In the case of 'pure chance without inheritance' the function has to be organized as a sequence of substitutions where there is no connection between two succeeding generations , . Every generation is generated by random selection from the set of all possible states given as the set of all binary strings with length , written as or abbreviated .
The probability to select a certain string is with classical probability assuming equal conditions for every possible value . According to Bernouli's Ars conjectandi (1713) ([27]:IV.4, cited in [104]:29), holds the law, that increasing the number of events towards infinity will produce enough events for each kind of possible value. This means every possible value will approach the limit (with going towards infinity).
With regard to the fundamental question whether this procedure can 'in principle' find all possible values one can say that random selection can select any possible value out of . The open question is the minimal size for a set in the beginning of such a process that all important values can be included simultaneously.
Within the fact that in principle all possible values can be found there remains the question of the quality of the intermediate sets . Measuring the quality of each intermediate set one can define the velocity of approaching the solution set by measuring the number of necessary selections.
To answer these questions some empirical results will be given first. We will start with the following examples ( := number of individuals in the set ; := number of selections with n elements each): (i) n=1,N=32, (ii) n=32, N=32, (iii) n=1, N=32x32.
In these examples the function 'freqCheckRep(l,n,N,K)' computes for a population with many members of binary strings with length and -many events with -many repetitions the frequency of each individual converted into the percentage of the ideal values. From each run with events the mean values is computed as well as the half of the mean difference between the maximum and the meinimum of these frequencies.
This is illustrated in the example below for a population with member and events. The 1st row shows the decimal value of a binary string between [0,31]. The 2nd row shows the frequency of a value during events. The 3rd row shows the percentage of these frequencies compared to the ideal values according to Bernoulli's formula. Thus in this one run with events we get a mean values of and the half distance of .
-->L=5, n=1, N=32, [DIST2,STD, MEAN, FREQ,FX, POPX]=frequCheck(l,n,N) 0. 1. 100. 1. 2. 200. 2. 1. 100. 3. 0. 0. 4. 0. 0. 5. 0. 0. 6. 0. 0. 7. 2. 200. 8. 1. 100. 9. 2. 200. 10. 1. 100. 11. 0. 0. 12. 3. 300. 13. 3. 300. 14. 2. 200. 15. 2. 200. 16. 0. 0. 17. 0. 0. 18. 0. 0. 19. 1. 100. 20. 1. 100. 21. 1. 100. 22. 0. 0. 23. 0. 0. 24. 0. 0. 25. 0. 0. 26. 3. 300. 27. 0. 0. 28. 2. 200. 29. 0. 0. 30. 1. 100. 31. 3. 300. FREQ = 300. 4. 200. 6. 100. 8. 0. 14. MEAN = 150. STD = 107.76318 DIST2 = 150.
If one repeats such a run -many times then one gets the following results:
-->l=5, n=1, N=32, K=10,[MEANX, DIST2X, DIST20, MEAN0]=freqCheckRep(l,n,N,K) MEAN0 = 167.5 DIST20 = 200. DIST2X = 150. 150. 150. 150. 200. 150. 200. 200. 150. 200. MEANX = 150. 150. 150. 150. 175. 150. 200. 200. 150. 200.
This example shows that the values in case of are vary 'unstable' covering a range with deviations of 150 - 200% above and below the 'mean'.
Increasing the numbers either with regard to the events or shows some improvements.
-->l=5, n=32, N=32, K=10,[MEANX, DIST2X, DIST20, MEAN0]=freqCheckRep(l,n,N,K) MEAN0 = 100.65639 DIST20 = 32.8125 DIST2X = 37.5 40.625 31.25 37.5 35.9375 29.6875 39.0625 29.6875 46.875 32.8125 MEANX = 100.86806 103.32031 98.125 104.91071 101.75781 97.65625 98.784722 97.851563 101.15132 102.13816 -->l=5, n=1, N=32*32, K=10,[MEANX, DIST2X, DIST20, MEAN0]=freqCheckRep(l,n,N,K) MEAN0 = 101.38233 DIST20 = 32.8125 DIST2X = 28.125 31.25 35.9375 28.125 34.375 39.0625 34.375 34.375 31.25 32.8125 MEANX = 102.91667 101.38889 102.43056 103.95833 100.78125 99.826389 101.5625 102.02206 99.804688 99.131944
While the mean values are in these examples close to the 'ideal' of 100% the deviation with about 30% above and below the mean values is still vary high. The interesting question is how 'fast' the deviations will approach zero or some . For this we have another empirical experiment.
Gerd Doeben-Henisch 2012-03-31