C++ fast random number generator [duplicate] - c++

This question already has answers here:
C++ generating random numbers
(11 answers)
Closed 8 years ago.
I'm sorry if this is kind of a dumb question, but I'm new to c++, and honestly can't find the answer;
When I use rand(), of course I have to first use srand().
At first i'd just import <ctime> and do srand(time()), and this worked. But if I called rand() more than once a second - how often time() changes - then I'd get the same answer. So for instance;
#include <iostream>
#include <ctime>
#include <cstdlib>
using namespace std;
int main()
{
bool x = true;
while(x == true)
{
int num = 1;
srand(time(NULL));
num = rand();
cout<<num%10<<endl;
}
}
Might produce something like, 6666666666777777777700000000003333333333
Which is no good for my purposes - I'd prefer something like 163509284749301935766.

You should only seed the random number generator once. Right now you are seeding it in the loop and using time(NULL) just means the seed changes once per second which gives you the bad output you have described.
DO this instead:
int main()
{
bool x = true;
int num = 1;
srand(time(NULL));
while(x == true)
{
num = rand();
cout<<num%10<<endl;
}
}
And if you really care about the random numbers generated you might want to use something other than rand(). The reason is that rand() has poor statistical properties for pseudo random number generation, it is often implemented as a Linear congruential generator. If you need high quality randomness then you should prefer something else such as the new c++ random number generators http://en.cppreference.com/w/cpp/numeric/random.
In fact there's even a report on depreciating the old rand() to try to push people to use the newer c++ standard library random functions.
In this particular case you take a modulus which causes a few subtle problems:
num = rand();
cout<<num%10<<endl;
Even if rand() was perfect if the modulus here isn't a divisor of the maximum value returned by rand() you will get a non-uniform distribution as a result. Here's a quick explanation, say rand() returned values in the range of [0,25] then taking the modulus would do the following.
before after modulus
[0-9] [0-9]
[10-19] [0-9]
[20-25] [0-5]
You'll see that you are more likely to get [0-5] than [6-9] which means you now no longer have a uniform number being generated. Note that this small range is for educational purposes only, the maximum value of rand() is mandated by the standard to be at least 32767. However it illustrates an important point, the larger the maximum generated number the better.
This uniformity of distribution problem aside the modulus has the particularly insidious effect of decreasing the quality of the pseudo-randomness even further for some implementations.
Using std::uniform_int_distribution avoids many problems so I would recommend changing your existing code to use the new library. Doing so would look like this:
#include <iostream>
#include <random>
using namespace std;
int main()
{
std::default_random_engine generator;
generator.seed( /* your seed for the RNG goes here */ );
std::uniform_int_distribution<int> distribution(0,9);//note the min and max parameters are inclusive here
while(true)
{
cout << distribution(generator) << endl;
}
}

A function f() that would generate random numbers in range [low,high] can be easily, robustly and safely defined with c++11 library facilities :
#include <random>
#include <iostream>
int f(int low, int high)
{
std::random_device rd;
std::mt19937 gen(rd());
std::uniform_int_distribution<> dis(low, high);
return dis(gen);
}
std::uniform_int_distribution will give you a random number in your range (eg (0,9) ) this way
On the rationale behind refraining from using the old rand you can check this

Related

C++ Password Generator Generates Same Password [duplicate]

In this rather basic C++ code snippet involving random number generation:
include <iostream>
using namespace std;
int main() {
cout << (rand() % 100);
return 0;
}
Why am I always getting an output of 41? I'm trying to get it to output some random number between 0 and 100. Maybe I'm not understanding something about how the rand function works?
You need to change the seed.
int main() {
srand(time(NULL));
cout << (rand() % 101);
return 0;
}
This srand thing also works for C.
See also:
http://xkcd.com/221/
For what its worth you are also only generating numbers between 0 and 99 (inclusive). If you wanted to generate values between 0 and 100 you would need.
rand() % 101
in addition to calling srand() as mentioned by others.
srand() seeds the random number generator. Without a seed, the generator is unable to generate the numbers you are looking for. As long as one's need for random numbers is not security-critical (e.g. any sort of cryptography), common practice is to use the system time as a seed by using the time() function from the <ctime> library as such: srand(time(0)). This will seed the random number generator with the system time expressed as a Unix timestamp (i.e. the number of seconds since the date 1/1/1970). You can then use rand() to generate a pseudo-random number.
Here is a quote from a duplicate question:
The reason is that a random number generated from the rand() function isn't
actually random. It simply is a transformation. Wikipedia gives a better
explanation of the meaning of pseudorandom number generator: deterministic
random bit generator. Every time you call rand() it takes the seed and/or the
last random number(s) generated (the C standard doesn't specify the algorithm
used, though C++11 has facilities for specifying some popular algorithms), runs
a mathematical operation on those numbers, and returns the result. So if the
seed state is the same each time (as it is if you don't call srand with a truly
random number), then you will always get the same 'random' numbers out.
If you want to know more, you can read the following:
http://www.dreamincode.net/forums/topic/24225-random-number-generation-102/
http://www.dreamincode.net/forums/topic/29294-making-pseudo-random-number-generators-more-random/
You are not seeding the number.
Use This:
#include <iostream>
#include <ctime>
using namespace std;
int main()
{
srand(static_cast<unsigned int>(time(0)));
cout << (rand() % 100) << endl;
return 0;
}
You only need to seed it once though. Basically don't seed it every random number.
random functions like borland complier
using namespace std;
int sys_random(int min, int max) {
return (rand() % (max - min+1) + min);
}
void sys_randomize() {
srand(time(0));
}
"srand(time(NULL));" as 1st line at "main()" won't help you if you're using "rand()" at static init. somewhere. You better create "struct rnd_init { rnd_init() { srand (time (nullptr)); } }" named whatever suits you, as a static var at the scope where "rand()" is being used: at some constructor, or whatever.

Clarification on calling srand [duplicate]

This question already has answers here:
How often should I call srand() in a C++ application?
(4 answers)
What does 'seeding' mean?
(4 answers)
Closed 3 years ago.
I'm wondering why it's advantageous to seed srand at the beginning of the program, instead of where I use it.
I generate pseudo-random numbers when seeding srand at the beginning of my program but I get the numbers all the same when I seed srand inside the function I call to generate the numbers
#include <iostream>
#include <ctime>
using namespace std;
int rng()
{
const int SIZE = 10;
int rng[10];
srand(time(NULL));
for (int i = 0; i < 10; i++)
{
rng[i] = rand() % 128 + 1;
return rng[i];
}
}
int main()
{
int array;
//srand(time(NULL)); If i put it here i get actual random numbers
cout << "Welcome to the program";
cout << "\nthis is your rng\n";
for (int i = 0; i < 10; i++)
{
array = rng();
cout << array << endl;
}
return 0;
}
When I run the program all of the numbers are the same, but when I delete the seeding from in the rng function and uncomment the srand in the main module the numbers are pseudo-random which is what I want. Im wondering why though. I've looked into it and heard that im seeding srand with a time and when I run that function the loop iterates so fast that all of the numbers are generated with the same seed value so they're all the same, but I'm wondering what's the difference from that and having srand(time(NULL)) in main because either way doesn't the function generate the numbers so fast they'll be at the same seed value anyway? It doesn't appear that way because of the different output but im curious, why?
time returns number of seconds since 1.1.1970 so calling it repeatedly during one second will indeed return same values. It doesn't matter exactly where you put srand as long as it's before all rand calls and it should only be called once per program as it's global and obviously resets the random sequence. So if you use it only where you need it, you risk that when some other part of the code will need it too and calls srand again, it will interfere with your rand calls. It's not necessary to call it at all but then the seed will always be the same. It's good for debugging to have an option to set the seed deterministicly.
That said, don't use it, just don't.
As you observed time is not a good seed generator and rand is not even good random number generator, certainly not for floats and x mod n. Use <random> library. It has std::random_device which can generate true random numbers = good seeds. Sadly it's not required to. std::mt19937 is go-to RNG which together with std::XX_YY_distributions should be more than enough for everything but the most extreme need for randomness. It's also thread-safe because you control access to the generator and how it's used.

How To Randomly Pick A Variable Using Rand()

I am trying to make a text based fighter in C++, this is one of the first things that I have done. So far I have this:
//Text Based Fighter
#include <iostream>
#include <stdlib.h> //srand, rand
#include <string>
using namespace std;
int main() {
//Player
int playerHealth = 100;
int attack1;
int attack2;
int attack3;
string attack;
int npc1;
int npc2;
cout << "Do you want to attack " << rand()[npc1,npc2];
//varname = rand() % 10 + 1;
return 0;
}
What I am wanting it to do is randomly pick between npc1 and npc2, thank you.
Also any comments on how I am writing my code would be appreciated, I have only started a couple of days ago thank you, if you need any more detail please feel free to ask, thank you.
You can just use an array of an arbitrary number of variables to choose from:
int attack[n]; //For some int-constant n
attack[rand() % n]; //choose a random attack-variable, use it
For just 2 choices you can take the remainder from 2 with a ternary expression:
int choice = rand() % 2 == 0 ? npc1 : npc2;
If you have more than 2 choices, or even if you don't, you can make an array with those and index into it.
int npc_choices[2];
int choice = npc_choices[rand() % 2];
If the number of choices is not a power of 2 you will likely introduce a very small bias into the selection with the modulo % operator. If you're not working on anything with statistical significance or with a huge number of choices I wouldn't worry about it.
It’s easy to make mistakes when generating pseudo-random numbers. For example, in some cases using rand() % RANGE can lead to a subtly-wrong distribution of numbers. (See this reference for examples of the problem.)
This may not matter if what you are doing is trivial.
If you want high-quality pseudo-random numbers, there are ways to fix rand() (see above reference), but modern C++ also provides <random> and uniform_int_distribution.
Here’s an example, simulating throwing a 6-sided die, adapted from examples in Boost and the C++ Reference:
#include <iostream>
#include <random>
std::random_device rd;
std::mt19937 gen(rd());
int roll_die() {
std::uniform_int_distribution<> dist(1, 6);
return dist(gen);
}
int main() {
std::cout << roll_die() << std::endl;
}
The part that says dist(1, 6) could be changed to dist(0, 1) to produce output in the range [0, 1] (inclusive) with a uniform distribution.
If you have just two choices in C++11 you can use std::bernoulli_distribution and here is an overly simplified sample:
#include <iostream>
#include <random>
int main()
{
std::random_device rd;
std::mt19937 gen(rd());
// give "true"1/2 of the time
// give "false" 1/2 of the time
std::bernoulli_distribution d(0.5);
int npcs[2] = {100, 101};
int index = d(gen) ? 0 : 1;
std::cout << "Do you want to attack " << npcs[index] ;
}
using an array is more flexible since it expands easily to more than two choices and then you would need to use std::uniform_int_distribution to choose between [0,N].
In the long run using rand() is not a good idea, although in many simple cases it may work fine. As Pete mentions as long you understand the limitations of rand() you can use it and the C FAQ has a good section on it, How can I get random integers in a certain range?.

C/C++ algorithm to produce same pseudo-random number sequences from same seed on different platforms? [duplicate]

This question already has answers here:
Consistent pseudo-random numbers across platforms
(5 answers)
Closed 9 years ago.
The title says it all, I am looking for something preferably stand-alone because I don't want to add more libraries.
Performance should be good since I need it in a tight high-performance loop. I guess that will come at a cost of the degree of randomness.
Any particular pseudo-random number generation algorithm will behave like this. The problem with rand is that it's not specified how it is implemented. Different implementations will behave in different ways and even have varying qualities.
However, C++11 provides the new <random> standard library header that contains lots of great random number generation facilities. The random number engines defined within are well-defined and, given the same seed, will always produce the same set of numbers.
For example, a popular high quality random number engine is std::mt19937, which is the Mersenne twister algorithm configured in a specific way. No matter which machine, you're on, the following will always produce the same set of real numbers between 0 and 1:
std::mt19937 engine(0); // Fixed seed of 0
std::uniform_real_distribution<> dist;
for (int i = 0; i < 100; i++) {
std::cout << dist(engine) << std::endl;
}
Here's a Mersenne Twister
Here is another another PRNG implementation in C.
You may find a collection of PRNG here.
Here's the simple classic PRNG:
#include <iostream>
using namespace std;
unsigned int PRNG()
{
// our initial starting seed is 5323
static unsigned int nSeed = 5323;
// Take the current seed and generate a new value from it
// Due to our use of large constants and overflow, it would be
// very hard for someone to predict what the next number is
// going to be from the previous one.
nSeed = (8253729 * nSeed + 2396403);
// Take the seed and return a value between 0 and 32767
return nSeed % 32767;
}
int main()
{
// Print 100 random numbers
for (int nCount=0; nCount < 100; ++nCount)
{
cout << PRNG() << "\t";
// If we've printed 5 numbers, start a new column
if ((nCount+1) % 5 == 0)
cout << endl;
}
}

C++11/Boost Random libraries, starting the generation at a specified point in the cycle

Using the C++11 #include<random> or Boost C++ Boost.Random is it possible to use one seed to start the random number generator at an arbitrary sequence in that you can choose?.
In other words I want to be able to specific where in the sequence the number generator starts, while using the same seed.
For example if i'm using the mt19937 generator with a length of cycle 2^19937-1 I would like to start generating random numbers at a user specified position in the length of the cycle. Say I pick 1000, the generator will start at the 1000th position in the length of the cycle.
Yes. There is a member discard(unsigned long long z) that does this for you. For example:
#include <random>
#include <cassert>
int main()
{
std::mt19937 e1(6492);
std::mt19937 e2(6492);
const int N = 1000;
for (int i = 0; i < N; ++i)
e1();
e2.discard(N);
assert(e1() == e2());
}
This program should not assert.