i < nBytes; ++i) {
n[i] = ~n[i];
}
}
void NegateByte (unsigned char* n, unsigned int nBytes)
{
// Flip each bit:
FlipByte(n, nBytes);
unsigned char* one = (unsigned char*)malloc(nBytes);
unsigned char* orig = (unsigned char*)malloc(nBytes);
one[0] = 0x1;
orig[0] = n[0];
for (unsigned int i = 1; i < nBytes; ++i) {
one[i] = 0x0;
orig[i] = n[i];
}
// Add binary representation of 1
AddByte(orig, nBytes, one, nBytes, n, nBytes);
free(one);
free(orig);
}
void AddBit(unsigned char* a, unsigned char* b, unsigned char* c,
unsigned char* result) {
*result = ((*a + *b + *c) & 0x1);
*c = (((*a + *b + *c) >> 1) & 0x1);
}
void SetBit(unsigned char* bytes, unsigned int byte, unsigned int bit,
bool val)
{
// shift desired bit into LSB position, and AND with 00000001
if (val) {
// OR with 00001000
bytes[byte] |= (0x01 << bit);
}
else{ // (!val), meaning we want to set to 0
// AND with 11110111
bytes[byte] &= ~(0x01 << bit);
}
}
void BalanceNumBytes (unsigned char* & a, unsigned int & aBytes,
unsigned char* & b, unsigned int & bBytes,
unsigned int & nBytes)
{
if (aBytes > bBytes) {
nBytes = aBytes;
b = (unsigned char*)realloc(b, nBytes);
bBytes = nBytes;
b[nBytes-1] = ((b[0] >> 7) & 0x1) ? 0xFF : 0x00;
} else if (bBytes > aBytes) {
nBytes = bBytes;
a = (unsigned char*)realloc(a, nBytes);
aBytes = nBytes;
a[nBytes-1] = ((a[0] >> 7) & 0x1) ? 0xFF : 0x00;
} else {
nBytes = aBytes;
}
}
The first thing to notice is that signed vs. unsigned doesn't matter to the generated bit pattern in two's complement. All that changes is the interpretation of the result.
The second thing to notice is that an addition has carried if the result is less than either input when done with unsigned arithmetic.
void AddByte(unsigned char* & a, unsigned int & aBytes,
unsigned char* & b, unsigned int & bBytes,
unsigned char* & sum, unsigned int & nBytes)
{
// Ensure that both of our addends have the same length in memory:
BalanceNumBytes(a, aBytes, b, bBytes, nBytes);
unsigned char carry = 0;
for (int j = 0; j < nbytes; ++j) { // need to reverse the loop for big-endian
result[j] = a[j] + b[j];
unsigned char newcarry = (result[j] < a[j] || (unsigned char)(result[j]+carry) < a[j]);
result[j] += carry;
carry = newcarry;
}
}
I have a list of N 64-bit integers whose bits represent small sets. Each integer has at most k bits set to 1. Given a bit mask, I would like to find the first element in the list that matches the mask, i.e. element & mask == element.
Example:
If my list is:
index abcdef
0 001100
1 001010
2 001000
3 000100
4 000010
5 000001
6 010000
7 100000
8 000000
and my mask is 111000, the first element matching the mask is at index 2.
Method 1:
Linear search through the entire list. This takes O(N) time and O(1) space.
Method 2:
Precompute a tree of all possible masks, and at each node keep the answer for that mask. This takes O(1) time for the query, but takes O(2^64) space.
Question:
How can I find the first element matching the mask faster than O(N), while still using a reasonable amount of space? I can afford to spend polynomial time in precomputation, because there will be a lot of queries. The key is that k is small. In my application, k <= 5 and N is in the thousands. The mask has many 1s; you can assume that it is drawn uniformly from the space of 64-bit integers.
Update:
Here is an example data set and a simple benchmark program that runs on Linux: http://up.thirld.com/binmask.tar.gz. For large.in, N=3779 and k=3. The first line is N, followed by N unsigned 64-bit ints representing the elements. Compile with make. Run with ./benchmark.e >large.out to create the true output, which you can then diff against. (Masks are generated randomly, but the random seed is fixed.) Then replace the find_first() function with your implementation.
The simple linear search is much faster than I expected. This is because k is small, and so for a random mask, a match is found very quickly on average.
A suffix tree (on bits) will do the trick, with the original priority at the leaf nodes:
000000 -> 8
1 -> 5
10 -> 4
100 -> 3
1000 -> 2
10 -> 1
100 -> 0
10000 -> 6
100000 -> 7
where if the bit is set in the mask, you search both arms, and if not, you search only the 0 arm; your answer is the minimum number you encounter at a leaf node.
You can improve this (marginally) by traversing the bits not in order but by maximum discriminability; in your example, note that 3 elements have bit 2 set, so you would create
2:0 0:0 1:0 3:0 4:0 5:0 -> 8
5:1 -> 5
4:1 5:0 -> 4
3:1 4:0 5:0 -> 3
1:1 3:0 4:0 5:0 -> 6
0:1 1:0 3:0 4:0 5:0 -> 7
2:1 0:0 1:0 3:0 4:0 5:0 -> 2
4:1 5:0 -> 1
3:1 4:0 5:0 -> 0
In your example mask this doesn't help (since you have to traverse both the bit2==0 and bit2==1 sides since your mask is set in bit 2), but on average it will improve the results (but at a cost of setup and more complex data structure). If some bits are much more likely to be set than others, this could be a huge win. If they're pretty close to random within the element list, then this doesn't help at all.
If you're stuck with essentially random bits set, you should get about (1-5/64)^32 benefit from the suffix tree approach on average (13x speedup), which might be better than the difference in efficiency due to using more complex operations (but don't count on it--bit masks are fast). If you have a nonrandom distribution of bits in your list, then you could do almost arbitrarily well.
This is the bitwise Kd-tree. It typically needs less than 64 visits per lookup operation. Currently, the selection of the bit (dimension) to pivot on is random.
#include <limits.h>
#include <time.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
typedef unsigned long long Thing;
typedef unsigned long Number;
unsigned thing_ffs(Thing mask);
Thing rand_mask(unsigned bitcnt);
#define WANT_RANDOM 31
#define WANT_BITS 3
#define BITSPERTHING (CHAR_BIT*sizeof(Thing))
#define NONUMBER ((Number)-1)
struct node {
Thing value;
Number num;
Number nul;
Number one;
char pivot;
} *nodes = NULL;
unsigned nodecount=0;
unsigned itercount=0;
struct node * nodes_read( unsigned *sizp, char *filename);
Number *find_ptr_to_insert(Number *ptr, Thing value, Thing mask);
unsigned grab_matches(Number *result, Number num, Thing mask);
void initialise_stuff(void);
int main (int argc, char **argv)
{
Thing mask;
Number num;
unsigned idx;
srand (time(NULL));
nodes = nodes_read( &nodecount, argv[1]);
fprintf( stdout, "Nodecount=%u\n", nodecount );
initialise_stuff();
#if WANT_RANDOM
mask = nodes[nodecount/2].value | nodes[nodecount/3].value ;
#else
mask = 0x38;
#endif
fprintf( stdout, "\n#### Search mask=%llx\n", (unsigned long long) mask );
itercount = 0;
num = NONUMBER;
idx = grab_matches(&num,0, mask);
fprintf( stdout, "Itercount=%u\n", itercount );
fprintf(stdout, "KdTree search %16llx\n", (unsigned long long) mask );
fprintf(stdout, "Count=%u Result:\n", idx);
idx = num;
if (idx >= nodecount) idx = nodecount-1;
fprintf( stdout, "num=%4u Value=%16llx\n"
,(unsigned) nodes[idx].num
,(unsigned long long) nodes[idx].value
);
fprintf( stdout, "\nLinear search %16llx\n", (unsigned long long) mask );
for (idx = 0; idx < nodecount; idx++) {
if ((nodes[idx].value & mask) == nodes[idx].value) break;
}
fprintf(stdout, "Cnt=%u\n", idx);
if (idx >= nodecount) idx = nodecount-1;
fprintf(stdout, "Num=%4u Value=%16llx\n"
, (unsigned) nodes[idx].num
, (unsigned long long) nodes[idx].value );
return 0;
}
void initialise_stuff(void)
{
unsigned num;
Number root, *ptr;
root = 0;
for (num=0; num < nodecount; num++) {
nodes[num].num = num;
nodes[num].one = NONUMBER;
nodes[num].nul = NONUMBER;
nodes[num].pivot = -1;
}
nodes[num-1].value = 0; /* last node is guaranteed to match anything */
root = 0;
for (num=1; num < nodecount; num++) {
ptr = find_ptr_to_insert (&root, nodes[num].value, 0ull );
if (*ptr == NONUMBER) *ptr = num;
else fprintf(stderr, "Found %u for %u\n"
, (unsigned)*ptr, (unsigned) num );
}
}
Thing rand_mask(unsigned bitcnt)
{struct node * nodes_read( unsigned *sizp, char *filename)
{
struct node *ptr;
unsigned size,used;
FILE *fp;
if (!filename) {
size = (WANT_RANDOM+0) ? WANT_RANDOM : 9;
ptr = malloc (size * sizeof *ptr);
#if (!WANT_RANDOM)
ptr[0].value = 0x0c;
ptr[1].value = 0x0a;
ptr[2].value = 0x08;
ptr[3].value = 0x04;
ptr[4].value = 0x02;
ptr[5].value = 0x01;
ptr[6].value = 0x10;
ptr[7].value = 0x20;
ptr[8].value = 0x00;
#else
for (used=0; used < size; used++) {
ptr[used].value = rand_mask(WANT_BITS);
}
#endif /* WANT_RANDOM */
*sizp = size;
return ptr;
}
fp = fopen( filename, "r" );
if (!fp) return NULL;
fscanf(fp,"%u\n", &size );
fprintf(stderr, "Size=%u\n", size);
ptr = malloc (size * sizeof *ptr);
for (used = 0; used < size; used++) {
fscanf(fp,"%llu\n", &ptr[used].value );
}
fclose( fp );
*sizp = used;
return ptr;
}
Thing value = 0;
unsigned bit, cnt;
for (cnt=0; cnt < bitcnt; cnt++) {
bit = 54321*rand();
bit %= BITSPERTHING;
value |= 1ull << bit;
}
return value;
}
Number *find_ptr_to_insert(Number *ptr, Thing value, Thing done)
{
Number num=NONUMBER;
while ( *ptr != NONUMBER) {
Thing wrong;
num = *ptr;
wrong = (nodes[num].value ^ value) & ~done;
if (nodes[num].pivot < 0) { /* This node is terminal */
/* choose one of the wrong bits for a pivot .
** For this bit (nodevalue==1 && searchmask==0 )
*/
if (!wrong) wrong = ~done ;
nodes[num].pivot = thing_ffs( wrong );
}
ptr = (wrong & 1ull << nodes[num].pivot) ? &nodes[num].nul : &nodes[num].one;
/* Once this bit has been tested, it can be masked off. */
done |= 1ull << nodes[num].pivot ;
}
return ptr;
}
unsigned grab_matches(Number *result, Number num, Thing mask)
{
Thing wrong;
unsigned count;
for (count=0; num < *result; ) {
itercount++;
wrong = nodes[num].value & ~mask;
if (!wrong) { /* we have a match */
if (num < *result) { *result = num; count++; }
/* This is cheap pruning: the break will omit both subtrees from the results.
** But because we already have a result, and the subtrees have higher numbers
** than our current num, we can ignore them. */
break;
}
if (nodes[num].pivot < 0) { /* This node is terminal */
break;
}
if (mask & 1ull << nodes[num].pivot) {
/* avoid recursion if there is only one non-empty subtree */
if (nodes[num].nul >= *result) { num = nodes[num].one; continue; }
if (nodes[num].one >= *result) { num = nodes[num].nul; continue; }
count += grab_matches(result, nodes[num].nul, mask);
count += grab_matches(result, nodes[num].one, mask);
break;
}
mask |= 1ull << nodes[num].pivot;
num = (wrong & 1ull << nodes[num].pivot) ? nodes[num].nul : nodes[num].one;
}
return count;
}
unsigned thing_ffs(Thing mask)
{
unsigned bit;
#if 1
if (!mask) return (unsigned)-1;
for ( bit=random() % BITSPERTHING; 1 ; bit += 5, bit %= BITSPERTHING) {
if (mask & 1ull << bit ) return bit;
}
#elif 0
for (bit =0; bit < BITSPERTHING; bit++ ) {
if (mask & 1ull <<bit) return bit;
}
#else
mask &= (mask-1); // Kernighan-trick
for (bit =0; bit < BITSPERTHING; bit++ ) {
mask >>=1;
if (!mask) return bit;
}
#endif
return 0xffffffff;
}
struct node * nodes_read( unsigned *sizp, char *filename)
{
struct node *ptr;
unsigned size,used;
FILE *fp;
if (!filename) {
size = (WANT_RANDOM+0) ? WANT_RANDOM : 9;
ptr = malloc (size * sizeof *ptr);
#if (!WANT_RANDOM)
ptr[0].value = 0x0c;
ptr[1].value = 0x0a;
ptr[2].value = 0x08;
ptr[3].value = 0x04;
ptr[4].value = 0x02;
ptr[5].value = 0x01;
ptr[6].value = 0x10;
ptr[7].value = 0x20;
ptr[8].value = 0x00;
#else
for (used=0; used < size; used++) {
ptr[used].value = rand_mask(WANT_BITS);
}
#endif /* WANT_RANDOM */
*sizp = size;
return ptr;
}
fp = fopen( filename, "r" );
if (!fp) return NULL;
fscanf(fp,"%u\n", &size );
fprintf(stderr, "Size=%u\n", size);
ptr = malloc (size * sizeof *ptr);
for (used = 0; used < size; used++) {
fscanf(fp,"%llu\n", &ptr[used].value );
}
fclose( fp );
*sizp = used;
return ptr;
}
UPDATE:
I experimented a bit with the pivot-selection, favouring bits with the highest discriminatory value ("information content"). This involves:
making a histogram of the usage of bits (can be done while initialising)
while building the tree: choosing the one with frequency closest to 1/2 in the remaining subtrees.
The result: the random pivot selection performed better.
Construct a a binary tree as follows:
Every level corresponds to a bit
It corresponding bit is on go right, otherwise left
This way insert every number in the database.
Now, for searching: if the corresponding bit in the mask is 1, traverse both children. If it is 0, traverse only the left node. Essentially keep traversing the tree until you hit the leaf node (BTW, 0 is a hit for every mask!).
This tree will have O(N) space requirements.
Eg of tree for 1 (001), 2(010) and 5 (101)
root
/ \
0 1
/ \ |
0 1 0
| | |
1 0 1
(1) (2) (5)
With precomputed bitmasks. Formally is is still O(N), since the and-mask operations are O(N). The final pass is also O(N), because it needs to find the lowest bit set, but that could be sped up, too.
#include <limits.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
/* For demonstration purposes.
** In reality, this should be an unsigned long long */
typedef unsigned char Thing;
#define BITSPERTHING (CHAR_BIT*sizeof (Thing))
#define COUNTOF(a) (sizeof a / sizeof a[0])
Thing data[] =
/****** index abcdef */
{ 0x0c /* 0 001100 */
, 0x0a /* 1 001010 */
, 0x08 /* 2 001000 */
, 0x04 /* 3 000100 */
, 0x02 /* 4 000010 */
, 0x01 /* 5 000001 */
, 0x10 /* 6 010000 */
, 0x20 /* 7 100000 */
, 0x00 /* 8 000000 */
};
/* Note: this is for demonstration purposes.
** Normally, one should choose a machine wide unsigned int
** for bitmask arrays.
*/
struct bitmap {
char data[ 1+COUNTOF (data)/ CHAR_BIT ];
} nulmaps [ BITSPERTHING ];
#define BITSET(a,i) (a)[(i) / CHAR_BIT ] |= (1u << ((i)%CHAR_BIT) )
#define BITTEST(a,i) ((a)[(i) / CHAR_BIT ] & (1u << ((i)%CHAR_BIT) ))
void init_tabs(void);
void map_empty(struct bitmap *dst);
void map_full(struct bitmap *dst);
void map_and2(struct bitmap *dst, struct bitmap *src);
int main (void)
{
Thing mask;
struct bitmap result;
unsigned ibit;
mask = 0x38;
init_tabs();
map_full(&result);
for (ibit = 0; ibit < BITSPERTHING; ibit++) {
/* bit in mask is 1, so bit at this position is in fact a don't care */
if (mask & (1u <<ibit)) continue;
/* bit in mask is 0, so we can only select items with a 0 at this bitpos */
map_and2(&result, &nulmaps[ibit] );
}
/* This is not the fastest way to find the lowest 1 bit */
for (ibit = 0; ibit < COUNTOF (data); ibit++) {
if (!BITTEST(result.data, ibit) ) continue;
fprintf(stdout, " %u", ibit);
}
fprintf( stdout, "\n" );
return 0;
}
void init_tabs(void)
{
unsigned ibit, ithing;
/* 1 bits in data that dont overlap with 1 bits in the searchmask are showstoppers.
** So, for each bitpos, we precompute a bitmask of all *entrynumbers* from data[], that contain 0 in bitpos.
*/
memset(nulmaps, 0 , sizeof nulmaps);
for (ithing=0; ithing < COUNTOF(data); ithing++) {
for (ibit=0; ibit < BITSPERTHING; ibit++) {
if ( data[ithing] & (1u << ibit) ) continue;
BITSET(nulmaps[ibit].data, ithing);
}
}
}
/* Logical And of two bitmask arrays; simular to dst &= src */
void map_and2(struct bitmap *dst, struct bitmap *src)
{
unsigned idx;
for (idx = 0; idx < COUNTOF(dst->data); idx++) {
dst->data[idx] &= src->data[idx] ;
}
}
void map_empty(struct bitmap *dst)
{
memset(dst->data, 0 , sizeof dst->data);
}
void map_full(struct bitmap *dst)
{
unsigned idx;
/* NOTE this loop sets too many bits to the left of COUNTOF(data) */
for (idx = 0; idx < COUNTOF(dst->data); idx++) {
dst->data[idx] = ~0;
}
}