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Prolog is a programming language that has been in existence since 1970. It is more like SQL than a traditional programming language, as it is declarative rather than imperative. In other words, you give a series of facts and clauses and then make queries. Prolog consists of terms, which can be atoms, numbers, variables, or a compound term. An atom is a general name with no defined value, for example, x, red, USA, etc. Numbers are either integers or floating points. A variable begins with an uppercase letter or an underscore, for example, X, Color, etc. A compound term may be a functor (similar to a function in a regular programming language) or a list of items.
A prolog program consists of facts and rules and is started with a query. A rule is something that is always true if the conditions are matched.
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An example of a Prolog assignment problem solved by our experts

Question
There are 3 people A, B, and C and they live in houses numbered 1, 2, and 3 (numbered from left to right). We know the following facts:
a) A’s house is to the right of the house painted yellow.
b) B lives in the red house.
c) The greenhouse is not number 3.
Who lives in which house and what color is each house?
NB: This task should be completed using a swi-prolog (use swipe at the command line to execute it).
Solution
This type of problem can easily be solved in your head but the task is to write a program to solve it. You need to solve the problem using generate and test. This is the least sophisticated technique, as it uses brute force to try all the possible combinations. However, it is sufficient to solve this program in a reasonable time.
Here is how the Prolog code for the program should look like.
%—— DIRECTIVES …
%
:- set_prolog_flag(toplevel_print_anon, false).
%
:- set_prolog_flag(toplevel_print_options,[max_depth(100)]).
%
%—— PEOPLE AND COLOURS …
%
people([a, b, c]).
cols([green, red, yellow]).
%
%—— THE PREDICATE is_element/2 …
%
is_element(E,[E|_]).
is_element(E,[_|T]) :- is_element(E,T).
%
The is_element allows you to check if the first value is contained in the second value.
I?- is element(9,[1,9,8,4]).
true.
?- is element(9,[2,0,1,8]).
false.
?- is element(E,[1,9,8,4]).
E = 1 ;
E = 9 ;
E = 8 ;
E = 4 ;
Task 1:
Write a predicate that generates all the pairs of p(Person, Color) called allhousepairs.
?- allhousepairs(Ps)
Ps = [p(a,green),p(a,red),p(a,yellow),p(b,green),p(b,red),p(b,yellow),p(c,green),p(c,red),p(c,yellow)]
Task 2:
Write a predicate that generates all the triplets (p(Person1, Color1), p(Person2, Color2), p(Person3, Color3)) called all_triplets.
?- all_triplets(Ts)
Ts = [ (p(a,green),p(b,red),p(c,yellow)), (p(a,green),p(b,yellow),p(c,red)),
…..]
Task 3:
Define a predicate that examines a triplet to find which person lives in which colored house or which color house is occupied by a person.
?- lives pattern((p(b,yellow),p(c,red),p(a,green)), yellow, Person).
Person = b .
?- lives pattern((p(b,yellow),p(c,red),p(a,green)), red, Person).
Person = c .
?- lives pattern((p(b,yellow),p(c,red),p(a,green)), Colour, b).
Colour = yellow .
Task 4:
Define a predicate house_info1 to check the first condition a) A’s house is to the right of the house painted yellow.
?- house info1((p(c,red),p(b,yellow),p(a,green))).
true.
?- house info1((p(a,red),p(c,yellow),p(b,green))).
False
Task 5:
Define a predicate house_info2 to check the second condition b) B lives in the red house.
?- house_info2((p(b,red),p(c,yellow),p(a,green))).
true.
?- house_info2((p(a,red),p(c,yellow),p(b,green))).
False.
Task 6:
Define a predicate house_info3 to check the final condition c) The green house is not number 3.
?- house info3((p(a,red),p(c,green),p(b,yellow))).
true.
?- house info3((p(a,red),p(c,yellow),p(b,green))).
False.
Task 7:
Define a predicate house_info that checks each of the previous 3 predicates.
?- house info((p(c,yellow),p(a,green),p(b,red))).
true.
?- house info((p(c,red),p(b,yellow),p(a,green))).
False.
Task 8:
Now use house_info to verify there is only 1 possible solution.
The finished solution is the type of Prolog homework help we can offer,
%—— DIRECTIVES …
%
:- set_prolog_flag(toplevel_print_anon, false).
%
:- set_prolog_flag(toplevel_print_options,[max_depth(100)]).
%
%—— PEOPLE AND COLOURS …
%
people([a, b, c]).
cols([green, red, yellow]).
%
%—— THE PREDICATE is_element/2 …
%
is_element(E,[E|_]).
is_element(E,[_|T]) :- is_element(E,T).
%
% Task 1
% The goal succeeds if PS is a list of all p(P, C) where P is an element of People and C is an element of Colors,
% where People and Colors are the names of people and colors.
allhousepairs(Ps):- people(People), cols(Colors),
bagof(p(Person, Color), (is_element(Person, People), is_element(Color, Colors)), Ps).
% Task 2
% The goal succeeds if Ts is a list of all triplets A, B, C where
% A, B, and C are pairs of people and colors, and no color and person repeats.
all_triplets(Ts):- allhousepairs(AllPairs),
bagof((p(P1, C1), p(P2, C2), p(P3, C3)),
(
is_element(p(P1, C1), AllPairs), is_element(p(P2, C2), AllPairs), is_element(p(P3, C3), AllPairs),
not(P1 = P2), not(P1 = P3), not(P2 = P3),
not(C1 = C2), not(C1 = C3), not(C2 = C3)
),
Ts
).
% Task 3
% The goal succeeds id the patter contains p(Person, Colour) at any position.
lives_pattern(((p(Person,Colour), _, _)), Colour, Person).
lives_pattern((_, (p(Person,Colour), _)), Colour, Person).
lives_pattern((_, _, (p(Person,Colour))), Colour, Person).
% Task 4
% The goal succeeds id p(a, _) is to the right top(_, yellow),
% i.e. if a’s house is to the right of the yellow house.
house_info1((_, p(_, yellow), p(a, _))).
house_info1((p(_, yellow), p(a, _), _)).
house_info1((p(_, yellow), _, p(a, _))).
% Task 5
% The goal succeeds id b’s house is red in the pattern.
house_info2(Pattern):- lives_pattern(Pattern, red, b).
% Task 6
% The goal succeeds if the third house is not green.
house_info3((_, _, p(_, Color3))):- Color3 \= green.
% Task 7
% The pattern satisfies all the conditions if it satisfies each
% rule house_info1, house_info2, house_info3.
house_info(Pattern):- house_info1(Pattern), house_info2(Pattern), house_info3(Pattern).
% Task 8
% Run this:
% (here we find all the patterns that assign persons and colors to houses
% and satisfy all the conditions)
% bagof(Pattern, (all_triplets(Ts), is_element(Pattern, Ts), house_info(Pattern)), Patterns).
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