Boolean Normal Forms, Shellability, and Reliability Computations
Summary (1 min read)
R, and π(B) < π(S), or
- More interestingly, the authors can now prove the following result.
- Let us mention here that in the special cases of aligned and regular functions, there are more efficient dualization algorithms known in the literature (see, e.g., [5, 9, 10, 24] ), which run in O(n 2 m) time.
- None of those procedures, however, seem to be extendable for the class of shellable functions.
- In the previous section, the authors have established that every positive function can be represented by a shellable DNF (Theorem 3.3).
- In fact, there exist positive Boolean functions in 2n variables which have only n prime implicants but for which every shellable DNF representation involves at least 2 n − 1 terms.
5. The LE property for DNFs.
- The proof of Theorem 5.1 establishes that testing the LE property is already NP-complete for DNFs of degree 5 or more (if the authors call degree of a DNF the number of literals in its longest term).
- This implies, in particular, that the LE property can be tested in polynomial time for DNFs of degree 2.
- The complexity of this problem remains open for DNFs of degree 3 or 4.
Did you find this useful? Give us your feedback
Citations
2 citations
Cites background from "Boolean Normal Forms, Shellability,..."
...Though the length of terms is not considered in [4], the proof shows the NP-completeness for k=5....
[...]
...paper, we improve the proof of [4] and prove that it is NPcomplete for k=4....
[...]
...It is shown in [4] that it is NP-complete to decide if a function has lexico-exchange...
[...]
...NP-completeness of testing lexico-exchange property, which is equivalent to ordered tree-shellability, is proved in [4]....
[...]
References
42,654 citations
40,020 citations
12,336 citations
4,258 citations
3,876 citations
"Boolean Normal Forms, Shellability,..." refers background in this paper
...In the context of reliability theory, the prime implicants of f represent the minimal cutsets of the system under study, namely, the minimal subsets of elements whose failure causes the whole system to fail (see [3, 25])....
[...]
...One of the fundamental issues in reliability is to compute the probability that a positive Boolean function (describing the state—operating or failed—of a complex system) take value 1 when each variable (representing the state of individual components) takes value 0 or 1 randomly and independently of the value of the other variables (see, for instance, [3, 25])....
[...]
Related Papers (5)
Frequently Asked Questions (8)
Q2. What is the simplest way to show that j JR′?
If j ∈ JR,π(B), let R ∈ R be such that π(R) < π(B) and {j} = R\\B. Since R ⊆ Bj and Bi ∈ R′, the authors deduce j = i, and thus {j} = R \\ Bi, which implies j ∈ JR′,π′(Bi).
Q3. How can the authors obtain a DNF representation of fd?
If f = ∨ I∈I ∧j∈I xj , then fd = ∧ I∈I ∨ j∈I xj (by De Morgan’s laws), and a DNF representation of fd can be obtained by applying the distributive laws to the latter expression.
Q4. What is the LE property for a DNF?
The authors say that Ψ has the LE property with respect to a permutation σ of (x1, x2, . . . , xn), or that σ is an LE order for Ψ, ifΨσ(x1, . . . , xn) = ∨ I∈I ∧ j∈I σ(xj)has the LE property with respect to (x1, x2, . . . , xn).
Q5. What is the leftmost implicant of f?
The leftmost implicants of the function f(x1, . . . , x4) = x1x2∨x3x4 are the sets {1, 2}, {1, 3, 4}, {2, 3, 4}, and {3, 4}, listed here in lexicographic order.
Q6. What is the meaning of the term fd?
In the context of reliability theory, the prime implicants of fd represent the minimal cutsets of the system under study, namely, the minimal subsets of elements whose failure causes the whole system to fail (see [3, 25]).
Q7. What is the simplest way to explain the ODNF?
Let us consider an arbitrary DNFΨ(x1, . . . , xn) = m∨ k=1 ∧ j∈Ik xj ∧ j∈Jk x j .(3.3)The authors say that DNF Ψ is a shelled ODNF if Ψ is orthogonal and Ψ is of the form ΨI,π (see (3.2)), where The author= {I1, . . . , Im} and π is a shelling order of I.
Q8. What is the first proof of the DNF?
As a first proof, let us consider the DNFΦ = ∨ I∈I ∧ j∈I xj ,where The authordenotes the family of all implicants of the function f , and let π be a permutation ordering these implicants in a nonincreasing order by their cardinality.