PI and randomness of digits after decimal point. Priority of Russia  
Tags: rng, probability, logarithm, integral
Published: 2020-03-14 06:56:03

PI and randomness of digits after decimal point. Priority of Russia 

On March 14, on day "3.14", PI celebrates its day 

Importance of topic: research on distribution of PI digits and irrational numbers opens up possibility of studying continuous data in real time, 
when incoming data is counted without pauses for a separate input of data array

Understanding patterns of distributions human and natural and fake and machine actually apply results of 4 types of research:

1. Creation of randomness
2. Overcoming randomness
3. Falsification of randomness
4. Overcoming falsification of randomness

Gap of Education in Russia and CIS and USSR: 
we don't study logarithm and integral in lower grades
and later consider simplest allegedly difficult
as well as algorithms for fast calculations are not studied 

In addition fast computing algorithms are not taught in school, 
although algorithms are clear in many programming languages 
without use of quantum computers

But this work highlights public pyramid 
spectral integral logarithmic binomial criterion


Using 55,000 digits of PI after decimal point,
first in a Word compatible program replace special characters
numbers are translated into a column and then in an Excel compatible 
program digits are divided into binary attributes: 
small \ large and even \ odd 

Results: average for both binary distributions: about 0.5
and binary divisions correspond to theoretical probability 

Table is applied:
![Danilin Table 1](https://habrastorage.org/webt/gw/lw/-e/gwlw-ezfehiqjvao3biiqpdspts.png)

Where are main formulas: 
Cell-Formula-Explanation

C1 =AVERAGE(D1:D55000)
Average value of sequence numbers

C2 =AVERAGE(B1:B55000)
Average distribution value 1

D1 =IF(B1<C$2;0;1)
If number is less than average, then 0, otherwise 1

D2 =IF(B2<C$2;0;1)
If number is less than 1, then 0, otherwise 1, and so on.

E2 =IF(D2=D1;E1+1;0)
If distribution features are same, then counter of same attributes is +1,
otherwise counter is reset to zero

F2 =IF(E3=0;E2;" ")
If counter is reset to zero, largest counter is fixed

H2 =COUNTIF(F$1:F$55000;G2)
Number of signs 1 in a row and so on

H12 =H2/H3
Ratio of nearest number of attributes

J1 =IF(B1/2=ROUND(B1/2);0;1)
If number is even, then 0, otherwise 1

J2 =IF(B2/2=ROUND(B2/2);0;1)
If number is even, then 0, otherwise 1, etc 

K2 =IF(J2=J1;K1+1;0)
If distribution features are same, then counter of same attributes is +1,
otherwise counter is reset to zero

L2 =IF(K3=0;K2;" ")
If counter is reset to zero, largest counter is fixed

N2 =COUNTIF(L$1:L$55000;M2)
Number of signs 1 in a row and so on

N12 =H2/H3
Ratio of nearest number of attributes

![Danilin Line11](https://habrastorage.org/webt/sn/fs/9g/snfs9gyyk2y0bwylbvwaxicjism.png)

Other control functions can be programmed in table 
You can create graphs of values of any cells in table 

Continuation of table explores random permutations of sequence

![Danilin Table22](https://habrastorage.org/webt/ec/5-/jb/ec5-jbjbfeqmfmi25a_uirj7fxg.png)

Column Q - for a random permutation of integers up to 10^6,
to avoid repeating random events;
Column R - initially a copy of column B, later modified;
Columns T...AE - same as C...N columns"

Cell-Formula-Explanation

Q1 =RANDBETWEEN(0;1000000)
For a random permutation

Q2 =RANDBETWEEN(0;1000000)
Random for permutation, etc.

Preliminary conclusions: predominance of spectrum of repetitive features is found, what is typical for natural sequences, for example, by typing manually 3000 digits, 1st spectrum of repeated features will be higher than theoretical value 

![Danilin Line22](https://habrastorage.org/webt/hm/qp/gu/hmqpguloxh1vb3w8nuhwsjih2ie.png)

Using capabilities table for a permutation of elements of sequence and shuffled, spectra take on theoretical values as would be synthesized by RNG and CSPRNG and CPRNG 

Spectra follow formula of Danilin: N = LOG(1-c)/LOG(1-p) by principle:
when C=P=0.5; N = 1 = log0.5/log0.5 = log(1-1/2)/log(1-1/2) = 1
for C=0.25; P=0.5; N = 2 = log0.75/log0.5 = log(1-1/4)/log(1-1/2) = 2, etc 

{% youtube YIJmgOTnkTU %}

![Danilin22](https://habrastorage.org/webt/08/l1/fp/08l1fp4ytxhytzfen4x9lkdm7ww.gif)

Program for distribution spectra of random number 
of consecutive identical features less \ more and even \ odd

Practical distributions correspond to theoretical ones
so random sequence is qualitative
and it is possible to study patterns of different sequences

Binomial Logarithmic Integral Pyramidal Distribution
BLIP distribution of Random numbers

 <source lang="vb">'dablip.bas
	
	RANDOMIZE TIMER
	tb = TIMER: s = 0
	OPEN "dablip.txt" FOR OUTPUT AS #2
	n = VAL(MID$(TIME$, 7, 2))*10 ^ 5
	DIM b(n), d(n), e(n), f(n)
	DIM j(n), k(n), m(n), p(16), q(16)
	LOCATE 1, 1: PRINT " THEORY Average BIG EVEN "
	
	FOR i = 2 TO n-1
	 b(i) = INT(RND*900)+100: s = s+b(i): m = s/i
	
	 IF b(i) < m THEN d(i) = 0 ELSE d(i) = 1
	 IF (b(i) MOD 2) = 0 THEN j(i) = 0 ELSE j(i) = 1
	
	 IF d(i) = d(i-1) THEN e(i) = e(i-1)+1 ELSE e(i) = 0
	 IF e(i) = 0 THEN f(i) = e(i-1) ELSE f(i) = 12
	 IF f(i) > 12 THEN f(i) = 12
	
	 IF j(i) = j(i-1) THEN k(i) = k(i-1)+1 ELSE k(i) = 0
	 IF k(i) = 0 THEN m(i) = k(i-1) ELSE m(i) = 12
	 IF m(i) > 12 THEN m(i) = 12
	
	 p(f(i)) = p(f(i))+1: q(m(i)) = q(m(i))+1
	
	 IF (i MOD 1000) = 0 THEN LOCATE 3, 1: PRINT i, " from ", n, INT(100*i/n); " %",
	NEXT
	
	LOCATE 3, 1: FOR t = 1 TO 12
	 PRINT INT(n/(2^(t+1))), INT((p(t-1)+q(t-1))/2), p(t-1), q(t-1)
	NEXT
	
	te = TIMER
	PRINT: PRINT te-tb; "second", INT(n/(te-tb)); " in second "
	PRINT n, " elements ",
	
	PRINT #2, te-tb; "second", INT(n/(te-tb)); " in second "
	PRINT #2, n, " elements ",: PRINT #2,
	
	PRINT #2,: PRINT #2, " THEORY Average BIG EVEN ": PRINT #2,
	FOR t = 1 TO 12
	 PRINT #2, INT(n/(2^(t+1))), INT((p(t-1)+q(t-1))/2), p(t-1), q(t-1)
	NEXT
</source>
Feature of program: index of indixes p(f(i)) & q(m(i))

Results:

 40 second 139'555 in second 
 5'600'000 elements 

 	THEORY	Average	BIG		EVEN

	1400000	1400610	1399595	1401625
	700000	700026	700122	699931
	350000	349716	349508	349925
	175000	174823	174892	174755
	87500	87424	87564	87285
	43750	43837	43931	43744
	21875	22028	21983	22074
	10937	10850	10865	10835
	5468	5481	5496	5466
	2734	2755	2732	2778
	1367	1388	1396	1380
	687		687		687		687

I think random have problems with parity:
parity of random changes too sharply

For self-study:

1. Uniformity of each of digits of PI
2. Methods for approximate calculation of PI
3. Learn 8 digits of PI after decimal point
4. Find formulas for calculating digits of PI
5. Remember circle length formulas