random even number

Gerb68

Hi
I'm trying very hard to get a random even number from either Math.random() or the nextInt() methods but I cant seem to get anywhere.

I have resorted to to getting a random number from a group of 5 even numbers but still this does not work...

int num2=(int)(Math.random()*10)%25+1;
int[] numbers = new int[]{2,4,6,8,10};

for(int i=0; i < numbers.length; i++)
{
if (numbers == num2)
//if (numbers%2 == 0 && num2 == numbers) I also tried this
{
System.out.println(num2 + " is even number.");
}
if (numbers != num2)
{
System.out.println(num2 + " is odd number.");
}
}

Any ideas on this would help alot and be much appreciated.
Thanx

Ger

John_Mc

Gerb68 wrote: »

Hi
I'm trying very hard to get a random even number from either Math.random() or the nextInt() methods but I cant seem to get anywhere.

I have resorted to to getting a random number from a group of 5 even numbers but still this does not work...

int num2=(int)(Math.random()*10)%25+1;
int[] numbers = new int[]{2,4,6,8,10};

for(int i=0; i < numbers.length; i++)
{
if (numbers == num2)
//if (numbers%2 == 0 && num2 == numbers) I also tried this
{
System.out.println(num2 + " is even number.");
}
if (numbers != num2)
{
System.out.println(num2 + " is odd number.");
}
}

Any ideas on this would help alot and be much appreciated.
Thanx

Ger

Here's how you can do it in .NET C#. Minimum number is 2 and Max is 1000.

Random random = new Random();
int randomNumber =1;
while(randomNumber % 2 != 0)
{
randomNumber = random.Next(2,1000);
}
//randomNumber is even here

srsly78

Just get random int and divide by 2.

John_Mc

srsly78 wrote: »

Just get random int and divide by 2.

Which is essentially what I've done inside a While loop so that it keeps getting a new random int until the modulus 2 is even.

srsly78

Oops I meant multiply by 2

MagicRon

John_Mc wrote: »

Here's how you can do it in .NET C#. Minimum number is 2 and Max is 1000.

Random random = new Random();
int randomNumber =1;
while(randomNumber % 2 != 0)
{
randomNumber = random.Next(2,1000);
}
//randomNumber is even here

"Infinite" Loop Alert...

Giblet

If number != even, add 1

k.p.h

Get number

If (Number % 2 > 0 )

Number + 1 = even number

Mod operator gives the remainder,As John_Mc has shown if the remainder is greater than 0 the number is not even. Use that as a test until you get the number you want.

MagicRon

Giblet wrote: »

If number != even, add 1

if number != even, return 4;

Giblet

Guaranteed random, every time.

ressem

int num2=(int)(Math.random()*10)%25+1;

Not sure what you are trying to do with the mod 25 here. Math.random picks numbers equal or greater than 0, and less than 1.

So
int num=(int)(Math.random()*5)*2;
gives numbers 0,2,4,6,8. Add another 2 to it if you want 2,4,6,8,10.

seamus

You don't need any while loop

(Set of all integers) * 2 === (Set of all even numbers)

k.p.h

seamus wrote: »

You don't need any while loop

(Set of all integers) * 2 === (Set of all even numbers)

Neat..

seamus

k.p.h wrote: »

Neat..

Yes, one of the few things I remember from set theory because it's cool.

Actualy srsly78 got there before me

To flesh it out a little more, if the OP is looking to get a random number between 2 and 10 (inclusive), then all you have to do is get a random number between 1 and 5 (inclusive) and multiply it by 2.

The Math.random() function returns a random floating-point number between 0 and 1 (but not zero or 1). The typical way to turn this into an integer is to multiply the number by 10 (this makes it a random floating-point number between 0 and 9), then use Math.floor() to "chop off" the digits after the decimal point.

So this is

Math.floor(Math.random() * 10)

So now we have a random number between 0 and 9. But we need five digits, not ten.

So now we can use modulus to "reduce" it to a number lower than 5. Or simply % 5. If the number is 9, we now have 4. If the number is 2, it stays 2.

So
Math.floor(Math.random() * 10) % 5

gives us a random number between 0 and 4. But we want between 1 & 5. So we add one. Then to get between 2 and 10, we multiply by two.

So the total line is

(((Math.floor(Math.random() * 10) % 5) + 1) * 2)

You can also multiply the 2 in, and get ((Math.floor(Math.random() * 20) % 10) + 2) but for the sake of documenting a piece of code, I would stick with the first one

ressem

(but not zero or 1)

Zero is allowed. Something of the order of 1 in 10^14 chance. i.e every Murphy's law day.

(((Math.floor(Math.random() * 10) % 5) + 1) * 2)

or
=2+(int)(Math.random()*5)*2;

The (int) cast does the truncation, and is above multiplication on the precedence table. So 2+ [0-4] * 2

Or is there something I'm missing?

longhalloween

Why don't you try this?

#include <stdio.h>
#include <time.h>

int main (){
srand((unsigned)time(NULL));

int randomNum;
int check = 0x0001;
randomNum = rand()%1000 + 2;
if (randomNum & check ==1)
randomNum +=1;


printf("%d", randomNum);

system("pause");
return 0 ;

}

Basically the random number is ANDed with 0x0001, if the last digit is an odd number, 1 is added, if it's even, it's just printed out.

yawha

Or just set the value to itself XORed with 1.