Different behaviour of rng() and rand() in parfor-loop
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    JonasBGood
 on 11 Feb 2014
  
    
    
    
    
    Commented: Walter Roberson
      
      
 on 31 Jan 2023
            Hey folks,
I experienced a different output of the standard function rand() in a parfor-loop although I am using the same seed-value. Example:
>> parfor i = 1:1, rng(123), rand(1,1), end
ans =
    0.2751
>> for i = 1:1, rng(123), rand(1,1), end
ans =
    0.6965
Is this a bug? I would expect exactly the same output of rand() in parfor.
Thanks & greetings Jonas
------------------------------------------------
ps: here are my detailed version information, I use an Intel i7-2600 processor.
----------------------------------------------------------------------------------------------------
MATLAB Version: 8.2.0.701 (R2013b)
MATLAB License Number: ••••••
Operating System: Microsoft Windows 7 Version 6.1 (Build 7601: Service Pack 1)
Java Version: Java 1.7.0_11-b21 with Oracle Corporation Java HotSpot™ 64-Bit Server VM mixed mode
----------------------------------------------------------------------------------------------------
MATLAB Version 8.2 (R2013b)
DIPUM Toolbox Version 1.1.4
DIPUM Toolbox Version 1.1.3
Image Processing Toolbox Version 8.3 (R2013b)
Optimization Toolbox Version 6.4 (R2013b)
Parallel Computing Toolbox Version 6.3 (R2013b)
2 Comments
  John Fox
 on 20 Jul 2017
				I had the exact same problem. My for loops gave a different answer than my parfor loops. The reason is
As described in Control Random Number Streams, each worker in a cluster has an independent random number generator stream. By default, therefore, each worker in a pool, and each iteration in a parfor-loop has a unique, independent set of random numbers. Subsequent runs of the parfor-loop generate different numbers.
 I fixed this with rng(123,'twister'). At least this worked for me.
  John Fox
 on 20 Jul 2017
				Notice what happens when you add 'twister' to the parfor loop
>> for i = 1:1, rng(123), rand(1,1), end
ans =
    0.6965
Which is the same answer you got.
>> parfor i = 1:1, rng(123,'twister'), rand(1,1), end
ans =
    0.6965
It is the same answer when you add 'twister' to the parfor loop
Accepted Answer
  Edric Ellis
    
      
 on 12 Feb 2014
        The client and the workers are set up to use different random generators, so this is expected. This is covered in the documentation. Note that "rng(seed)" changes only the seed but not the underlying generator type.
1 Comment
  Walter Roberson
      
      
 on 31 Jan 2023
				More Answers (0)
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