Tag Archives: Non-Circular Satisfiability

Serializability of Database Histories

This is another problem with a cool elegant reduction once you get past the baggage you need to know to understand the problem.  This database section seems to be full of problems like these.

The problem: Serializability of Database Histories.  This is problem SR33 in the appendix.

The description: We have a set V of variables in our database, and a set T of transactions, where each transaction i has a read operation (Ri) that reads some subset of V, and a write operation Wi that writes some (possibly different) subset of V.  We’re also given a “history” H of T, which permutes the order of all reads and writes maintaining the property that for all i, Ri comes before Wi in the history.  Think of this as a set of parallel transactions that reach a central database.  H is the order the database processes these operations.

Can we find a serial history H’ of T, with the following properties:

  • Each Ri occurs immediately before its corresponding Wi.
  • A “live transaction” is a transaction (Ri, Wi) where either Wi is the last time a variable is written before the Rj of some other live transaction or the last time the variable is written at all.  The set of live transactions in H and H’ needs to be the same.
  • For each pair of live transactions (Ri, Wi) and (Rj, Wj), for any variable v in Wi∩Rj, Wi is the last write set to contain v before Rj in H if and only if Wi is the last write set to contain v before Rj in H’.  The paper says that this means transaction j “reads from” (or “reads v from”) transaction i.

Example: The paper by Papadimitriou, Bernstein, and Rothnie that has the reduction has a good simple example of a non-serializable history:

H= <R1, R2, W2, W1>, where R1 and W2 access a variable x, and R2 and W1 access a variable y.  Both transactions are live since they both write their variables for the last time.  Notice that neither transaction reads any variable.  But the two possible candidates for H’ are: <R1, W1, R2, W2> (where R2 reads the y written by W1) and <R2, W2, R1, W1> (where R1 reads the x written by W2), so neither H’ candidate has the same set of transactions reading variables from each other.

Reduction: Is from Non-Circular Satisfiability.  Given a formula, they generate a “polygraph” of a database history.  A polygraph (N,A,B) is a directed graph (N,A) along with a set B of “bipaths” (paths that are 2 edges long).  If a bipath{(v,u), (u,w)} is in B, then the edge (w,v) is in A.  So, if a bipath exists in B from v to w, then an edge in A exists from w back to v.  This means that we can view a polygraph (N,A,B) as a family of directed graphs.  Each directed graph in the family has the same vertices and an edge set A’ that is a superset of A and contains at least one edge in each bipath in B. They define an acyclic polygraph as a polygraph (represented as a family of directed graphs) where at least one directed graph in the family is acyclic.

In the paper, they relate database histories to polygraphs by letting the vertex set N bet a set of live transactions.  We build edges in A (u,v) from transactions that write a variable (vertex u)  to transactions that read the same variable (vertex v).  If some other vertex w also has that variable in their read set then the bipath {(v,w), (w,u)} exists in B.  So edges (u,v) in A mean that u “happens before” v since u writes a variable that v reads.  A bipath {(v,w), (w,u)} means that w also reads the same variable, so must happen before u or after v.  They show that a history is serializable if and only if the polygraph for the history is acyclic.

So, given a formula, they build a polygraph that is acyclic if and only if the formula is satisfiable.  The polygraph will have 3 vertices (aj, bj, and cj) for each variable xj in the formula.  Each a vertex connects by an edge in A to its corresponding B vertex.  We also have a bipath in B from the b vertex through the corresponding c vertex back to the a vertex.

Each literal Cik  (literal #k of clause i) generates two vertices yik and zik.  We add edges in A from each yik to zi(k+1)mod 3  (in other words, the y vertex of each clause connects to the “next” z vertex, wrapping around if necessary).  If literal Cik is a positive occurrence of variable Xj, we add edges (cj, yik) and (bj, zik) to A, and the bipath {(zik, yik), (yik, bj)} to B.  If the literal is negative, we instead add (zik, cj) to A and {(aj, zik), (zik, yik)} to B.

If the polygraph is acyclic (and thus the history it represents is serializable), then there is some acyclic digraph in the family of directed graphs related to the polygraph.  So the bipath {(bj, cj), (cj, aj)} will have either the first edge from b-c (which we will represent as “false”) or will have the second edge from c-a (which we will represent as “true”).  (The directed graph can’t have both because its edges are a superset of A, which means it has the edge (aj, bj) and taking both halves of he bipath will cause a cycle).

If our acyclic directed graph has the “false” (b-c) version of the edge for a literal, then it also has to have the z-y edge of the bipath associated with the literal (otherwise there is a cycle).  If all of the literals in a clause were set to false, this would cause a cycle between these bipath edges and the y-z edges we added in A for each clause.  So at least one literal per clause must be true, which gives us a way to satisfy the formula.

If the formula is satisfiable, then build the acyclic digraph that starts with all of A, and takes the bipath edges corresponding to the truth value of each variable, as defined above.  This implies ways you need to take the edges from the bipaths for the literals, to avoid cycles.  The only way now for the graph to be acyclic is for there to be a cycle of x’s and y’s in the edges and bipath edges.  But that would imply that we’ve set all of the literals in a clause to false. Since we know that the clause can be made true (since the original formula is satisfiable), we know that a way exists to make the directed graph acyclic.

Difficulty: 7.  It takes a lot of baggage to get to the actual reduction here, but once you do, I think it’s pretty easy and cool to see how the cycles arise from the definitions of the graphs and from the formula.

Non-Circular Satisfiability

The next reduction uses a new variant of Satisfiability, so I figured it would be worth making a separate post for it.

The problem: Non-Circular Satisfiability.  This problem is not in the appendix.

The description: A CNF-Satisfiability clause is mixed if it contains both variables and their negations.  A Satisfiability formula is non-circular if each variable occurs in a mixed clause at most once.  Given a non-circular satisfiability formula, is it satisfiable?

Example: This is actually a more general version of Monotone Satisfiability– in Monotone Sat, no mixed clauses can exist.  So all Monotone formulas are also Non-Circular.

Notice that we’re not necessarily restricting things to 3Sat clauses- we can have more or less than 3 variables per clause.  So here is a (satisfiable) instance:

F = (x1 ∨ x2) ∧ (~x1 ∨ ~x2) ∧ (~x1 ∨ x2)

This formula has just one mixed clause (the third one) and is satisfiable if x1 is false and x2 is true.  The formula becomes unsatisfiable if we add the clause (x1 ∨ ~x2), but then we would have both variables be in a second mixed clause.

Reduction: The easy way is to just do this by restriction from Montone Sat.  (Given a monotone sat formula, it’s automatically non-circular, so we’re done).  But if we want an actual reduction, we can use the paper by Papadimitriou, Bernstein, and Rothnie which has the next two reductions in G&J and uses this problem for the first one.  They also show the reduction for this problem in their paper and reduce from regular CNF-SAT.  So we have a formula F.  For each variable x in F, suppose x appears m times in F. Create m new variables x1 through xm.  The first instance of x in F will be replaced by x1, the second by ~x2, the third by x3, and so on down.  We then need to basically add the rules x1 ≡~x2 ≡ x3 …  (forcing each new literal we replaced x with to all have the same truth values).  in CNF form, that’s equivalent to the clauses: (x1 ∨ x2) ∧ (~x2 ∨ ~x3) ∧ (x3 ∨ x4) .. and so on down.  This means our new formula is equivalent to F (and is satisfiable when F is).  Each of these new clauses is non-mixed.  The only other occurrence of each xi variable is the one time it replaces x in the original F, which may or may not be a mixed clause, but either way, that means each variable appears in at most one mixed clause.

Difficulty: 3.  This is about as hard as the Monotone Sat reduction.