I've implemented a signal processing FFT algorithm in Python using np.fft (too easy). Now I'm working on doing this in C using an integer algorithm. After some research, I found that one of the most popular integer FFT libraries in C on the internet is the one by Roberts, Slaney, and Bouras which can be found in many places including the second entry here.fft,
There are a few concepts I don't understand and am hoping for guidance.
Specifically, the example script included in the library linked above separates the input signal into two bins, real and imaginary, by putting all of the even indexes of the signal in the first half and the odd indexes in the second half of the signal.
for (i=0; i<N; i++){
x[i] = AMPLITUDE*cos(i*FREQUENCY*(2*3.1415926535)/N);
if (i & 0x01) // only odd index
fx[(N+i)>>1] = x[i]; // N+i >> 1 is len(input)+i/2
else // only even index
fx[i>>1] = x[i];
}
fix_fftr(fx, log2N, 0);
The signal array has not changed length but now contains two of almost the same signal. Then FFT driver function (fix_fftr) takes the entire input signal as an argument and does the exact same thing
if (inverse)
scale = fix_fft(fr, fi, m-1, inverse);
for (int i=1; i<n; i+=2) {
tt = f[n+i-1]; // even index
f[n+i-1] = f[i]; // odd index into the second half
f[i] = tt; // even index into the first half
}
if (!inverse)
scale = fix_fft(fr, fi, m-1, inverse);
return scale;
whats the reason for this?
The first part is computing the twiddle factors, which are constants for a given length FFT and independent of the data.
The second part appears to be part of the data shuffling based on recursive bit-reversed addressing, which is a component within an in-place FFT.
Related
I want to generate all permutations of 32 card deck, I represent cards as numbers 0-7, so I don´t care about color of the card. The game is very simple (divide deck into two gropus, compare two cards, add both cards to group of bigger card). I have already code this part of game, but deck is now generating randomly, and I want to look to all possibilities of cards, and make some statistics about it. How can I code this card generating? I totaly don´t know, how to code it.
Because I was just studying Aaron Williams 2009 paper "Loopless Generation of Multiset Permutations by Prefix Shifts", I'll contribute a version of his algorithm, which precisely solves this problem. I believe it to be faster than the standard C++ next_permutation which is usually cited for this problem, because it doesn't rely on searching the input vector for the pivot point. But more extensive benchmarking would be required to produce a definitive answer; it is quite possible that it ends up moving more data around.
Williams' implementation of the algorithm avoids data movement by storing the permutation in a linked list, which allows the "prefix shift" (rotate a prefix of the vector by one position) to be implemented by just modifying two next pointers. That makes the algorithm loopless.
My version here differs in a couple of ways.
First, it uses an ordinary array to store the values, which means that the shift does require a loop. On the other hand, it avoids having to implement a linked-list datatype, and many operations are faster on arrays.
Second, it uses suffix shifts rather than prefix shifts; in effect, it produces the reverse of each permutation (compared with Williams' implementation). I did that because it simplifies the description of the starting condition.
Finally, it just does one permutation step. One of the great things about Williams' algorithm is that the state of the permutation sequence can be encapsulated in a single index value (as well as the permutation itself, of course). This implementation returns the state to be provided to the next call. (Since the state variable will be 0 at the end, the return value doubles as a termination indicator.)
Here's the code:
/* Do a single permutation of v in reverse coolex order, using
* a modification of Aaron Williams' loopless shift prefix algorithm.
* v must have length n. It may have repeated elements; the permutations
* generated will be unique.
* For the first call, v must be sorted into non-descending order and the
* third parameter must be 1. For subsequent calls, the third parameter must
* be the return value of the previous call. When the return value is 0,
* all permutations have been generated.
*/
unsigned multipermute_step(int* v, unsigned n, unsigned state) {
int old_end = v[n - 1];
unsigned pivot = state < 2 || v[state - 2] > v[state] ? state - 1 : state - 2;
int new_end = v[pivot];
for (; pivot < n - 1; ++pivot) v[pivot] = v[pivot + 1];
v[pivot] = new_end;
return new_end < old_end ? n - 1 : state - 1;
}
In case that comment was unclear, you could use the following to produce all shuffles of a deck of 4*k cards without regard to suit:
unsigned n = 4 * k;
int v[n];
for (unsigned i = 0; i < k; ++i)
for (unsigned j = 0; j < 4; ++j)
v[4 * i + j] = i;
unsigned state = 1;
do {
/* process the permutation */
} while ((state = multipermute_step(v, n, state);
Actually trying to do that for k == 8 will take a while, since there are 32!/(4!)8 possible shuffles. That's about 2.39*1024. But I did do all the shuffles of decks of 16 cards in 0.3 seconds, and I estimate that I could have done 20 cards in half an hour.
I've got an assignment - compare 2 matrix multiplications - in the default way, and multiplication after transposition of second matrix, we should point the difference which method is faster. I've written something like this below, but time and time2 are nearly equal to each other. In one case the first method is faster, I run the multiplication with the same size of matrix, and in another one the second method is faster. Is something done wrong? Should I change something in my code?
clock_t start = clock();
int sum;
for(int i=0; i<size; ++i) {
for(int j=0; j<size; ++j) {
sum = 0;
for(int k=0; k<size; ++k) {
sum = sum + (m1[i][k] * m2[k][j]);
}
score[i][j] = sum;
}
}
clock_t end = clock();
double time = (end-start)/(double)CLOCKS_PER_SEC;
for(int i=0; i<size; ++i) {
for(int j=0; j<size; ++j) {
int temp = m2[i][j];
m2[i][j] = m2[j][i];
m2[j][i] = temp;
}
}
clock_t start2 = clock();
int sum2;
for(int i=0; i<size; ++i) {
for(int j=0; j<size; ++j) {
sum2 = 0;
for(int k=0; k<size; ++k) {
sum2 = sum2 + (m1[k][i] * m2[k][j]);
}
score[i][j] = sum2;
}
}
clock_t end2 = clock();
double time2 = (end2-start2)/(double)CLOCKS_PER_SEC;
You have multiple severe issues with your code and/or your understanding. Let me try to explain.
Matrix multiplication is bottlenecked by the rate at which the processor can load and store the values to memory. Most current architectures use cache to help with this. Data is moved from memory to cache and from cache to memory in blocks. To maximize the benefit of caching, you want to make sure you will use all the data in that block. To do that, you make sure you access the data sequentially in memory.
In C, multi-dimensional arrays are specified in row-major order. It means that the rightmost index is consecutive in memory; i.e. that a[i][k] and a[i][k+1] are consecutive in memory.
Depending on the architecture, the time it takes for the processor to wait (and do nothing) for the data to be moved from RAM to cache (and vice versa), may or may not be included in the CPU time (that e.g. clock() measures, albeit at a very poor resolution). For this kind of measurement ("microbenchmark"), it is much better to measure and report both CPU and real (or wall clock) time used; especially so if the microbenchmark is run on different machines, to get a better idea of the practical impact of the change.
There will be a lot of variation, so typically, you measure the time taken by a few hundred repeats (each repeat possibly making more than one operation; enough to be easily measured), storing the duration of each, and report their median. Why median, and not minimum, maximum, average? Because there will always be occasional glitches (unreasonable measurement due to an external event, or something), which typically yield a much higher value than normal; this makes the maximum uninteresting, and skews the average (mean) unless removed. The minimum is typically an over-optimistic case, where everything just happened to go perfectly; that rarely occurs in practice, so is only a curiosity, not of practical interest. The median time, on the other hand, gives you a practical measurement: you can expect 50% of all runs of your test case to take no more than the median time measured.
On POSIXy systems (Linux, Mac, BSDs), you should use clock_gettime() to measure the time. The struct timespec format has nanosecond precision (1 second = 1,000,000,000 nanoseconds), but resolution may be smaller (i.e., the clocks change by more than 1 nanosecond, whenever they change). I personally use
#define _POSIX_C_SOURCE 200809L
#include <time.h>
static struct timespec cpu_start, wall_start;
double cpu_seconds, wall_seconds;
void timing_start(void)
{
clock_gettime(CLOCK_REALTIME, &wall_start);
clock_gettime(CLOCK_THREAD_CPUTIME_ID, &cpu_start);
}
void timing_stop(void)
{
struct timespec cpu_end, wall_end;
clock_gettime(CLOCK_REALTIME, &wall_end);
clock_gettime(CLOCK_THREAD_CPUTIME_ID, &cpu_end);
wall_seconds = (double)(wall_end.tv_sec - wall_start.tv_sec)
+ (double)(wall_end.tv_nsec - wall_start.tv_nsec) / 1000000000.0;
cpu_seconds = (double)(cpu_end.tv_sec - cpu_start.tv_sec)
+ (double)(cpu_end.tv_nsec - cpu_start.tv_nsec) / 1000000000.0;
}
You call timing_start() before the operation, and timing_stop() after the operation; then, cpu_seconds contains the amount of CPU time taken and wall_seconds the real wall clock time taken (both in seconds, use e.g. %.9f to print all meaningful decimals).
The above won't work on Windows, because Microsoft does not want your C code to be portable to other systems. It prefers to develop their own "standard" instead. (Those C11 "safe" _s() I/O function variants are a stupid sham, compared to e.g. POSIX getline(), or the state of wide character support on all systems except Windows.)
Matrix multiplication is
c[r][c] = a[r][0] * b[0][c]
+ a[r][1] * b[1][c]
: :
+ a[r][L] * b[L][c]
where a has L+1 columns, and b has L+1 rows.
In order to make the summation loop use consecutive elements, we need to transpose b. If B[c][r] = b[r][c], then
c[r][c] = a[r][0] * B[c][0]
+ a[r][1] * B[c][1]
: :
+ a[r][L] * B[c][L]
Note that it suffices that a and B are consecutive in memory, but separate (possibly "far" away from each other), for the processor to utilize cache efficiently in such cases.
OP uses a simple loop, similar to the following pseudocode, to transpose b:
For r in rows:
For c in columns:
temporary = b[r][c]
b[r][c] = b[c][r]
b[c][r] = temporary
End For
End For
The problem above is that each element participates in a swap twice. For example, if b has 10 rows and columns, r = 3, c = 5 swaps b[3][5] and b[5][3], but then later, r = 5, c = 3 swaps b[5][3] and b[3][5] again! Essentially, the double loop ends up restoring the matrix to the original order; it does not do a transpose.
Consider the following entries and the actual transpose:
b[0][0] b[0][1] b[0][2] b[0][0] b[1][0] b[2][0]
b[1][0] b[1][1] b[1][2] ⇔ b[0][1] b[1][1] b[2][1]
b[2][0] b[2][1] b[2][2] b[0][2] b[1][2] b[2][2]
The diagonal entries are not swapped. You only need to do the swap in the upper triangular portion (where c > r) or in the lower triangular portion (where r > c), to swap all entries, because each swap swaps one entry from the upper triangular to the lower triangular, and vice versa.
So, to recap:
Is something done wrong?
Yes. Your transpose does nothing. You haven't understood the reason why one would want to transpose the second matrix. Your time measurement relies on a low-precision CPU time, which may not reflect the time taken by moving data between RAM and CPU cache. In the second test case, with m2 "transposed" (except it isn't, because you swap each element pair twice, returning them back to the way they were), your innermost loop is over the leftmost array index, which means it calculates the wrong result. (Moreover, because consecutive iterations of the innermost loop accesses items far from each other in memory, it is anti-optimized: it uses the pattern that is worst in terms of speed.)
All the above may sound harsh, but it isn't intended to be, at all. I do not know you, and I am not trying to evaluate you; I am only pointing out the errors in this particular answer, in your current understanding, and only in the hopes that it helps you, and anyone else encountering this question in similar circumstances, to learn.
Goal: Implement the diagram shown below in OpenCL. The main thing needed from the OpenCl kernel is to multiply the coefficient array and temp array and then accumilate all those values into one at the end. (That is probably the most time intensive operation, parallelism would be really helpful here).
I am using a helper function for the kernel that does the multiplication and addition (I am hoping this function will be parallel as well).
Description of the picture:
One at a time, the values are passed into the array (temp array) which is the same size as the coefficient array. Now every time a single value is passed into this array, the temp array is multiplied with the coefficient array in parallel and the values of each index are then concatenated into one single element. This will continue until the input array reaches it's final element.
What happens with my code?
For 60 elements from the input, it takes over 8000 ms!! and I have a total of 1.2 million inputs that still have to be passed in. I know for a fact that there is a way better solution to do what I am attempting. Here is my code below.
Here are some things that I know are wrong with he code for sure. When I try to multiply the coefficient values with the temp array, it crashes. This is because of the global_id. All I want this line to do is simply multiply the two arrays in parallel.
I tried to figure out why it was taking so long to do the FIFO function, so I started commenting lines out. I first started by commenting everything except the first for loop of the FIFO function. As a result this took 50 ms. Then when I uncommented the next loop, it jumped to 8000ms. So the delay would have to do with the transfer of data.
Is there a register shift that I could use in OpenCl? Perhaps use some logical shifting method for integer arrays? (I know there is a '>>' operator).
float constant temp[58];
float constant tempArrayForShift[58];
float constant multipliedResult[58];
float fifo(float inputValue, float *coefficients, int sizeOfCoeff) {
//take array of 58 elements (or same size as number of coefficients)
//shift all elements to the right one
//bring next element into index 0 from input
//multiply the coefficient array with the array thats the same size of coefficients and accumilate
//store into one output value of the output array
//repeat till input array has reached the end
int globalId = get_global_id(0);
float output = 0.0f;
//Shift everything down from 1 to 57
//takes about 50ms here
for(int i=1; i<58; i++){
tempArrayForShift[i] = temp[i];
}
//Input the new value passed from main kernel. Rest of values were shifted over so element is written at index 0.
tempArrayForShift[0] = inputValue;
//Takes about 8000ms with this loop included
//Write values back into temp array
for(int i=0; i<58; i++){
temp[i] = tempArrayForShift[i];
}
//all 58 elements of the coefficient array and temp array are multiplied at the same time and stored in a new array
//I am 100% sure this line is crashing the program.
//multipliedResult[globalId] = coefficients[globalId] * temp[globalId];
//Sum the temp array with each other. Temp array consists of coefficients*fifo buffer
for (int i = 0; i < 58; i ++) {
// output = multipliedResult[i] + output;
}
//Returned summed value of temp array
return output;
}
__kernel void lowpass(__global float *Array, __global float *coefficients, __global float *Output) {
//Initialize the temporary array values to 0
for (int i = 0; i < 58; i ++) {
temp[i] = 0;
tempArrayForShift[i] = 0;
multipliedResult[i] = 0;
}
//fifo adds one element in and calls the fifo function. ALL I NEED TO DO IS SEND ONE VALUE AT A TIME HERE.
for (int i = 0; i < 60; i ++) {
Output[i] = fifo(Array[i], coefficients, 58);
}
}
I have had this problem with OpenCl for a long time. I am not sure how to implement parallel and sequential instructions together.
Another alternative I was thinking about
In the main cpp file, I was thinking of implementing the fifo buffer there and having the kernel do the multiplication and addition. But this would mean I would have to call the kernel 1000+ times in a loop. Would this be the better solution? Or would it just be completely inefficient.
To get good performance out of GPU, you need to parallelize your work to many threads. In your code you are just using a single thread and a GPU is very slow per thread but can be very fast, if many threads are running at the same time. In this case you can use a single thread for each output value. You do not actually need to shift values through a array: For every output value a window of 58 values is considered, you can just grab these values from memory, multiply them with the coefficients and write back the result.
A simple implementation would be (launch with as many threads as output values):
__kernel void lowpass(__global float *Array, __global float *coefficients, __global float *Output)
{
int globalId = get_global_id(0);
float sum=0.0f;
for (int i=0; i< 58; i++)
{
float tmp=0;
if (globalId+i > 56)
{
tmp=Array[i+globalId-57]*coefficient[57-i];
}
sum += tmp;
}
output[globalId]=sum;
}
This is not perfect, as the memory access patterns it generates are not optimal for GPUs. The Cache will likely help a bit, but there is clearly a lot of room for optimization, as the values are reused several times. The operation you are trying to perform is called convolution (1D). NVidia has an 2D example called oclConvolutionSeparable in their GPU Computing SDK, that shows an optimized version. You adapt use their convolutionRows kernel for a 1D convolution.
Here's another kernel you can try out. There are a lot of synchronization points (barriers), but this should perform fairly well. The 65-item work group is not very optimal.
the steps:
init local values to 0
copy coefficients to local variable
looping over the output elements to compute:
shift existing elements (work items > 0 only)
copy new element (work item 0 only)
compute dot product
5a. multiplication - one per work item
5b. reduction loop to compute sum
copy dot product to output (WI 0 only)
final barrier
the code:
__kernel void lowpass(__global float *Array, __constant float *coefficients, __global float *Output, __local float *localArray, __local float *localSums){
int globalId = get_global_id(0);
int localId = get_local_id(0);
int localSize = get_local_size(0);
//1 init local values to 0
localArray[localId] = 0.0f
//2 copy coefficients to local
//don't bother with this id __constant is working for you
//requires another local to be passed in: localCoeff
//localCoeff[localId] = coefficients[localId];
//barrier for both steps 1 and 2
barrier(CLK_LOCAL_MEM_FENCE);
float tmp;
for(int i = 0; i< outputSize; i++)
{
//3 shift elements (+barrier)
if(localId > 0){
tmp = localArray[localId -1]
}
barrier(CLK_LOCAL_MEM_FENCE);
localArray[localId] = tmp
//4 copy new element (work item 0 only, + barrier)
if(localId == 0){
localArray[0] = Array[i];
}
barrier(CLK_LOCAL_MEM_FENCE);
//5 compute dot product
//5a multiply + barrier
localSums[localId] = localArray[localId] * coefficients[localId];
barrier(CLK_LOCAL_MEM_FENCE);
//5b reduction loop + barrier
for(int j = 1; j < localSize; j <<= 1) {
int mask = (j << 1) - 1;
if ((localId & mask) == 0) {
localSums[local_index] += localSums[localId +j]
}
barrier(CLK_LOCAL_MEM_FENCE);
}
//6 copy dot product (WI 0 only)
if(localId == 0){
Output[i] = localSums[0];
}
//7 barrier
//only needed if there is more code after the loop.
//the barrier in #3 covers this in the case where the loop continues
//barrier(CLK_LOCAL_MEM_FENCE);
}
}
What about more work groups?
This is slightly simplified to allow a single 1x65 work group computer the entire 1.2M Output. To allow multiple work groups, you could use / get_num_groups(0) to calculate the amount of work each group should do (workAmount), and adjust the i for-loop:
for (i = workAmount * get_group_id(0); i< (workAmount * (get_group_id(0)+1) -1); i++)
Step #1 must be changed as well to initialize to the correct starting state for localArray, rather than all 0s.
//1 init local values
if(groupId == 0){
localArray[localId] = 0.0f
}else{
localArray[localSize - localId] = Array[workAmount - localId];
}
These two changes should allow you to use a more optimal number of work groups; I suggest some multiple of the number of compute units on the device. Try to keep the amount of work for each group in the thousands though. Play around with this, sometimes what seems optimal on a high-level will be detrimental to the kernel when it's running.
Advantages
At almost every point in this kernel, the work items have something to do. The only time fewer than 100% of the items are working is during the reduction loop in step 5b. Read more here about why that is a good thing.
Disadvantages
The barriers will slow down the kernel just by the nature of what barriers do: the pause a work item until the others reach that point. Maybe there is a way you could implement this with fewer barriers, but I still feel this is optimal because of the problem you are trying to solve.
There isn't room for more work items per group, and 65 is not a very optimal size. Ideally, you should try to use a power of 2, or a multiple of 64. This won't be a huge issue though, because there are a lot of barriers in the kernel which makes them all wait fairly regularly.
I have an array of doubles which is the result of the FFT applied on an array, that contains the audio data of a Wav audio file in which i have added a 1000Hz tone.
I obtained this array thought the DREALFT defined in "Numerical Recipes".(I must use it).
(The original array has a length that is power of two.)
Mine array has this structure:
array[0] = first real valued component of the complex transform
array[1] = last real valued component of the complex transform
array[2] = real part of the second element
array[3] = imaginary part of the second element
etc......
Now, i know that this array represent the frequency domain.
I want to determine and kill the 1000Hz frequency.
I have tried this formula for finding the index of the array which should contain the 1000Hz frequency:
index = 1000. * NElements /44100;
Also, since I assume that this index refers to an array with real values only, i have determined the correct(?) position in my array, that contains imaginary values too:
int correctIndex=2;
for(k=0;k<index;k++){
correctIndex+=2;
}
(I know that surely there is a way easier but it is the first that came to mind)
Then, i find this value: 16275892957.123705, which i suppose to be the real part of the 1000Hz frequency.(Sorry if this is an imprecise affermation but at the moment I do not care to know more about it)
So i have tried to suppress it:
array[index]=-copy[index]*0.1f;
I don't know exactly why i used this formula but is the only one that gives some results, in fact the 1000hz tone appears to decrease slightly.
This is the part of the code in question:
double *copy = malloc( nCampioni * sizeof(double));
int nSamples;
/*...Fill copy with audio data...*/
/*...Apply ZERO PADDING and reach the length of 8388608 samples,
or rather 8388608 double values...*/
/*Apply the FFT (Sure this works)*/
drealft(copy - 1, nSamples, 1);
/*I determine the REAL(?) array index*/
i= 1000. * nSamples /44100;
/*I determine MINE(?) array index*/
int j=2;
for(k=0;k<i;k++){
j+=2;
}
/*I reduce the array value, AND some other values aroud it as an attempt*/
for(i=-12;i<12;i+=2){
copy[j-i]=-copy[i-j]*0.1f;
printf("%d\n",j-i);
}
/*Apply the inverse FFT*/
drealft(copy - 1, nSamples, -1);
/*...Write the audio data on the file...*/
NOTE: for simplicity I omitted the part where I get an array of double from an array of int16_t
How can i determine and totally kill the 1000Hz frequency?
Thank you!
As Oli Charlesworth writes, because your target frequency is not exactly one of the FFT bins (your index, TargetFrequency * NumberOfElements / SamplingRate, is not exactly an integer), the energy of the target frequency will be spread across all bins. For a start, you can eliminate some of the frequency by zeroing the bin closest to the target frequency. This will of course affect other frequencies somewhat too, since it is slightly off target. To better suppress the target frequency, you will need to consider a more sophisticated filter.
However, for educational purposes: To suppress the frequency corresponding to a bin, simply set that bin to zero. You must set both the real and the imaginary components of the bin to zero, which you can do with:
copy[index*2 + 0] = 0;
copy[index*2 + 1] = 1;
Some notes about this:
You had this code to calculate the position in the array:
int correctIndex = 2;
for (k = 0; k < index; k++) {
correctIndex += 2;
}
That is equivalent to:
correctIndex = 2*(index+1);
I believe you want 2*index, not 2*(index+1). So you were likely reducing the wrong bin.
At one point in your question, you wrote array[index] = -copy[index]*0.1f;. I do not know what array is. You appeared to be working in place in copy. I also do not know why you multiplied by 1/10. If you want to eliminate a frequency, just set it to zero. Multiplying it by 1/10 only reduces it to 10% of its original magnitude.
I understand that you must pass copy-1 to drealft because the Numerical Recipes code uses one-based indexing. However, the C standard does not support the way you are doing it. The behavior of the expression copy-1 is not defined by the standard. It will work in most C implementations. However, to write supported portable code, you should do this instead:
// Allocate one extra element.
double *memory = malloc((nCampioni+1) * sizeof *memory);
// Make a pointer that is convenient for your work.
double *copy = memory+1;
…
// Pass the necessary base address to drealft.
drealft(memory, nSamples, 1);
// Suppress a frequency.
copy[index*2 + 0] = 0;
copy[index*2 + 1] = 0;
…
// Free the memory.
free(memory);
One experiment I suggest you consider is to initialize an array with just a sine wave at the desired frequency:
for (i = 0; i < nSamples; ++i)
copy[i] = sin(TwoPi * Frequency / SampleRate * i);
(TwoPi is of course 2*3.1415926535897932384626433.) Then apply drealft and look at the results. You will see that much of the energy is at a peak in the closest bin to the target frequency, but much of it has also spread to other bins. Clearly, zeroing a single bin and performing the inverse FFT cannot eliminate all of the frequency. Also, you should see that the peak is in the same bin you calculated for index. If it is not, something is wrong.
I just finished a homework problem for Computer Science 1 (yes, it's homework, but hear me out!). Now, the assignment is 100% complete and working, so I don't need help on it. My question involves the efficiency of an algorithm I'm using (we aren't graded on algorithmic efficiency yet, I'm just really curious).
The function I'm about to present currently uses a modified version of the linear search algorithm (that I came up with, all by myself!) in order to check how many numbers on a given lottery ticket match the winning numbers, assuming that both the numbers on the ticket and the numbers drawn are in ascending order. I was wondering, is there any way to make this algorithm more efficient?
/*
* Function: ticketCheck
*
* #param struct ticket
* #param array winningNums[6]
*
* Takes in a ticket, counts how many numbers
* in the ticket match, and returns the number
* of matches.
*
* Uses a modified linear search algorithm,
* in which the index of the successor to the
* last matched number is used as the index of
* the first number tested for the next ticket value.
*
* #return int numMatches
*/
int ticketCheck( struct ticket ticket, int winningNums[6] )
{
int numMatches = 0;
int offset = 0;
int i;
int j;
for( i = 0; i < 6; i++ )
{
for( j = 0 + offset; j < 6; j++ )
{
if( ticket.ticketNum[i] == winningNums[j] )
{
numMatches++;
offset = j + 1;
break;
}
if( ticket.ticketNum[i] < winningNums[j] )
{
i++;
j--;
continue;
}
}
}
return numMatches;
}
It's more or less there, but not quite. In most situations, it's O(n), but it's O(n^2) if every ticketNum is greater than every winningNum. (This is because the inner j loop doesn't break when j==6 like it should, but runs the next i iteration instead.)
You want your algorithm to increment either i or j at each step, and to terminate when i==6 or j==6. [Your algorithm almost satisfies this, as stated above.] As a result, you only need one loop:
for (i=0,j=0; i<6 && j<6; /* no increment step here */) {
if (ticketNum[i] == winningNum[j]) {
numMatches++;
i++;
j++;
}
else if (ticketNum[i] < winningNum[j]) {
/* ticketNum[i] won't match any winningNum, discard it */
i++;
}
else { /* ticketNum[i] > winningNum[j] */
/* discard winningNum[j] similarly */
j++;
}
}
Clearly this is O(n); at each stage, it either increments i or j, so the most steps it can do is 2*n-1. This has almost the same behaviour as your algorithm, but is easier to follow and easier to see that it's correct.
You're basically looking for the size of the intersection of two sets. Given that most lottos use around 50 balls (or so), you could store the numbers as bits that are set in an unsigned long long. Finding the common numbers is then a simple matter of ANDing the two together: commonNums = TicketNums & winningNums;.
Finding the size of the intersection is a matter of counting the one bits in the resulting number, a subject that's been covered previously (though in this case, you'd use 64-bit numbers, or a pair of 32-bit numbers, instead of a single 32-bit number).
Yes, there is something faster, but probably using more memory. Make an array full of 0 in the size of the possible numbers, put a 1 on every drawn number. For every ticket number add the value at the index of that number.
int NumsArray[MAX_NUMBER+1];
memset(NumsArray, 0, sizeof NumsArray);
for( i = 0; i < 6; i++ )
NumsArray[winningNums[i]] = 1;
for( i = 0; i < 6; i++ )
numMatches += NumsArray[ticket.ticketNum[i]];
12 loop rounds instead of up to 36
The surrounding code left as an exercise.
EDIT: It also has the advantage of not needing to sort both set of values.
This is really only a minor change on a scale like this, but if the second loop reaches a number bigger than the current ticket number, it is already allowed to brake. Furthermore, if your seconds traverses numbers lower than your ticket number, it may update the offset even if no match is found within that iteration.
PS:
Not to forget, general results on efficiency make more sense, if we take the number of balls or the size of the ticket to be variable. Otherwise it is too much dependent of the machine.
If instead of comparing the arrays of lottery numbers you were to create two bit arrays of flags -- each flag being set if it's index is in that array -- then you could perform a bitwise and on the two bit arrays (the lottery ticket and the winning number sets) and produce another bit array whose bits were flags for matching numbers only. Then count the bits set.
For many lotteries 64 bits would be enough, so a uint64_t should be big enough to cover this. Also, some architectures have instructions to count the bits set in a register, which some compilers might be able to recognize and optimize for.
The efficiency of this algorithm is based both on the range of lottery numbers (M) and the number of lottery numbers per ticket (N). The setting if the flags is O(N), while the and-ing of the two bit arrays and counting of the bits could be O(M), depending on if your M (lotto number range) is larger than the size that the target cpu can preform these operations on directly. Most likely, though, M will be small and its impact will likely be less than that of N on the performance.