Alphabet | Probability (per 2000) \(P_{1}\) | Weighting \(\log_{1.5}(256-P_{1})\) (3 s.f.) |
---|---|---|

E | 254 | 1.7 |

T | 182 | 10.6 |

A | 164 | 11.2 |

O | 150 | 11.5 |

I | 140 | 11.7 |

N | 134 | 11.8 |

S | 126 | 12.0 |

H | 122 | 12.1 |

R | 120 | 12.1 |

D | 86 | 12.7 |

L | 80 | 12.8 |

U | 56 | 13.1 |

C | 56 | 13.1 |

M | 48 | 13.2 |

W | 48 | 13.2 |

F | 44 | 13.2 |

G | 40 | 13.3 |

Y | 40 | 13.3 |

P | 38 | 13.3 |

B | 30 | 13.4 |

V | 20 | 13.5 |

K | 16 | 13.5 |

J | 4 | 13.6 |

X | 4 | 13.6 |

Q | 2 | 13.7 |

Z | 2 | 13.7 |

Using these weightings we can define a simple weighting system that can convert standard english letters (or any other language that uses these 26 symbols) into SKI strings.

SKI Combinator | Weighting \(W_{s}\) |
---|---|

S | 3.3 \((W_{3})\) |

K | 2.2 \((W_{2})\) |

I | 1.1 \((W_{1})\) |

Until I can devise a better weighting system that will supersede this very rudimentary weighting system this will suffice.

Now to select a string we can use the equation \(W_{s} = (W_{3})^{a} + (W_{2})^{b} + (W_{1})^{c} - 3\) . By taking the floor function of the solutions for the equation (namely a, b and c) will represent the number combinators that are available to form the string.

**Example**

Take A. From the first table, A has a weighting of 1.7 . So 1.7 = 3.3^a + 2.2^b +1.1^c -3 . One possible solution is a=0 b=1 and c=4. So A can be written as KIII or IKII or IIKI or IIIK. Thus the value of the individual letters and words will be the length of the longest sequence that can be beta reduced. Therefore the value of A is 4.

Note, this is largely incomplete and woefully undefined as of the time of writing. If there any mistakes please don't hesitate to point them out. Any suggestions will be welcome. : )