I'm trying to write a program that calculates the value of propositional logic. I can do it for 4 to 6 variables but for 100 I have no idea.
a propositional formula with n = 100 variables A1, A2, . . . , specified. The formula is in the Conjunctive normal form. A clause C of the formula is replaced by a subset
TC ⊆ {1, 2, . . . , n} ∪ {−1, −2, . . . , −n}
coded as follows:
• Ai occurs in C if and only if i ∈ TC.
• ¬Ai occurs in C if and only if −i ∈ TC.
An occupancy is sought
b : {A1, A2, . . . , An} → {true, false},
such that the number τ (b) of clauses in the formula that are true under assignment b is as large as possible.
does anyone have an idea for an iterative procedure
for this optimization problem
How do i choose the starting assignment b0?
How do i get an assignment bi+1 with τ (bi) < τ (bi+1) from an existing assignment bi?
When do i stop the iteration?
Thank you
Related
I have 4 non negative real variable that are A, B, C and X. Based on the current problem that I have, I notice that the variable X must belong to the interval of [B,C] and the relation will be a bunch of if-else conditions like this:
If A < B:
x = B
elseif A > C:
x = C
elseif B<=A<=C:
x = A
As you can see, it quite difficult to reformulate as a Mixed Integer Programming problem with corresponding decision variable (d1, d2 and d3). I have try reading some instructions regarding if-then formulation using big M method at this site:
https://www.math.cuhk.edu.hk/course_builder/1415/math3220/L2%20(without%20solution).pdf but it seem that this problem is more challenging than their tutorial.
Could you kindly provide me with a formulation for this situation ?
Thank you very much !
I'm looking for a tool for determining whether a given set of linear equations/inequalities (A) entails another given set of linear equations/inequalities (B). The return value should be either 'true' or 'false'.
To illustrate, let's look at possible instances of A and B and the expected return value of the algorithm:
A: {Z=3,Y=Z+2, X < Y} ;
B: {X<5} ;
Expected result: true
A: {Z=3,Y=Z+2, X < Y} ;
B: {X<10} ;
Expected result: true
A: {Z=3,Y=Z+2, X < Y} ;
B: {X=3} ;
Expected result: false
A: {X<=Y,X>=Y} ;
B: {X=Y} ;
Expected result: true
A: {X<=Y,X>=Y} ;
B: {X=Y, X>Z+2} ;
Expected result: false
Typically A contains up to 10 equations/inequalities, and B contains 1 or 2. All of them are linear and relatively simple. We may even assume that all variables are integers.
This task - of determining whether A entails B - is part of a bigger system and therefore I'm looking for tools/source code/packages that already implemented something like that and I can use.
Things I started to look at:
Theorem provers with algebra - Otter, EQP and Z3 (Vampire is currently unavailable for download).
Coq formal proof management system.
Linear Programming.
However, my experience with these tools is very limited and so far I didn't find a promising direction. Any guidelines and ideas from people more experienced than me will be highly appreciated.
Thank you for your time!
I think I found a working solution. The problem can be rephrased as an assignment problem and then it can be solved by theorem provers such as Z3 and with some work probably also by linear programming solvers.
To illustrate, let's look at the first example I gave above:
A: {Z=3, Y=Z+2, X<Y} ;
B: {X<5} ;
Determining whether A entails B is equivalent to determining whether it is impossible that A is true while B is false. This is a simple simple logical equivalence. In our case, it means that rather than checking whether the condition in B follows from the ones in A, we can check whether there is no assignment of X, Y and Z that satisfies the conditions in A and not in B.
Now, when phrased as an assignment problem, a theorem prover such as Z3 can be called for the task. The following code checks if the conditions in A are satisfiable while the one in B is not:
(declare-fun x () Int)
(declare-fun y () Int)
(declare-fun z () Int)
(assert (= z 3))
(assert (= y (+ z 2)))
(assert (< x y))
(assert (not (< x 5)))
(check-sat)
(get-model)
(exit)
Z3 reports that there is no model that satisfies these conditions, thus it is not possible that B doesn't follow from A, and therefore we can conclude that B follows from A.
I'm in a Formal languages class and have a grammar quiz coming up. I'm assuming something like this will appear.
Consider the alphabet ∑ = {a, b, c}. Construct a grammar that generates the language L = {bab^nabc^na^p : n ≥ 0, p ≥ 1}. Assume that the start variable is S.
It was a very long time since I worked with formal languages for the last time, so, please, forgive my rustyness, but this would be the language: We divide S to a prefix variable (A) and a suffix variable (B). Then, we handle the prefix and the suffix separately, both of them have a possible rule of further recursion, and an end sign of empty, where no occurrence is required and the constant where at least an occurrence is required.
{bab^nabc^na^p : n ≥ 0, p ≥ 1}
S -> ASB
A -> babAabc
A -> {empty}
B -> Ba
B -> a
Given that fib(n)=fib(n-1)+fib(n-2) for n>1 and given that fib(0)=a, fib(1)=b (some a, b >0), which of the following is true?
fib(n) is
Select one or more:
a. O(n^2)
b. O(2^n)
c. O((1-sqrt 5)/2)^n)
d. O(n)
e. Answer depends on a and b.
f. O((1+sqrt 5)/2)^n)
Solving the Fibonacci sequence I got that:
fib(n)= 1/(sqrt 5) ((1+sqrt 5)/2)^n - 1/(sqrt 5) ((1-sqrt 5)/2)^n
But what would be the time complexity in this case? Would that mean the answers are c and f?
From your closed form of your formula, the term 1 / (sqrt 5) ((1 - sqrt 5) / 2)^n has limit 0 as n grows to infinity (|(1 - sqrt 5) / 2| < 1). Therefore we can ignore this term. Also since in time complexity theory we don't care about muliplication constants the following is true:
fib(n) = Θ(φ^n)
where φ = (1 + sqrt 5) / 2 a.k.a. the golden ratio constant.
So it's an exponential function and we can exclude a, d, e. We can exclude c since as was said it has limit 0. But answer b is also correct because φ < 2 and O expresses an upper bound.
Finally, the correct answers are:
b, f
Θ(φ^n) is correct when a=1 and b=1 or a=1 and b=2 . The value of φ depends on a and b.
For computing fib(n-1) and fib(n-2) if we compute them recursively complexity is exponential, but if we save two last values and use them, complexity is O(n) and not depends on a and b.
I've read about it in a book but it wasn't explained at all. I also never saw it in a program. Is part of Prolog syntax? What's it for? Do you use it?
It represents implication. The righthand side is only executed if the lefthand side is true. Thus, if you have this code,
implication(X) :-
(X = a ->
write('Argument a received.'), nl
; X = b ->
write('Argument b received.'), nl
;
write('Received unknown argument.'), nl
).
Then it will write different things depending on it argument:
?- implication(a).
Argument a received.
true.
?- implication(b).
Argument b received.
true.
?- implication(c).
Received unknown argument.
true.
(link to documentation.)
It's a local version of the cut, see for example the section on control predicated in the SWI manual.
It is mostly used to implement if-then-else by (condition -> true-branch ; false-branch). Once the condition succeeds there is no backtracking from the true branch back into the condition or into the false branch, but backtracking out of the if-then-else is still possible:
?- member(X,[1,2,3]), (X=1 -> Y=a ; X=2 -> Y=b ; Y=c).
X = 1,
Y = a ;
X = 2,
Y = b ;
X = 3,
Y = c.
?- member(X,[1,2,3]), (X=1, !, Y=a ; X=2 -> Y=b ; Y=c).
X = 1,
Y = a.
Therefore it is called a local cut.
It is possible to avoid using it by writing something more wordy. If I rewrite Stephan's predicate:
implication(X) :-
(
X = a,
write('Argument a received.'), nl
;
X = b,
write('Argument b received.'), nl
;
X \= a,
X \= b,
write('Received unknown argument.'), nl
).
(Yeah I don't think there is any problem with using it, but my boss was paranoid about it for some reason, so we always used the above approach.)
With either version, you need to be careful that you are covering all cases you intend to cover, especially if you have many branches.
ETA: I am not sure if this is completely equivalent to Stephan's, because of backtracking if you have implication(X). But I don't have a Prolog interpreter right now to check.