| p Statistics: Chain Distance |
| by JVSchmidt |
| General | |
|
When testing the difference of sums we knew that this test is not very sensitive due to the combinatorical proberty of the sum,
generally: n + m = (n-x) + (m+x). But we can treat any digit position as an coordinate value and ask about the distance of to chains. Let us define the DISTANCE OF TWO CHAINS (DOC) as: DOC = SUM i=1,L ABS(Xi - Yi) where L = length of chains X and Y Xi,Yi = i-th digit of chain X respectively Y Here is an example for p with L=5. First sequence X=14159 First sequence Y=26535 DOC = abs(1-2) + abs(4-6) + abs(1-5) + abs(5-3) + abs(9-5) = = 1 + 2 + 4 + 2 + 4 = 13 The recursive law for the expected distribution can easily be found. Let w(L,d) be the probability that two chains of length L have a DOC = d. Then w(L+1,d) = w(L,d) / 10 + sum w(L,d-i) x 2*(10-i)/100 where sums taken for i=1 to 9. Starting condition is: w(1,d) = 2*(10-d)/100 for d=1-9 w(1,0) = 1/10 We understand that the DOC-value can be an indicator for possible correlations between digits of neighbouring sequences. | |
| Result's Overview | |
| Digits analyzed: 4.2 * 10 9 Analysis started at digit: 1 Ellapsed computer time for one class: 3 min 40 sec | |
| Length of chains L | Number of examined pairs of chains K = N/(2*L) | Chi2 | Number of statistical relevant classes |
| 2 | 1.050.000.000 | 13,0688 | 19 |
| 5 | 420.000.000 | 41,7334 | 46 |
| 10 | 210.000.000 | 76,3766 | 72 |
| 20 | 105.000.000 | 108,3159 | 105 |
| 40 | 52.500.0000 | 161,6699 | 148 |
| 70 | 30.000.0000 | 214,09242 | 191 |
| 80 | 26.250.000 | 212,11474 | 203 |