Inductively Defining Sets of Strings - string
CS student slogging through a logic class. This question has me befuddled
Inductively Defining Sets of Strings
Find an inductive definition for the following set of strings:
S = {apbcr | p is a natural number, and r is a natural number greater than 0} *the p and r are superscripts*
I'd suggest the following definition:
Related
Is L = {ww^Ru | w, u ∈ {0,1}+} regular language?
let L = {wwRu | w, u ∈ {0,1}+}. Is L regular language ? Note that w, u cannot be empty.
I've tried to prove it is not regular language by the pumping lemma, but I failed when w = 0^p1^p, 01^p, (01)^p. Once I take y = 0^p or 1^p, xyyz will be 00.../11.../01^n0... etc.
And I cannot draw its DFA/NFA or write its regular expression to prove it is regular language.
So is L regular or not ? How can I prove it ?
The language is not regular, and we can prove it using the Myhill-Nerode theorem. Consider the sequence of strings 01, 0101, ..., (01)^n, ... First, notice that none of these strings are in the language. Any prefix of any of these strings which has even length is of the form (01)^2m for some m, and therefore just a shorter string in the sequence; splitting such a prefix in two either has both substrings start with 0 and end with 1, or else it has the first substring start and end with 0 and the second start and end with 1. In either case, these strings are not of the form w(w^R)u for any w or u. Next, notice that the shortest possible string which we can append to any of these strings, to produce a string in the language, is always the reverse of itself followed by either 0 or 1. That is, to turn 01 into a string in the language, we must append 100 or 101; there are no shorter strings we can append to 01 to get a string in the language. The same holds true for 0101: 10100 and 10101 are the shortest possible strings that take 0101 to a string in L. And so on for each string of the form (01)^n. This means that each string of the form (01)^n is distinguishable with respect to the target language w(w^R)u. The Myhill-Nerode theorem tells us that a minimal DFA for a regular language has exactly as many states as there are equivalence classes under the indistinguishability relation. Because we have infinitely many distinguishable strings with respect to our language, a minimal DFA for this language must have infinitely many states. But, a DFA cannot have infinitely many states; this is a contradiction. This means that our language cannot be regular.
The language is REGULAR: L = 00(0+1)+ + 11(0+1)+ + 0(11)+0(0+1)+ + 1(00)+1(0+1)+
alloy predicate calculus post production
As I'm new to alloy, this is most likely a simple question. I've been through the on-line tutorials and am now reading the Software Abstractions, revised edition. On page 34 there is an example at the bottom of the page:
r' = {b:B, a:A, c:C | a->b->c in r}
where the text says that this defines a new relation of B->A->C. I don't see how an explicit order for r' is achieved by this statement.
It's the property of set comprehension
{a: A | somePredicate1[a]} is of type A and returns a set containing all atoms for which somePredicate1 holds;
{a: A, b: B | somePredicate2[a, b]} is of type A->B and returns a relation containing all a->b tuples for which somePredicate2 holds;
and so on
The syntax of set comprehension basically consists of two parts (1) type declaration (before the | character), and (2) predicate which must hold for every element in the returned set.
Is "set" the default multiplicity?
Are these two equivalent:
r: A -> B
r: A set -> set B
That is, is set the default multiplicity?
If yes, then I will quibble with the definition of the arrow operator in the Software Abstractions book. The book says on page 55:
The arrow product (or just product) p->q of two relations p and q is
the relation you get by taking every combination of a tuple from p and
a tuple from q and concatenating them.
I interpret that definition to mean the only valid instance for p->q is one that has every possible combination of tuples from p with tuples from q. But that's not right (I think). Any instance containing mappings between p and q is valid. For example, on page 56 is this example,
Name = {(N0), (N1)}
Addr = {(D0), (D1)}
The book says this is a valid relation for Name->Addr
{(N0, D0), (N0, D1), (N1, D0), (N1, D1)}
But that's not the only valid relation, right? For example, this is a valid relation:
{(N0, D0), (N1, D1)}
Is that right?
The declaration r : A->B means r is a subset of A->B. The expression A->B has just one value, which is the cross product of A and B. The declaration results in a set of possible values for r, which would include both the example given in the book that you cite, and the example that you ask about.
design NFA which accepts specific length of strings
Im looking forward to design a FA which accepts some kind of string that accept some A and B.
First a string which the number of A is five more times higher than B.
i mean L={w∈{A,B}* and (nA(W)-nB(W) mod 5=0)
And also a FA which accept different number of character in a string:
L={A^n B^m C^k | n,k>0 and m>3}
I design some FAs But they did not work perfectly on this complicated strings.
Any help on how should i design this ?
Unfortunately, your questions are confusing as the english text doesn't agree with the mathematical formula. I will try to answer to these four questions then:
A language which consists of string over {a,b} that the number of a (= #a(w))
is five times as the number of b ( #b(w)),
L = { w in {a,b}* : #a(w)>#b(w) and #a(w)=#b(w)mod5 }
This cannot be done by an NFA. The proof is simple by using the pumping lemma (P.L) with the string a^pb^5p, where p is the constant of P.L.
For the language: L={w∈{A,B}* : (nA(W)-nB(W)) mod 5=0} that you wrote,
you can do it with an DFA that consists of a cycle of 5 states.
The transitions are, if you read a go clockwise if you read b go counter-clocwise. Choose at random one state to be initial state and the same state will be the final state.
For the language L={A^n B^m C^k | n,k>0 and m>3}, it should be easy to find out
if you read L as L=A(A)* B(B)* c^4(C)*
For the language that accepts different number of character in the string (let's say over a,b). The language should be R={ w in {a,b}* : #a(w) not equal #b(w)}
This language again it cannot be recognized by an NFA. If this language was regular (recognzied by an NFA) so would be this language:
L=a*b* intersection (R complement). The language L is {a^n b^n/ n non-negative integer}.
The language L is the first example of most books when they speak about languages that are non-regular.
Hopefully, you will find this answer helpful.
Translate list comprehension to Prolog
I have a list comprehension in Haskell that I want to translate to Prolog. The point of the list comprehension is rotating a 4 by 4 grid: rotate :: [Int] -> [Int] rotate grid = [ grid !! (a + 4 * b) | a <- [0..3], b <- [0..3] ] Now in Prolog, I translated it like this: rotateGrid([T0,T1,T2,T3,T4,T5,T6,T7,T8,T9,T10,T11,T12,T13,T14,T15], [T0,T4,T8,T12,T1,T5,T9,T13,T2,T6,T10,T14,T3,T7,T11,T15]). Can we do better?
We can use findall/3 for list comprehensions (Cf. the SWI-Prolog Documentation). E.g., ?- findall(X, between(1,10,X), Xs). Xs = [1,2,3,4,5,6,7,8,9,10] Xs is a list holding all values that can unify with X when X is a number between 1 and 10. This is roughly equivalent to the Haskell expression let Xs = [x | x <- [1..10]](1). You can read a findall/3 statement thus: "find all values of [First Argument] such that [Conditions in Second Argument] hold, and put those values in the list, [Third Argument]". I've used findall/3 to write a predicate rotate_grid(+Grid, ?RotatedGrid). Here is a list of the approximate Haskell-Prolog equivalences I used in the predicate; each line shows the relation between the value that the Haskell expression will evaluate to and the Prolog variable with the same value: a <- [0..3] = A in between(0, 3, A) b <- [0..3] = B in between(0, 3, B) (a + 4 * d) = X in X is A + 4 * D <Grid> !! <Index> = Element in nth0(Index, Grid, Element) Then we simply need to find all the values of Element: rotate_grid(Grid, RotatedGrid) :- findall( Element, ( between(0,3,A), between(0,3,B), Index is A + 4 * B, nth0(Index, Grid, Element) ), RotatedGrid ). To verify that this produces the right transformation, I down-cased the Prolog code from the question and posed the following query: ?- rotate_grid([t0,t1,t2,t3,t4,t5,t6,t7,t8,t9,t10,t11,t12,t13,t14,t15], [t0,t4,t8,t12,t1,t5,t9,t13,t2,t6,t10,t14,t3,t7,t11,t15]). | true. Footnotes: (1): between/3 isn't actually the analogue of [m..n], since the latter returns a list of values from m to n where between(M,N,X) will instantiate X with each value between M and N (inclusive) on backtracking. To get a list of numbers in SWI-Prolog, we can use numlist(M,N,Ns). So a stricter analogue for x <- [1.10] would be the conjunction member(X, Ns), numlist(1, 10, Ns).
You want a permutation of a list. The concrete elements are not considered. Therefore, you can generalize your Haskell signature to rotate :: [x] -> [x] This is already a very valuable hint for Prolog: the list's elements will not be considered - elements will not even be compared. So a Prolog solution should be able to handle variables directly, like so: ?- rotateGrid(L,R). L = [_A,_B,_C,_D,_E,_F,_G,_H,_I,_J,_K,_L,_M,_N,_O,_P], R = [_A,_E,_I,_M,_B,_F,_J,_N,_C,_G,_K,_O,_D,_H,_L,_P]. And your original definition handles this perfectly. Your version using list comprehensions suggests itself to be realized via backtracking, certain precautions have to be taken. Using findall/3, as suggested by #aBathologist will rename variables: ?- length(L,16),rotate_grid(L,R). L = [_A,_B,_C,_D,_E,_F,_G,_H,_I,_J,_K,_L,_M,_N,_O,_P], R = [_Q,_R,_S,_T,_U,_V,_W,_X,_Y,_Z,_A1,_B1,_C1,_D1,_E1,_F1]. The built-in predicate bagof/3 addresses this problem. Note that we have to declare all local, existential variables explicitly: rotate_grid2(Grid, RotatedGrid) :- bagof( Element, A^B^Index^ % declaration of existential variables ( between(0,3,A), between(0,3,B), Index is A + 4 * B, nth0(Index, Grid, Element) ), RotatedGrid). For lists that are shorter than 16 elements, the Haskell version produces a clean error, but here we get pretty random results: ?- L=[1,2,3,4],rotate_grid(L,R). L = [1,2,3,4], R = [1,2,3,4]. ?- L=[1,2,3,4,5],rotate_grid(L,R). L = [1,2,3,4,5], R = [1,5,2,3,4]. This is due to the unclear separation between the part that enumerates and "generates" a concrete element. The cleanest way is to add length(Grid, 16) prior to the goal bagof/3. List comprehensions in Prolog Currently, only B-Prolog offers a form of list comprehensions: R#=[E: A in 0..3,B in 0..3,[E,I],(I is A+4*B,nth0(I,L,E))]. However, it does not address the second problem: | ?- L = [1,2,3], R#=[E: A in 0..3,B in 0..3,[E,I],(I is A+4*B,nth0(I,L,E))]. L = [1,2,3] R = [1,2,3] yes
Use a loop predicate foreach/4
If the comprehension should retain variables, which is for example important in constraint programming, a Prolog system could offer a predicate foreach/4. This predicate is the DCG buddy of foreach/2.
Here is how variables are not retained via findall/3, the
result R contains fresh variables according to the ISO
core semantics of findall/3:
Welcome to SWI-Prolog (threaded, 64 bits, version 7.7.1)
SWI-Prolog comes with ABSOLUTELY NO WARRANTY. This is free software.
?- functor(L,foo,5), findall(X,
(between(1,5,N), M is 6-N, arg(M,L,X)), R).
L = foo(_5140, _5142, _5144, _5146, _5148),
R = [_5210, _5204, _5198, _5192, _5186].
And here is how variables can be retained via foreach/4,
the resulting list has the same variables as the compound
we started with:
Jekejeke Prolog 3, Runtime Library 1.3.0
(c) 1985-2018, XLOG Technologies GmbH, Switzerland
?- [user].
helper(N,L) --> [X], {M is 6-N, arg(M,L,X)}.
Yes
?- functor(L,foo,5), foreach(between(1,5,N),helper(N,L),R,[]).
L = foo(_A,_G,_M,_S,_Y),
R = [_Y,_S,_M,_G,_A]
Using foreach/4 instead of bagof/3 might seem a little bit over the top. foreach/4 will probably only show its full potential when implementing Picat loops, since it can build up constraints, what bagof/3 cannot do.
foreach/4 is an implementation without the full materialization of all solution that are then backtracked. It shares with bagof/3 the reconstruct of variables, but still allows backtracking in the conjunction of the closures.