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Download Computational Complexity of Solving Equation Systems by Przemyslaw Broniek PDF

By Przemyslaw Broniek

This quantity considers the computational complexity of opting for even if a approach of equations over a set algebra A has an answer. It examines intimately the 2 difficulties this results in: SysTermSat(A) and SysPolSat(A), during which equations are equipped out of phrases or polynomials, respectively. The booklet characterizes these algebras for which SysPolSat may be solved in a polynomial time. to this point, reports and their results haven't coated algebras that generate a range admitting kind 1 within the experience of Tame Congruence concept. seeing that unary algebras admit simply sort 1, this e-book specializes in those algebras to take on the most challenge. It discusses numerous points of unary algebras and proves that the Constraint delight challenge for relational constructions is polynomially corresponding to SysTermSat over unary algebras. The book’s ultimate chapters speak about partial characterizations, current conclusions, and describe the issues which are nonetheless open.

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Example text

Now take a valuation s : X → {0, 1} satisfying formula F to produce a valuation v : V → A. Since F contains the sets of the form S A (x) for each x ∈ V , it is enough to valuate x by some element from A to satisfy all of the S A (x)’s. To determine v(xi ) note that since s is a Positive- 1- in- 3- Sat solution, we know that for each i exactly one of s(X 1i ), s(X 2i ), s(X 3i ), say s(X αi ), takes the true value. We set v(xi ) to an arbitrarily j chosen element of Aα . Whenever s(X αi ) = s(X β ) this valuation v satisfies: f α (v(xi )) = α j ⇔ s(X αi ) = 1 ⇔ s(X β ) = 1 ⇔ f β (v(x j )) = β.

Thanks to our numeration of variables in V we know that X ij is placed in jth position in a constraint Ci , thus s(X ij ) = ε(ai ) j = t j (ai ) = t j (v(xi )). Analogously s(X lk ) = ε(ak )l = tl (ak ) = tl (v(xk )). Since s satisfies the constraint (R= , (X ij , X lk )), we get s(X ij ) = s(X lk ). Finally t j (v(xi )) = s(X ij ) = s(X lk ) = tl (v(xk )), as required. Conversely, take a solution v : V → A of J . 16, v(xi ) ∈ C i which gives ε(v(xi )) ∈ RC i . We define a valuation s : V → D by putting s(X ij ) = ε(v(xi )) j = t j (v(xi )).

Conversely, take a solution v : V → A of the instance F . First of all, since v satisfies all equations in the sets S A (x), we get v(x) ∈ A for all x ∈ V . To see that s : X → {0, 1} determined by: s(X αi ) = 1, 0, if v(xi ) ∈ Aα , otherwise, j is well defined suppose that X αi = X β . • Case 1: α = β. In this case the equation f α (xi ) ≈ f α (x j ) belongs to F . Since v is its solution, we get f α (v(xi )) = f α (v(x j )). Thus v(xi ) ∈ Aα ⇔ v(x j ) ∈ Aα , j so that s(X αi ) = s(X α ), as required.

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