rand() between m and n

Discussion in 'C++' started by Gary Wessle, Aug 1, 2006.

  1. Gary Wessle

    Gary Wessle Guest

    Hi

    I need help to generate some random numbers between 2 and 8.

    #include <cstdlib>
    using std::rand;

    the following was out of my range,

    int main() {

    for (int i = 0; i < 50; i++){
    int x = (int(rand())/444489786)*8;
    cout << x << '\t' << endl;
    }
    }

    it can be any quality random number.


    thanks
     
    Gary Wessle, Aug 1, 2006
    #1
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  2. Gary Wessle

    Marco Wahl Guest

    Gary Wessle <> writes:
    > I need help to generate some random numbers between 2 and 8.


    So you need random numbers out of 3, 4, 5, 6, 7, right?

    > the following was out of my range,


    Did you get all zeros?

    > int main() {
    >
    > for (int i = 0; i < 50; i++){
    > int x = (int(rand())/444489786)*8;
    > cout << x << '\t' << endl;
    > }
    > }
    >
    > it can be any quality random number.


    So the rand function should suffice.

    Converting the result of the rand function to float
    allows you to normalize the rand result into the
    interval [0, 1] as floating point number. Then you can
    multiply with five (you want to pick a number from a
    range of five numbers, don't you?) and then round and
    add three to the result.

    For some code example see e.g.

    http://cplus.about.com/od/advancedtutorials/l/aa041303c.htm


    HTH
    --
    Marco Wahl
    http://visenso.com
     
    Marco Wahl, Aug 1, 2006
    #2
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  3. > I need help to generate some random numbers between 2 and 8.
    >
    > #include <cstdlib>
    > using std::rand;
    >
    > the following was out of my range,
    >
    > int main() {
    >
    > for (int i = 0; i < 50; i++){
    > int x = (int(rand())/444489786)*8;
    > cout << x << '\t' << endl;
    > }
    > }


    // return random number elm[from; upto]
    int rnd_range( int from, int upto)
    {
    return (rand() % (upto - from + 1)) + from;
    }
     
    Gernot Frisch, Aug 1, 2006
    #3
  4. Gary Wessle

    Chris Theis Guest

    "Gernot Frisch" <> wrote in message
    news:...
    >
    >> I need help to generate some random numbers between 2 and 8.
    >>
    >> #include <cstdlib>
    >> using std::rand;
    >>
    >> the following was out of my range,
    >>
    >> int main() {
    >>
    >> for (int i = 0; i < 50; i++){
    >> int x = (int(rand())/444489786)*8;
    >> cout << x << '\t' << endl;
    >> }
    >> }

    >
    > // return random number elm[from; upto]
    > int rnd_range( int from, int upto)
    > {
    > return (rand() % (upto - from + 1)) + from;
    > }
    >


    This works absolutely fine, but still the other approach using a random
    number between r out of [0,1) ( x = Min + r * (Max-Min) ) is favorable
    because it doesn't tamper with the "randomness" of the generator and has no
    influence on the period of the generated sequence.

    Cheers
    Chris
     
    Chris Theis, Aug 1, 2006
    #4
  5. Gary Wessle

    Gary Wessle Guest

    "Chris Theis" <> writes:

    > "Gernot Frisch" <> wrote in message
    > news:...
    > >
    > >> I need help to generate some random numbers between 2 and 8.
    > >>
    > >> #include <cstdlib>
    > >> using std::rand;
    > >>
    > >> the following was out of my range,
    > >>
    > >> int main() {
    > >>
    > >> for (int i = 0; i < 50; i++){
    > >> int x = (int(rand())/444489786)*8;
    > >> cout << x << '\t' << endl;
    > >> }
    > >> }

    > >
    > > // return random number elm[from; upto]
    > > int rnd_range( int from, int upto)
    > > {
    > > return (rand() % (upto - from + 1)) + from;
    > > }
    > >

    >
    > This works absolutely fine, but still the other approach using a random
    > number between r out of [0,1) ( x = Min + r * (Max-Min) ) is favorable



    do you mean
    int x = from + rand() * (to-from) ?

    > because it doesn't tamper with the "randomness" of the generator and has no
    > influence on the period of the generated sequence.
    > Cheers
    > Chris



    I am not sure what wrong I am doing, followed the link provided by
    Marco, which was
    http://cplus.about.com/od/advancedtutorials/l/aa041303c.htm

    the following puts out zeros on my screen. only zeros no mater how
    many times I run it.

    #include <ctime>
    using std::time;


    int main() {

    int x;
    const int N = 100;
    srand( static_cast<unsigned> (time( NULL )) );
    for (int i = 0; i < 4; i++) {
    x = static_cast<int> ( N*rand())/(RAND_MAX+1) ;
    cout << x << endl;
    }
    }
     
    Gary Wessle, Aug 1, 2006
    #5
  6. Gary Wessle

    Marcus Kwok Guest

    Gary Wessle <> wrote:
    > I am not sure what wrong I am doing, followed the link provided by
    > Marco, which was
    > http://cplus.about.com/od/advancedtutorials/l/aa041303c.htm
    >
    > the following puts out zeros on my screen. only zeros no mater how
    > many times I run it.


    #include <iostream>

    > #include <ctime>
    > using std::time;


    using std::cout;

    > int main() {
    >
    > int x;
    > const int N = 100;
    > srand( static_cast<unsigned> (time( NULL )) );
    > for (int i = 0; i < 4; i++) {
    > x = static_cast<int> ( N*rand())/(RAND_MAX+1) ;


    Let's rearrange your spacing:

    x = static_cast<int>(N*rand()) / (RAND_MAX+1);

    So, you are taking the value of N*rand(), and then casting it to an int.
    Then you are dividing it by (RAND_MAX+1), which is also an int.
    Therefore you have an int/int, so the division truncates, resulting in a
    zero.

    I would try casting the result of N*rand() to a double, then performing
    the (floating-point) division, then casting THAT result to an int:

    x = static_cast<int>(static_cast<double>(N*rand()) / (RAND_MAX+1));

    Of course, you could break this up into smaller steps.

    > cout << x << endl;
    > }
    > }


    --
    Marcus Kwok
    Replace 'invalid' with 'net' to reply
     
    Marcus Kwok, Aug 1, 2006
    #6
  7. Gary Wessle

    Gary Wessle Guest

    lid (Marcus Kwok) writes:

    > Gary Wessle <> wrote:
    > > I am not sure what wrong I am doing, followed the link provided by
    > > Marco, which was
    > > http://cplus.about.com/od/advancedtutorials/l/aa041303c.htm
    > >
    > > the following puts out zeros on my screen. only zeros no mater how
    > > many times I run it.

    >
    > #include <iostream>
    >
    > > #include <ctime>
    > > using std::time;

    >
    > using std::cout;
    >
    > > int main() {
    > >
    > > int x;
    > > const int N = 100;
    > > srand( static_cast<unsigned> (time( NULL )) );
    > > for (int i = 0; i < 4; i++) {
    > > x = static_cast<int> ( N*rand())/(RAND_MAX+1) ;

    >
    > Let's rearrange your spacing:
    >
    > x = static_cast<int>(N*rand()) / (RAND_MAX+1);
    >
    > So, you are taking the value of N*rand(), and then casting it to an int.
    > Then you are dividing it by (RAND_MAX+1), which is also an int.
    > Therefore you have an int/int, so the division truncates, resulting in a
    > zero.
    >
    > I would try casting the result of N*rand() to a double, then performing
    > the (floating-point) division, then casting THAT result to an int:
    >
    > x = static_cast<int>(static_cast<double>(N*rand()) / (RAND_MAX+1));
    >
    > Of course, you could break this up into smaller steps.
    >
    > > cout << x << endl;
    > > }
    > > }

    >
    > --
    > Marcus Kwok
    > Replace 'invalid' with 'net' to reply



    do you mean this, it still puts out all zeros, could you try it on
    your machine and report back? thanks


    ****************
    ****************
    #include <iostream>
    #include <cstdlib>

    using namespace std;

    int main() {

    srand( static_cast<unsigned> (time( NULL )) );
    int N = 8;
    int x;
    for (unsigned i=0; i<5; i++){
    x = static_cast<int>(static_cast<double>(N*rand()) / (RAND_MAX+1));
    cout << x << '\t';
    }
    }
    ****************
    ****************
     
    Gary Wessle, Aug 2, 2006
    #7
  8. Gary Wessle

    Marco Wahl Guest

    Gary Wessle <> writes:

    > do you mean this, it still puts out all zeros, could you try it on
    > your machine and report back? thanks
    >
    >
    > #include <iostream>
    > #include <cstdlib>
    >
    > using namespace std;
    >
    > int main() {
    >
    > srand( static_cast<unsigned> (time( NULL )) );
    > int N = 8;
    > int x;
    > for (unsigned i=0; i<5; i++){
    > x = static_cast<int>(static_cast<double>(N*rand()) / (RAND_MAX+1));
    > cout << x << '\t';
    > }
    > }


    I checked and also get all 0s. The output looks more interesting when using the line

    x = static_cast<int>(N * static_cast<double>(rand()) / (static_cast<double>(RAND_MAX) + 1));

    There is an overflow on my machine at least when calculating RAND_MAX+1.


    HTH
    --
    Marco Wahl
    http://visenso.com
     
    Marco Wahl, Aug 2, 2006
    #8
  9. Gary Wessle

    Guest

    Hello all,
    why you are going far away?
    first use srand() to randomize radom number, srand get a parameter as
    seed, like:
    srand(3); or you can use srand(time(0));
    then use this method to have randome number in a certain range:
    x=minum number +rand()% maximum number;
    for example: x=1+rand()%6; gives you a random number between 1 to 6;
    Also you mist use srand() before using rand, you can use it at the
    first of main()
    Hope I could help you;
     
    , Aug 2, 2006
    #9
  10. Gary Wessle

    Gary Wessle Guest

    the thing is if you need to generate say 5 random numbers between 2
    and 8 then this will not work, run this

    for (int i= 0; i< 5; i++){
    srand( time( 0));
    int x= 2+ rand() %8;
    cout << x << '\n';
    }
     
    Gary Wessle, Aug 2, 2006
    #10
  11. Gary Wessle

    Kai-Uwe Bux Guest

    Gary Wessle wrote:

    > the thing is if you need to generate say 5 random numbers between 2
    > and 8 then this will not work, run this
    >
    > for (int i= 0; i< 5; i++){
    > srand( time( 0));
    > int x= 2+ rand() %8;


    int x = 2 + rand() % 6;

    > cout << x << '\n';
    > }



    Best

    Kai-Uwe Bux
     
    Kai-Uwe Bux, Aug 2, 2006
    #11
  12. Gary Wessle

    Gary Wessle Guest

    Marco Wahl <> writes:

    > Gary Wessle <> writes:
    >
    > > do you mean this, it still puts out all zeros, could you try it on
    > > your machine and report back? thanks
    > >
    > >
    > > #include <iostream>
    > > #include <cstdlib>
    > >
    > > using namespace std;
    > >
    > > int main() {
    > >
    > > srand( static_cast<unsigned> (time( NULL )) );
    > > int N = 8;
    > > int x;
    > > for (unsigned i=0; i<5; i++){
    > > x = static_cast<int>(static_cast<double>(N*rand()) / (RAND_MAX+1));
    > > cout << x << '\t';
    > > }
    > > }

    >
    > I checked and also get all 0s. The output looks more interesting when using the line
    >
    > x = static_cast<int>(N * static_cast<double>(rand()) / (static_cast<double>(RAND_MAX) + 1));
    >
    > There is an overflow on my machine at least when calculating RAND_MAX+1.
    >
    >
    > HTH
    > --
    > Marco Wahl
    > http://visenso.com


    here is the final code for future readers.

    ****************************************************************
    int randam(int from, int to) {
    int f = from;
    int t = to;
    int a = to-from+1;
    return static_cast<int> (a * static_cast<double>(rand())/
    (static_cast<double>(RAND_MAX) + 1))+from;
    }

    int main() {

    // print out 5 random numbers between 2 and 8
    srand( static_cast<unsigned> (time( NULL )) );
    for(int i=0; i<5; i++) cout << randam(2,8) << " ";

    }
     
    Gary Wessle, Aug 2, 2006
    #12
  13. Gary Wessle

    Chris Theis Guest

    "Gary Wessle" <> wrote in message
    news:...
    > Marco Wahl <> writes:
    >
    >> Gary Wessle <> writes:
    >>
    >> > do you mean this, it still puts out all zeros, could you try it on
    >> > your machine and report back? thanks
    >> >
    >> >
    >> > #include <iostream>
    >> > #include <cstdlib>
    >> >
    >> > using namespace std;
    >> >
    >> > int main() {
    >> >
    >> > srand( static_cast<unsigned> (time( NULL )) );
    >> > int N = 8;
    >> > int x;
    >> > for (unsigned i=0; i<5; i++){
    >> > x = static_cast<int>(static_cast<double>(N*rand()) /
    >> > (RAND_MAX+1));
    >> > cout << x << '\t';
    >> > }
    >> > }

    >>
    >> I checked and also get all 0s. The output looks more interesting when
    >> using the line
    >>
    >> x = static_cast<int>(N * static_cast<double>(rand()) /
    >> (static_cast<double>(RAND_MAX) + 1));
    >>
    >> There is an overflow on my machine at least when calculating RAND_MAX+1.
    >>
    >>
    >> HTH
    >> --
    >> Marco Wahl
    >> http://visenso.com

    >
    > here is the final code for future readers.
    >
    > ****************************************************************
    > int randam(int from, int to) {
    > int f = from;
    > int t = to;
    > int a = to-from+1;
    > return static_cast<int> (a * static_cast<double>(rand())/
    > (static_cast<double>(RAND_MAX) + 1))+from;
    > }
    >
    > int main() {
    >
    > // print out 5 random numbers between 2 and 8
    > srand( static_cast<unsigned> (time( NULL )) );
    > for(int i=0; i<5; i++) cout << randam(2,8) << " ";
    >
    > }


    Hi Gary,

    > do you mean int x = from + rand() * (to-from) ?

    Yes, that's what I mean. It a simple scaling and adding an offset to the
    result of a random number in the [0,1).

    The code above should give you what you are looking for. In a more compact
    form you can write it like this:

    return static_cast<int>( from + ( (to - from) * (rand() / (RAND_MAX +
    1.0)) ));

    But be careful - the statement above is different than writing

    return static_cast<int>( from + ( (to - from) * (rand() / (RAND_MAX +
    1)) ));

    The "beast" that you were running into was the integer division of rand() by
    another much larger integer. The result of this integer devision is always 0
    and that's why you obtained zeros (or the offset from). However, by using
    1.0 instead of 1 in the calculation you implicitly convert the divisor to a
    float and consequently the division is performed and results in a float.

    HTH
    Chris
     
    Chris Theis, Aug 2, 2006
    #13
  14. Gary Wessle

    Chris Theis Guest

    <> wrote in message
    news:...
    > Hello all,
    > why you are going far away?
    > first use srand() to randomize radom number, srand get a parameter as
    > seed, like:
    > srand(3); or you can use srand(time(0));
    > then use this method to have randome number in a certain range:
    > x=minum number +rand()% maximum number;


    Yes, that surely works but brings back my original comment. Using the modulo
    operation you interfere with the randomness and don't make use of the full
    period of the random generator. Anyway, it depends on your application
    whether this matters or not.

    Cheers
    Chris
     
    Chris Theis, Aug 2, 2006
    #14
  15. Gary Wessle

    Kai-Uwe Bux Guest

    Chris Theis wrote:

    >
    > <> wrote in message
    > news:...
    >> Hello all,
    >> why you are going far away?
    >> first use srand() to randomize radom number, srand get a parameter as
    >> seed, like:
    >> srand(3); or you can use srand(time(0));
    >> then use this method to have randome number in a certain range:
    >> x=minum number +rand()% maximum number;

    >
    > Yes, that surely works but brings back my original comment. Using the
    > modulo operation you interfere with the randomness


    How, and why?

    > and don't make use of the full period of the random generator.


    Huh? I see that that can happen. However, I do not see why the alternative
    of chopping [0,RAND_MAX] into n intervals of equal length it a priory
    better.

    As far as I can see, any kind of mapping N values to n < N values can tamper
    with the period or other measures of randomness. Whether a particular
    mapping fares better or worse depends on the underlying random number
    generator.


    > Anyway, it depends on your application whether this matters or not.


    True, and I think up-thread it was stated that quality does not matter.


    Best

    Kai-Uwe Bux
     
    Kai-Uwe Bux, Aug 2, 2006
    #15
  16. Gary Wessle wrote:
    > Hi
    >
    > I need help to generate some random numbers between 2 and 8.
    >
    > #include <cstdlib>
    > using std::rand;
    >
    > the following was out of my range,
    >
    > int main() {
    >
    > for (int i = 0; i < 50; i++){
    > int x = (int(rand())/444489786)*8;
    > cout << x << '\t' << endl;
    > }
    > }
    >
    > it can be any quality random number.
    >
    >
    > thanks


    This is one way of doing it:


    #include <iostream>
    #include <cstdlib>
    #include <ctime>

    using namespace std;


    template<typename T>
    class Random {
    public:
    Random(T min_, T max_);

    T operator()(T min_, T max_); ///< Set new range for random numbers,
    and return a random number.

    T operator()(); ///< Return a random number.

    private:
    T _min, _max;
    T _delta;
    };


    template<typename T>
    inline Random<T>::Random(T min_, T max_)
    :_min(min_),
    _max(max_){
    srand(time(0));
    srand(time(0));
    _delta = RAND_MAX / (_max - _min);}


    template<typename T>
    inline T Random<T>::eek:perator()(T min_, T max_){
    _min = min_;
    _max = max_;
    _delta = RAND_MAX / (_max - _min);
    return(operator()());}


    template<typename T>
    inline T Random<T>::eek:perator()(){
    return((rand() / _delta) + _min);}



    int main(){
    Random<float> randFloat(-10, -20);

    cerr << "5 random floating point numbers between -10 and -20: ";
    for(int i = 0; i < 5; i++)
    cerr << fixed << randFloat() << " ";
    cerr << endl;

    randFloat(-100, 100);

    cerr << "5 random floating point numbers between -100 and 100: ";
    for(int i = 0; i < 5; i++)
    cerr << fixed << randFloat() << " ";
    cerr << endl;


    Random<int> randInt(20000, 10000);

    cerr << "5 random integers between 20000 and 10000: ";
    for(int i = 0; i < 5; i++)
    cerr << fixed << randInt() << " ";
    cerr << endl;

    randInt(-10, 10);

    cerr << "5 random integers between -10 and 10: ";
    for(int i = 0; i < 5; i++)
    cerr << fixed << randInt() << " ";
    cerr << endl;
    return 0;}


    lars@ibmr52:~/programming/c$ g++ -g -Wall random.cpp -o random &&
    ../random
    5 random floating point numbers between -10 and -20: -15.719680
    -10.247149 -10.654663 -16.673820 -13.213199
    5 random floating point numbers between -100 and 100: 63.869286
    -77.031998 -25.940384 -4.096813 -49.285671
    5 random integers between 20000 and 10000: 14281 19753 19346 13327
    16787
    5 random integers between -10 and 10: 6 -8 -3 -1 -5


    --
    mvh, Lars Rune Nøstdal
    http://lars.nostdal.org/
     
    =?iso-8859-1?q?Lars_Rune_N=F8stdal?=, Aug 2, 2006
    #16
  17. Gary Wessle

    Marcus Kwok Guest

    Gary Wessle <> wrote:
    > do you mean this, it still puts out all zeros, could you try it on
    > your machine and report back? thanks
    >
    >
    > ****************
    > ****************
    > #include <iostream>
    > #include <cstdlib>
    >
    > using namespace std;
    >
    > int main() {
    >
    > srand( static_cast<unsigned> (time( NULL )) );
    > int N = 8;
    > int x;
    > for (unsigned i=0; i<5; i++){
    > x = static_cast<int>(static_cast<double>(N*rand()) / (RAND_MAX+1));
    > cout << x << '\t';
    > }
    > }
    > ****************
    > ****************


    Interesting, on my machine (Windows XP, VS .NET 2003), after I added
    #include <ctime>, my output for repeated runs was (using spaces instead
    of tabs):

    2 5 1 1 1
    2 6 6 7 2
    2 6 2 2 7

    etc.

    --
    Marcus Kwok
    Replace 'invalid' with 'net' to reply
     
    Marcus Kwok, Aug 2, 2006
    #17
  18. Gary Wessle

    Chris Theis Guest

    "Kai-Uwe Bux" <> wrote in message
    news:eaqde5$q22$...
    > Chris Theis wrote:
    >
    >>
    >> <> wrote in message
    >> news:...
    >>> Hello all,
    >>> why you are going far away?
    >>> first use srand() to randomize radom number, srand get a parameter as
    >>> seed, like:
    >>> srand(3); or you can use srand(time(0));
    >>> then use this method to have randome number in a certain range:
    >>> x=minum number +rand()% maximum number;

    >>
    >> Yes, that surely works but brings back my original comment. Using the
    >> modulo operation you interfere with the randomness

    >
    > How, and why?
    >
    >> and don't make use of the full period of the random generator.

    >
    > Huh? I see that that can happen. However, I do not see why the alternative
    > of chopping [0,RAND_MAX] into n intervals of equal length it a priory
    > better.
    >


    Well we're getting into a hot topic here which has been discussed
    extensively. On one hand using modulo you might skew the results towards
    lower values and on the other hand you can have an effect on the randomness.
    The reason is that the modulo operation uses low order bits and depending on
    the type of random generator the high-order bits might show better
    randomness. However, this is something which depends on the implementation
    of rand and is the subject of endless discussions! For a thorough discussion
    on this including the mathematics you might refer to Knuth's book The Art of
    computer programming Vol.2 and/or Numerical Recipes.

    > As far as I can see, any kind of mapping N values to n < N values can
    > tamper
    > with the period or other measures of randomness. Whether a particular
    > mapping fares better or worse depends on the underlying random number
    > generator.
    >


    Yes and no. Of course you will never get good results if your random
    generator is crap. However, the simple scaling and offsetting does not
    interfere with any property of the RNG and therefore does not affect the
    "randomness". In opposition to this the modulo mapping can affect the
    randomness because you're introducing a limitation.

    >
    >> Anyway, it depends on your application whether this matters or not.

    >
    > True, and I think up-thread it was stated that quality does not matter.
    >
    >
    > Best
    >
    > Kai-Uwe Bux


    Cheers
    Chris
     
    Chris Theis, Aug 2, 2006
    #18
  19. Gary Wessle

    Pete Becker Guest

    Chris Theis wrote:
    > "Kai-Uwe Bux" <> wrote in message
    > news:eaqde5$q22$...
    >
    >>
    >>Huh? I see that that can happen. However, I do not see why the alternative
    >>of chopping [0,RAND_MAX] into n intervals of equal length it a priory
    >>better.
    >>

    >
    >
    > Well we're getting into a hot topic here which has been discussed
    > extensively. On one hand using modulo you might skew the results towards
    > lower values and on the other hand you can have an effect on the randomness.
    > The reason is that the modulo operation uses low order bits


    That's not the reason. Consider a drastically simplified case: a rand()
    function that produces values between 0 and 5 inclusive, and an attempt
    to generate values from 0 to 4 inclusive. Using the remainder operator
    (C and C++ do not have a modulus operator, but for positive arguments
    remainder and modulus do the same thing), the generated value 0, 1, 2,
    3, and 4 each map to themselves, and the generated value 5 maps to 0. So
    you get twice as many 0's in the output.

    When you use floating point, you're still mapping six values into five
    slots, and one of the slots will come up twice as often as the others.
    Try it.

    In general, this problem occurs whenever the number of values in the
    range returned by the generator is not an exact multiple of the number
    of values in the target range. The solution, regardless of how you
    implement the range conversion, is to discard some values from the
    generator, so that you end up with a range whose size is a multiple of
    the size of the target range. In the case of the simple example above,
    throw away every 5 that you get from the generator.

    > and depending on
    > the type of random generator the high-order bits might show better
    > randomness. However, this is something which depends on the implementation
    > of rand and is the subject of endless discussions! For a thorough discussion
    > on this including the mathematics you might refer to Knuth's book The Art of
    > computer programming Vol.2 and/or Numerical Recipes.


    That is, indeed, one of the assertions made about random number
    generators. The mismatch in range sizes is far more significant.

    >
    >
    >>As far as I can see, any kind of mapping N values to n < N values can
    >>tamper
    >>with the period or other measures of randomness. Whether a particular
    >>mapping fares better or worse depends on the underlying random number
    >>generator.
    >>

    >
    >
    > Yes and no. Of course you will never get good results if your random
    > generator is crap. However, the simple scaling and offsetting does not
    > interfere with any property of the RNG and therefore does not affect the
    > "randomness".


    It does affect the randomness of the resulting sequence, as shown above.
     
    Pete Becker, Aug 2, 2006
    #19
  20. Gary Wessle

    Pete Becker Guest

    Chris Theis wrote:

    > "Gary Wessle" <> wrote in message
    > news:...
    >
    >>
    >>here is the final code for future readers.
    >>
    >>****************************************************************
    >>int randam(int from, int to) {
    >> int f = from;
    >> int t = to;
    >> int a = to-from+1;
    >> return static_cast<int> (a * static_cast<double>(rand())/
    >> (static_cast<double>(RAND_MAX) + 1))+from;
    >>}
    >>
    >>int main() {
    >>
    >>// print out 5 random numbers between 2 and 8
    >> srand( static_cast<unsigned> (time( NULL )) );
    >> for(int i=0; i<5; i++) cout << randam(2,8) << " ";
    >>
    >>}

    >
    >
    > The code above should give you what you are looking for. In a more compact
    > form you can write it like this:
    >


    As I pointed out in another message, this doesn't compensate for
    mismatched range sizes. The bias is less obvious than when using
    remainainder because remainder puts the excess values down at the low
    end of the target range, while the floating-point approach distributes
    them more evenly. But try that code with a custom rand function that
    generates values between 0 and 8, inclusive. One of the values will turn
    up twice as often as the others. (I haven't done the analysis to figure
    out which one.)

    To avoid this bias, compute the largest multiple of the number of values
    in the target range (m * (to - from + 1)) that is less than or equal to
    the number of values in the generator's range (loosely, RAND_MAX+1, but
    beware of overlows). Each time you call rand, check whether the value is
    greater than or equal to this largest multiple; if it is, call rand again.
     
    Pete Becker, Aug 2, 2006
    #20
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