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  • Score the best grades with our online Prolog assignment help service

Score the best grades with our online Prolog assignment help service

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.
Students who are just getting started with Prolog may find creating programs in this language being a daunting task. This is because algorithms are specified differently than in normal programming languages.

But your dream career as a Prolog programmer highly depends on how you perform in this language. It is, therefore, important that you gain extensive knowledge of it and seek professional guidance on areas you find problematic. You can get exceptional online Prolog assignment help services from our competent experts and become the best student in your class. serves students who are stuck with Prolog assignments. Our Prolog assignment solutions are prepared according to the instructions provided by the student, simplifying each complex area, and making it easy to understand. If you are willing to pay for Prolog assignments, do not look any further as we are the best company for the job. Just upload your assignment details and leave the rest to us.

Topics covered by our online Prolog homework help service

Truth be told, there is not a single topic in Prolog that our experts cannot provide assistance with. We make sure of this by performing several tests on them to determine how good they are at delivering Prolog homework solutions to students.

Only the most qualified and experienced individuals join our team and are authorized to prepare students’ assignments. All our experts have vast knowledge and experience in delivering quality assistance in the following Prolog concepts:

  • Disjunctions and queries: If your concern is to decide whether a query is true or false, our experts can help you with that. Additionally, they will guide you on all the theories and concepts related to disjunctions and queries.
  • Structuring logic programs: If you are facing difficulties using Prolog syntax to develop program structures, you can avail professional guidance from our experts.
  • Programming techniques: There are various techniques used to develop programs in Prolog. Students need to get familiar with all of them in order to shine in the language and deliver assignments that garner them decent grades. Our experts assist students who cannot get their heads around the various Prolog programming techniques by simplifying this area in a manner that is easy for them to understand.
There are many other topics with which our experts have provided quality assistance through our online Prolog homework help service. However, the above three are the most problematic ones as most of the homework help requests usually come from these. We have a team of top-notch online prolog experts who are highly qualified and experienced. 

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Being a reputed provider of help with Prolog homework in the UK and other countries, we have been handling every request made by students diligently, enabling them to achieve their academic goals.

An example of a Prolog assignment problem solved by our experts

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).

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.
:- set_prolog_flag(toplevel_print_anon, false).
:- set_prolog_flag(toplevel_print_options,[max_depth(100)]).
people([a, b, c]).
cols([green, red, yellow]).
%—— THE PREDICATE is_element/2 …
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]).
?- is element(9,[2,0,1,8]).
?- 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))).
?- house info1((p(a,red),p(c,yellow),p(b,green))).

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))).
?- house_info2((p(a,red),p(c,yellow),p(b,green))).

Task 6:
Define a predicate house_info3 to check the final condition c) The greenhouse is not number 3.
?- house info3((p(a,red),p(c,green),p(b,yellow))).
?- house info3((p(a,red),p(c,yellow),p(b,green))).

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))).
?- house info((p(c,red),p(b,yellow),p(a,green))).

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,
:- set_prolog_flag(toplevel_print_anon, false).
:- set_prolog_flag(toplevel_print_options,[max_depth(100)]).
people([a, b, c]).
cols([green, red, yellow]).
%—— THE PREDICATE is_element/2 …
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)

% Task 3
% The goal succeeds if 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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