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Luis A. Afonso
Posts:
4,715
From:
LIsbon (Portugal)
Registered:
2/16/05


PI on duty to check a PRNG
Posted:
Mar 1, 2013 9:31 AM


PI on duty to check a PRNG
Suppose we are estimate by a random simulation / Monte Carlo procedure pi=3.14159? . The Probability to have at most a difference relative to the true pi value is given by the DKW formula: en.wikipedia.org/wiki/DvoretzkyKieferWolfowitz inequality Table of values x and their probabilities, N=16E06 trials. ___x<3.1412_____p=0.00 ___x=3.1412_____p=0.01 _____3.1413_______0.13 _____3.1414_______0.60 _____3.1415_______(1.51) _____3.1416_______(2.00) _____3.1417_______(1.39) ____ 3.1418_______0.51 _____3.1419_______0.10 _____3.1420_______0.01 >3.1420__________ 0.00
Therefore we must observe our empirical results inside the interval [3.1412, 3.1420] if the simulations are working properly.
Experience setting:
In the Cartesian plane (orthogonal axis, x, y) let be the square with vertices on (0, 0), (1, 0), (1, 1), (0, 1) and C= (0.5, 0.5) the centre of the circumference (radius= 0.5) tangent to the square sides. When a point W= (randomX, randomY) falls inside the circumference with area pi*0.5*0.5 one has: _____(x  .5)^2 + (y  .5)^2 <= 0.25, condition to which a success is counted.
Outputs:
_3.1417 , 3.1416 , 3.1416 , 3.1416 , 3.1417 , 3.1416 , _3.1422*, 3.1415 , 3.1416 , 3.1417 , 3.1415 , 3.1414 , _3.1417 , 3.1417 , 3.1417
The stared value is anomalous. . . The PRNG´s performance is not as good as it was desired. It should be used only for illustrative or pedagogical, not research, purposes. Luis A. Afonso
REM "minus" CLS PRINT " (x.5)^2+(y.5)^2 <= .25 " PRINT all = 1.6E+07 FOR rpt = 1 TO 10 inn = 0 COLOR 13 FOR j = 1 TO all RANDOMIZE TIMER LOCATE 4, 50: PRINT USING "##########"; all  j u = (RND  .5) ^ 2 + (RND  .5) ^ 2 IF u <= .5 ^ 2 THEN inn = inn + 1 LOCATE 10 + rpt, 40 PRINT USING "#.#### "; 4 * inn / j NEXT j NEXT rpt END



