Problem set 4 - Emmanuel Briand

Mar 19, 2015 - variables Xi that makes the formula True ? This is an instance of the. 2-SAT (2-satisfiability) problem. Associate to the formula the fol-.
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Problem set 4 March 19, 2015

Emmanuel Briand. Universidad de Sevilla. 2014–2015. Discrete Mathematics. Grado Ingeniería Informática.

Various topics Problem 1. Let G be a graph with 56 edges, whose complementary graph has 80 edges. How many vertices does G have? Problem 2. How many edges does the complete graph on n vertices (often denoted Kn ) have? same question for the complete biparatite graph on m + n vertices Km,n (whose vertices are 1, 2, . . . , m and 10 , 20 ,. . . , n0 and edges are all pairs {i, j0 }). Problem 3. A leaf of a tree is a vertex with degree 1. How many leaves does a tree with 1000 edges have, at most? At least? Problem 4. How many articulation points dos a tree with 1000 edges has at most? At least? Problem 5. Let G be a graph with 26 vertices, 21 edges and no cycle. How many connected components does G have?

2-SAT and strongly connected components Consider the formula:

( X1 ∨ X2 ) ∧ ( X3 ∨ X4 ) ∧ ( X1 ∨ X4 ) ∧ ( X1 ∨ X3 ) ∧ ( X2 ∨ X4 )

(1)

where X stands for the negation of X. This formula is a conjunction of disjunction of two variables (ior negations of variables). Does there exist an assignation of boolean values (True/False) to the variables Xi that makes the formula True ? This is an instance of the 2-SAT (2-satisfiability) problem. Associate to the formula the following digraph: its vertices are the variables Xi and their negations Xi . For each disjunction A ∨ B, add two edges: one is A → B (that represents the implication A ⇒ B that is equivalent to A ∨ B), the other is B → A (that represents the implication B ⇒ A, also equivalent to A ∨ B). This digraph is the implication graph of the formula. Paths in this graph are implications derived from the formula. In particular, it is necessary, for making the formula true, to assign the same boolean value to all vertices that belong to a given strongly connected component. Actually, the following theorem holds: Theorem 1. The formula is satisfiable if and only if no strongly connected component contains at the same time a variable Xi and its negation. Problem 6. Draw the implication graph for Formula (1). Determine whether or not it is satisfiable. If so, find an assignment of values that makes it true. Do the same for the formulas:

( X1 ∨ X2 ) ∧ ( X3 ∨ X4 ) ∧ ( X1 ∨ X4 ) ∧ ( X1 ∨ X3 ) ∧ ( X2 ∨ X4 ) ∧ ( X1 ∨ X3 )

problem set 4

and

( X1 ∨ X3 ) ∧ ( X1 ∨ X4 ) ∧ ( X2 ∨ X4 ) ∧ ( X0 ∨ X5 ) ∧ ( X1 ∨ X5 ) ∧ ( X2 ∨ X5 ) ∧ ( X3 ∨ X6 ) ∧ ( X4 ∨ X6 ) ∧ ( X5 ∨ X6 ) ∧ ( X7 ∨ X2 ) ∧ ( X7 ∨ X3 )

2