We all know good old Math.random(). It returns a random floating point number between 0 and 1.
What I can't seem to find any evidence about is if zero or one is exclusive or inclusive.
I know that if they are inclusive, the probability of hitting either one of these values is seriously low.
But I can't help but wonder if I should wasting an if statement looking for it or not.
In my current scenario zero is not a problem, but one is.
var __rand:uint = Math.floor( Math.random() * myArray.length );
var result:String = myArray[__rand];
if the 1 in Math.random() is exclusive, then I will know that will NEVER be 1, and therefore __rand could never equal myArray.length and should always be below it.. But just wasn't sure if I should waste time in some performance critical code if I should account for it.
PS: The code above is NOT the performance critical code, just an example
Basically, just 2 simple questions.
1) Is returning one impossible or possible.
2) If possible, is it worth accounting for it.
As per the docs:
Returns a pseudo-random number n, where 0 <= n < 1. The number
returned is calculated in an undisclosed manner, and is
"pseudo-random" because the calculation inevitably contains some
element of non-randomness.
So it can be 0 but not 1. You don't have to worry about index out of bounds.
By the way, if this was really performance critical code, you are better off casting the value as int or uint rather than using Math.floor (see this performance test).
Math.random will return a number between 0 and (1 exclusive). Never will return a 1.
Related
I'm trying to solve this recursion:
T(n)=log(T(n-1))+1 :n>2
T(n)=O(1) :n=2
I'm getting an answer of O(1), but I feel I'm missing something.
I'd be happy to get some help.
Thanks in advance!
Your recursion relation converges to 1.0 for any value >= 1.0
Your answer of O(1) is quite correct. You may be a little thrown by the recursion relation being expressed in such a straightforward fashion of time, rather than being given the algorithm?
Let me try again. Also, perhaps we're both a little confused. I answered at the single-call level; perhaps you need the overall answer (more likely, now that I think about it).
First, let's take a single call. If n=2, it's constant time. If n>2 ... this is where I'm not quite familiar with the notation. Does this denote time for a single call, or for the entire recursion sequence descending to n=2? I think it has to be for the single call, due to practical considerations. This makes my earlier answer incorrect.
Look at the call for n=3. This expands and solves as
T(3) = log(T(2)) + 1
T(3) = log(1) + 1
T(3) = 0 + 1 = 1
By induction T(n) = 1 for n >= 2. As it turns out, even if the base case -- T(2) -- has a value other than 1, so long as it's finite and more than 1/ (whatever base we're using for logs), the series will converge to 1, and each call will be in constant time.
Thus, to solve T(n), we have n-2 calls, each of which is T(1). This gives an overall complexity of O(n).
Is that more clear?
I have a CUDA program whose kernel basically does the following.
I provide a list of n points in cartesian coordinates e.g. (x_i,y_i) in a plane of dimension dim_x * dim_y. I invoke the kernel accordingly.
For every point on this plane (x_p,y_p) I calculate by a formula the time it would take for each of those n points to reach there; given those n points are moving with a certain velocity.
I order those times in increasing order t_0,t_1,...t_n where the precision of t_i is set to 1. i.e. If t'_i=2.3453 then I would only use t_i=2.3.
Assuming the times are generated from a normal distribution I simulate the 3 quickest times to find the percentage of time those 3 points reached earliest. Hence suppose prob_0 = 0.76,prob_1=0.20 and prob_2=0.04 by a random experiment. Since t_0 reaches first most amongst the three, I also return the original index (before sorting of times) of the point. Say idx_0 = 5 (An integer).
Hence for every point on this plane I get a pair (prob,idx).
Suppose n/2 of those points are of one kind and the rest are of other. A sample image generated looks as follows.
Especially when precision of the time was set to 1 I noticed that the number of unique 3 tuples of time (t_0,t_1,t_2) was just 2.5% of the total data points i.e. number of points on the plane. This meant that most of the times the kernel was uselessly simulating when it could just use the values from previous simulations. Hence I could use a dictionary having key as 3-tuple of times and value as index and prob. Since as far as I know and tested, STL can't be accessed inside a kernel, I constructed an array of floats of size 201000000. This choice was by experimentation since none of the top 3 times exceeded 20 seconds. Hence t_0 could take any value from {0.0,0.1,0.2,...,20.0} thus having 201 choices. I could construct a key for such a dictionary like the following
Key = t_o * 10^6 + t_1 * 10^3 + t_2
As far as the value is concerned I could make it as (prob+idx). Since idx is an integer and 0.0<=prob<=1.0, I could retrieve both of those values later by
prob=dict[key]-floor(dict[key])
idx = floor(dict[key])
So now my kernel looks like the following
__global__ my_kernel(float* points,float* dict,float *p,float *i,size_t w,...){
unsigned int col = blockIdx.y*blockDim.y + threadIdx.y;
unsigned int row = blockIdx.x*blockDim.x + threadIdx.x;
//Calculate time taken for each of the points to reach a particular point on the plane
//Order the times in increasing order t_0,t_1,...,t_n
//Calculate Key = t_o * 10^6 + t_1 * 10^3 + t_2
if(dict[key]>0.0){
prob=dict[key]-floor(dict[key])
idx = floor(dict[key])
}
else{
//Simulate and find prob and idx
dict[key]=(prob+idx)
}
p[row*width+col]=prob;
i[row*width+col]=idx;
}
The result is quite similar to the original program for most points but for some it is wrong.
I am quite sure that this is due to race condition. Notice that dict was initialized with all zeroes. The basic idea would be to make the data structure "read many write once" in a particular location of the dict.
I am aware that there might be much more optimized ways of solving this problem rather than allocating so much memory. Please let me know in that case. But I would really like to understand why this particular solution is failing. In particular I would like to know how to use atomicAdd in this setting. I have failed to use it.
Unless your simulation in the else branch is very long (~100s of floating-point operations), a lookup table in global memory is likely to be slower than running the computation. Global memory access is very expensive!
In any case, there is no way to save time by "skipping work" using conditional branching. The Single Instruction, Multiple Thread architecture of a GPU means that the instructions for both sides of the branch will be executed serially, unless all of the threads in a block follow the same branch.
edit:
The fact that you are seeing a performance increase as a result of introducing the conditional branch and you didn't have any problems with deadlock suggests that all the threads in each block are always taking the same branch. I suspect that once dict starts getting populated, the performance increase will go away.
Perhaps I have misunderstood something, but if you want to calculate the probability of an event x, assuming a normal distribution and given the mean mu and standard deviation sigma, there is no need to generate a load of random numbers and approximate a Gaussian curve. You can directly calculate the probability:
p = exp(-((x - mu) * (x - mu) / (2.0f * sigma * sigma))) /
(sigma * sqrt(2.0f * M_PI));
What does the term "STEP" means in bullet physics?
What does the function stepSimulation() and its parameters mean?
I have read the documentation but i could not get hold of anything.
Any valid explanation would be of great help.
I know I'm late, but I thought the accepted answer was only marginally better than the documentation's description.
timeStep: The amount of seconds, not milliseconds, passed since the last call to stepSimulation.
maxSubSteps: Should generally stay at one so Bullet interpolates current values on its own. A value of zero implies a variable tick rate, meaning Bullet advances the simulation exactly timeStep seconds instead of interpolating. This feature is buggy and not recommended. A value greater than one must always satisfy the equation timeStep < maxSubSteps * fixedTimeStep or you're losing time in the simulation.
fixedTimeStep: Inversely proportional to the simulation's resolution. Resolution increases as this value decreases. Keep in mind that a higher resolution means it takes more CPU.
btDynamicsWorld::stepSimulation(
btScalar timeStep,
int maxSubSteps=1,
btScalar fixedTimeStep=btScalar(1.)/btScalar(60.));
timeStep - time passed after last simulation.
Internally simulation is done for some internal constant steps. fixedTimeStep
fixedTimeStep ~~~ 0.01666666 = 1/60
if timeStep is 0.1 then it will include 6 (timeStep / fixedTimeStep) internal simulations.
To make glider movements BulletPhysics interpolate final step results according reminder after division (timeStep / fixedTimeStep)
timeStep - the amount of time in seconds to step the simulation by. Typically you're going to be passing it the time since you last called it.
maxSubSteps - the maximum number of steps that Bullet is allowed to take each time you call it.
fixedTimeStep - regulates resolution of the simulation. If your balls penetrates your walls instead of colliding with them try to decrease it.
Here i would like to address the issue in Proxy's answer about special meaning of value 1 for maxSubSteps. There is only one special value, that is 0 and you most likely don't want to use it because then simulation will go with non-constant time step. All other values are the same. Let's have a look at the actual code:
if (maxSubSteps)
{
m_localTime += timeStep;
...
if (m_localTime >= fixedTimeStep)
{
numSimulationSubSteps = int(m_localTime / fixedTimeStep);
m_localTime -= numSimulationSubSteps * fixedTimeStep;
}
}
...
if (numSimulationSubSteps)
{
//clamp the number of substeps, to prevent simulation grinding spiralling down to a halt
int clampedSimulationSteps = (numSimulationSubSteps > maxSubSteps) ? maxSubSteps : numSimulationSubSteps;
...
for (int i = 0; i < clampedSimulationSteps; i++)
{
internalSingleStepSimulation(fixedTimeStep);
synchronizeMotionStates();
}
}
So, there is nothing special about maxSubSteps equal to 1. You should really abide this formula timeStep < maxSubSteps * fixedTimeStep if you don't want to lose time.
I am attempting to get a random bearing, from 0 to 359.9.
SET bearing = FLOOR((RAND() * 359.9));
I may call the procedure that runs this request within the same while loop, immediately one after the next. Unfortunately, the randomization seems to be anything but unique. e.g.
Results
358.07
359.15
357.85
I understand how randomization works, and I know because of my quick calls to the same function, the ticks used to generate the random number are very close to one another.
In any other situation, I would wait a few milliseconds in between calls or reinit my Random object (such as in C#), which would greatly vary my randomness. However, I don't want to wait in this situation.
How can I increase randomness without waiting?
I understand how randomization works, and I know because of my quick calls to the same function, the ticks used to generate the random number are very close to one another.
That's not quite right. Where folks get into trouble is when they re-seed a random number generator repeatedly with the current time, and because they do it very quickly the time is the same and they end up re-seeding the RNG with the same seed. This results in the RNG spitting out the same sequence of numbers each time it is re-seeded.
Importantly, by "the same" I mean exactly the same. An RNG is either going to return an identical sequence or a completely different one. A "close" seed won't result in a "similar" sequence. You will either get an identical sequence or a totally different one.
The correct solution to this is not to stagger your re-seeds, but actually to stop re-seeding the RNG. You only need to seed an RNG once.
Anyways, that is neither here nor there. MySQL's RAND() function does not require explicit seeding. When you call RAND() without arguments the seeding is taken care of for you meaning you can call it repeatedly without issue. There's no time-based limitation with how often you can call it.
Actually your SQL looks fine as is. There's something missing from your post, in fact. Since you're calling FLOOR() the result you get should always be an integer. There's no way you'll get a fractional result from that assignment. You should see integral results like this:
187
274
89
345
That's what I got from running SELECT FLOOR(RAND() * 359.9) repeatedly.
Also, for what it's worth RAND() will never return 1.0. Its range is 0 ≤ RAND() < 1.0. You are safe using 360 vs. 359.9:
SET bearing = FLOOR(RAND() * 360);
What is the best way to constrain the values of a PRNG to a smaller range? If you use modulus and the old max number is not evenly divisible by the new max number you bias toward the 0 through (old_max - new_max - 1). I assume the best way would be something like this (this is floating point, not integer math)
random_num = PRNG() / max_orginal_range * max_smaller_range
But something in my gut makes me question that method (maybe floating point implementation and representation differences?).
The random number generator will produce consistent results across hardware and software platforms, and the constraint needs to as well.
I was right to doubt the pseudocode above (but not for the reasons I was thinking). MichaelGG's answer got me thinking about the problem in a different way. I can model it using smaller numbers and test every outcome. So, let's assume we have a PRNG that produces a random number between 0 and 31 and you want the smaller range to be 0 to 9. If you use modulus you bias toward 0, 1, 2, and 3. If you use the pseudocode above you bias toward 0, 2, 5, and 7. I don't think there can be a good way to map one set into the other. The best that I have come up with so far is to regenerate the random numbers that are greater than old_max/new_max, but that has deep problems as well (reducing the period, time to generate new numbers until one is in the right range, etc.).
I think I may have naively approached this problem. It may be time to start some serious research into the literature (someone has to have tackled this before).
I know this might not be a particularly helpful answer, but I think the best way would be to conceive of a few different methods, then trying them out a few million times, and check the result sets.
When in doubt, try it yourself.
EDIT
It should be noted that many languages (like C#) have built in limiting in their functions
int maximumvalue = 20;
Random rand = new Random();
rand.Next(maximumvalue);
And whenever possible, you should use those rather than any code you would write yourself. Don't Reinvent The Wheel.
This problem is akin to rolling a k-sided die given only a p-sided die, without wasting randomness.
In this sense, by Lemma 3 in "Simulating a dice with a dice" by B. Kloeckner, this waste is inevitable unless "every prime number dividing k also divides p". Thus, for example, if p is a power of 2 (and any block of random bits is the same as rolling a die with a power of 2 number of faces) and k has prime factors other than 2, the best you can do is get arbitrarily close to no waste of randomness, such as by batching multiple rolls of the p-sided die until p^n is "close enough" to a power of k.
Let me also go over some of your concerns about regenerating random numbers:
"Reducing the period": Besides batching of bits, this concern can be dealt with in several ways:
Use a PRNG with a bigger "period" (maximum cycle length).
Add a Bays–Durham shuffle to the PRNG's implementation.
Use a "true" random number generator; this is not trivial.
Employ randomness extraction, which is discussed in Devroye and Gravel 2015-2020 and in my Note on Randomness Extraction. However, randomness extraction is pretty involved.
Ignore the problem, especially if it isn't a security application or serious simulation.
"Time to generate new numbers until one is in the right range": If you want unbiased random numbers, then any algorithm that does so will generally have to run forever in the worst case. Again, by Lemma 3, the algorithm will run forever in the worst case unless "every prime number dividing k also divides p", which is not the case if, say, k is 10 and p is 32.
See also the question: How to generate a random integer in the range [0,n] from a stream of random bits without wasting bits?, especially my answer there.
If PRNG() is generating uniformly distributed random numbers then the above looks good. In fact (if you want to scale the mean etc.) the above should be fine for all purposes. I guess you need to ask what the error associated with the original PRNG() is, and whether further manipulating will add to that substantially.
If in doubt, generate an appropriately sized sample set, and look at the results in Excel or similar (to check your mean / std.dev etc. for what you'd expect)
If you have access to a PRNG function (say, random()) that'll generate numbers in the range 0 <= x < 1, can you not just do:
random_num = (int) (random() * max_range);
to give you numbers in the range 0 to max_range?
Here's how the CLR's Random class works when limited (as per Reflector):
long num = maxValue - minValue;
if (num <= 0x7fffffffL) {
return (((int) (this.Sample() * num)) + minValue);
}
return (((int) ((long) (this.GetSampleForLargeRange() * num))) + minValue);
Even if you're given a positive int, it's not hard to get it to a double. Just multiply the random int by (1/maxint). Going from a 32-bit int to a double should provide adequate precision. (I haven't actually tested a PRNG like this, so I might be missing something with floats.)
Psuedo random number generators are essentially producing a random series of 1s and 0s, which when appended to each other, are an infinitely large number in base two. each time you consume a bit from you're prng, you are dividing that number by two and keeping the modulus. You can do this forever without wasting a single bit.
If you need a number in the range [0, N), then you need the same, but instead of base two, you need base N. It's basically trivial to convert the bases. Consume the number of bits you need, return the remainder of those bits back to your prng to be used next time a number is needed.