Rand.Int

[AK.E]

In my program I've been trying to get one variable to generate a number any from 2-14 and 28-36.

Turing:

var randomNumber : int := 0 | | % some code goes here | loop randomNumber := Rand.Int (2. 14) end loop

I also want the variable to generate anything from 28 - 36.
Is there a way to do this? I don't want any numbers from 15 - 27.
If there is more information that you need just ask.

[Teti]

Its hard to help without more code to see how it will work with your program.

[ericfourfour]

You can try getting a random number between 2 and 22. If it is greater than 14, add 14. I'm sure there is a formula you can use as well.

[agnivohneb]

You can try something like this

code:

var randomNumber : int := 0     |     | % some code goes here     | loop     randomNumber := Rand.Int (2, 36)     exit when randomNumber <= 14 or randomNumber >= 28 end loop

I have a question for you how did you get it to say turing then write the code in the turing colors

[[Gandalf]]

Or you could use two Rand.Int() calls, which would be slightly more efficient then the above while loop. Not a big deal though.

agnivohneb, use [syntax="Turing"]code goes here[/syntax].

[Cervantes]

Use syntax tags. [ syntax="turing"] code [ /syntax]


agnivohneb's method would work, but it's a bad way to do it. On a bad day, it could take a long time to generate a random number between 2 and 36 that is <= 14 or >= 28. It's possible we could just keep drawing numbers between 15 and 27 for a long time.

ericfourfour's method, however, is good. We have to generate exactly one random number every time. No more.

[Cervantes]

Gandalf @ Wed Jan 24, 2007 5:15 pm wrote:

Or you could use two Rand.Int() calls, which would be slightly more efficient then the above while loop. Not a big deal though.

2-14 has 13 possible values in the range, but 28-36 has 9 possible values. Thus, if you did something like this:

code:

if Rand.Int (0, 1) = 0 then     put Rand.Int (2, 14) else     put Rand.Int (28, 36) end if

The chances of getting a 28 is higher than the chances of getting a 2. You'd have to do correction for that by changing the Rand.Int (0,1) at the start to be more like Rand.Int (1, 13+9) then checking whether that is <= 13 or not. So, yeah: that's needlessly complicated. ericfourfour's method.

edit: [Gandalf], your name breaks the quote system. It's the use of [ and ].