Merge branch 'master' of gl.kwarc.info:teaching/AI

Frederik/README.md

+Readme
+===
+
+AI 1 - 2019/2020
+---
+
+* 2019-10-28: `prolog/2019-10-28-tutorial.pl` (simple prolog exercises)
+* 2019-10-31 (substition for Katja): `prolog/2019-10-31-tutorial.pl` (simple prolog exercises)
+* 2019-11-04: `prolog/2019-11-04-trees.pl` (some simple tree exercises)
+* 2019-11-11: `2019-11-11-nary-trees.pl` (tree exercise in preparation for tree-search homework)

Frederik/prolog/2019-10-28-tutorial.pl

+% This is the prolog file we created during the tutorial.
+%
+% Remark: inspired by this problem list: http://www.ic.unicamp.br/~meidanis/courses/mc336/2009s2/prolog/problemas/
+
+
+% INTRODUCTION
+
+country(germany).
+
+geolocation(X) :- country(X).
+
+
+bigger(russia, germany).
+bigger(germany, belgium).
+bigger(A, B) :-
+    bigger(A, C), bigger(C, B).
+
+
+% LIST EXERCISES
+
+firstelement([A|_], A).
+
+
+lastelement([A], A).
+lastelement([_|A], B) :-
+    lastelement(A, B).
+
+
+secondlast([H,_], H).
+secondlast([_|A], B) :-
+    secondlast(A, B).
+
+
+second([], []).
+second([A], [A]).
+second([A,_|B], [A|Restlist]) :-
+        second(B, Restlist).
+
+
+prepend(A, B, [A|B]).
+
+
+duplicate([], []).
+duplicate([A], [A,A]).
+duplicate([H|T], [H,H|T2]) :-
+        duplicate(T, T2).
+
+
+% INSERTION SORT
+
+insert(Element, [], [Element]).
+
+insert(Element, [Hs|Ts], [Element,Hs|Ts]):-
+    Element =< Hs.
+
+insert(Element, [Hs|Ts], [Hs|Tl]) :-
+    Element >= Hs, 
+    insert(Element, Ts, Tl).
+
+inssort([], []).
+inssort([H|T], Result) :-
+            inssort(T, X),
+            insert(H, X, Result).
+

Frederik/prolog/2019-10-31-tutorial.pl

+% INTRODUCTION
+
+animal(elephant).
+
+bigger(elephant, bird).
+bigger(elephant, cow).
+bigger(cow, mouse).
+
+bigger(X, Y) :-
+    bigger(X, A), bigger(A, Y).
+
+% Note: The above definition for a transitive relation
+% results in loop, which wasn't resolved during the tutorial.
+% If you want to know more about it, take a look at this answer:
+% https://stackoverflow.com/a/28615704
+
+
+% SIMPLE LIST TASKS
+% inspired from http://www.ic.unicamp.br/~meidanis/courses/mc336/2009s2/prolog/problemas/
+
+getfirst([X|_], X).
+
+getlast([X], X).
+getlast([_|R], X) :-
+    getlast(R, X).
+
+getsecondlast([X,_], X).
+getsecondlast([_|R], X) :-
+    getsecondlast(R, X).
+
+getlength([], 0).
+getlength([X|R], L) :-
+    getlength(R, N), L is N+1.
+
+second([], []).
+second([X], [X]).
+second([X,Y|R], [X|S]) :-
+    second(R, S).
+
+fourth(X, Y) :-
+    second(X, Z), second(Z, Y).
+
+prepend(X,SomeList,[X|SomeList]).
+
+duplicate([], []).
+duplicate([X|R], [X,X|S]) :-
+    duplicate(R, S).
+
+
+% INSERTION SORT
+
+insert(X, [], [X]).
+insert(X, [H|R], [H|S]) :-
+    X >= H, insert(X, R, S).
+insert(X, [H|R], [X,H|R]) :-
+    X =< H.
+
+inssort([], []).
+inssort([H|T], L) :-
+    inssort(T, TSorted),
+    write('inserting '),
+    write(H),
+    write('\n'),
+    insert(H, TSorted, L).
+

Frederik/prolog/2019-11-04-trees.pl

+% Check that a tree is indeed a binary tree
+% (as specified in the homework)
+isbinary(nil).
+isbinary(tree(N, X, Y)) :-
+    integer(N),
+    isbinary(X),
+    isbinary(Y).
+
+% Helper predicate: get the maximum of two integers
+% e.g max(2,7,X) results in X=7.
+%
+% ! is the cut operator.
+% It tells prolog not to backtrack anymore.
+% In this case, we want to ignore the second case,
+% if the first case is applicable.
+max(X,Y,X) :- X >= Y, !.
+max(_,Y,Y).
+
+% Get the largest node in a tree
+% (assuming it only contains natural numbers)
+treemax(nil, 0).
+treemax(tree(N, X, Y), Result) :-
+    treemax(X, XRes),
+    treemax(Y, YRes),
+    max(XRes, YRes, XYRes),
+    max(XYRes, N, Result).
+
+% Get the depth of a tree
+treedepth(nil, 0).
+treedepth(tree(_, X, Y), Depth) :-
+    treedepth(X, XDepth),
+    treedepth(Y, YDepth),
+    max(XDepth, YDepth, D),
+    Depth is D + 1.
+
+% helper predicate: concatenate two lists
+concatenate([], L, L).
+concatenate([H|T], L, [H|TL]) :-
+    concatenate(T, L, TL).
+
+% get the leaves of a binary tree
+leaves(nil, []).
+leaves(tree(N, nil, nil), [N]) :- !.
+leaves(tree(N, X, Y), Result) :-
+    leaves(X, XL),
+    leaves(Y, YL),
+    concatenate(XL, YL, Result).
+
+% compute the result of a formula (which you can think of as a tree).
+% Example tree: plus(times(2,6), 7)
+compute(SomeInt, SomeInt) :-
+    integer(SomeInt).
+compute(times(X, Y), R) :-
+    compute(X, XRes),
+    compute(Y, YRes),
+    R is XRes * YRes.
+compute(plus(X, Y), R) :-
+    compute(X, XRes),
+    compute(Y, YRes),
+    R is XRes + YRes.
+

Frederik/prolog/2019-11-11-nary-trees.pl

+% The tree definition from the homework assignment.
+istree(tree(Value,Children)) :-
+    string(Value),subtrees(Children).
+subtrees([]).
+subtrees([(Cost,T)|Rest]) :-
+    number(Cost),istree(T), subtrees(Rest).
+
+
+% An example tree.
+% If you want to call e.g. addweights with this tree, you can do it with the following trick:
+% tree1(X), addweights(X, Sum).
+tree1(tree("A", [(2, tree("B", [(1, tree("E", [])), (2, tree("F", []))])),
+                 (5, tree("C", [(2, tree("G", []))])),
+                 (2, tree("D", [(1, tree("X", [(1, tree("Y", []))]))]))])).
+
+% Add up all the weights of the edges in the tree.
+% (we called it cw in the tutorial)
+addweights(tree(_, []), 0).
+addweights(tree(_, [(X, T)|Rest]), S) :-
+   addweights(T, S1),
+   addweights(tree("asdf", Rest), S2),
+   S is X + S1 + S2.
+
+% Also adds up all the weights, but in a different way:
+% In the previous attempt, we always expected a tree as argument and constructed trees on demand.
+% In this approach we'll take a list of (Cost, Tree) pairs instead.
+% This is not as pretty, but a useful approach for some tree search algorithms.
+addweights2([], 0).
+addweights2([(X, tree(_, Children))|Rest], S) :-
+    addweights2(Rest, S1),
+    addweights2(Children, S2),
+    S is S1 + S2 + X.
+addweights2(T, S) :-
+    istree(T),
+    addweights2([(0, T)], S).

Leon/tut1/examples.pl

+peter.
+
+%?- peter.
+
+likes(peter, mary).
+likes(peter, sophie).
+cool(X) :- likes(peter,X).
+
+%?- trace, cool(sophie).
+
+%?- trace, cool(X).
+
+has_wheels(mybike, 2).
+has_wheels(mytricycle, 3).
+has_wheels(mytoycar, 4).
+has_wheels(mycar, 4).
+has_motor(mybike).
+has_motor(mycar).
+car(X) :- has_wheels(X, 4), has_motor(X).
+
+%?- trace, car(Y).
+
+
+% Cool cool cool, but how can we code with this?
+% The answers is for the most part...RECURSION!
+
+nat(zero).
+nat(s(X)) :- nat(X).
+
+% add(X, Y, Z) -> X + Y = Z
+add(X, zero, X).
+add(X, s(Y), s(Z)) :- add(X, Y, Z).
+
+% We see, there is no "return" in Prolog, only predicates.
+% For every predicate please state in a comment which parameters
+% are input and which are output.
+
+
+%% List comprehension:
+% list e.g. [1,2,3] or [jamie,cersei], empty list = []
+% pattern matching with lists:
+% [Head|Tail], where Head is an element and Tail is a List!
+% Also works with first 2 elements [X1, X2 | T]
+
+append([],L2,L2).
+append([H|T],L2,[H|Res]):-
+	append(T,L2,Res).
+
+
+
+
+
+% A practical example
+
+% Signature for our problem:
+% countN(N : int, L : List[int]) -> int
+% counts the occurrences of N in L and returns the result.
+% With Prolog-like lists: countN(N, [H | T]) -> int
+
+% L = [2, 3, 5, 2, 7, 1, 2, 2, 5, 9]
+% countN(3, L) -> 1
+% countN(2, L) -> 4
+% countN(8, L) -> 0
+% countN(5, L) -> 2
+
+% Delcarative Approach
+% countN(N, L) -> res
+% res = 0
+% for e in L:
+%    if e == N:
+%      res += 1
+% return res
+
+% Functional / Recursive Approach
+% Base Cases:
+% countN(N, []) -> return 0
+% Recursive Case:
+% countN(N, [H | T]) ->
+%  if N == H
+%    return countN(N, [T]) + 1
+%  else
+%    return countN(N, [T])
+
+
+% ... in prolog we can't "return", so we need another variable in the predicate
+% countN(N, L, RES) the number of occurrences of N in list L
+% will be "returned/safed" in RES (Will be in relation with RES)
+
+% Base Cases:
+countN(_, [], 0).
+%?- trace, countN(2, [], X).
+% Recursive Case:
+countN(N, [N | T], RES) :- countN(N, T, OLDRES), RES is OLDRES + 1.
+countN(N, [M | T], RES) :- not(N=M), countN(N, T, RES).
+% Point out: "N" in first line list compr. "is" can be very useful
+%?- trace, countN(2, [2], X).