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This is the first time I'm trying random numbers with C (I miss C#). Here is my code:
int i, j = 0;
for(i = 0; i <= 10; i++) {
j = rand();
printf("j = %d\n", j);
}
with this code, I get the same sequence every time I run the code. But it generates different random sequences if I add srand(/*somevalue/*) before the for loop. Can anyone explain why?
You have to seed it. Seeding it with the time is a good idea:
srand()
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main ()
{
srand ( time(NULL) );
printf ("Random Number: %d\n", rand() %100);
return 0;
}
You get the same sequence because rand() is automatically seeded with the a value of 1 if you do not call srand().
Edit
Due to comments
rand() will return a number between 0 and RAND_MAX (defined in the standard library). Using the modulo operator (%) gives the remainder of the division rand() / 100. This will force the random number to be within the range 0-99. For example, to get a random number in the range of 0-999 we would apply rand() % 1000.
rand() returns pseudo-random numbers. It generates numbers based on a given algorithm.
The starting point of that algorithm is always the same, so you'll see the same sequence generated for each invocation. This is handy when you need to verify the behavior and consistency of your program.
You can set the "seed" of the random generator with the srand function(only call srand once in a program) One common way to get different sequences from the rand() generator is to set the seed to the current time or the id of the process:
srand(time(NULL)); or srand(getpid()); at the start of the program.
Generating real randomness is very very hard for a computer, but for practical non-crypto related work, an algorithm that tries to evenly distribute the generated sequences works fine.
To quote from man rand :
The srand() function sets its argument
as the seed for a new sequence of
pseudo-random integers to be returned
by rand(). These sequences are
repeatable by calling srand() with the
same seed value.
If no seed value is provided, the
rand() function is automatically
seeded with a value of 1.
So, with no seed value, rand() assumes the seed as 1 (every time in your case) and with the same seed value, rand() will produce the same sequence of numbers.
There's a lot of answers here, but no-one seems to have really explained why it is that rand() always generates the same sequence given the same seed - or even what the seed is really doing. So here goes.
The rand() function maintains an internal state. Conceptually, you could think of this as a global variable of some type called rand_state. Each time you call rand(), it does two things. It uses the existing state to calculate a new state, and it uses the new state to calculate a number to return to you:
state_t rand_state = INITIAL_STATE;
state_t calculate_next_state(state_t s);
int calculate_return_value(state_t s);
int rand(void)
{
rand_state = calculate_next_state(rand_state);
return calculate_return_value(rand_state);
}
Now you can see that each time you call rand(), it's going to make rand_state move one step along a pre-determined path. The random values you see are just based on where you are along that path, so they're going to follow a pre-determined sequence too.
Now here's where srand() comes in. It lets you jump to a different point on the path:
state_t generate_random_state(unsigned int seed);
void srand(unsigned int seed)
{
rand_state = generate_random_state(seed);
}
The exact details of state_t, calculate_next_state(), calculate_return_value() and generate_random_state() can vary from platform to platform, but they're usually quite simple.
You can see from this that every time your program starts, rand_state is going to start off at INITIAL_STATE (which is equivalent to generate_random_state(1)) - which is why you always get the same sequence if you don't use srand().
If I remember the quote from Knuth's seminal work "The Art of Computer Programming" at the beginning of the chapter on Random Number Generation, it goes like this:
"Anyone who attempts to generate random numbers by mathematical means is, technically speaking, in a state of sin".
Simply put, the stock random number generators are algorithms, mathematical and 100% predictable. This is actually a good thing in a lot of situations, where a repeatable sequence of "random" numbers is desirable - for example for certain statistical exercises, where you don't want the "wobble" in results that truly random data introduces thanks to clustering effects.
Although grabbing bits of "random" data from the computer's hardware is a popular second alternative, it's not truly random either - although the more complex the operating environment, the more possibilities for randomness - or at least unpredictability.
Truly random data generators tend to look to outside sources. Radioactive decay is a favorite, as is the behavior of quasars. Anything whose roots are in quantum effects is effectively random - much to Einstein's annoyance.
Random number generators are not actually random, they like most software is completely predictable. What rand does is create a different pseudo-random number each time it is called One which appears to be random. In order to use it properly you need to give it a different starting point.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main ()
{
/* initialize random seed: */
srand ( time(NULL) );
printf("random number %d\n",rand());
printf("random number %d\n",rand());
printf("random number %d\n",rand());
printf("random number %d\n",rand());
return 0;
}
This is from http://www.acm.uiuc.edu/webmonkeys/book/c_guide/2.13.html#rand:
Declaration:
void srand(unsigned int seed);
This function seeds the random number generator used by the function rand. Seeding srand with the same seed will cause rand to return the same sequence of pseudo-random numbers. If srand is not called, rand acts as if srand(1) has been called.
rand() returns the next (pseudo) random number in a series. What's happening is you have the same series each time its run (default '1'). To seed a new series, you have to call srand() before you start calling rand().
If you want something random every time, you might try:
srand (time (0));
Rand does not get you a random number. It gives you the next number in a sequence generated by a pseudorandom number generator. To get a different sequence every time you start your program, you have to seed the algorithm by calling srand.
A (very bad) way to do it is by passing it the current time:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main() {
srand(time(NULL));
int i, j = 0;
for(i = 0; i <= 10; i++) {
j = rand();
printf("j = %d\n", j);
}
return 0;
}
Why this is a bad way? Because a pseudorandom number generator is as good as its seed, and the seed must be unpredictable. That is why you may need a better source of entropy, like reading from /dev/urandom.
call srand(sameSeed) before calling rand(). More details here.
Seeding the rand()
void srand (unsigned int seed)
This function establishes seed as the seed for a new series of pseudo-random numbers. If you call rand before a seed has been established with srand, it uses the value 1 as a default seed.
To produce a different pseudo-random series each time your program is run, do srand (time (0))
None of you guys are answering his question.
with this code i get the same sequance everytime the code but it generates random sequences if i add srand(/somevalue/) before the for loop . can someone explain why ?
From what my professor has told me, it is used if you want to make sure your code is running properly and to see if there is something wrong or if you can change something.
I had an interesting problem today. As part of practice with my fluency with objective-c, as I sit in math class, I write programs for each problem done on the board in hopes to increase my math and programming capabilities.
However, today I encountered a problem. One of the questions was something like "Find the greatest prime number that 10564245 (<-- example number) is divisible by"
So, I went in and made the program. I got as far as doing the loop for values to check, and then began to code the part where it checks the reminder, and if it's 0, it logs it, and if it isn't, it skips it.
However, since the number is too big to be an int, it had to be a double. When I tried to plug the number in, the program gave me errors when I wanted to use the % operator with the double. Is there any way to find remainders if you have a very large number?
Thanks
ERROR: Invalid operands to binary expression
EDIT: SOLVED!
I took a little from each answer. We have the fmod capability, but I ended up using long instead of int, I don't know why I didn't think of the origionally
Use fmod().
#include <stdio.h>
#include <math.h>
int main(int argc, char *argv[]) {
double a = 77879878978942312315687897;
double b = 10;
printf("%.f",fmod(a, b));
}
Double values aren't precise for integer values and so probably shouldn't be used for this sort of thing. Instead, you can use a long into or even a long long int to do your calculations
What is a quick and easy way to 'checksum' an array of floating point numbers, while allowing for a specified small amount of inaccuracy?
e.g. I have two algorithms which should (in theory, with infinite precision) output the same array. But they work differently, and so floating point errors will accumulate differently, though the array lengths should be exactly the same. I'd like a quick and easy way to test if the arrays seem to be the same. I could of course compare the numbers pairwise, and report the maximum error; but one algorithm is in C++ and the other is in Mathematica and I don't want the bother of writing out the numbers to a file or pasting them from one system to another. That's why I want a simple checksum.
I could simply add up all the numbers in the array. If the array length is N, and I can tolerate an error of 0.0001 in each number, then I would check if abs(sum1-sum2)<0.0001*N. But this simplistic 'checksum' is not robust, e.g. to an error of +10 in one entry and -10 in another. (And anyway, probability theory says that the error probably grows like sqrt(N), not like N.) Of course, any checksum is a low-dimensional summary of a chunk of data so it will miss some errors, if not most... but simple checksums are nonetheless useful for finding non-malicious bug-type errors.
Or I could create a two-dimensional checksum, [sum(x[n]), sum(abs(x[n]))]. But is the best I can do, i.e. is there a different function I might use that would be "more orthogonal" to the sum(x[n])? And if I used some arbitrary functions, e.g. [sum(f1(x[n])), sum(f2(x[n]))], then how should my 'raw error tolerance' translate into 'checksum error tolerance'?
I'm programming in C++, but I'm happy to see answers in any language.
i have a feeling that what you want may be possible via something like gray codes. if you could translate your values into gray codes and use some kind of checksum that was able to correct n bits you could detect whether or not the two arrays were the same except for n-1 bits of error, right? (each bit of error means a number is "off by one", where the mapping would be such that this was a variation in the least significant digit).
but the exact details are beyond me - particularly for floating point values.
i don't know if it helps, but what gray codes solve is the problem of pathological rounding. rounding sounds like it will solve the problem - a naive solution might round and then checksum. but simple rounding always has pathological cases - for example, if we use floor, then 0.9999999 and 1 are distinct. a gray code approach seems to address that, since neighbouring values are always single bit away, so a bit-based checksum will accurately reflect "distance".
[update:] more exactly, what you want is a checksum that gives an estimate of the hamming distance between your gray-encoded sequences (and the gray encoded part is easy if you just care about 0.0001 since you can multiple everything by 10000 and use integers).
and it seems like such checksums do exist: Any error-correcting code can be used for error detection. A code with minimum Hamming distance, d, can detect up to d − 1 errors in a code word. Using minimum-distance-based error-correcting codes for error detection can be suitable if a strict limit on the minimum number of errors to be detected is desired.
so, just in case it's not clear:
multiple by minimum error to get integers
convert to gray code equivalent
use an error detecting code with a minimum hamming distance larger than the error you can tolerate.
but i am still not sure that's right. you still get the pathological rounding in the conversion from float to integer. so it seems like you need a minimum hamming distance that is 1 + len(data) (worst case, with a rounding error on each value). is that feasible? probably not for large arrays.
maybe ask again with better tags/description now that a general direction is possible? or just add tags now? we need someone who does this for a living. [i added a couple of tags]
I've spent a while looking for a deterministic answer, and been unable to find one. If there is a good answer, it's likely to require heavy-duty mathematical skills (functional analysis).
I'm pretty sure there is no solution based on "discretize in some cunning way, then apply a discrete checksum", e.g. "discretize into strings of 0/1/?, where ? means wildcard". Any discretization will have the property that two floating-point numbers very close to each other can end up with different discrete codes, and then the discrete checksum won't tell us what we want to know.
However, a very simple randomized scheme should work fine. Generate a pseudorandom string S from the alphabet {+1,-1}, and compute csx=sum(X_i*S_i) and csy=sum(Y_i*S_i), where X and Y are my original arrays of floating point numbers. If we model the errors as independent Normal random variables with mean 0, then it's easy to compute the distribution of csx-csy. We could do this for several strings S, and then do a hypothesis test that the mean error is 0. The number of strings S needed for the test is fixed, it doesn't grow linearly in the size of the arrays, so it satisfies my need for a "low-dimensional summary". This method also gives an estimate of the standard deviation of the error, which may be handy.
Try this:
#include <complex>
#include <cmath>
#include <iostream>
// PARAMETERS
const size_t no_freqs = 3;
const double freqs[no_freqs] = {0.05, 0.16, 0.39}; // (for example)
int main() {
std::complex<double> spectral_amplitude[no_freqs];
for (size_t i = 0; i < no_freqs; ++i) spectral_amplitude[i] = 0.0;
size_t n_data = 0;
{
std::complex<double> datum;
while (std::cin >> datum) {
for (size_t i = 0; i < no_freqs; ++i) {
spectral_amplitude[i] += datum * std::exp(
std::complex<double>(0.0, 1.0) * freqs[i] * double(n_data)
);
}
++n_data;
}
}
std::cout << "Fuzzy checksum:\n";
for (size_t i = 0; i < no_freqs; ++i) {
std::cout << real(spectral_amplitude[i]) << "\n";
std::cout << imag(spectral_amplitude[i]) << "\n";
}
std::cout << "\n";
return 0;
}
It returns just a few, arbitrary points of a Fourier transform of the entire data set. These make a fuzzy checksum, so to speak.
How about computing a standard integer checksum on the data obtained by zeroing the least significant digits of the data, the ones that you don't care about?
I have a simple c program for printing n Fibonacci numbers and I would like to compile it to ELF object file. Instead of setting the number of fibonacci numbers (n) directly in my c code, I would like to set them in the registers since I am simulating it for an ARM processor.How can I do that?
Here is the code snippet
#include <stdio.h>
#include <stdlib.h>
#define ITERATIONS 3
static float fib(float i) {
return (i>1) ? fib(i-1) + fib(i-2) : i;
}
int main(int argc, char **argv) {
float i;
printf("starting...\n");
for(i=0; i<ITERATIONS; i++) {
printf("fib(%f) = %f\n", i, fib(i));
}
printf("finishing...\n");
return 0;
}
I would like to set the ITERATIONS counter in my Registers rather than in the code.
Thanks in advance
The register keyword can be used to suggest to the compiler that it uses a registers for the iterator and the number of iterations:
register float i;
register int numIterations = ITERATIONS;
but that will not help much. First of all, the compiler may or may not use your suggestion. Next, values will still need to be placed on the stack for the call to fib(), and, finally, depending on what functions you call within your loop, code in the procedure are calling could save your register contents in the stack frame at procedure entry, and restore them as part of the code implementing the procedure return.
If you really need to make every instruction count, then you will need to write machine code (using an assembly language). That way, you have direct control over your register usage. Assembly language programming is not for the faint of heart. Assembly language development is several times slower than using higher level languages, your risk of inserting bugs is greater, and they are much more difficult to track down. High level languages were developed for a reason, and the C language was developed to help write Unix. The minicomputers that ran the first Unix systems were extremely slow, but the reason C was used instead of assembly was that even then, it was more important to have code that took less time to code, had fewer bugs, and was easier to debug than assembler.
If you want to try this, here are the answers to a previous question on stackoverflow about resources for ARM programming that might be helpful.
One tactic you might take is to isolate your performance-critical code into a procedure, write the procedure in C, the capture the generated assembly language representation. Then rewrite the assembler to be more efficient. Test thoroughly, and get at least one other set of eyeballs to look the resulting code over.
Good Luck!
Make ITERATIONS a variable rather than a literal constant, then you can set its value directly in your debugger/simulator's watch or locals window just before the loop executes.
Alternatively as it appears you have stdio support, why not just accept the value via console input?
How to do numerical integration (what numerical method, and what tricks to use) for one-dimensional integration over infinite range, where one or more functions in the integrand are 1d quantum harmonic oscillator wave functions. Among others I want to calculate matrix elements of some function in the harmonic oscillator basis:
phin(x) = Nn Hn(x) exp(-x2/2)
where Hn(x) is Hermite polynomial
Vm,n = \int_{-infinity}^{infinity} phim(x) V(x) phin(x) dx
Also in the case where there are quantum harmonic wavefunctions with different widths.
The problem is that wavefunctions phin(x) have oscillatory behaviour, which is a problem for large n, and algorithm like adaptive Gauss-Kronrod quadrature from GSL (GNU Scientific Library) take long to calculate, and have large errors.
An incomplete answer, since I'm a little short on time at the moment; if others can't complete the picture, I can supply more details later.
Apply orthogonality of the wavefunctions whenever and wherever possible. This should significantly cut down the amount of computation.
Do analytically whatever you can. Lift constants, split integrals by parts, whatever. Isolate the region of interest; most wavefunctions are band-limited, and reducing the area of interest will do a lot to save work.
For the quadrature itself, you probably want to split the wavefunctions into three pieces and integrate each separately: the oscillatory bit in the center plus the exponentially-decaying tails on either side. If the wavefunction is odd, you get lucky and the tails will cancel each other, meaning you only have to worry about the center. For even wavefunctions, you only have to integrate one and double it (hooray for symmetry!). Otherwise, integrate the tails using a high order Gauss-Laguerre quadrature rule. You might have to calculate the rules yourself; I don't know if tables list good Gauss-Laguerre rules, as they're not used too often. You probably also want to check the error behavior as the number of nodes in the rule goes up; it's been a long time since I used Gauss-Laguerre rules and I don't remember if they exhibit Runge's phenomenon. Integrate the center part using whatever method you like; Gauss-Kronrod is a solid choice, of course, but there's also Fejer quadrature (which sometimes scales better to high numbers of nodes, which might work nicer on an oscillatory integrand) and even the trapezoidal rule (which exhibits stunning accuracy with certain oscillatory functions). Pick one and try it out; if results are poor, give another method a shot.
Hardest question ever on SO? Hardly :)
I'd recommend a few other things:
Try transforming the function onto a finite domain to make the integration more manageable.
Use symmetry where possible - break it up into the sum of two integrals from negative infinity to zero and zero to infinity and see if the function is symmetry or anti-symmetric. It could make your computing easier.
Look into Gauss-Laguerre quadrature and see if it can help you.
The WKB approximation?
I am not going to explain or qualify any of this right now. This code is written as is and probably incorrect. I am not even sure if it is the code I was looking for, I just remember that years ago I did this problem and upon searching my archives I found this. You will need to plot the output yourself, some instruction is provided. I will say that the integration over infinite range is a problem that I addressed and upon execution of the code it states the round off error at 'infinity' (which numerically just means large).
// compile g++ base.cc -lm
#include <iostream>
#include <cstdlib>
#include <fstream>
#include <math.h>
using namespace std;
int main ()
{
double xmax,dfx,dx,x,hbar,k,dE,E,E_0,m,psi_0,psi_1,psi_2;
double w,num;
int n,temp,parity,order;
double last;
double propogator(double E,int parity);
double eigen(double E,int parity);
double f(double x, double psi, double dpsi);
double g(double x, double psi, double dpsi);
double rk4(double x, double psi, double dpsi, double E);
ofstream datas ("test.dat");
E_0= 1.602189*pow(10.0,-19.0);// ev joules conversion
dE=E_0*.001;
//w^2=k/m v=1/2 k x^2 V=??? = E_0/xmax x^2 k-->
//w=sqrt( (2*E_0)/(m*xmax) );
//E=(0+.5)*hbar*w;
cout << "Enter what energy level your looking for, as an (0,1,2...) INTEGER: ";
cin >> order;
E=0;
for (n=0; n<=order; n++)
{
parity=0;
//if its even parity is 1 (true)
temp=n;
if ( (n%2)==0 ) {parity=1; }
cout << "Energy " << n << " has these parameters: ";
E=eigen(E,parity);
if (n==order)
{
propogator(E,parity);
cout <<" The postive values of the wave function were written to sho.dat \n";
cout <<" In order to plot the data should be reflected about the y-axis \n";
cout <<" evenly for even energy levels and oddly for odd energy levels\n";
}
E=E+dE;
}
}
double propogator(double E,int parity)
{
ofstream datas ("sho.dat") ;
double hbar =1.054*pow(10.0,-34.0);
double m =9.109534*pow(10.0,-31.0);
double E_0= 1.602189*pow(10.0,-19.0);
double dx =pow(10.0,-10);
double xmax= 100*pow(10.0,-10.0)+dx;
double dE=E_0*.001;
double last=1;
double x=dx;
double psi_2=0.0;
double psi_0=0.0;
double psi_1=1.0;
// cout <<parity << " parity passsed \n";
psi_0=0.0;
psi_1=1.0;
if (parity==1)
{
psi_0=1.0;
psi_1=m*(1.0/(hbar*hbar))* dx*dx*(0-E)+1 ;
}
do
{
datas << x << "\t" << psi_0 << "\n";
psi_2=(2.0*m*(dx/hbar)*(dx/hbar)*(E_0*(x/xmax)*(x/xmax)-E)+2.0)*psi_1-psi_0;
//cout << psi_1 << "=psi_1\n";
psi_0=psi_1;
psi_1=psi_2;
x=x+dx;
} while ( x<= xmax);
//I return 666 as a dummy value sometimes to check the function has run
return 666;
}
double eigen(double E,int parity)
{
double hbar =1.054*pow(10.0,-34.0);
double m =9.109534*pow(10.0,-31.0);
double E_0= 1.602189*pow(10.0,-19.0);
double dx =pow(10.0,-10);
double xmax= 100*pow(10.0,-10.0)+dx;
double dE=E_0*.001;
double last=1;
double x=dx;
double psi_2=0.0;
double psi_0=0.0;
double psi_1=1.0;
do
{
psi_0=0.0;
psi_1=1.0;
if (parity==1)
{double psi_0=1.0; double psi_1=m*(1.0/(hbar*hbar))* dx*dx*(0-E)+1 ;}
x=dx;
do
{
psi_2=(2.0*m*(dx/hbar)*(dx/hbar)*(E_0*(x/xmax)*(x/xmax)-E)+2.0)*psi_1-psi_0;
psi_0=psi_1;
psi_1=psi_2;
x=x+dx;
} while ( x<= xmax);
if ( sqrt(psi_2*psi_2)<=1.0*pow(10.0,-3.0))
{
cout << E << " is an eigen energy and " << psi_2 << " is psi of 'infinity' \n";
return E;
}
else
{
if ( (last >0.0 && psi_2<0.0) ||( psi_2>0.0 && last<0.0) )
{
E=E-dE;
dE=dE/10.0;
}
}
last=psi_2;
E=E+dE;
} while (E<=E_0);
}
If this code seems correct, wrong, interesting or you do have specific questions ask and I will answer them.
I am a student majoring in physics, and I also encountered the problem. These days I keep thinking about this question and get my own answer. I think it may help you solve this question.
1.In gsl, there are functions can help you integrate the oscillatory function--qawo & qawf. Maybe you can set a value, a. And the integration can be separated into tow parts, [0,a] and [a,pos_infinity]. In the first interval, you can use any gsl integration function you want, and in the second interval, you can use qawo or qawf.
2.Or you can integrate the function to a upper limit, b, that is integrated in [0,b]. So the integration can be calculated using a gauss legendry method, and this is provided in gsl. Although there maybe some difference between the real value and the computed value, but if you set b properly, the difference can be neglected. As long as the difference is less than the accuracy you want. And this method using the gsl function is only called once and can use many times, because the return value is point and its corresponding weight, and integration is only the sum of f(xi)*wi, for more details you can search gauss legendre quadrature on wikipedia. Multiple and addition operation is much faster than integration.
3.There is also a function which can calculate the infinity area integration--qagi, you can search it in the gsl-user's guide. But this is called everytime you need to calculate the integration, and this may cause some time consuming, but I'm not sure how long will it use in you program.
I suggest NO.2 choice I offered.
If you are going to work with Harmonic oscillator functions less than n = 100 you might want to try:
http://www.mymathlib.com/quadrature/gauss_hermite.html
The program computes an integral via gauss-hermite quadrature with 100 zeroes and weights (the zeroes of H_100). Once you go over Hermite_100 the integrals are not as accurate.
Using this integration method I wrote a program calculating exactly what you want to calculate and it works fairly well. Also, there might be a way to go beyond n=100 by using the asymptotic form of the Hermite-polynomial zeroes but I haven't looked into it.