Advice on a simple random number generation problem.

Question

I have a simple integer vector, with only 4 values. I want to loop through the vector, assigning either 0 or 1 to each value. I want it to be random to the point that its different every time.

I thought the following would suffice:

  for (int i = 0; i < (int)numberVect.size(); i++)
  {
      numberVect[i] = rand() % 2;
  }

However, what I found was that every time I would close and re-open the program, the exact same values would be assigned.

So the first time through it would be 1,0,0,0, then if I ran the loop again without closing the program it would be 0,1,0,1. which seems perfect.

HOWEVER, after closing the program and restarting it up again, I would find the first sequence would again be 1,0,0,0 and the second would again be 0,1,0,1.

Also I have a second question in relation to this problem:

Can you suggest a way to make sure that AT LEAST one of the values in the vector is a 1? while still allowing the random generation to work seamlessly?

Thanks a lot and I hope you guys can help :D

possible duplicate of [why do i always get the same sequence of random numbers with rand() ?](http://stackoverflow.com/questions/1108780/why-do-i-always-get-the-same-sequence-of-random-numbers-with-rand) — GWW, Jul 28 '11 at 18:57
Note that `rand()` `srand()` is pretty c-ish. It works just fine for lots of applications, but gets awkward when you need more complicated random-number-generation. I suggest you rather use the c++0x / tr1 random facilities. — leftaroundabout, Jul 28 '11 at 19:09
Why do you cast the result of `size` to `int`, rather than making the counter of type `size_t`?? — Kerrek SB, Jul 28 '11 at 21:08

score 6 · Answer 1 · edited May 23 '17 at 12:11

Question 1

You need to seed your random number generator with srand.

The sequence of pseudo-random numbers generated depends on the seed value (and are thus reproducible with the same seed value). I use the the current time as the seed value:

srand(time(NULL));

Question 2

There are only 16 possible sequences for you to choose from (4 bits).

Out of these, you want to randomly select one with at least 1 high bit. Therefore you just need to randomly select a number from 1 to 15 and then convert it to binary!

There are two ways of doing this:

Method 1:

int R = rand() % 15 + 1;

Method 2:

float r = static_cast<float>(rand())/RAND_MAX;
int R = static_cast<int>( ceil(1 + (15-1)*r) );

Once you have R, convert it to binary. Here's a sample program with Method 1:

// Seed the random number generator
srand(static_cast<unsigned int>(time(NULL)));

// Get the random number from 1 - 15
int R = (rand()%15)+1;
cout<<R<<endl;

// Convert to binary
vector<int> numberVect(4);  
for (int i = static_cast<int>(numberVect.size())-1; i >=0 ; i--)
{
    numberVect[i] = R%2;
    R/=2;
}

// Display
cout<<numberVect[0]<<","<<numberVect[1]<<","<<numberVect[2]<<","<<numberVect[3]<<endl;

Note: `static_cast<int>(x)` is the C++ way of doing `(int)x` if you didn't know.

thank you so much, would never have thought of something like this. — Qas Piano, Jul 29 '11 at 01:52

score 2 · Answer 2 · answered Jul 28 '11 at 18:55

2

You should seed your random generator with the current time at program start to avoid this behaviour.

answered Jul 28 '11 at 18:55

Hyperboreus

31,997
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score 1 · Answer 3 · answered Jul 28 '11 at 18:56

1

You need to seed your random numbers. Using the time is common practice.

answered Jul 28 '11 at 18:56

Isaac Fife

1,659
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score 1 · Accepted Answer · answered Jul 28 '11 at 18:59

1

You need to seed the random generator, before using it. For that you need srand()

I usually do that with passing current time to srand() as:

#include <time.h>

srand(time(NULL)); //seeding!

//now use
int whatever = rand();

Now everytime, you run this code you'll get different random sequence,even if there is one second of difference between two consecutive runs.

answered Jul 28 '11 at 18:59

Nawaz

353,942
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851

thanks so much, thanks to everyone who answered, this answer was the most dummy proof (and i am a dummy ;)) truly i am grateful. has now idea about seeding the rand func :) THANKS!!!! – Qas Piano Jul 28 '11 at 19:44

score 0 · Answer 5 · answered Jul 28 '11 at 18:57

Do this in your program before you generate any random numbers:

  srand(time(NULL));

The random number generator isn't "truly random", it needs to start somewhere, and typically people will use the current system time as a good "kinda random" start point, guaranteeing that each run of the program will be different.

Your second question: can you make sure that at least one of the values is a "1" while still being random? Well, no, of course not, then it wouldn't be random. :) You can post-process, though, if you're willing to forgo some of the "randomness"-- after you generate the vector, check to see if there's at least one "1" value. If not, set one of them to "1". :)

score 0 · Answer 6 · answered Jul 28 '11 at 18:57

You have to initialize the random seed. Include stdlib.h and time.h, then call this before calling rand() for the first time:

srand(time(NULL));

This code initializes the random seed with the current system time, in seconds, so you always get different results (unless you start the program twice during one second.)

score 0 · Answer 7 · answered Jul 28 '11 at 18:58

You're getting the same results each time you run the program because you aren't seeding rand. It's usually recommended to seed based on time:

srand(time(NULL));

To make sure at least 1 number is 1, you could use a bool to indicate if you've gotten a 1 yet. If at the end of the array, you don't have a 1, do a final call to rand with a mod of the vector size, and set that value to 1.