Generating uniform random numbers in Lua

2020-01-28 09:03发布

问题:

I am working on programming a Markov chain in Lua, and one element of this requires me to uniformly generate random numbers. Here is a simplified example to illustrate my question:

example = function(x)
    local r = math.random(1,10)
    print(r)
    return x[r]
end

exampleArray = {"a","b","c","d","e","f","g","h","i","j"}

print(example(exampleArray))

My issue is that when I re-run this program multiple times (mash F5) the exact same random number is generated resulting in the example function selecting the exact same array element. However, if I include many calls to the example function within the single program by repeating the print line at the end many times I get suitable random results.

This is not my intention as a proper Markov pseudo-random text generator should be able to run the same program with the same inputs multiple times and output different pseudo-random text every time. I have tried resetting the seed using math.randomseed(os.time()) and this makes it so the random number distribution is no longer uniform. My goal is to be able to re-run the above program and receive a randomly selected number every time.

回答1:

You need to run math.randomseed() once before using math.random(), like this:

math.randomseed(os.time())

From your comment that you saw the first number is still the same. This is caused by the implementation of random generator in some platforms.

The solution is to pop some random numbers before using them for real:

math.randomseed(os.time())
math.random(); math.random(); math.random()

Note that the standard C library random() is usually not so uniformly random, a better solution is to use a better random generator if your platform provides one.

Reference: Lua Math Library



回答2:

Standard C random numbers generator used in Lua isn't guananteed to be good for simulation. The words "Markov chain" suggest that you may need a better one. Here's a generator widely used for Monte-Carlo calculations:

local A1, A2 = 727595, 798405  -- 5^17=D20*A1+A2
local D20, D40 = 1048576, 1099511627776  -- 2^20, 2^40
local X1, X2 = 0, 1
function rand()
    local U = X2*A2
    local V = (X1*A2 + X2*A1) % D20
    V = (V*D20 + U) % D40
    X1 = math.floor(V/D20)
    X2 = V - X1*D20
    return V/D40
end

It generates a number between 0 and 1, so r = math.floor(rand()*10) + 1 would go into your example. (That's multiplicative random number generator with period 2^38, multiplier 5^17 and modulo 2^40, original Pascal code by http://osmf.sscc.ru/~smp/)



回答3:

math.randomseed(os.clock()*100000000000)
for i=1,3 do
    math.random(10000, 65000)
end

Always results in new random number. Changing seed value will ensure randomness, Dont follow os.time() cuz its epoch time and changes after one second but os.clock() dont have same value at any close instance. Cheers!