I have a problem with Shuffling this array with Arduino software:
int questionNumberArray[10]={0,1,2,3,4,5,6,7,8,9};
Does anyone know a build in function or a way to shuffle the values in the array without any repeating?
The simplest way would be this little for loop:
int questionNumberArray[] = {0,1,2,3,4,5,6,7,8,9};
const size_t n = sizeof(questionNumberArray) / sizeof(questionNumberArray[0]);
for (size_t i = 0; i < n - 1; i++)
{
size_t j = random(0, n - i);
int t = questionNumberArray[i];
questionNumberArray[i] = questionNumberArray[j];
questionNumberArray[j] = t;
}
Let's break it line by line, shall we?
int questionNumberArray[] = {0,1,2,3,4,5,6,7,8,9};
You don't need to put number of cells if you initialize an array like that. Just leave the brackets empty like I did.
const size_t n = sizeof(questionNumberArray) / sizeof(questionNumberArray[0]);
I decided to store number of cells in n constant. Operator sizeof gives you number of bytes taken by your array and number of bytes taken by one cell. You divide first number by the second and you have size of your array.
for (size_t i = 0; i < n - 1; i++)
Please note, that range of the loop is n - 1. We don't want i to ever have value of last index.
size_t j = random(0, n - i);
We declare variable j that points to some random cell with index greater than i. That is why we never wanted i to have n - 1 value - because then j would be out of bound. We get random number with Arduino's random function: https://www.arduino.cc/en/Reference/Random
int t = questionNumberArray[i];
questionNumberArray[i] = questionNumberArray[j];
questionNumberArray[j] = t;
Simple swap of two values. It's possible to do it without temporary t variable, but the code is less readable then.
In my case the result was as follows:
questionNumberArray[0] = 0
questionNumberArray[1] = 9
questionNumberArray[2] = 7
questionNumberArray[3] = 4
questionNumberArray[4] = 6
questionNumberArray[5] = 5
questionNumberArray[6] = 1
questionNumberArray[7] = 8
questionNumberArray[8] = 2
questionNumberArray[9] = 3
Related
I have the following problem to solve: given a number N and 1<=k<=N, count the number of possible sums of (1,...,k) which add to N. There may be equal factors (e.g. if N=3 and k=2, (1,1,1) is a valid sum), but permutations must not be counted (e.g., if N=3 and k=2, count (1,2) and (2,1) as a single solution). I have implemented the recursive Python code below but I'd like to find a better solution (maybe with dynamic programming? ). It seems similar to the triple step problem, but with the extra constraint of not counting permutations.
def find_num_sums_aux(n, min_k, max_k):
# base case
if n == 0:
return 1
count = 0
# due to lower bound min_k, we evaluate only ordered solutions and prevent permutations
for i in range(min_k, max_k+1):
if n-i>=0:
count += find_num_sums_aux(n-i, i, max_k)
return count
def find_num_sums(n, k):
count = find_num_sums_aux(n,1,k)
return count
This is a standard problem in dynamic programming (subset sum problem).
Lets define the function f(i,j) which gives the number of ways you can get the sum j using a subset of the numbers (1...i), then the result to your problem will be f(k,n).
for each number x of the range (1...i), x might be a part of the sum j or might not, so we need to count these two possibilities.
Note: f(i,0) = 1 for any i, which means that you can get the sum = 0 in one way and this way is by not taking any number from the range (1...i).
Here is the code written in C++:
int n = 10;
int k = 7;
int f[8][11];
//initializing the array with zeroes
for (int i = 0; i <= k; i++)
for (int j = 0; j <= n; j++)
f[i][j] = 0;
f[0][0] = 1;
for (int i = 1; i <= k; i++) {
for (int j = 0; j <= n; j++) {
if (j == 0)
f[i][j] = 1;
else {
f[i][j] = f[i - 1][j];//without adding i to the sum j
if (j - i >= 0)
f[i][j] = f[i][j] + f[i - 1][j - i];//adding i to the sum j
}
}
}
cout << f[k][n] << endl;//print f(k,n)
Update
To handle the case where we can repeat the elements like (1,1,1) will give you the sum 3, you just need to allow picking the same element multiple times by changing the following line of code:
f[i][j] = f[i][j] + f[i - 1][j - i];//adding i to the sum
To this:
f[i][j] = f[i][j] + f[i][j - i];
Below is a piece of C code run from R used to compare each row of a matrix to a vector. The number of identical values is stored in the first column of a two-column matrix.
I know it can easily be done in R (as done to check the results), but this is a first step for a more complex use case.
When openmp is not used, it works ok. When openmp is used, it give correlated (0.99) but inconsistent results.
Question1: What am I doing wrong?
Question2: I use a double for loop to fill the output matrix (ret) with zeros. What would be a better solution?
Also, inconsistencies were observed when the code was used in a package. I tried to make the code reproducible using inline, but it does not recognize the openmp statements (I tried to include 'omp.h', in the parameters of cfunction, ...).
Question3: How can we make this code work with inline?
I'm (too?) far outside my comfort zone on this topic.
library(inline)
compare <- cfunction(c(x = "integer", vec = "integer"), "
const int I = nrows(x), J = ncols(x);
SEXP ret;
PROTECT(ret = allocMatrix(INTSXP, I, 2));
int *ptx = INTEGER(x), *ptvec = INTEGER(vec), *ptret = INTEGER(ret);
for (int i=0; i<I; i++)
for (int j=0; j<2; j++)
ptret[j * I + i] = 0;
int i, j;
#pragma omp parallel for default(none) shared(ptx, ptvec, ptret) private(i,j)
for (j=0; j<J; j++)
for (i=0; i<I; i++)
if (ptx[i + I * j] == ptvec[j]) {++ptret[i];}
UNPROTECT(1);
return ret;
")
N = 3e3
M = 1e4
m = matrix(sample(c(-1:1), N*M, replace = TRUE), nc = M)
v = sample(-1:1, M, replace = TRUE)
cc = compare(m, v)
cr = rowSums(t(t(m) == v))
all.equal(cc[,1], cr)
Thanks to the comments above, I reconsidered the data race issue.
IIUC, my loop was parallelized on j (the columns). Then, each thread had its own value of i (the rows), but possible identical values across threads, that were then trying to increment ptret[i] at the same time.
To avoid this, I now loop on i first, so that only a single thread will increment each row.
Then, I realized that I could move the zero-initialization of ptret within the first loop.
It seems to work. I get identical results, increased CPU usage, and 3-4x speedup on my laptop.
I guess that solves questions 1 and 2. I will have a closer look at the inline/openmp problem.
Code below, fwiw.
#include <omp.h>
#include <R.h>
#include <Rinternals.h>
#include <stdio.h>
SEXP c_compare(SEXP x, SEXP vec)
{
const int I = nrows(x), J = ncols(x);
SEXP ret;
PROTECT(ret = allocMatrix(INTSXP, I, 2));
int *ptx = INTEGER(x), *ptvec = INTEGER(vec), *ptret = INTEGER(ret);
int i, j;
#pragma omp parallel for default(none) shared(ptx, ptvec, ptret) private(i, j)
for (i = 0; i < I; i++) {
// init ptret to zero
ptret[i] = 0;
ptret[I + i] = 0;
for (j = 0; j < J; j++)
if (ptx[i + I * j] == ptvec[j]) {
++ptret[i];
}
}
UNPROTECT(1);
return ret;
}
int j = 10;
void f() {
int *i;
i = &j;
*i = 7;
i = (int *) malloc(sizeof(int));
*i = j;
j = j + 5;
printf("%d %d", *i, j);
}
Write down the two values that the function f will output
I am not able to fully understand how pointers work
So this is my interpretation of the code
int j = 10;
J is assigned to value 10 in memory and has a address lets say 200
int *i;
declaring a pointer
i = &j;
i value is now 200
*i = 7
value of *i is 7
i = (int *) malloc(sizeof(int))
do not really understand what the above code is doing, but I think it assigns the variable i to the size of integer in array?
*i = j
Does *i point to address of j or the value of j
j = j + 5
the value of j (which I don't know what is) + 5
Thank you
First off, I'd recommend either reading a good C book or looking for a tutorial online to get some basics right.
int *i;
i = &j;
This makes pointer variable i to point to the same memory region where variable j resides (so your guess is right).
*i = 7;
Now, the location pointed to by i is updated to read 7. So, j is no longer 10, it's now 7 as *i and j are the same value.
i = (int *) malloc(sizeof(int));
Now, you allocated new piece of heap memory sufficient to hold a single int value. i no longer points to the location of j.
*i = j;
Set the value stored in location pointed to by i to have the same value as j, in this case 7.
j = j + 5;
Increment j by 5. Since i no longer points to &j the value stored in *i is unaffected.
printf("%d %d", *i, j);
Print value of *i and j.
As previously demonstrated i points to a memory location holding a value of 7 and j is 7 + 5 = 12.
I would like to make real time audio processing with Qt and display the spectrum using FFTW3.
What I've done in steps:
I capture any sound from computer device and fill it into the buffer.
I assign sound samples to double array
I compute the fundamental frequency.
when I'm display the fundamental frequency and Magnetitude when the microphone is on but no signal(silence) , the fundamental frequency is not what I expected , the code don't always return zero , sometimes the code returns 1500Hz,2000hz as frequency
and when the microphone is off (mute) the code don't return zero as fundamamental frequency but returns a number between 0 and 9000Hz. Any help woulbd be appreciated
here is my code
QByteArray *buffer;
QAudioInput *audioInput;
audioInput = new QAudioInput(format, this);
//Check the number of samples in input buffer
qint64 len = audioInput->bytesReady();
//Limit sample size
if(len > 4096)
len = 4096;
//Read sound samples from input device to buffer
qint64 l = input->read(buffer.data(), len);
int input_size= BufferSize;
int output_size = input_size; //input_size/2+1;
fftw_plan p3;
double in[output_size];
fftw_complex out[output_size];
short *outdata = (short*)m_buffer.data();// assign sample into short array
int data_size = size_t(outdata);
int data_size1 = sizeof(outdata);
int count = 0;
double w = 0;
for(int i(chanelNumber); i < output_size/2; i= i + 2) //fill array in
{
w= 0.5 * (1 - cos(2*M_PI*i/output_size)); // Hann Windows
double x = 0;
if(i < data_size){
x = outdata[i];
}
if(count < output_size){
in[count] = x;// fill Array In with sample from buffer
count++;
}
}
for(int i=count; i<output_size; i++){
in[i] = 0;
}
p3 = fftw_plan_dft_r2c_1d(output_size, in, out, FFTW_ESTIMATE);// create Plan
fftw_execute(p3);// FFT
for (int i = 0; i < (output_size/2); i++) {
long peak=0;
double Amplitudemax=0;
double r1 = out[i][0] * out[i][0];
double im1 = out[i][3] * out[i][4];
double t1 = r1 + im1;
//double t = 20*log(sqrt(t1));
double t = sqrt(t1)/(double)(output_size/2);
double f = (double)i*8000 / ((double)output_size/2);
if(Magnitude > AmplitudeMax)
{
AmplitudeMax = Magnitude;
Peak =2* i;
}
}
fftw_destroy_plan(p3);
return Peak*(static_cast<double>(8000)/output_Size);
What you think is silence might contain some small amount of noise. The FFT of random noise will also appear random, and thus have a random magnitude peak. But it is possible that noise might come from equipment or electronics in the environment (fans, flyback transformers, etc.), or the power supply to your ADC or mic, thus showing some frequency biases.
If the noise level is low enough, normally one checks the level of the magnitude peak, compares it against a threshold, and cuts off frequency estimation reporting below this threshold.
Hi all, I have an array of length N, and I'd like to divide it as best as possible between 'size' processors. N/size has a remainder, e.g. 1000 array elements divided by 7 processes, or 14 processes by 3 processes.
I'm aware of at least a couple of ways of work sharing in MPI, such as:
for (i=rank; i<N;i+=size){ a[i] = DO_SOME_WORK }
However, this does not divide the array into contiguous chunks, which I'd like to do as I believe is faster for IO reasons.
Another one I'm aware of is:
int count = N / size;
int start = rank * count;
int stop = start + count;
// now perform the loop
int nloops = 0;
for (int i=start; i<stop; ++i)
{
a[i] = DO_SOME_WORK;
}
However, with this method, for my first example we get 1000/7 = 142 = count. And so the last rank starts at 852 and ends at 994. The last 6 lines are ignored.
Would be best solution to append something like this to the previous code?
int remainder = N%size;
int start = N-remainder;
if (rank == 0){
for (i=start;i<N;i++){
a[i] = DO_SOME_WORK;
}
This seems messy, and if its the best solution I'm surprised I haven't seen it elsewhere.
Thanks for any help!
If I had N tasks (e.g., array elements) and size workers (e.g., MPI ranks), I would go as follows:
int count = N / size;
int remainder = N % size;
int start, stop;
if (rank < remainder) {
// The first 'remainder' ranks get 'count + 1' tasks each
start = rank * (count + 1);
stop = start + count;
} else {
// The remaining 'size - remainder' ranks get 'count' task each
start = rank * count + remainder;
stop = start + (count - 1);
}
for (int i = start; i <= stop; ++i) { a[i] = DO_SOME_WORK(); }
That is how it works:
/*
# ranks: remainder size - remainder
/------------------------------------\ /-----------------------------\
rank: 0 1 remainder-1 size-1
+---------+---------+-......-+---------+-------+-------+-.....-+-------+
tasks: | count+1 | count+1 | ...... | count+1 | count | count | ..... | count |
+---------+---------+-......-+---------+-------+-------+-.....-+-------+
^ ^ ^ ^
| | | |
task #: rank * (count+1) | rank * count + remainder |
| |
task #: rank * (count+1) + count rank * count + remainder + count - 1
\------------------------------------/
# tasks: remainder * count + remainder
*/
Here's a closed-form solution.
Let N = array length and P = number of processors.
From j = 0 to P-1,
Starting point of array on processor j = floor(N * j / P)
Length of array on processor j = floor(N * (j + 1) / P) – floor(N * j / P)
Consider your "1000 steps and 7 processes" example.
simple division won't work because integer division (in C) gives you the floor, and you are left with some remainder: i.e. 1000 / 7 is 142, and there will be 6 doodads hanging out
ceiling division has the opposite problem: ceil(1000/7) is 143, but then the last processor overruns the array, or ends up with less to do than the others.
You are asking for a scheme to evenly distribute the remainder over processors. Some processes should have 142, others 143. There must be a more formal approach but considering the attention this question's gotten in the last six months maybe not.
Here's my approach. Every process needs to do this algorithm, and just pick out the answer it needs for itself.
#include <mpi.h>
#include <stdio.h>
#include <stdlib.h>
int main (int argc, char ** argv)
{
#define NR_ITEMS 1000
int i, rank, nprocs;;
int *bins;
MPI_Init(&argc, &argv);
MPI_Comm_rank(MPI_COMM_WORLD, &rank);
MPI_Comm_size(MPI_COMM_WORLD, &nprocs);
bins = calloc(nprocs, sizeof(int));
int nr_alloced = 0;
for (i=0; i<nprocs; i++) {
remainder = NR_ITEMS - nr_alloced;
buckets = (nprocs - i);
/* if you want the "big" buckets up front, do ceiling division */
bins[i] = remainder / buckets;
nr_alloced += bins[i];
}
if (rank == 0)
for (i=0; i<nprocs; i++) printf("%d ", bins[i]);
MPI_Finalize();
return 0;
}
I know this is long sense gone but a simple way to do this is to give each process the floor of the (number of items) / (number of processes) + (1 if process_num < num_items mod num_procs). In python, an array with work counts:
# Number of items
NI=128
# Number of processes
NP=20
# Items per process
[NI/NP + (1 if P < NI%NP else 0)for P in range(0,NP)]
Improving off of #Alexander's answer: make use of min to condense the logic.
int count = N / size;
int remainder = N % size;
int start = rank * count + min(rank, remainder);
int stop = (rank + 1) * count + min(rank + 1, remainder);
for (int i = start; i < stop; ++i) { a[i] = DO_SOME_WORK(); }
I think that the best solution is to write yourself a little function for splitting work across processes evenly enough. Here's some pseudo-code, I'm sure you can write C (is that C in your question ?) better than I can.
function split_evenly_enough(num_steps, num_processes)
return = repmat(0, num_processes) ! pseudo-Matlab for an array of num_processes 0s
steps_per_process = ceiling(num_steps/num_processes)
return = steps_per_process - 1 ! set all elements of the return vector to this number
return(1:mod(num_steps, num_processes)) = steps_per_process ! some processes have 1 more step
end
How about this?
int* distribute(int total, int processes) {
int* distribution = new int[processes];
int last = processes - 1;
int remaining = total;
int process = 0;
while (remaining != 0) {
++distribution[process];
--remaining;
if (process != last) {
++process;
}
else {
process = 0;
}
}
return distribution;
}
The idea is that you assign an element to the first process, then an element to the second process, then an element to the third process, and so on, jumping back to the first process whenever the last one is reached.
This method works even when the number of processes is greater than the number of elements. It uses only very simple operations and should therefore be very fast.
I had a similar problem, and here is my non optimum solution with Python and mpi4py API. An optimum solution would take into account how the processors are laid out, here extra work is ditributed to lower ranks. The uneven workload only differ by one task, so it should not be a big deal in general.
from mpi4py import MPI
import sys
def get_start_end(comm,N):
"""
Distribute N consecutive things (rows of a matrix , blocks of a 1D array)
as evenly as possible over a given communicator.
Uneven workload (differs by 1 at most) is on the initial ranks.
Parameters
----------
comm: MPI communicator
N: int
Total number of things to be distributed.
Returns
----------
rstart: index of first local row
rend: 1 + index of last row
Notes
----------
Index is zero based.
"""
P = comm.size
rank = comm.rank
rstart = 0
rend = N
if P >= N:
if rank < N:
rstart = rank
rend = rank + 1
else:
rstart = 0
rend = 0
else:
n = N//P # Integer division PEP-238
remainder = N%P
rstart = n * rank
rend = n * (rank+1)
if remainder:
if rank >= remainder:
rstart += remainder
rend += remainder
else:
rstart += rank
rend += rank + 1
return rstart, rend
if __name__ == '__main__':
comm = MPI.COMM_WORLD
n = int(sys.argv[1])
print(comm.rank,get_start_end(comm,n))