prolog代写: simulates a non-deterministic finite state automaton

%%%% project

Your code must work on swipl prolog, which can be down loaded to your computer.
See the attached file, fsa_sim_handout.pl.
Note the name contains underscores ( _ ), not dashes ( – ).
This file defines predicates for simulating a Non-deterministic Finite State Automaton.
This assignment is to fill in the definitions of predicates state_struct, next_state_name, and accepts,
so that they work as the comments indicate that they should.
The predicates demo1 and demo2 (defined in the handout file) show how to call this code interactively.

%%%% simulates a non-deterministic finite state
%%%% automaton.

%%% ================ representations

%%% input character string is represented as a list of symbols, e.g., [a, b, a, c]

%%% A state is represented using the functor state, whuch takes 3 arguments:
%%% – name, a symbol: the name of this state
%%% – transition list: a list of transition structures
%%% – accepting: the symbol yes or no

%%% a transition structure is represented using the functor transition,
%%% which takes 2 arguments:
%%% – character, which is a character or epsilon
%%% – name of resulting state.

%%% There is no special structure or functor for an fsa. It is just represented as a list of state structures, the
%%% first state structure in the list is the starting state of the fsa

%%% e.g., this fsa:

% ___ 0 ___ 0
% | | 1 | |
% \-> even <—————-odd <-/
% \ ^ |
% \—————-| | 1
% 1 V
% State even is the foo ————> bar
% start state epsilon
% Even and bar are
% accepting states.
%
%%% is represented as follows:
%%% [state(even, [transition(0, even), transition(1, odd)], yes),
%%% state(odd,[transition(1, foo), transition(0, odd), transition(1, even)], no),
%%% state(foo, [transition(epsilon, bar)], no)
%%% state(bar, [ ], yes)]

%%% ================ predicates

%%% state_struct(+Name, +States, -State_struct)
%%% Name is a state name,
%%% States a list of state structures, and
%%% State_Struct the state structure from States with name Name

% define state_struct here

%%% next_state_name(+State_struct, +Characters, -Next_name, -Next_chars)
%%% find (the name of) a next state to transition to
%%% State_struct is the current state,
%%% Characters is the list of characters currently left in the input,
%%% Next_name is the name of the next state, and Next_chars is the
%%% remaining list of input chars

% define next_state_name here

% accepts(+State_struct, +Fsa, +Chars) true if Fsa accepts character
% string Chars when starting from the state represented by State_struct

% define accepts here

%%% run(Fsa, Chars) Fsa is a list of state stuctures representing
%%% a Fsa (so that the first state in the list is the start state of
%%% the fsa), Chars is a list of symbols and numbers representing
%%% the input to the fsa. run succeeds if and only if the fsa
%%% would accept the sequence of characters

run([State1 | States], Chars):- accepts(State1,[State1 | States], Chars).

% demo1 and demo2 demonstrate calls to run. They each specify a
% automaton and allow you to run that automaton with different input
% strings, by calling, e.g., demo1([1, 0]).
%

% demo1 and demo2 are here to show you how to call your code. They
% are NOT intended to provide sufficient test data for your code. Among
% other things, they do not test epsilon transitions.
% demo1 shows a deterministic finite state automaton.
% demo1([ ]) should return true, as should demo1([1,1]), demo1([0,1,1]),
% demo1([1,1,0,0,1,0,1]), etc (any string with an even number of 1’s)
% demo1([1]) should return false, as should demo1([1,0,1,1]) as should any
% string with an odd number of 1’s
demo1(Chars):- run([state(even,[ transition(0, even),transition(1, odd)], yes),
state(odd,[transition(0, odd), transition(1, even)], no)],
Chars).

% demo2 shows non-determinism: there are transitions from state odd
% both to even and to other for the character 1. both demo2([1,1,0]). and demo2([1,1,2]).
% should return true.
demo2(Chars):- run([state(even,[ transition(0, even),transition(1, odd)], yes),
state(odd,[transition(1, even), transition(1, other), transition(0, odd)], no),
state(other,[transition(2, otherb)],no),
state(otherb,[ ], yes)],
Chars).