if you need biased numbers you should look into bell curves and statistics.
the most efficent way if your range is small could be to use a lookup table with the correct distribution. just squaring the numbers may seem like a good idea, lets see.
normal random distribution of 0-9
0 1 2 3 4 5 6 7 8 9
distribution squared
0 1 4 9 16 25 36 49 64 81
distribution normalized (ie num/maxNum*range)
0 0.111 0.444 1 1.778 2.778 4 5.444 7.111 9
hmm, seems the distribution just got spread away from the center (ie clustered below and above the mean).
ie that wont work.
Random number between 2 bounds.
a person,
if you dont start with 0-9 and instead with 0-1, i believe, squring it will work....
hmm.... guess not...
wait.. would this not work?
seems to me that now instead of the mean bieng 0.45 in the caes of the top 10 #s, it is now 0.285, and now all biased#s that started below .707.... are now rounded down
please explain y this method doesnt work...
lol
ive changed my mind on weather this method works or not around 4 times durring the time of this post...
someone has probobly already cleared the whole thing up
*sigh*
tazzel3d ~ dwiel
if you dont start with 0-9 and instead with 0-1, i believe, squring it will work....
rand .0 .1 .2 .3 .4 .5 .6 .7 .8 .9 bias .0 .01 .04 .09 .16 .25 .36 .49 .64 .81 hmm.... guess not...
wait.. would this not work?
seems to me that now instead of the mean bieng 0.45 in the caes of the top 10 #s, it is now 0.285, and now all biased#s that started below .707.... are now rounded down
please explain y this method doesnt work...
lol
ive changed my mind on weather this method works or not around 4 times durring the time of this post...
someone has probobly already cleared the whole thing up
*sigh*
tazzel3d ~ dwiel
quote:Original post by a person
if you need biased numbers you should look into bell curves and statistics.
the most efficent way if your range is small could be to use a lookup table with the correct distribution. just squaring the numbers may seem like a good idea, lets see.
normal random distribution of 0-9
0 1 2 3 4 5 6 7 8 9
distribution squared
0 1 4 9 16 25 36 49 64 81
distribution normalized (ie num/maxNum*range)
0 0.111 0.444 1 1.778 2.778 4 5.444 7.111 9
hmm, seems the distribution just got spread away from the center (ie clustered below and above the mean).
ie that wont work.
Huh? Care to explain how you can say that this distribution is "clustered below and above the mean"? The first derivative of this function keeps increasing (obviously), so it''s only "clustered" below the mean. The gap between two numbers is 2n, so it grows increasingly sparse.
Cédric
If I''m reading this right, I believe your mean will increase exponentially with the range you input which I don''t think you want. Hmm.. I may have to leave for a while to think of a solution here....
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Ahah !
Late nite genius strikes again.
CRandom::CRandom()
{
stored = 0;
}
void CRandom::GetRandom()
{
if(!stored)
{
srand((unsigned)(time(NULL)));
}
stored = rand();
}
float CRandom::GetNumInRange(float upper, float lower)
{
float retval;
float diff = upper-lower;
GetRandom();
retval = fabs(diff*sin(stored));
return retval;
}
okay, why sin() ?
It returns between 1 and -1.
I abs it, and its between 1 and 0.
Seed it with rand(), multiply it times the difference, kaboom; a slow, more-or-less-random tween bounds solution.
Bugle4d
Late nite genius strikes again.
CRandom::CRandom()
{
stored = 0;
}
void CRandom::GetRandom()
{
if(!stored)
{
srand((unsigned)(time(NULL)));
}
stored = rand();
}
float CRandom::GetNumInRange(float upper, float lower)
{
float retval;
float diff = upper-lower;
GetRandom();
retval = fabs(diff*sin(stored));
return retval;
}
okay, why sin() ?
It returns between 1 and -1.
I abs it, and its between 1 and 0.
Seed it with rand(), multiply it times the difference, kaboom; a slow, more-or-less-random tween bounds solution.
Bugle4d
So now im kinda wondering how you would find the % of #s based on the bias function, f(x) = x*x, and how to graph the bias somehow so I could see the frequency of each #. This would be a good way to test the bias function to see if it correctly biases your #s
Tazzel3d ~ dwiel
P.S. I prolly just made no sence what so ever, but please bear with me.. lol
all I want to do is graphically represent the amount of bias an equation gives.
thanx
Tazzel3d ~ dwiel
P.S. I prolly just made no sence what so ever, but please bear with me.. lol
all I want to do is graphically represent the amount of bias an equation gives.
thanx
Like I said, the first derivative gives you a pretty good idea; the lower it is, the more numbers there is around that point.
So
f(x) = x²
f''(x) = 2x
That''s a linear graph, so there are more numbers close to 0² than there are close to 100². It also works for
f(x) = x
There is probably a more correct mathematical way of representing the ''density'' of a function, but I don''t know. Maybe with integrals...
Cédric
So
f(x) = x²
f''(x) = 2x
That''s a linear graph, so there are more numbers close to 0² than there are close to 100². It also works for
f(x) = x
There is probably a more correct mathematical way of representing the ''density'' of a function, but I don''t know. Maybe with integrals...
Cédric
I guess you might be able to just graph the function.. or its inverse it you wanted to see the ''density'' function. Will this work acuratly? If so, it makes it much easier to generate the bias function...
just an idea
Tazzel3d ~ dwiel
just an idea
Tazzel3d ~ dwiel
float randomnum = ScaleNumber(BiasNumber(RANDOM_FLOAT, min, max);
- Tazzel3d
there a " missing here? BiasNumber takes 1 value no?
Dont speak C, so forgive me if im wrong.
and what is RAND_MAX?
surely you just want a random number between 0 and 1, if you divide it by anything you risk affecting the squaring thing, cause you squash the data.
if you take your random float (0 to 1), square it, times it by the range (max-min), and add on min, then thats what you want.
note if your range is large, eg. 100,000 then you'll have 5 0's at the end of your number.
anyway, i've knocked up a quick program in Blitz, and it demonstrates the squaring thing to work lovely, its not much, but it does seem to show that the ditribution is shaped like the inverse tho.
fatray
Edit: put in the blitz code.
edit: then tidied it up a bit, and threw in more coments for any non-blitzers
[edited by - fatray on August 2, 2002 7:34:04 PM]
[edited by - fatray on August 2, 2002 7:35:00 PM]
[edited by - fatray on August 2, 2002 7:41:14 PM]
- Tazzel3d
there a " missing here? BiasNumber takes 1 value no?
Dont speak C, so forgive me if im wrong.
and what is RAND_MAX?
surely you just want a random number between 0 and 1, if you divide it by anything you risk affecting the squaring thing, cause you squash the data.
if you take your random float (0 to 1), square it, times it by the range (max-min), and add on min, then thats what you want.
note if your range is large, eg. 100,000 then you'll have 5 0's at the end of your number.
anyway, i've knocked up a quick program in Blitz, and it demonstrates the squaring thing to work lovely, its not much, but it does seem to show that the ditribution is shaped like the inverse tho.
fatray
;bias random number generator.Graphics 640,480,16,2Const Number_of_numbers=50000Dim number#(Number_of_numbers) ;an array of number_of_numbers +1 floats, ;(0-n_o_n inclusive), # is blitz for floatDim frequency(500) ; an array of 500 integers, used for drawing my bar chartFor a=1 To Number_of_numbers ;each number = random float between 0 and 1 number(a) = Rnd#(0,1)Nextbar_chart_calc()render()WaitKey()For a=1 To Number_of_numbers ;square each number, to hopefully bias the ;numbers towards the origin number(a) = number(a) * number(a) Nextbar_chart_calc()render()WaitKey()EndFunction bar_chart_calc() ;we'll be drawing a bar chart with frequency along the Y, so we need to count what interval the number lies in. ;we'll have the intervals 0.002 wide, (range/500) class_width#=0.002 For a=0 To 500 frequency(a)=0 Next For a=1 To Number_of_numbers ;class is an int ;just need to scale up the range by 500 class=number(a)*500 frequency(class)=frequency(class)+1 NextEnd FunctionFunction render() Cls Color 200,0,0 For a= 0 To 500 Line a+70,400,a+70,400-frequency(a)*1 Next Color 200,200,200 Text 320,460,"Any key to cont.",1,1End Function Edit: put in the blitz code.
edit: then tidied it up a bit, and threw in more coments for any non-blitzers
[edited by - fatray on August 2, 2002 7:34:04 PM]
[edited by - fatray on August 2, 2002 7:35:00 PM]
[edited by - fatray on August 2, 2002 7:41:14 PM]
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