I have a habit of trying out correctness about some logical statements with worlfram alpha by generating truth table for them. For example, I can try if this:
((¬x→y)∧(¬x→¬y))→x
is correct or not by geerating truth table for ((¬x→y)∧(¬x→¬y)) which turns out to be the same as x column in the same truth table. Hence the above is correct. However is there any way I can check same for biconditionals involving nested existential and universal quantifiers and predicates? For example can I somehow verify rules of this kind?:
(∀x)(∀y)ϕ(x,y)⇔(∀y)(∀x)ϕ(x,y)
Update
I am able to do following check ∀x,y(x∨y) as follows:
Resolve[ForAll[{x,y}, x or y]]
which correctly returns False as (x∨y) does not hold for all x and y.
So now I thought I can do something similar to obtain True for following (which is a general fact): ¬(∀x)ϕ(x)⇔(∃x)¬ϕ(x). I tried this:
Resolve[ForAll[x,(not ForAll[x, x]) xnor (exists[x,not x])]]
But it did not work. Note that ⇔ is nothing but xnor. So how do I do this especially something like following also correctly returns True:
Resolve[not ForAll[x, x]]
which stands for ¬∀x(x).
Related
I have written an ASP program with an optimization condition in the end. When I compile it, even though I get the correct result, in the summary I get a negative value for the optimization (i.e. -3).
Do you know why this happens?
The optimization code part looks like this:
number_of_leaves(N) :- #count{X : leaf(X)} = N.
#maximize {N : number_of_leaves(N)}.
In clingo only #minimize is actually implemented. #maximize is realized by using #minimize - just with negated values. Therefore the "maximum" is negated as well - it shows -3 when the value should be 3.
I am trying to solve a problem involving the equating of sums of exponentials.
This is how I would do it hardcoded:
#NLconstraint(m, exp(x[25])==exp(x[14])+exp(x[18]))
This works fine with the rest of the code. However, when I try to do it for an arbitrary set of equations like the above I get an error. Here's my code:
#NLconstraint(m,[k=1:length(LHSSum)],sum(exp.(LHSSum[k][i]) for i=1:length(LHSSum[k]))==sum(exp.(RHSSum[k][i]) for i=1:length(RHSSum[k])))
where LHSSum and RHSSum are arrays containing arrays of the elements that need to be exponentiated and then summed over. That is LHSSum[1]=[x[1],x[2],x[3],...,x[n]]. Where x[i] are variables of type JuMP.Variable. Note that length(LHSSum)=length(RHSSum).
The error returned is:
LoadError: exp is not defined for type Variable. Are you trying to build a nonlinear problem? Make sure you use #NLconstraint/#NLobjective.
So a simple solution would be to simply do all the exponentiating and summing outside of the #NLconstraint function, so the input would be a scalar. However, this too presents a problem since exp(x) is not defined since x is of type JuMP.variable, whereas exp expects something of type real. This is strange since I am able to calculate exponentials just fine when the function is called within an #NLconstraint(). I.e. when I code this line#NLconstraint(m,exp(x)==exp(z)+exp(y)) instead of the earlier line, no errors are thrown.
Another thing I thought to do would be a Taylor Series expansion, but this too presents a problem since it goes into #NLconstraint land for powers greater than 2, and then I get stuck with the same vectorization problem.
So I feel stuck, I feel like if JuMP would allow for the vectorized evaluation of #NLconstraint like it does for #constraint, this would not even be an issue. Another fix would be if JuMP implements it's own exp function to allow for the exponentiation of JuMP.Variable type. However, as it is I don't see a way to solve this problem in general using the JuMP framework. Do any of you have any solutions to this problem? Any clever workarounds that I am missing?
I'm confused why i isn't used in the expressions you wrote. Do you mean:
#NLconstraint(m, [k = 1:length(LHSSum)],
sum(exp(LHSSum[k][i]) for i in 1:length(LHSSum[k]))
==
sum(exp(RHSSum[k][i]) for i in 1:length(RHSSum[k])))
I'm a beginner in Prolog and I am dealing with a problem that might seem stupid to you, but I really can't understand what I'm doing wrong! Ok, I have this file fruits.pl and inside that I have something like this:
fruit(apple,small,sweet).
fruit(lemon,small,nosweet).
fruit(melon,big,sweet).
I have already (inside that file made a coexist(X,Y) atom that checks if two fruits can be put together in a plate. It works fine! But now I can't create a suggest(X) that takes as a parameter a fruit and returns a list of fruits that can be put together in the same plate.
The thing is I was trying to make something like that
suggest(X) :- findall(Y,fruit(Y,_,_), List), coexist(X,Y).
What do you think? Every time I try to run this in swi prolog there is a warning 'singleton variable' and when I press
suggest(apple).
then it says false..
sorry for my english :/
Predicates in Prolog do not return anything. You have goals that are satisfied or not and you can interpret that as returning true or false.
Your predicate suggest(X) should contain another parameter that will be bound to the list of fruits that go together with X. An option would be: suggest(X, List) which describes the following relation: List represents all the fruits that go together with X. Then, you could ask:
?- suggest(apple, List).
List = [pear, cherry].
The goal findall(Y, ... , ...) uses the Y variable internally and Y is still unbound after the goal is satisfied. So, you should move coexist(X,Y) inside the second argument of findall/3 which is the goal that is satisfied in all possible ways. Th rule below works only if X is instantiated (suggest(+X, -List)).
suggest(X, List) :- findall(Y, (fruit(Y,_,_), coexist(X, Y)), List).
You can read this as follows: "List represents all fruits Y that coexist with X".
When you try to define a predicate in Prolog, first of all pretend that you have written that predicate already and start with imagining how you would use it. That is, what queries you would like to pose.
To me, it looks as if coexist/2 already describes what you want. BTW, may_coexist/2 might be a more descriptive name. Why do you want this in a separate list? And why using fruit/3 at all? But for the sake of the question let's assume that this makes sense. So essentially you would have now a relation fruit_compatible/2:
fruit_compatible(F, G) :-
fruit(F, _, _),
may_coexist(F, G),
fruit(G, _, _). % maybe you want to add this?
And now, let's assume you want this list too. So you would have a relation fruit_suggestions/2. How to use it?
?- fruit_suggestions(apple, L).
L = [cherry,pear].
or ... should it be rather L = [pear,cherry]? Or both?
?- fruit_suggestions(lemon, L).
L = [orange].
So every time I want a suggestion I have to think of a fruit. Always thinking: what fruit should it be? Fortunately there is a less demanding way in Prolog: Simply use a variable instead of the fruit! Now we should get all suggestions at once!
?- fruit_suggestions(F, L).
F = apple, L = [cherry, pear]
; F = lemon, L = [orange]
; F = cromulon, L = [embiggy, mushfruit].
So we need to implement it such that it will behave that way. findall/3 alone does not solve this. And implementing it manually is far from trivial. But there is setof/3 which handles variables in exactly that manner. Many of the tiny nitty-gritty design decisions have already been made, like that the list will be sorted ascendingly.
fruit_suggestions(F, L) :-
setof(G, fruit_compatible(F, G), L).
Edit: Due to the discussion below, here would be a solution that also permits empty lists. Note that this sounds trivial but it is not. To see this, consider the query:
?- fruit_suggestions(F, []).
What does it mean? What should F be? Also things that are no fruits at all? In that case we would have to produce solutions for everything. Like F = badger ; F = 42 ; .... Most probably this does not make much sense. What might be intended is those fruits that are incompatible with everything. To this end, we need to add a new rule:
fruit_suggestions(F, []) :-
setof(t,X^Y^fruit(F,X,Y),_),
\+ fruit_compatible(F, _).
fruit_suggestions(F, L) :-
setof(G, fruit_compatible(F, G), L).
I have got a term from which I want to get set of variables name.
Eg. input: my_m(aa,b,B,C,max(D,C),D)
output: [B,C,D] (no need to be ordered as order of appearance in input)
(That would call like set_variable_name(Input,Output).)
I can simply get [B,C,D,C,D] from the input, but don't know how to implement set (only one appearance in output). I've tried something like storing in rbtrees but that failed, because of
only_one([],T,T) :- !.
only_one([X|XS],B,C) :- rb_in(X,X,B), !, only_one(XS,B,C).
only_one([X|XS],B,C) :- rb_insert(B,X,X,U), only_one(XS,U,C).
it returns tree with only one node and unification like B=C, C=D.... I think I get it why - because of unification of X while questioning rb_in(..).
So, how to store only once that name of variable? Or is that fundamentally wrong idea because we are using logic programming? If you want to know why I need this, it's because we are asked to implement A* algorithm in Prolog and this is one part of making search space.
You can use sort/2, which also removes duplicates.
In algebra if I make the statement x + y = 3, the variables I used will hold the values either 2 and 1 or 1 and 2. I know that assignment in programming is not the same thing, but I got to wondering. If I wanted to represent the value of, say, a quantumly weird particle, I would want my variable to have two values at the same time and to have it resolve into one or the other later. Or maybe I'm just dreaming?
Is it possible to say something like i = 3 or 2;?
This is one of the features planned for Perl 6 (junctions), with syntax that should look like my $a = 1|2|3;
If ever implemented, it would work intuitively, like $a==1 being true at the same time as $a==2. Also, for example, $a+1 would give you a value of 2|3|4.
This feature is actually available in Perl5 as well through Perl6::Junction and Quantum::Superpositions modules, but without the syntax sugar (through 'functions' all and any).
At least for comparison (b < any(1,2,3)) it was also available in Microsoft Cω experimental language, however it was not documented anywhere (I just tried it when I was looking at Cω and it just worked).
You can't do this with native types, but there's nothing stopping you from creating a variable object (presuming you are using an OO language) which has a range of values or even a probability density function rather than an actual value.
You will also need to define all the mathematical operators between your variables and your variables and native scalars. Same goes for the equality and assignment operators.
numpy arrays do something similar for vectors and matrices.
That's also the kind of thing you can do in Prolog. You define rules that constraint your variables and then let Prolog resolve them ...
It takes some time to get used to it, but it is wonderful for certain problems once you know how to use it ...
Damien Conways Quantum::Superpositions might do what you want,
https://metacpan.org/pod/Quantum::Superpositions
You might need your crack-pipe however.
What you're asking seems to be how to implement a Fuzzy Logic system. These have been around for some time and you can undoubtedly pick up a library for the common programming languages quite easily.
You could use a struct and handle the operations manualy. Otherwise, no a variable only has 1 value at a time.
A variable is nothing more than an address into memory. That means a variable describes exactly one place in memory (length depending on the type). So as long as we have no "quantum memory" (and we dont have it, and it doesnt look like we will have it in near future), the answer is a NO.
If you want to program and to modell this behaviour, your way would be to use a an array (with length equal to the number of max. multiple values). With this comes the increased runtime, hence the computations must be done on each of the values (e.g. x+y, must compute with 2 different values x1+y1, x2+y2, x1+y2 and x2+y1).
In Perl , you can .
If you use Scalar::Util , you can have a var take 2 values . One if it's used in string context , and another if it's used in a numerical context .