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I am surprising to notice that it is somehow difficult to obtain a correct fit of interaction function from gam().
To be more specific, I want to estimate an additive function:
y=m_1(x)+m_2(z)+m_{12}(x,z)+u,
where m_1(x)=x^2, m_2(z)=z^2,m_{12}(x,z)=xz. The following code generate this model:
test1 <- function(x,z,sx=1,sz=1) {
#--m1(x) function
m.x<-x^2
m.x<-m.x-mean(m.x)
#--m2(z) function
m.z<-z^2
m.z<-m.z-mean(m.z)
#--m12(x,z) function
m.xz<-x*z
m.xz<-m.xz-mean(m.xz)
m<-m.x+m.z+m.xz
return(list(m=m,m.x=m.x,m.z=m.z,m.xz=m.xz))
}
n <- 1000
a=0
b=2
x <- runif(n,a,b)/20
z <- runif(n,a,b)
u <- rnorm(n,0,0.5)
model<-test1(x,z)
y <- model$m + u
So I use gam() by fitting the model as
b3 <- gam(y~ ti(x) + ti(z) + ti(x,z))
vis.gam(b3);title("tensor anova")
#---extracting basis matrix
B.f3<-model.matrix.gam(b3)
#---extracting series estimator
b3.hat<-b3$coefficients
Question: when I plot the estimated function by gam()above against its true function, I end up with
par(mfrow=c(1,3))
#---m1(x)
B.x<-B.f3[,c(2:5)]
b.x.hat<-b3.hat[c(2:5)]
plot(x,B.x%*%b.x.hat)
points(x,model$m.x,col='red')
legend('topleft',c('Estimate','True'),lty=c(1,1),col=c('black','red'))
#---m2(z)
B.z<-B.f3[,c(6:9)]
b.z.hat<-b3.hat[c(6:9)]
plot(z,B.z%*%b.z.hat)
points(z,model$m.z,col='red')
legend('topleft',c('Estimate','True'),lty=c(1,1),col=c('black','red'))
#---m12(x,z)
B.xz<-B.f3[,-c(1:9)]
b.xz.hat<-b3.hat[-c(1:9)]
plot(x,B.xz%*%b.xz.hat)
points(x,model$m.xz,col='red')
legend('topleft',c('Estimate','True'),lty=c(1,1),col=c('black','red'))
However, the function estimate of m_1(x) is largely different from x^2, and the interaction function estimate m_{12}(x,z) is also largely different from xz defined in test1 above. The results are the same if I use predict(b3).
I really can't figure it out. Can anybody help me out by explaining why the results end up with this? Greatly appreciate it!
First, the problem of the above issue is not due to the package, of course. It is closely related to the identification conditions of the smooth functions. One common practice is to impose the assumptions that E(mj(.))=0 for all individual function j=1,...,d, and E(m_ij(x_i,x_j)|x_i)=E(m_ij(x_i,x_j)|x_j)=0 for i not equal to j. Those conditions require one to employ centered basis function in series estimator, which has been done already in GAM package. However, in my case above, function m(x,z)=x*z defined in test1 does not satisfy the above identification assumptions, since the integral of x*z with respect to either x or z is not zero when x and z have range from zero to two.
Furthermore, series estimator allows the individual and interaction function to be identified if one impose m(0)=0 or m(0,x_j)=m(x_i,0)=0. This can be readily achieved if we center the basis function around zero. I have tried both cases, and they work well whenever DGP satisfies the identification conditions.
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 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
Sorry guys I cannot provide the pic here...I was unable to upload the pic...so.. I will give the transition table of the problem.
(S/I)....a...b.....c.......e(elipson)
p>.......{p}.....{q}...{r} ..¤(phi)
q>.......{q} ....{r} ..¤.... {p}
r(final)>..{r}...¤....{p}....{q}
Here ¤ is phai
p is starting state
And r is final state
My doubt is...Will e-closure of final state {r} have starting state {p}......,even if starting state will not have a direct reach through elipson to final state
.....but...final state reach the starting state through elipson to state {q} and then to starting state {p}
In my book it is given that
e-closure (r)={r,q}
But my question is why it is not ....{p,q,r}...while final state {r} is reaching starting state {p} as well...
ϵ-closure(s) is a set of NFA states reachable from NFA state s on
ϵ-transitions alone.
You were correct in thinking the same. Please follow the definition mentioned above.
So, ϵ-closure(r) = set of NFA states reachable from NFA state r on ϵ-transitions alone = {p,q,r}. Hence, your book has incorrectly computed it.
The answer has to be {p,q,r}.
NOTE that state {r} is included because a state will always remain on itself at ϵ-transition. State {p} was possible because p is reachable from NFA state q on ϵ-transition, which was reachable from NFA state r on ϵ-transition.
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Given an arbitrary propositional formula PHI (linear constraints on some variables), what is the best way to determine the (approximate) upper and lower bound for each variable?
Some variables may be unbounded. In this case, an algorithm should conclude that there's no upper/lower bound for those variables.
e.g., PHI = (x=3 AND y>=1). The upper and lower bound for x are both 3. The lower bound for y is 1, and y does not have an upper bound.
This is a very simple example, but is there a solution in general (perhaps using SMT solver)?
This assumes the SAT/UNSAT domain of each variable is continuous.
Use an SMT solver to check for a solution to the formula. If there's no solution then that means no upper/lower bounds, so stop.
For each variable in the formula, conduct two binary searches over the domain of the variable, one searching for the lower bound, the other for the upper bound. The starting value in the search for each variable is the value for the variable in the solution found in step 1. Use the SMT solver to probe each search value for satisfiability and methodically bracket the bounds for each variable.
Pseudo code for the search functions, assuming integer domain variables:
lower_bound(variable, start, formula)
{
lo = -inf;
hi = start;
last_sat = start;
incr = 1;
do {
variable = (lo + hi) / 2;
if (SMT(formula) == UNSAT) {
lo = variable + incr;
} else {
last_sat = variable;
hi = variable - incr;
}
} while (lo <= hi);
return last_sat;
}
and
upper_bound(variable, start, formula)
{
lo = start;
hi = +inf;
last_sat = start;
do {
variable = (lo + hi) / 2;
if (SMT(formula) == SAT) {
last_sat = variable;
lo = variable + incr;
} else {
hi = variable - incr;
}
} while (lo <= hi);
return last_sat;
}
-inf/+inf are the smallest/largest values representable in the domain of each variable. If lower_bound returns -inf then the variable has no lower bound. If upper_bound returns +inf then the variable has no upper bound.
In practice most of such optimization problems require some sort of iterate-until-maximum/minimum kind of external driver on top of the SMT solver. Quantified approaches are also possible that can leverage the particular capabilities of the SMT solvers, but in practice such constraints end up being too hard for the underlying solver. See this discussion in particular: How to optimize a piece of code in Z3? (PI_NON_NESTED_ARITH_WEIGHT related)
Having said that, most high-level language bindings include the necessary machinery to simplify your life. For instance, if you use the Haskell SBV library to script z3, you can have:
Prelude> import Data.SBV
Prelude Data.SBV> maximize Quantified head 2 (\[x, y] -> x.==3 &&& y.>=1)
Just [3,1]
Prelude Data.SBV> maximize Quantified (head . tail) 2 (\[x, y] -> x.==3 &&& y.>=1)
Nothing
Prelude Data.SBV> minimize Quantified head 2 (\[x, y] -> x.==3 &&& y.>=1)
Just [3,1]
Prelude Data.SBV> minimize Quantified (head . tail) 2 (\[x, y] -> x.==3 &&& y.>=1)
Just [3,1]
The first result states x=3, y=1 maximizes x with respect to the predicate x==3 && y>=1.
The second result says there is no value that maximizes y with respect to the same predicate.
Third call says x=3,y=1 minimizes the predicate with respect to x.
Fourth call says x=3,y=1 minimizes the predicate with respect to y.
(See http://hackage.haskell.org/packages/archive/sbv/0.9.24/doc/html/Data-SBV.html#g:34 for details.)
You can also use the option "Iterative" (instead of Quantified) to have the library do the optimization iteratively instead of using quantifiers. These two techniques are not equivalent as the latter can get stuck in a local minima/maxima, but iterative approaches might solve problems where the quantified version is way too much to handle for the SMT solver.