CS 5104 take-home quiz - 1 - April 20, 2006

Solution to CS 5104 Take-home quiz 1. (5 points) Show that the ≤P relation is a transitive relation on languages. That is, show

that if L1 ≤P L2 and L2 ≤P L3, then L1 ≤P L3. Let M1 be a polynomial-time TM that reduces L1 to L2 Let M2 be a polynomial-time TM that reduces L2 to L3 Construct M = On input w: Run M1 on w to get w′ Run M2 on w′ to get w′′ Output w′′ The runtime of M is polynomial in size |w|. w ∈ L1 ⇔ w′ ∈ L2 ⇔ w′′ ∈ L3 Therefore, M is a polynomial-time TM that reduces L1 to L3

2. (5 points) Fill out the table described in the polynomial time algorithm for context-

free language recognition from Theorem 7.16 for string w = baba and CFG G: S → RT R → TR | a T → TR | b

last symbol

b a b a b { T } { R, T } { S } { S, R, T } a { R } { S } { S } b { T } { R, T }

first symbol

a { R }

CS 5104 take-home quiz - 2 - April 20, 2006

3. (15 points) A subset of the nodes of graph G is a dominating set if every node of G is either in the subset or is adjacent to some node in the subset. Let

DOM-SET = { 〈G, k〉 | G has a dominating set with k nodes } Show that it is NP-complete by giving a reduction from VERTEX-COVER.

I. DOM-SET ∈ NP A. Certificate: a set of vertices that dominate G B. Algorithm: M = “On input 〈G, c〉: for each vertex v in G check if v ∈ c or ∃ x ∈ c such that (v, x) is in G C. Runtime: Polynomial in V[G] + E[G] II. DOM-SET is NP-hard (we show VERTEX-COVER ≤P DOM-SET A. Algorithm: R: 〈G, k〉 → 〈G′, k′〉 R = “On input 〈G, k〉: set k′ = k copy all edges and their vertices from G into G′ // note: this does not copy vertices with no edges for each edge (x, y) in G add a vertex zx,y to G′ add edges (x, zx,y) and (y, zx,y) to G′

B. Runtime: Clearly polynomial C. Correctness: Show: 〈G, k〉 ∈ VERTEX-COVER ⇔ 〈G′, k′〉 ∈ DOM-SET (⇒) Assume G has a vertex cover V of size k ⇒ V dominates all vertices that have at least one edge, and V dominates all new vertices zx,y added to G′ ⇒ V dominates G′ (⇐) Assume that G′ has a dominating set D of size k ⇒ Let D′ = (D ∩ V[G]) ∪ X where X ⊆ V[G] and x ∈ X if 1. (x, y) ∈ E[G] 2. x, y ∉ D ∩ V[G] 3. y ∉ X

4. zx,y ∈ D (zx,y ∈ V[G′] – V[G] by R’s construction of G′) ⇒ D′ ≤ k (X has ≤ one element for every element of V[G′] – V[G]) and for all edges (u, v) ∈ G,

Case 1: u ∈ D ∩ V[G] or v ∈ D ∩ V[G] ⇒ D′ covers (u, v) Case 2: u,v ∉ D ∩ V[G] ⇒ zu,v ∈ D ⇒ u or v ∈ X ⇒ D′ covers (u,v)