rand() between m and n

**Marco Wahl** · Aug 1 '06, 08:05 AM

Re: rand() between m and n

Gary Wessle <phddas@yahoo.c omwrites:

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

**Gernot Frisch** · Aug 1 '06, 09:05 AM

Re: rand() between m and n

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;
}

**Chris Theis** · Aug 1 '06, 03:35 PM

Re: rand() between m and n

"Gernot Frisch" <Me@Privacy.net wrote in message
news:4j8kd7F6qo 0lU1@individual .net...

>

>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

**Gary Wessle** · Aug 1 '06, 04:45 PM

Re: rand() between m and n

"Chris Theis" <christian.thei s@nospam.cern.c hwrites:

"Gernot Frisch" <Me@Privacy.net wrote in message
news:4j8kd7F6qo 0lU1@individual .net...

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<uns igned(time( NULL )) );
for (int i = 0; i < 4; i++) {
x = static_cast<int ( N*rand())/(RAND_MAX+1) ;
cout << x << endl;
}
}

**Marcus Kwok** · Aug 1 '06, 06:35 PM

Re: rand() between m and n

Gary Wessle <phddas@yahoo.c omwrote:

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<uns igned(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<d ouble>(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

**Gary Wessle** · Aug 2 '06, 01:15 AM

Re: rand() between m and n

ricecake@gehenn om.invalid (Marcus Kwok) writes:

Gary Wessle <phddas@yahoo.c omwrote:

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<uns igned(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<d ouble>(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<uns igned(time( NULL )) );
int N = 8;
int x;
for (unsigned i=0; i<5; i++){
x = static_cast<int >(static_cast<d ouble>(N*rand() ) / (RAND_MAX+1));
cout << x << '\t';
}
}
*************** *
*************** *

**Marco Wahl** · Aug 2 '06, 07:35 AM

Re: rand() between m and n

Gary Wessle <phddas@yahoo.c omwrites:

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<uns igned(time( NULL )) );
int N = 8;
int x;
for (unsigned i=0; i<5; i++){
x = static_cast<int >(static_cast<d ouble>(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<dou ble>(rand()) / (static_cast<do uble>(RAND_MAX) + 1));

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

HTH
--
Marco Wahl

http://visenso.com

**arash.koushkestani@gmail.com** · Aug 2 '06, 08:15 AM

Re: rand() between m and n

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;

**Gary Wessle** · Aug 2 '06, 08:55 AM

Re: rand() between m and n

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';
}

**Kai-Uwe Bux** · Aug 2 '06, 09:25 AM

Re: rand() between m and n

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

**Gary Wessle** · Aug 2 '06, 09:25 AM

Re: rand() between m and n

Marco Wahl <mw@visenso.dew rites:

Gary Wessle <phddas@yahoo.c omwrites:
>

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<uns igned(time( NULL )) );
int N = 8;
int x;
for (unsigned i=0; i<5; i++){
x = static_cast<int >(static_cast<d ouble>(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<dou ble>(rand()) / (static_cast<do uble>(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<dou ble>(rand())/
(static_cast<do uble>(RAND_MAX) + 1))+from;
}

int main() {

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

}

**Chris Theis** · Aug 2 '06, 01:35 PM

Re: rand() between m and n

"Gary Wessle" <phddas@yahoo.c omwrote in message
news:874pwveb1o .fsf@localhost. localdomain...

Marco Wahl <mw@visenso.dew rites:
>

>Gary Wessle <phddas@yahoo.c omwrites:
>>

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<uns igned(time( NULL )) );
int N = 8;
int x;
for (unsigned i=0; i<5; i++){
x = static_cast<int >(static_cast<d ouble>(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<dou ble>(rand()) /
>(static_cast<d ouble>(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<dou ble>(rand())/
(static_cast<do uble>(RAND_MAX) + 1))+from;
}
>
int main() {
>
// print out 5 random numbers between 2 and 8
srand( static_cast<uns igned(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 '06, 01:35 PM

Re: rand() between m and n

<arash.koushkes tani@gmail.comw rote in message
news:1154507129 .015574.198850@ h48g2000cwc.goo glegroups.com.. .

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

**Kai-Uwe Bux** · Aug 2 '06, 02:35 PM

Re: rand() between m and n

Chris Theis wrote:

>
<arash.koushkes tani@gmail.comw rote in message
news:1154507129 .015574.198850@ h48g2000cwc.goo glegroups.com.. .

>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

rand() between m and n

rand() between m and n

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