the regular languages are closed under the operation:
init(L) = the set of the strings w such that for some x, wx is in L.
EDIT :
x can be any string, a character or empty string
How can I prove that ?
OK, misread the quesion on the first time, now I get it. It still trivial. Looking at the automate what you searching is a partion of the automate into two state sets S1 and S2, so that just one transition is between them (and if its from S1->S2 S1 contains of course the start node, and S2 the end node). Such exist always (exception empty language), in case there is no such node you can add one, so w is just a set containing the empty word, which is of course also regular (as well as the empty language case).
Unless I'm misunderstanding, the answer is that you can't. Because it's not true.
First, let's consider the language L = {aa, bb, cc} and alphabet {a, b, c}
So, init(L) = {a, b, c}. However, each of the elemnts in init(L) are not in L.
Edit: If we are concatenating empty characters, then init(L) = {a, b, c, aa, bb, cc}. Which is still not equal to L.
A language is regular iff there's a finite-state automaton that recognizes it. So, suppose L is a regular language and let A be an automaton that recognizes it. Now, say that a state of A is "good" if there's some set of possible transitions starting there and ending in the "accept" state. Define a new automaton A' in which all transitions to "good" states are replaced by direct transitions to the accept state. Then the language recognized by A' is exactly init(L).
I think it's a new DFA B, that makes all the state of A(the original DFA) that can reach to the final states of A the final state of B.
Related
I'm looking at an exam question which says
'Explain how a correctly formed arithmetic expression over variables a, b, c that contains additions, multiplication and brackets can be recognised by a DFA with a counter. (Such a DFA can increment and decrement the counter on each transition as well as test it for zero).'
I'm not quite sure I understand what it's actually asking. If someone could provide a hint that would be greatly appreciated. (i.e. what the multiplication of a and b would be for example)
If all you want is just a hint, then I think this should be enough:
Some strings represent valid arithmetic expressions such as
a + b
(a + b)*(c + d)
and some are invalid such as
a + - b
a b
)
()
(a + b)*)c+d(
"can be recognised by a DFA with a counter" here means that you can build a Deterministic finite automaton such that it stops at one of the "accept states" after processing a sting if and only if that string represents a valid arithmetic expression
P.S. Additional hint: the counting part is important here because a simple DFA can't even recognize whether an arbitrary string of brackets is properly matched.
I'm working on a problem from the Languages and Machines: An Introduction to the Theory of Computer Science (3rd Edition) in Chapter 2 Example 6.
I need help finding the answer of:
Recursive definition of set strings over {a,b} that contains one b and even number of a's before the first b?
When looking for a recursive definition, start by identifying the base cases and then look for the recursive steps - like you're doing induction. What are the smallest strings in this language? Well, any string must have a b. Is b alone a string in the language? Why yes it is, since there are zero as that come before it and zero is an even number.
Rule 1: b is in L.
Now, given any string in the language, how can we get more strings? Well, we can apparently add any number of as to the end of the string and get another string in the language. In fact, we can get all such strings from b if we simply allow you to add one more a to the end of a string in the language. From x in L, we therefore recover xa, xaa, ..., xa^n, ... = xa*.
Rule 2: if x is in L then xa is in L.
Finally, what can we do to the beginning of strings in our language? The number of as must be even. So far, rules 1 and 2 only allow us to construct strings that have zero as before the b. We should be able to get two, four, six, etc., all the even numbers, of as. A rule that lets us add two as to any string in our language will let us add ever more as to the beginning while maintaining the evenness property we require. Starting with x in L, we recover aax, aaaax, ..., (aa)^(2n)x, ... = (aa)*x.
Rule 3: if x is in L, then aax is in L.
Optionally, you may add the sometimes implicitly understood rule that only those things allowed by the aforementioned rules are allowed. Otherwise, technically anything is allowed since we haven't explicitly disallowed anything yet.
Rule 4: Nothing is in L unless by virtue of some combination of the rules 1, 2 and/or 3 above.
I am a trying to prove that every regular language is decidable.
So in order to prove that I am trying to show that I can move from deterministic finite automaton (DFA) to a Turing decidable machine.
So I am not sure how to construct a Turing machine that simulates the original automate (DFA).
The states (in the automate and the Turing machine ) will be similar off course.. but I am not sure how to continue..
Thanks in advance.
Shiran
Wouldn't this work : the input tape carries the word to be recognised followed by some letter # from outside the input alphabet. For each state q of the DFA, you have a Turing machine state q with transitions:
input letter -> tape operation, next state, write symbol
c for each input letter -> move right, the state the DFA reaches from state q with letter c, write c
# -> STOP ACCEPT or STOP REJECT, depending on whether q is final in the DFA.
A problem I'm working on with Prolog is to see if a train can travel from one destination to the next. There are two rules.
A train can travel through on or more intermediary from one destination to the next.
Ex: San Francisco to Los Angeles
Los Angeles to Irvine
Irvine to San Diego
This gives a route from San Francisco to San Diego.
A train can travel to and from a destination. So if a train can travel from San Francisco to Los Angeles, it can travel from Los Angeles to San Francisco.
This is the code I currently have.
nonStopTrain(sandiego,oceanside).
nonStopTrain(lasvegas,sandiego).
nonStopTrain(sanfrancisco,bakersfield).
nonStopTrain(bakersfield,sandiego).
nonStopTrain(oceanside,losangeles).
nonStopTrain(portland,sanfrancisco).
nonStopTrain(seattle,portland).
canTravel(From, To) :- nonStopTrain(From, To); nonStopTrain(To, From).
canTravel(From, To) :-
canTravel(From, Through), canTravel(Through, To).
The problem is the ability to travel bidirectionally. When I run this program, I keep running back and fourth between the same places, and I'm not exactly sure why.
The problem with a naive solution is that there are an infinite number of ways to get from point A to point B if you don't eliminate cycles. Suppose I want to go from Seattle to San Francisco. Without handling cycles, we're going to get each of these as a unique solution:
seattle -> portland -> seattle -> portland -> sanfrancisco
seattle -> portland -> seattle -> portland -> seattle -> portland -> sanfrancisco
seattle -> (portland -> seattle) x N -> sanfrancisco
There's no limit to the number of times you can double back on yourself, so there's effectively an infinite number of solutions once you have as little as three nodes connected. In practice you don't want any solutions where you double back on yourself, but Prolog doesn't know that and there's no intuitive and naive way to prevent it.
One of the better ways forward is to simply keep track of where you've been. To do that we're going to need to make the predicate take an extra argument. First I've also introduced a helper predicate.
connectedDirectly(From, To) :- nonStopTrain(From, To) ; nonStopTrain(To, From).
Having this separated out will reduce the desire to call canTravel recursively when we really just want to attach one more leg to the journey. Now for canTravel:
canTravel(From, To) :- canTravel(From, To, []).
This is an arity 2 rule that maps onto our new arity 3 rule. The list of places we've been is always empty initially. Now we need a base case:
canTravel(From, To, _) :- connectedDirectly(From, To).
This should be obvious. Now the inductive case is a little different, because we need to do two things: find a new leg to attach to the journey, make sure we haven't been through that leg before, and then recur, adding the new leg to the list of places we've been. Finally, we want to ensure we don't get large cycles where we end up where we started, so we add a rule to the end to make sure we don't.
canTravel(From, To, Visited) :-
connectedDirectly(From, Through),
\+ memberchk(Through, Visited),
canTravel(Through, To, [Through|Visited]),
From \= To.
Now if you run it you'll find you get 98 solutions and all the solutions are symmetric:
?- forall(canTravel(X, Y), (write(X-Y), nl)).
sandiego-oceanside
lasvegas-sandiego
sanfrancisco-bakersfield
... etc.
So, happily, we were able to avoid going for a breadth-first search solution.
Edit
I have apparently confused the situation by overloading the name canTravel for two separate predicates. In Prolog, a predicate is uniquely defined by the name and the arity, much like overloading in C++ or Java where the "effective method" is determined by the number of arguments and the name, not just the name.
Your intuition is correct—the empty list in canTravel(From, To) :- canTravel(From, To, []) is establishing an initial binding for the list of visited places. It's not exactly allocating storage so much as establishing a base case.
There are really two uses of canTravel inside itself. One of them is calling canTravel/3 from canTravel/2. In this case, canTravel/3 is really sort of like a helper, doing the actual work of canTravel/2, but with an internal variable that we are initializing to the empty list. The other use is canTravel/3 from within canTravel/3, and for that we're both using it to achieve the same goal: recursion, Prolog's primary "looping" construction.
The third argument in canTravel(From, To, _) :- connectedDirectly(From, To) is what makes this clause part of canTravel/3. This is the base case of the recursion, so it doesn't need to consider the places we've visited so far (although the inductive case will prevent a circular journey). We could also check it here, but it turns out to be more expensive and have no effect on the resultset:
canTravel(From, To, Visited) :- connectedDirectly(From, To), \+ memberchk(To, Visited).
I concluded that if it was adding expense and complexity without changing the answers we could omit the check, which reduces the base case to the original one with the anonymous third variable.
It may make more sense to see this without the overloading, in which case it looks like this:
canTravel(From, To) :- canTravel_loop(From, To, []).
canTravel_loop(From, To, _) :- connectedDirectly(From, To).
canTravel_loop(From, To, Visited) :-
connectedDirectly(From, Through),
\+ memberchk(Through, Visited),
canTravel_loop(Through, To, [Through|Visited]),
From \= To.
Edit 2
Regarding the "bar operator," your intuition is once again correct. :) I'm using it here to prepend an item to a list. What's confusing you is that in Prolog, with unification, most expressions express relationships rather than procedures. So depending on the context, [X|Xs] might be used to construct a new list (if you have X and XS in hand) or it might be used to break an implicit list into a head X and tail Xs. Look at all the ways I can use it just from the repl:
?- X = hello, Xs = [world, new, user], Y = [X|Xs].
Y = [hello, world, new, user].
This is basically how we're using it in canTravel: we have Through and we have Visited, so we're making a new list with Through first and Visited as the tail, and that's the third parameter to the recursive invocation. In procedural terms, we're just adding Through to a variable we're using in our loop.
But because this is Prolog, we're not limited to using things in one direction:
?- Y = [hello, world, new, user], Y = [X|Xs].
X = hello,
Xs = [world, new, user].
?- Y = [hello, world, new, user], [X|Xs] = Y.
X = hello,
Xs = [world, new, user].
Notice that Prolog didn't care which direction the assignment happened in, but it managed to "work backwards" to figure out what X and Xs should be using Y. This is one of the magical things about Prolog. (Note that in the examples in this session I'm omitting the variables which are echoed back because they obscure the point.)
In general, you want predicates that can solve for different parameters. For instance, member/2 can be used to test membership or to enumerate items. append/3 can build a new list from two old lists, or it can enumerate all the ways to split a list into two segments, or it can find a prefix or suffix given a list and a suffix or prefix.
As you get more used to this functionality you'll stop thinking of Prolog rules as being like functions in other languages and start to see them as relations: logical "truths" that exist between certain constructions. member/2 isn't written by trying to enumerate items or by seeking through a list looking for a particular value. It's implemented by saying: the relation member(Item, List) is true when the Item is the first thing in List:
member(Item, [Item|_]).
or else when Item is in the remainder of the list:
member(Item, [_|Tail]) :- member(Item, Tail).
This definition is sufficient for all the possible uses. If Item is not instantiated, it will be instantiated to the first item in the list, then the first item in the tail of that list, and so on. If Item is instantiated, it will be true if Item is the first item in the list or if it is the first item in the tail. Surprisingly, member/2 can even be used to generate lists that contain a value:
?- member(1, X).
X = [1|_G274] ;
X = [_G8, 1|_G12] ;
X = [_G8, _G11, 1|_G15] .
You can see what happened there: the _ in the second clause is being made into anonymous variables, so it's generating lists with the 1 in the first position, then the second, then the third, etc.
A lot of Prolog works like this. This one is also pretty surprising:
?- length(X, 3).
X = [_G273, _G276, _G279].
Hope this helps clarify things a bit more! :)
Do you have use some specific Prolog system?
Your program will work as intended without modifications (well, you have to add :- auto_table. as a first line of your program) in a system with tabling support, like B-Prolog.
I think adding a cut will stop your infinite recursion issue because once it finds an answer it won't keep backtracking forever:
canTravel(From, To) :- nonStopTrain(From, To); nonStopTrain(To, From).
canTravel(From, To) :-
canTravel(From, Through), canTravel(Through, To), !.
I have no doubt that there is a more correct solution than this though.
Ok, so i have this
edu_less(hs,college).
edu_less(college,masters).
edu_less(masters,phd).
I need to write a function to tell if something is less than the other. The predicate is
edu_le.
So if i put edu_le(hs,phd). it should return yes.
I came up with this.
edu_le(A,B) :- A = B.
edu_le(A,B) :- edu_less(A,B).
edu_le(A,B) :- edu_less(A,C), edu_le(C,B).
I really don't want it to go through everything and return multiple answers.
Is it possible to only return yes or no if it finds that it is in fact less than or equal to the 2nd argument?
So basically if i use the example edu_le(hs,phd) again, then because hs is less than college, and college is than masters, and masters is less than phd, then hs must be less than phd and it would say yes.
Sorry, really new to prolog, still trying to get the hang of this.
In the predicate definition
edu_le(A,B) :- A = B.
edu_le(A,B) :- edu_less(A,B).
edu_le(A,B) :- edu_less(A,C), edu_le(C,B).
the second clause is superfluous and causes repeated generation of answers. Use
edu_le(A,B) :- A = B.
edu_le(A,B) :- edu_less(A,C), edu_le(C,B).
This gives you one true answer, then no more answers (false) on backtracking. You can use a cut in the first clause, but then generating won't work anymore.
?- edu_le(hs,X).
X = hs ;
X = college ;
X = masters ;
X = phd ;
false.
becomes incomplete:
?- edu_le(hs,X).
X = hs.
As mat suggested, use once/1 instead. In a good Prolog implementation, this predicate works as if your program had cuts in strategic places, speeding up your program without disturbing its logical semantics.
The most practical way to write predicates like that is to use the cut (!). The cut causes further clauses not to be considered when backtracking. You would write your predicate as following:
edu_le(A,B) :- A = B, !.
edu_le(A,B) :- edu_less(A,B), !.
edu_le(A,B) :- edu_less(A,C), edu_le(C,B).
The last clause does not need a cut because there are no further clauses to consider anyway. The cut is placed after any tests to determine whether the clause should succeed.
Logic programming purists disapprove of the cut, because it makes the meaning of a predicate depend on the ordering of the clauses, which is unlike logic in mathematics.
!/0 also makes this program incomplete, consider for example the most general query with both versions:
?- edu_le(X, Y).
It is often better to use once/1 if you only want a single proof of a particular goal:
?- once(edu_le(hs, phd)).
I would suggest you NOT to follow the path proposed by Juho Östman and keep purity - otherwise, why should you use Prolog in first instance? If you are too lenient with sticking to the logical paradigm you obtain some unpleasing results. In this case, Juho's predicate is definitely different from yours, and I'll show you why.
First, just drop the useless edu_le(A,B) :- edu_less(A,B). rule, as larsmans suggests. You will obtain a less redundant version of your original predicate:
edu_le1(A, A).
edu_le1(A, B) :- edu_less(A, C), edu_le1(C, B).
It just behaves as edu_le, meaning: given an arbitrary query, it produces exactly the same answer, except for duplicates (edu_le1 has less). You may just be happy with it, but it still has some redundant answers that you may not like; e.g, under SWI:
?- edu_le1(hs, hs)
true ;
false.
Now you may say you do not like it because it still has the redundant false, but if you use Juho's predicate instead (without the useless rule):
edu_le2(A, A) :- !.
edu_le2(A, B) :- edu_less(A, C), edu_le2(C, B).
it's true that you eliminate the useless final false:
?- edu_le2(hs, hs)
true.
?-
but you lose more than that: You lose, as mat remarks, the possibility of generating all the solutions when one variable is not instantiated:
?- edu_le1(hs, B) %same, with more copies, for edu_le
B = hs ;
B = college ;
B = masters ;
B = phd ;
false.
?- edu_le2(hs, B)
B = hs. %bad!
?-
In other words, the latter predicate is NOT equivalent to the former: edu_le and edu_le1 have type edu_le(?A, ?B), while instead edu_le2 has type edu_le2(+A, +B) (see [1] for the meaning). Be sure: edu_le2 is less useful because it does less things, and thus can be reused in less contexts. This because the cut in edu_le2 is a red cut, i.e., a cut that changes the meaning of the predicate where it is introduced. You may nevertheless be content with it, given that you understand the difference between the two. It all depends on what you want to do with it.
If you want to get the best of the two worlds, you need to introduce in edu_le1 a proper green cut that lowers the redundancy when A and B are completely instantiated to terms. At the purpose, you must check that A and B are instantiated to the same term before cutting. You cannot do it with =, because = does not check, but unifies. The right operator is ==:
edu_le3(A, B) :- (A == B -> ! ; true), A = B.
edu_le3(A, B) :- edu_less(A, C), edu_le3(C, B).
Note that the additional cut in the first rule is active only when A and B happen to be the same term. Now that the cut is a proper green cut, the predicate works also in the most general cases as your original one:
?- edu_le3(A, A).
true.
?- edu_le3(A, B). %note that A and B are not the same term
A = B ;
A = hs,
B = college ;
A = hs,
B = masters ;
A = hs,
B = phd ;
A = college,
B = masters ;
A = college,
B = phd ;
A = masters,
B = phd ;
false.
?-
with Prolog backtracking through all the solutions.
I don't think there is some way to eliminate the last false without introducing too strong dependency on edu_lt. This because we must keep open the possibility that there is another edu_lt to explore, in the case you decide later to enrich it with more ground facts. So, in my opinion, this is the best you can have.
[1] SWI Prolog reference manual, section 4.1.