I'm trying to capture a string with a combination of a's and b's but always ending with b. In other words:
local patt = S'ab'^0 * P'b'
matching aaab and bbabb but not aaa or bba. The above however does not match anything. Is this because S'ab'^0 is greedy and matches the final b? I think so and can't think of any alternatives except perhaps resorting to lpeg.Cmt which seems like overkill. But maybe not, anyone know how to match such a pattern? I saw this question but the problem with the solution there is that it would stop at the first end marker (i.e. 'cat' there, 'b' here) and in my case I need to accept the middle 'b's.
P.S.
What I'm actually trying to do is match an expression whose outermost rule is a function call.
E.g.
func();
func(x)(y);
func_arr[z]();
all match but
exp;
func()[1];
4 + 5;
do not. The rest of my grammar works and I'm pretty sure this boils down to the same issue but for completeness, the grammar I'm working with looks something like:
top_expr = V'primary_expr' * V'postfix_op'^0 * V'func_call_op' * P';';
postfix_op = V'func_call_op' + V'index_op';
And similarly the V'postfix_op'^0 eats up the func_call_op I'm expecting at the end.
Yes, there is no backtracking, so you've correctly identified the problem. I think the solution is to list the valid postfix_op expressions; I'd change V'func_call_op' + V'index_op' to V'func_call_op'^0 * V'index_op' and also change the final V'func_call_op' to V'func_call_op'^1 to allow several function calls at the end.
Update: as suggested in the comments, the solution to the a/b problem would be (P'b'^0 * P'a')^0 * P'b'^1.
How about this?
local final = P'b' * P(-1)
local patt = (S'ab' - final)^0 * final
The pattern final is what we need at the end of the string.
The pattern patt matches the set 'ab' unless it is followed by the final sequence. Then it asserts that we have the final sequence. That stops the final 'b' from being eaten.
This doesn't guarantee that we get any a's (but neither would the pattern in the question have).
Sorry my answer comes too late but I think it's worth to give this question a more correct answer.
As I understand it, you just want a non-blind greedy match. But unfortunately the "official documentation" of LPeg only tells us how to use LPeg for blind greedy match (or repetition). But this pattern can be described by a parsing expression grammar. For rule S if you want to match as many E1 as you can followed by E2, you need to write
S <- E1 S / E2
The solution to a/b problem becomes
S <- [ab] S / 'b'
You might want to optimize the rule by inserting some a's in the first option
S <- [ab] 'a'* S / 'b'
which will reduce the recursions a lot. As for your real problem, here's my answser:
top_expr <- primary_expr p_and_f ';'
p_and_f <- postfix_op p_and_f / func_call_op
postfix_op <- func_call_op / index_op
Related
I am writing a parser for a query engine. My parser DCG query is not deterministic.
I will be using the parser in a relational manner, to both check and synthesize queries.
Is it appropriate for a parser DCG to not be deterministic?
In code:
If I want to be able to use query/2 both ways, does it require that
?- phrase(query, [q,u,e,r,y]).
true;
false.
or should I be able to obtain
?- phrase(query, [q,u,e,r,y]).
true.
nevertheless, given that the first snippet would require me to use it as such
?- bagof(X, phrase(query, [q,u,e,r,y]), [true]).
true.
when using it to check a formula?
The first question to ask yourself, is your grammar deterministic, or in the terminology of grammars, unambiguous. This is not asking if your DCG is deterministic, but if the grammar is unambiguous. That can be answered with basic parsing concepts, no use of DCG is needed to answer that question. In other words, is there only one way to parse a valid input. The standard book for this is "Compilers : principles, techniques, & tools" (WorldCat)
Now you are actually asking about three different uses for parsing.
A recognizer.
A parser.
A generator.
If your grammar is unambiguous then
For a recognizer the answer should only be true for valid input that can be parsed and false for invalid input.
For the parser it should be deterministic as there is only one way to parse the input. The difference between a parser and an recognizer is that a recognizer only returns true or false and a parser will return something more, typically an abstract syntax tree.
For the generator, it should be semi-deterministic so that it can generate multiple results.
Can all of this be done with one, DCG, yes. The three different ways are dependent upon how you use the input and output of the DCG.
Here is an example with a very simple grammar.
The grammar is just an infix binary expression with one operator and two possible operands. The operator is (+) and the operands are either (1) or (2).
expr(expr(Operand_1,Operator,Operand_2)) -->
operand(Operand_1),
operator(Operator),
operand(Operand_2).
operand(operand(1)) --> "1".
operand(operand(2)) --> "2".
operator(operator(+)) --> "+".
recognizer(Input) :-
string_codes(Input,Codes),
DCG = expr(_),
phrase(DCG,Codes,[]).
parser(Input,Ast) :-
string_codes(Input,Codes),
DCG = expr(Ast),
phrase(DCG,Codes,[]).
generator(Generated) :-
DCG = expr(_),
phrase(DCG,Codes,[]),
string_codes(Generated,Codes).
:- begin_tests(expr).
recognizer_test_case_success("1+1").
recognizer_test_case_success("1+2").
recognizer_test_case_success("2+1").
recognizer_test_case_success("2+2").
test(recognizer,[ forall(recognizer_test_case_success(Input)) ] ) :-
recognizer(Input).
recognizer_test_case_fail("2+3").
test(recognizer,[ forall(recognizer_test_case_fail(Input)), fail ] ) :-
recognizer(Input).
parser_test_case_success("1+1",expr(operand(1),operator(+),operand(1))).
parser_test_case_success("1+2",expr(operand(1),operator(+),operand(2))).
parser_test_case_success("2+1",expr(operand(2),operator(+),operand(1))).
parser_test_case_success("2+2",expr(operand(2),operator(+),operand(2))).
test(parser,[ forall(parser_test_case_success(Input,Expected_ast)) ] ) :-
parser(Input,Ast),
assertion( Ast == Expected_ast).
parser_test_case_fail("2+3").
test(parser,[ forall(parser_test_case_fail(Input)), fail ] ) :-
parser(Input,_).
test(generator,all(Generated == ["1+1","1+2","2+1","2+2"]) ) :-
generator(Generated).
:- end_tests(expr).
The grammar is unambiguous and has only 4 valid strings which are all unique.
The recognizer is deterministic and only returns true or false.
The parser is deterministic and returns a unique AST.
The generator is semi-deterministic and returns all 4 valid unique strings.
Example run of the test cases.
?- run_tests.
% PL-Unit: expr ........... done
% All 11 tests passed
true.
To expand a little on the comment by Daniel
As Daniel notes
1 + 2 + 3
can be parsed as
(1 + 2) + 3
or
1 + (2 + 3)
So 1+2+3 is an example as you said is specified by a recursive DCG and as I noted a common way out of the problem is to use parenthesizes to start a new context. What is meant by starting a new context is that it is like getting a new clean slate to start over again. If you are creating an AST, you just put the new context, items in between the parenthesizes, as a new subtree at the current node.
With regards to write_canonical/1, this is also helpful but be aware of left and right associativity of operators. See Associative property
e.g.
+ is left associative
?- write_canonical(1+2+3).
+(+(1,2),3)
true.
^ is right associative
?- write_canonical(2^3^4).
^(2,^(3,4))
true.
i.e.
2^3^4 = 2^(3^4) = 2^81 = 2417851639229258349412352
2^3^4 != (2^3)^4 = 8^4 = 4096
The point of this added info is to warn you that grammar design is full of hidden pitfalls and if you have not had a rigorous class in it and done some of it you could easily create a grammar that looks great and works great and then years latter is found to have a serious problem. While Python was not ambiguous AFAIK, it did have grammar issues, it had enough issues that when Python 3 was created, many of the issues were fixed. So Python 3 is not backward compatible with Python 2 (differences). Yes they have made changes and libraries to make it easier to use Python 2 code with Python 3, but the point is that the grammar could have used a bit more analysis when designed.
The only reason why code should be non-deterministic is that your question has multiple answers. In that case, you'd of course want your query to have multiple solutions. Even then, however, you'd like it to not leave a choice point after the last solution, if at all possible.
Here is what I mean:
"What is the smaller of two numbers?"
min_a(A, B, B) :- B < A.
min_a(A, B, A) :- A =< B.
So now you ask, "what is the smaller of 1 and 2" and the answer you expect is "1":
?- min_a(1, 2, Min).
Min = 1.
?- min_a(2, 1, Min).
Min = 1 ; % crap...
false.
?- min_a(2, 1, 2).
false.
?- min_a(2, 1, 1).
true ; % crap...
false.
So that's not bad code but I think it's still crap. This is why, for the smaller of two numbers, you'd use something like the min() function in SWI-Prolog.
Similarly, say you want to ask, "What are the even numbers between 1 and 10"; you write the query:
?- between(1, 10, X), X rem 2 =:= 0.
X = 2 ;
X = 4 ;
X = 6 ;
X = 8 ;
X = 10.
... and that's fine, but if you then ask for the numbers that are multiple of 3, you get:
?- between(1, 10, X), X rem 3 =:= 0.
X = 3 ;
X = 6 ;
X = 9 ;
false. % crap...
The "low-hanging fruit" are the cases where you as a programmer would see that there cannot be non-determinism, but for some reason your Prolog is not able to deduce that from the code you wrote. In most cases, you can do something about it.
On to your actual question. If you can, write your code so that there is non-determinism only if there are multiple answers to the question you'll be asking. When you use a DCG for both parsing and generating, this sometimes means you end up with two code paths. It feels clumsy but it is easier to write, to read, to understand, and probably to make efficient. As a word of caution, take a look at this question. I can't know that for sure, but the problems that OP is running into are almost certainly caused by unnecessary non-determinism. What probably happens with larger inputs is that a lot of choice points are left behind, there is a lot of memory that cannot be reclaimed, a lot of processing time going into book keeping, huge solution trees being traversed only to get (as expected) no solutions.... you get the point.
For examples of what I mean, you can take a look at the implementation of library(dcg/basics) in SWI-Prolog. Pay attention to several things:
The documentation is very explicit about what is deterministic, what isn't, and how non-determinism is supposed to be useful to the client code;
The use of cuts, where necessary, to get rid of choice points that are useless;
The implementation of number//1 (towards the bottom) that can "generate extract a number".
(Hint: use the primitives in this library when you write your own parser!)
I hope you find this unnecessarily long answer useful.
I have a fairly large expression that involves a lot of subexpressions of the form (100*A^3 + 200*A^2 + 100*A)*x or (-A^2 - A)*y or (100*A^2 + 100*A)*z
I know, but I don't know how to tell Maxima this, that it in this case is valid to make the approximation A+1 ~ A, thereby effectively removing anything but the highest power of A in each coefficient.
I'm now looking for functions, tools, or methods that I can use to guide Maxima in dropping various terms that aren't important.
I have attempted with subst, but that requires me to specify each and every factor separately, because:
subst([A+1=B], (A+2)*(A+1)*2);
subst([A+1=B], (A+2)*(A*2+2));
(%o1) 2*(A+2)*B
(%o2) (A+2)*(2*A+2)
(that is, I need to add one expression for each slightly different variant)
I tried with ratsimp, but that's too eager to change every occurrence:
ratsubst(B, A+1, A*(A+1)*2);
ratsubst(B, A+1, A*(A*2+2));
(%o3) 2*B^2-2*B
(%o4) 2*B^2-2*B
which isn't actually simpler, as I would have preferred the answer to have been given as 2*B^2.
In another answer, (https://stackoverflow.com/a/22695050/5999883) the functions let and letsimp were suggested for the task of substituting values, but I fail to get them to really do anything:
x:(A+1)*A;
let ( A+1, B );
letsimp(x);
(x)A*(A+1)
(%o6) A+1 --\> B
(%o7) A^2+A
Again, I'd like to approximate this expression to A^2 (B^2, whatever it's called).
I understand that this is, in general, a hard problem (is e.g. A^2 + 10^8*A still okay to approximate as A^2?) but I think that what I'm looking for is a function or method of calculation that would be a little bit smarter than subst and can recognize that the same substitution could be done in the expression A^2+A as in the expression 100*A^2+100*A or -A^2-A instead of making me create a list of three (or twenty) individual substitutions when calling subst. The "nice" part of the full expression that I'm working on is that each of these A factors are of the form k*A^n*(A+1)^m for various small integers n, m, so I never actually end up with the degenerate case mentioned above.
(I was briefly thinking of re-expressing my expression as a polynomial in A, but this will not work as the only valid approximation of the expression (A^3+A^2+A)*x + y is A^3*x + y -- I know nothing about the relative sizes of x and y.
I need a help with following:
flatten ([]) -> [];
flatten([H|T]) -> H ++ flatten(T).
Input List contain other lists with a different length
For example:
flatten([[1,2,3],[4,7],[9,9,9,9,9,9]]).
What is the time complexity of this function?
And why?
I got it to O(n) where n is a number of elements in the Input list.
For example:
flatten([[1,2,3],[4,7],[9,9,9,9,9,9]]) n=3
flatten([[1,2,3],[4,7],[9,9,9,9,9,9],[3,2,4],[1,4,6]]) n=5
Thanks for help.
First of all I'm not sure your code will work, at least not in the way standard library works. You could compare your function with lists:flatten/1 and maybe improve on your implementation. Try lists such as [a, [b, c]] and [[a], [b, [c]], [d]] as input and verify if you return what you expected.
Regarding complexity it is little tricky due to ++ operator and functional (immutable) nature of the language. All lists in Erlang are linked lists (not arrays like in C++), and you can not just add something to end of one without modifying it; before it was pointing to end of list, now you would like it to link to something else. And again, since it is not mutable language you have to make copy of whole list left of ++ operator, which increases complexity of this operator.
You could say that complexity of A ++ B is length(A), and it makes complexity of your function little bit greater. It would go something like length(FirstElement) + (lenght(FirstElement) + length(SecondElement)) + .... up to (without) last, which after some math magic could be simplified to (n -1) * 1/2 * k * k where n is number of elements, and k is average length of element. Or O(n^3).
If you are new to this it might seem little bit odd, but with some practice you can get hang of it. I would recommend going through few resources:
Good explanation of lists and how they are created
Documentation on list handling with DO and DO NOT parts
Short description of ++ operator myths and best practices
Chapter about recursion and tail-recursion with examples using ++ operator
I'm trying to teach myself Prolog. Below, I've written some code that I think should return all paths between nodes in an undirected graph... but it doesn't. I'm trying to understand why this particular code doesn't work (which I think differentiates this question from similar Prolog pathfinding posts). I'm running this in SWI-Prolog. Any clues?
% Define a directed graph (nodes may or may not be "room"s; edges are encoded by "leads_to" predicates).
room(kitchen).
room(living_room).
room(den).
room(stairs).
room(hall).
room(bathroom).
room(bedroom1).
room(bedroom2).
room(bedroom3).
room(studio).
leads_to(kitchen, living_room).
leads_to(living_room, stairs).
leads_to(living_room, den).
leads_to(stairs, hall).
leads_to(hall, bedroom1).
leads_to(hall, bedroom2).
leads_to(hall, bedroom3).
leads_to(hall, studio).
leads_to(living_room, outside). % Note "outside" is the only node that is not a "room"
leads_to(kitchen, outside).
% Define the indirection of the graph. This is what we'll work with.
neighbor(A,B) :- leads_to(A, B).
neighbor(A,B) :- leads_to(B, A).
Iff A --> B --> C --> D is a loop-free path, then
path(A, D, [B, C])
should be true. I.e., the third argument contains the intermediate nodes.
% Base Rule (R0)
path(X,Y,[]) :- neighbor(X,Y).
% Inductive Rule (R1)
path(X,Y,[Z|P]) :- not(X == Y), neighbor(X,Z), not(member(Z, P)), path(Z,Y,P).
Yet,
?- path(bedroom1, stairs, P).
is false. Why? Shouldn't we get a match to R1 with
X = bedroom1
Y = stairs
Z = hall
P = []
since,
?- neighbor(bedroom1, hall).
true.
?- not(member(hall, [])).
true.
?- path(hall, stairs, []).
true .
?
In fact, if I evaluate
?- path(A, B, P).
I get only the length-1 solutions.
Welcome to Prolog! The problem, essentially, is that when you get to not(member(Z, P)) in R1, P is still a pure variable, because the evaluation hasn't gotten to path(Z, Y, P) to define it yet. One of the surprising yet inspiring things about Prolog is that member(Ground, Var) will generate lists that contain Ground and unify them with Var:
?- member(a, X).
X = [a|_G890] ;
X = [_G889, a|_G893] ;
X = [_G889, _G892, a|_G896] .
This has the confusing side-effect that checking for a value in an uninstantiated list will always succeed, which is why not(member(Z, P)) will always fail, causing R1 to always fail. The fact that you get all the R0 solutions and none of the R1 solutions is a clue that something in R1 is causing it to always fail. After all, we know R0 works.
If you swap these two goals, you'll get the first result you want:
path(X,Y,[Z|P]) :- not(X == Y), neighbor(X,Z), path(Z,Y,P), not(member(Z, P)).
?- path(bedroom1, stairs, P).
P = [hall]
If you ask for another solution, you'll get a stack overflow. This is because after the change we're happily generating solutions with cycles as quickly as possible with path(Z,Y,P), only to discard them post-facto with not(member(Z, P)). (Incidentally, for a slight efficiency gain we can switch to memberchk/2 instead of member/2. Of course doing the wrong thing faster isn't much help. :)
I'd be inclined to convert this to a breadth-first search, which in Prolog would imply adding an "open set" argument to contain solutions you haven't tried yet, and at each node first trying something in the open set and then adding that node's possibilities to the end of the open set. When the open set is extinguished, you've tried every node you could get to. For some path finding problems it's a better solution than depth first search anyway. Another thing you could try is separating the path into a visited and future component, and only checking the visited component. As long as you aren't generating a cycle in the current step, you can be assured you aren't generating one at all, there's no need to worry about future steps.
The way you worded the question leads me to believe you don't want a complete solution, just a hint, so I think this is all you need. Let me know if that's not right.
Extremely just-started-yesterday new to F#.
What I want: To write code that parses the string "2 + 2" into (using as an example code from the tutorial project) Expr.Add(Expr.Num 2, Expr.Num 2) for evaluation. Some help to at least point me in the right direction or tell me it's too complex for my first F# project. (This is how I learn things: By bashing my head against stuff that's hard)
What I have: My best guess at code to extract the numbers. Probably horribly off base. Also, a lack of clue.
let script = "2 + 2";
let rec scriptParse xs =
match xs with
| [] -> (double)0
| y::ys -> (double)y
let split = (script.Split([|' '|]))
let f x = (split[x]) // "This code is not a function and cannot be applied."
let list = [ for x in 0..script.Length -> f x ]
let result = scriptParse
Thanks.
The immediate issue that you're running into is that split is an array of strings. To access an element of this array, the syntax is split.[x], not split[x] (which would apply split to the singleton list [x], assuming it were a function).
Here are a few other issues:
Your definition of list is probably wrong: x ranges up to the length of script, not the length of the array split. If you want to convert an array or other sequence to a list you can just use List.ofSeq or Seq.toList instead of an explicit list comprehension [...].
Your "casts" to double are a bit odd - that's not the right syntax for performing conversions in F#, although it will work in this case. double is a function, so the parentheses are unnecessary and what you are doing is really calling double 0 and double y. You should just use 0.0 for the first case, and in the second case, it's unclear what you are converting from.
In general, it would probably be better to do a bit more design up front to decide what your overall strategy will be, since it's not clear to me that you'll be able to piece together a working parser based on your current approach. There are several well known techniques for writing a parser - are you trying to use a particular approach?