% ----- STRUCTURE ----- % define families with genders and relationships % define easy and intermediate relationship predicates % specify on pedigree collapse determination using the Coefficient of Imbreeding (Wright, 1922) % (see SOURCES [4]) by using helper predicates and bundling them into an advanced logic. % define helper functions for the COI calculation % Sources

% ----- GOAL OF THIS PROGRAM ----- % This program sets itself the goal to find out the degree of pedigree collapse (incest) in families. % The coefficient of imbreeding (COI) is being measured with the formula by Wright, from 1922 (see SOURCES [4]). % Note: the pedigree collapse coefficient is normally not being used in science, due to misleading approximations. % For scientific purposes, the COI is normally being used. % The formula works as following:

% ----- USAGE - coi ----- % ?- coi(, , ). % should not be set with a value. % could be any Person in the knowledge base like charles_II_kingOfSpain or a variable. % with a float value in [0, 1]. % Recommended BaseRiskRate: 0, 0.03, 0.05. % --> coi(COI, charles_II_kingOfSpain, 0.03).

% ----- USAGE - cois to json ----- % ?- cois_to_json(, ), write_jsoned_cois(). % give a value, like 0.03 % let be the variable JSON % --> writes JSONed COIs into the console.

% ----- FORMULA ----- % FP = SUM[ 0.5 ^ (N1 + N2 + 1) * (1 + FCA)] % FP: Coefficient of Imbreeding of a Person P. % SUM: Sum of all results for each common ancestor of P. % N1: Number of generations between P and a found common ancestor of P's father (Parent1). % N2: same like N1, for the mother (Parent2). % FCA: Coefficient of Imbreeding for the found common ancetor himself.

% ----- EXAMPLE EQUATION ----- % Example: see the simple animal relationship example (see SOURCES [3], page 150): % animal_P has the same grandfather animal_c. % The equation results as following: % FP = 0.5 ^ (1 + 1 + 1) * (1 + 0) = 0.125 % N1: is 1, as one generation-difference of grandfather and father. % N1: is 1, as one generation-difference of grandfather and mother. % FCA: is 0, as no information about the ancestors of the grandfather was given. % Result: 0.125 --> 12.5% of the genes are duplicates

% ----- POSSIBLE FURTHER STEPS -----

• plot the solution for a given person or the average of many people with different base rates?
  • ask mates if anybody does grafviz in Prolog (visualization)
  • one person in dependency of the base rate [0, 10].
  • all people of a population, averaged. base rates [0, 10]?
• check out other methods of evaluating pedigree collapse