In this file (related to random number generation), there is following line:
private const int MSEED = 161803398;
which resembles golden ratio.
How come golden ratio plays a role in random number generation? Can you help me understand this from mathematical point of view?
NOTE: Naturally, I predict there will be many people saying: This is a programming question. However, I am asking for answer/clarification/insight from math point of view. Hope you can make this subtle distinction.
After some research, I found that the most likely origin of the code I linked to at the top is the following comment and code: (from the book NUMERICAL RECIPIES, pg. 198):
Finally, we give you Knuth's suggestion for a portable routine, which we have translated to the present conventions as RAN3. This is not based on the linear congruential method at all, but rather on a subtractive method. One might hope that its weaknesses, if any, are therefore of a highly different character from the weaknesses, if any, of RAN1 above [not given]. If you ever suspect trouble with one routine, it is a good idea to try the other in the same application. RAN3 has one nice feature: if your machine is poor on integer arithmetic, (i.e. is limited to 16-bit integers), substitution of the two "commented" lines for the one directly following them will render the routine entirely floating point.
FUNCTION RAN3(IDUM) Returns a uniform random deviate between 0.0 and 1.0. Set IDUM to any negative value to initialize or reinitialize the sequence. C IMPLICIT REAL*4(M) C PARAMETER (MBIG=4000000.,MSEED=1618033.,MZ=0.,FAC=2.5E-7) PARAMETER (MBIG=1000000000,MSEED=161803398,MZ=0,FAC=1.E-9) According to Knuth, any large MBIG, and any smaller (but still large) MSEED can be substituted for the above values. DIMENSION MA(55) Save MA. This value is special and should not be modified; see Knuth DATA IFF /0/ IF(IDUM.LT.0.OR.IFF.EQ.0)THEN Initialization IFF=1 MJ=MSEED-IABS(IDUM) Initialize MA(55) using the seed IDUM and the large number MSEED. MJ=MOD(MJ,MBIG) MA(55)=MJ MK=1 DO 11 I=1,54 Now initialize the rest of the table II=MOD(21*I,55) in a slightly random order MA(II)=MK with numbers that are not especially random MK=MJ-MK IF(MK.LT.MZ)MK=MK+MBIG MJ=MA(II) 11 CONTINUE DO 13 K=1,4 We randomize them by "warming up the generator" DO 12 I=1,55 MA(I)=MA(I)-MA(1+MOD(I+30,55)) IF(MA(I).LT.MZ)MA(I)=MA(I)+MBIG 12 CONTINUE 13 CONTINUE INEXT=0 Prepare indices for our first generated number INEXTP=31 The constant 31 is special; see Knuth IDUM=1 ENDIF INEXT=INEXT+1 Here is where we start, except on initialization. Increment INEXT, IF(INEXT.EQ.56)INEXT=1 wrapping around 56 to 1. INEXTP=INEXTP+1 Ditto for INEXTP IF(INEXTP.EQ.56)INEXTP=1 MJ=MA(INEXT)-MA(INEXTP) Now generate a new random number subtractively IF(MJ.LT.MZ)MJ=MJ+MBIG Be sure that it is in range MA(INEXT)=MJ and output the derived uniform deviate RAN3=MJ*FAC RETURN END