Math question about random range and fractions

Question

How do you say 2/3 of a number to give in random range? This is what I tried and it seemed to work at first, but this is a no go.

 var  power = //this is an inspector input number
 Slam = Random.Range(power/3 + power/3, power +1);

I am basically trying to say that I want slam to be a random number between 2/3s of power and total power.

Thanks

Answer 1

Best Answer

Answer by CHPedersen · May 26, 2011 at 07:05 AM

I just tested this (in C#), it worked fine for me and seems to do what you want:

 int test = 5;
 float Slam = Random.Range(0.6666f * test, test);

The reason it didn't work for you is possible due to your division getting truncated to an integer, if power is dynamically typed to an integer.

Notice that,

 int test = 5;
 Debug.Log(test / 3);

Outputs "1", because (int)(5/3) = 1.

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superventure · May 26, 2011 at 07:12 AM 0

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What about in Js? I tried .6666 and it's still giving me weird results.

CHPedersen · May 26, 2011 at 07:21 AM 0

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Can you elaborate on "weird results"? Try Debug.Log'ing both the precise input and the results, to see what is actually being produced. If you have 0.6666f*power taking place inside the call to Random.Range, then move it out and Debug.Log it first, to see what that calculation yields. That's the only way to debug these things - study each component one by one.

superventure · May 26, 2011 at 07:32 AM 0

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Well, I did suspect integers being a problem, but then I went and added '.0' to the end of the power variable. Numbers divisible by 3 work, others give me calculations under 1.0 ... Is there any way to round off the numbers we divide by in the equation so that integers can be used?

CHPedersen · May 26, 2011 at 07:47 AM 2

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Well, first of all, don't do those divisions. :) Division is actually computationally slower than multiplication, by more than 2x, so try to convert stuff into multiplications instead, especially in loops.

Second of all, I'm admittedly not a Js-programmer and I never code anything that's dynamically typed, but couldn't you simply write "var power : float;" to enforce that this is a floating point type? Then, any arithmetic operation you perform on it with a constant (such as multiplication by 0.6666) should also return a float without need of further casting.

superventure · May 26, 2011 at 08:22 AM 0

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The odd thing is and I just now realized, the '/3' works just fine except for in a specific onTrigger function. It refuses to calculate in any form of decency, and it is the exact same code! I also tried using float as well, but nothing changed. I think I do need a multiplication solution for this, I just don't know how to write it out.

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Answer 2

Answer by TowerOfBricks · May 26, 2011 at 07:57 PM

 var power : float = 100;

 slam : float = Random.Range (0.66666*power,power);
 Debug.Log ("A number in between 2/3 of "+power+" and "+power+" is for example "+slam);

That would work.

You could also do:

 slam : float = power-Random.value*power*0.3333;

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Math question about random range and fractions

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