TL;DR: How can I generate a graph while constraining it to be subisomorph to every graph in a positive list while being non-subisomorph to every graph in a negative list?
I have a list of directed heterogeneous attributed graphs labeled as positive or negative. I would like to find the smallest list of patterns(graphs with special values) such that:
Every input graph has a pattern that matches(= 'P is subisomorphic to G, and the mapped nodes have the same attribute values')
A positive pattern can only match a positive graph
A positive pattern does not match any negative graph
A negative pattern can only match a negative graph
A negative pattern does not match any negative graph
Exemple:
Input g1(+),g2(-),g3(+),g4(+),g5(-),g6(+)
Acceptable solution: p1(+),p2(+),p3(-) where p1(+) matches g1(+) and g4(+); p2(+) matches g3(+) and g6(+); and p3(-) matches g2(-) and g5(-)
Non acceptable solution: p1(+),p2(-) where p1(+) matches g1(+),g2(-),g3(+); p2(-) matches g4(+),g5(-),g6(+)
Currently, I'm able to generate graphs matching every graph in a list, but I can't manage to enforce the constraint 'A positive pattern does not match any negative graph'. I made a predicate 'matches', which takes as input a pattern and a graph, and uses a local array of variables 'mapping' to try and map nodes together. But when I try to use that predicate in a negative context, the following error is returned: MiniZinc: flattening error: free variable in non-positive context.
How can I bypass that limitation? I tried to code the opposite predicate 'not_matches' but I've not yet found how to specify 'for all node mapping, the isomorphism is invalid'. I also can't define the mapping outside the predicate, because a pattern can match a graph more than once and i need to be able to get all mappings.
Here is a reproductible exemple:
include "globals.mzn";
predicate p(array [1..5] of var 0..10:arr1, array [1..5] of 1..10:arr2)=
let{array [1..5] of var 1..5: mapping; constraint all_different(mapping)} in (forall(i in 1..5)(arr1[i]=0\/arr1[i]=arr2[mapping[i]]));
array [1..5] of var 0..10:arr;
constraint p(arr,[1,2,3,4,5]);
constraint p(arr,[1,2,3,4,6]);
constraint not p(arr,[1,2,3,5,6]);
solve satisfy;
For that exemple, the decision variable is an array and the predicate p is true if a mapping exists such that the values of the array are mapped together. One or more elements of the array can also be 0, used here as a wildcard.
[1,2,3,4,0] is an acceptable solution
[0,0,0,0,0] is not acceptable, it matches anything. And the solution should not match [1,2,3,5,6]
[1,2,3,4,7] is not acceptable, it doesn't match anything(as there is no 7 in the parameter arrays)
Thanks by advance! =)
Edit: Added non-acceptable solutions
It is probably good to note that MiniZinc's limitation is not coincidental. When the creation of a free variable is negated, rather then finding a valid assignment for the variable, instead the model would have to prove that no such valid assignment exists. This is a much harder problem that would bring MiniZinc into the field of quantified constraint programming. The only general solution (to still receive the same flattened constraint model) would be to iterate over all possible values for each variable and enforce the negated constraints. Since the number of possibilities quickly explodes and the chance of getting a good model is small, MiniZinc does not do this automatically and throws this error instead.
This technique would work in your case as well. In the not_matches version of your predicate, you can iterate over all possible permutations (the possible mappings) and enforce that they not correct (partial) mappings. This would be a correct way to enforce the constraint, but would quickly explode. I believe, however, that there is a different way to enforce this constraint that will work better.
My idea stems from the fact that, although the most natural way to describe a permutation from one array to the another is to actually create the assignment from the first to the second, when dealing with discrete variables, you can instead enforce that each has the exact same number of each possible value. As such a predicate that enforces X is a permutation of Y might be written as:
predicate is_perm(array[int] of var $$E: X, array[int] of var $$E: Y) =
let {
array[int] of int: vals = [i | i in (dom_array(X) union dom_array(Y))]
} in global_cardinality(X, vals) = global_cardinality(Y, vals);
Notably this predicate can be negated because it doesn't contain any free variables. All new variables (the resulting values of global_cardinality) are functionally defined. When negated, only the relation = has to be changed to !=.
In your model, we are not just considering full permutations, but rather partial permutations, and we use a dummy value otherwise. As such, the p predicate might also be written:
predicate p(array [int] of var 0..10: X, array [int] of var 1..10: Y) =
let {
set of int: vals = lb_array(Y)..ub_array(Y); % must not include dummy value
array[vals] of var int: countY = global_cardinality(Y, [i | i in vals]);
array[vals] of var int: countX = global_cardinality(X, [i | i in vals]);
} in forall(i in vals) (countX[i] <= countY[i]);
Again this predicate does not contain any free variables, and can be negated. In this case, the forall can be changed into a exist with a negated body.
There are a few things that we can still do to optimise p for this use case. First, it seems that global_cardinality is only defined for variables, but since Y is guaranteed par, we can rewrite it and have the correct counts during MiniZinc's compilation. Second, it can be seen that lb_array(Y)..ub_array(Y) gives the tighest possible set. In your example, this means that only slightly different versions of the global cardinality function are evaluated, that could have been
predicate p(array [1..5] of var 0..10: X, array [1..5] of 1..10: Y) =
let {
% CHANGE: Use declared values of Y to ensure CSE will reuse `global_cardinality` result values.
set of int: vals = 1..10; % do not include dummy value
% CHANGE: parameter evaluation of global_cardinality
array[vals] of int: countY = [count(j in index_set(Y)) (i = Y[j]) | i in vals];
array[vals] of var int: countX = global_cardinality(X, [i | i in 1..10]);
} in forall(i in vals) (countX[i] <= countY[i]);
Regarding the example. One approach might be to rewrite the not p(...) constraint to a specific not_p(...) constraint. But I'm how sure how that be formulated.
Here's an example but it's probably not correct:
predicate not_p(array [1..5] of var 0..10:arr1, array [1..5] of 1..10:arr2)=
let{
array [1..5] of var 1..5: mapping;
constraint all_different(mapping)
} in
exists(i in 1..5)(
arr1[i] != 0
/\
arr1[i] != arr2[mapping[i]]
);
This give 500 solutions such as
arr = [1, 0, 0, 0, 0];
----------
arr = [2, 0, 0, 0, 0];
----------
arr = [3, 0, 0, 0, 0];
...
----------
arr = [2, 0, 0, 3, 4];
----------
arr = [2, 0, 1, 3, 4];
----------
arr = [2, 1, 0, 3, 4];
Update
I added not before the exists loop.
I have understood that L={wxw^r|w,x belongs to {a,b}^* } is regular because it turns out to be the pattern of starting and ending with same symbol but I am not getting the proper explanation that how to say L={ww^rx|w,x belongs to {a,b}*} is regular language using DFA design.
Please help me in understanding this!
This is a trick question. The language L as you have specified is the language of the regular expression (a + b)*, that is, any string of a's and b's. The trick is that for any string y = s1.s2.s3...sk where si in {a, b}, we can write y = wxw^R where w is the empty string and x = y. Basically, the trick is that we can always choose w to be the empty string, and in that case we are left with L = {x | x in {a, b}^*}, clearly regular. Another way of thinking about it is this: can you find any string of a's and b's that is not in L? Is it not in L even if you take w to be the empty string?
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)+
Right now I'm learning for a test in cs and I have my problems with regular expressions. Here is example of a question I don’t understand.
We have a given alphabet Σ = {0, 1} and L1. L1 represent any word with odd number of 0(zeros) and exactly twice 1 (ones).
The prof. showed us an example how the solution should look like:
Σ = {a, b} with L1.
L1 represents any Word in which the sub-words aa or bb occurs.
L1 = (a ∪ b)* (aa ∪ bb)* (a ∪ b)*
thank you for your help
Try this:
(?mx)
^
(?=0*10*10*$) # check if the word consists of only 0s and exactly two 1s
(?:..){1,}. # check if the word length is odd
$
Demo on RegEx101. There you can find an exhaustive explanation of this regular expression.
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.