Generation of random numbers in batches

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NaN
NaN 2020-7-20
编辑： Adam Danz 2020-7-20
Hello,
Imagine the following code in Matlab
rng(1)
a = rand(1,100)
rng(1)
a2 = rand(1,200);
These are two random generators initialized with the same seed. The elements in a are equal to the 100 first elements in a2.
Now, I want to create a random sequence a3 that satisfies a2 = [a, a3].
How should I choose the seed of a3 to satisfy a2? is this even possible?
Best

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采纳的回答

NaN
NaN 2020-7-20
Found the answer.
Just need to store the state of the generator with s = rng and initialize with rng(s) the next generation
Thanks for your help

更多回答（3 个）

Steven Lord
Steven Lord 2020-7-20
The way to do what you're asking is to record the state of the random number generator. You can't really "work backwards".
>> rng(1)
>> a = rand(1, 100);
>> s = rng;
>> rng(1)
>> a2 = rand(1, 200);
>> rng(s) % Use the state recorded after generating the first 100
>> a3 = rand(1, 100);
>> isequal([a a3], a2)
ans =
logical
1
Adam Danz
Adam Danz 2020-7-20
编辑：Adam Danz 2020-7-20
The seeds below are chosen arbitrarily.
rng(1)
a = rand(1,100);
rng(2)
a3 = rand(1,100);
a2 = [a,a3]
You don't even have to change seeds if the a and a3 are generated consecutively.
rng(100)
a = rand(1,100);
a3 = rand(1,100);
a2 = [a,a3];
Bjorn Gustavsson
Bjorn Gustavsson 2020-7-20
if you know from the start how many point in each set just do something like:
a2 = randn(1,num_points_in_a2);
a = a2(1:min(num_points_in_a,numel(a2)));
a3 = a2((1+num_points_in_a):end);
If you dont know how many point you want beforehand just create and throw away the excess points you need - if it is as few as a couple of 100 elements bothering about avoiding might take more time than will be lost...

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