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(X_{i} - Y_{i})where L = length of chains X and Y X _{i},Y_{i} = 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) | Chi^{2}
| 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 |

Detailed results for this test you will find here: Details for Chain Distance Test (EXCEL file)

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