Programming Forums > Application Development > Java

kurt · Feb 12th, 2006, 4:14 AM

I'm trying to get a random double number with includes the low and high bound itself.

I've tried:

Math.random()*(highBound()-lowBound())+lowBound());

if i enter say 6 to 10 for low and high bound respectively, i can't get the number 6.0000 and 10.0000.

Is it possible to include both ends?

Arevos · Feb 12th, 2006, 4:32 AM

Math.random should return a number that is greater than or equal to 0, and less than 1. So your lower bound at least, will eventually turn up.

The problem is the 'eventually' part. Doubles are 64 bit types, so there's 2^64 representable values between 1 and 0; that's 1.8e19 in standard form. With this many choices, it'll be a while before you end up with exactly zero, especially considering that there are only 2.4e9 seconds in the average western human lifespan.

It's perfectly possible to include both ends, but statistically you won't see them unless you limit the number of values that Math.random can return. Perhaps you only need integers, or perhaps you only need numbers that go down to two decimal places.

ReggaetonKing · Feb 12th, 2006, 11:07 AM

highest = 10
lowest = 6
6 + (int) (Math.random() *10)

Polyphemus_ · Feb 12th, 2006, 11:27 AM

Quote:

Originally Posted by reggaeton_king

highest = 10
lowest = 6
6 + (int) (Math.random() *10)

Won't work, when Math.random() returns 1, you will get the number 16.

Use Math.random() * (10 - 6) instead.

ReggaetonKing · Feb 12th, 2006, 12:02 PM

yeaaaah! Sorry it only works when the lowest number is 1, sorry!

6 + (int) (Math.random() * 10 - 6)

Polyphemus_ · Feb 12th, 2006, 12:20 PM

Quote:

Originally Posted by reggaeton_king

yeaaaah! Sorry it only works when the lowest number is 1, sorry!

6 + (int) (Math.random() * 10 - 6)

Still wrong

(Toggle Plain Text)

6 + (int) (Math.random() * (10 - 6))

kurt · Feb 12th, 2006, 12:28 PM

Yeah, thanks guys.

For getting integers from 6 to 10, i have no problems.

It's decimal points from 6.00 to 10.00 that i can't get the both ends.

Since Arevos said that the numbers will eventually turn up, i guess there's no problem then.

I asked this because i thought that math.random returns a value between 0 to 1, not including 0 and 1. Now i'm confused.

ReggaetonKing · Feb 12th, 2006, 12:51 PM

put (int) in front of the statement, it will turn out with no demicals!

jaeusm · Feb 12th, 2006, 2:56 PM

Quote:

I asked this because i thought that math.random returns a value between 0 to 1, not including 0 and 1. Now i'm confused.

Math.random() returns a double greater than or equal to 0.0 and less than 1.0

ReggaetonKing · Feb 12th, 2006, 3:05 PM

it returns 0 - 0.99999999999999999999, better explanation

Feb 12th, 2006, 4:14 AM	#1
kurt Programmer Join Date: Oct 2005 Posts: 65 Rep Power: 3	Random double (inclusive of range) I'm trying to get a random double number with includes the low and high bound itself. I've tried: (Toggle Plain Text) Math.random()(highBound()-lowBound())+lowBound()); Math.random()(highBound()-lowBound())+lowBound()); if i enter say 6 to 10 for low and high bound respectively, i can't get the number 6.0000 and 10.0000. Is it possible to include both ends?

Feb 12th, 2006, 11:07 AM	#3
ReggaetonKing Sexy Programmer Join Date: Nov 2005 Location: New Jersey Posts: 891 Rep Power: 3	highest = 10 lowest = 6 6 + (int) (Math.random() *10) __________________ I would love to change the world, but they won't give me the source code!

Feb 12th, 2006, 12:02 PM	#5
ReggaetonKing Sexy Programmer Join Date: Nov 2005 Location: New Jersey Posts: 891 Rep Power: 3	yeaaaah! Sorry it only works when the lowest number is 1, sorry! 6 + (int) (Math.random() * 10 - 6) __________________ I would love to change the world, but they won't give me the source code!

Feb 12th, 2006, 12:51 PM	#8
ReggaetonKing Sexy Programmer Join Date: Nov 2005 Location: New Jersey Posts: 891 Rep Power: 3	put (int) in front of the statement, it will turn out with no demicals! __________________ I would love to change the world, but they won't give me the source code!

Feb 12th, 2006, 3:05 PM	#10
ReggaetonKing Sexy Programmer Join Date: Nov 2005 Location: New Jersey Posts: 891 Rep Power: 3	it returns 0 - 0.99999999999999999999, better explanation __________________ I would love to change the world, but they won't give me the source code!

Feb 12th, 2006, 4:32 AM	#2
Arevos Programming Guru Join Date: Aug 2005 Location: England Posts: 1,499 Rep Power: 4	Math.random should return a number that is greater than or equal to 0, and less than 1. So your lower bound at least, will eventually turn up. The problem is the 'eventually' part. Doubles are 64 bit types, so there's 2^64 representable values between 1 and 0; that's 1.8e19 in standard form. With this many choices, it'll be a while before you end up with exactly zero, especially considering that there are only 2.4e9 seconds in the average western human lifespan. It's perfectly possible to include both ends, but statistically you won't see them unless you limit the number of values that Math.random can return. Perhaps you only need integers, or perhaps you only need numbers that go down to two decimal places.

Feb 12th, 2006, 12:28 PM	#7
kurt Programmer Join Date: Oct 2005 Posts: 65 Rep Power: 3	Yeah, thanks guys. For getting integers from 6 to 10, i have no problems. It's decimal points from 6.00 to 10.00 that i can't get the both ends. Since Arevos said that the numbers will eventually turn up, i guess there's no problem then. I asked this because i thought that math.random returns a value between 0 to 1, not including 0 and 1. Now i'm confused.

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Programming Forums > Application Development > Java

Random double (inclusive of range)