Solutions to Quiz 1 (6.042J Mathematics for Computer Science)

Massachusetts Institute of Technology 6.042J/18.062J : Mathematics for Computer Science October 17 revised October 18, 125 minutes Solutions to Quiz 1 Problem 1 (15 points). Let G be an undirected graph. Let P (x, y) mean that there is a path from vertex x to vertex y. Express each of the following sentences in terms of P , quantifiers, logical connectives, and equality, using variables that range over the vertices of G. (Reminder: there is a zero­length path from any vertex to itself.) (a) (3 points) Vertices x and y are in the same connected component. Solution. P (x, y) � (b) (3 points) G has a vertex of degree zero. (Reminder: undirected graphs only have edges between distinct vertices, that is, no self­loops.) Solution. ∃x∀y. P (x, y) −→ x = y. � (c) (4 points) G has at least three connected components. Solution. ∃x, y, z. ¬P (x, y) ∧ ¬P (x, z) ∧ ¬P (y, z). � (d) (5 points) There is a positive­length simple path from x to y. Solution. There is a positive length simple path between two vertices iff the vertices are distinct and there is a path between them, so we can express this as x= � y ∧ P (x, y). � 2 Problem 2 (10 points). Classify each of the following binary relations as E: An equivalence relation. T: A Total order, P: A Partial order that is not total. S: A Symmetric relation that is not transitive. N: None of the above. (a) (2 points) The relation between times during a single day: x and y are at most twenty minutes apart. Solution. S: This relation is obviously symmetric. It is not transitive: 1:00R 1:15, and 1:15R 1:22, but ¬ (1:00R 1:22). � (b) (2 points) The relation between times during a single day: x is more than twenty minutes later than y. Solution. P: This relation is asymmetric and transitive. It is not a total ordering because some times are incomparable to each other. For example, 1:15 is incomparable to 1:22. � (c) (2 points) The relation between vertices in an arbitrary digraph: there is a path from v to w. Solution. N: This relation is transitive but won’t be a partial order unless the graph is acyclic. � (d) (2 points) The relation between vertices in an undirected graph: there is a path from v to w. Solution. E: In fact, the connected components of the graph are the equivalence classes of this relation. � (e) (2 points) The relation between Fall ’05 6.042 students: student s is older but also shorter than t. Solution. P: This relation is the product of the “older than” and the “shorter than” orders, and so is a partial order. It is not total because there is a younger student who is taller than an older student. � 3 Problem 3 (20 points). Let G0 = 1, G1 = 2, G2 = 4, and define Gn ::= Gn−1 + 2Gn−2 + Gn−3 (1) for n ≥ 3. Show by induction that Gn ≤ (2.2)n for all n ≥ 0. Solution. The proof is by strong induction with hypothesis P (n) := Gn ≤ (2.2)n . Proof. Base Cases n = 0: G0 ::= 1 = (2.2)0 . n = 1: G1 ::= 2 < 2.2 = (2.2)1 . n = 2: G2 ::= 4 < 4.84 = (2.2)2 . Inductive Step: Assume n ≥ 2 and P (k) for all k such that 0 ≤ k ≤ n. Gn+1 ::= Gn + 2Gn−1 + Gn−2 ≤ (2.2)n + 2(2.2)n−1 + (2.2)n−2 = (2.2)n−2 [(2.2)2 + 2(2.2) + 1] = (2.2)n−2 [(2.2 + 1)2 ] = (2.2)n−2 (3.2)2 = (2.2)n−2 (10.24) < (2.2)n−2 (10.648) = (2.2)n−2 (2.2)3 = (2.2)n+1 by (1) by induction hypothesis � Problem 4 (20 points). An intersection graph is an undirected graph whose vertices are sets and whose edges are specified by the rule that there is an edge between vertices A and B iff A = � B and A ∩ B �= ∅. (a) (1 point) Draw the intersection graph whose vertices are the sets {1, 2, 3} , {1, 9, 10} , {2, 4, 6, 8, 10} , {3, 4, 5} , {5, 6, 7} , {7, 8, 9} 4 Solution. � (b) (3 points) What is the chromatic number of the graph in part (a)? Solution. 4. The graph is isomorphic to the wheel, W5 . All wheels with an odd­length rim have chromatic number 4, as explained in lecture. � (c) (3 points) What is the largest k such that the graph in part (a) is k­connected? Solution. 3. � We now consider an arbitrary undirected graph, G. For any vertex, v, of G, let I(v) be the set of edges incident to v. (d) (3 points) Explain how to uniquely determine the vertex v given any two edges in I(v). Solution. Just find a vertex that the two edges have in common. This must be v. � (e) (10 points) An incidence­set is the set of edges incident to some vertex, that is, a set equal to I(v) for some vertex v of G. Prove that if G is a graph whose vertices all have degree greater than 1, then the function, I, is an isomorphism between G and the inter­ section graph whose vertices are the incidence­sets of G. Solution. Let H be the intersection graph whose vertices are the incidence­sets of G. By definition of H, the function, I, is a surjection onto the vertices of H. Moreover, since each incidence­set contains at least two edges, part (d) implies that given I(v), we can uniquely determine v. This means I is also an injection into the vertices of H, and therefore is a bijection from the vertices of G to the vertices of H. Also v—w is an edge of G iff iff iff [v—w ∈ I(v) and v—w ∈ I(w)] v �= w and I(v) ∩ I(w) �= ∅ I(v)—I(w) is an edge of H, 5 which proves that I is an isomorphism. Incidentally, if G happens to have any vertices of degree < 2, then it’s easy to assign special sets to these vertices in such a way that these special sets, along with the incidence sets of vertices of degree ≥ 2, are the vertices of an intersection graph that is isomorphic to G. This shows that every undirected graph is isomorphic to an intersection graph. � Problem 5 (10 points). Two banks only allow transactions that are multiples of 39 dollars or 57 dollars. Is there a series of transactions whose net result is a payment of 1 dollar from the first bank to the second bank? Briefly explain why or why not. Solution. Yes. Since 39 and 57 have no common prime factors, their gcd is 1, and so there are positive integers s, t such that s (39 ) − t (57 ) = 1. So the first bank should pay the second bank 39 dollars s times, and accept payment of 57 dollars form the second bank t times for a net transfer of 1 dollar to the second bank. � Problem 6 (15 points). Each year, Santa’s reindeer hold “Reindeer Games”, from which Rudolph is pointedly excluded. The Games consist of a sequence of matches, where one reindeer competes against another. Draws are not possible. On Christmas Eve, Santa produces a rank list of all his reindeer. If reindeer p lost a match to reindeer q, then p appears below q in Santa’s ranking, but if he has any choice because of unplayed matches, Santa can give higher rank to the reindeer he likes better. To prevent confusion, two reindeer may not play a match if either outcome would lead to a cycle of reindeer, where each lost to the next. Though it is only October, the 2004 Reindeer Games have already begun. We can describe the results so far with a binary relation, L, on the set of reindeer, where pLq means that reindeer p lost a match to reindeer q. Let L+ be the corresponding positive­length path relation1 . Note that L+ is a partial order, so we can regard a match loser as “smaller” than the winner. On the following page you’ll find a list of terms and a sequence of statements. Add the appropriate term to each statement. 1 Thus, reindeer p is related to reindeer q by L+ if p lost to q or if p lost to a reindeer who lost to q or if p lost to a reindeer who lost to a reindeer who lost to q, etc. 6 a strict partial order comparable elements an antichain a minimum element a maximum element Terms a weak partial order incomparable elements a maximal antichain a minimal element a maximal element a total order a chain a topological sort Statements (a) (1 point) An unbeaten reindeer is of the partial order L+ . Solution. a maximal element � (b) (1 point) A reindeer who has lost every match so far is of the partial order L+ . Solution. a minimal element � (c) (1 point) Two reindeer can not play a match if they are of L+ . Solution. comparable elements � (d) (1 point) A reindeer assured of first place in Santa’s ranking is of L+ . Solution. a maximum element � (e) (1 point) A sequence of reindeer which must appear in the same order in Santa’s rank list is . Solution. a chain � (f) (2 points) A set of reindeer such that any two could still play a match is . 7 Solution. an antichain � (g) (2 points) The fact that no reindeer loses a match to himself implies that L+ is . Solution. a strict partial order � (h) (2 points) Santa’s final ranking of his reindeer on Christmas Eve must be of L+ . Solution. a topological sort � (i) (2 points) No more matches are possible if and only if L+ is . Solution. a total order � (j) (2 points) Suppose that Santa has 11 reindeer. If no more matches can be played, what is the smallest possible number of matches already played? Solution. 10. Call the reindeer r1 , . . . , r11 . Then no more matches can be played if r1 lost to r2 , r2 lost to r3 , etc. If only ten matches have been played, then the corresponding digraph will not even be connected if we regard the edges as undirected. So there would be two disjoint sets of reindeer, S and T , such that no reindeer in S has played a match against a reindeer in T . Therefore, more matches can be played; in particular, any reindeer in S can play against any reindeer in T . � Problem 7 (10 points). A map is a connected planar graph with a planar drawing whose face boundaries are simple cycles. (a) (7 points) Prove that if a map has no simple cycle of length 3, then e ≤ 2v − 4, where v is the number of vertices and e is the number of edges in the graph. (2) 8 Solution. Since there is no simple cycle of length 3 and face boundaries are simple cycles, every face must be of length 4 or more. We know that in a planar drawing, each edge appears twice on the face boundaries, so � � � � 4 = 4f, (3) 2e = length(b) ≥ b∈ boundaries b∈ boundaries where f is the number of faces. By Euler’s formula, f = e − v + 2, so substituting for f in (3), yields 2e ≥ 4(e − v + 2), which simplifies to (2). � (b) (3 points) Prove that K3,3 is not a map. (K3,3 is the graph with six vertices and an edge from each of the first three vertices to each of the last three.) Solution. K3,3 is by definition bipartite, and hence has no odd­length simple cycle, and in particular, no length 3 cycle. (This could also be checked exhaustively.) However, it has 9 edges and 6 vertices, and since 9 > 8 = 2 · 6 − 4, it does not satisfy (2), and so cannot be a map. Incidentally, this implies that K3,3 is not planar. The reason is that connected planar graphs which are not maps are the ones with “dongles” or “crossbars,” as explained in lecture. But all edges in dongles and crossbars are cut edges. So a connected graph like K3,3 with no cut edges is planar iff it is a map. �