MA0301 ELEMENTARY DISCRETE MATHEMATICS NTNU, SPRING 2020 Solutions Set 12 Exercise 1. Let B be a Boolean algebra. For x; y; z 2 B find the dual expressions of i)(x +y ¯) · (¯z + y) ii) (1 + x) · y + x · y¯ · z iii)(x · y + 1) · (0 + x) · z Solution. Recall the definition of dual expression: it consists in the statement obtained by replacing + by ·, · by +, 1 by 0, and 0 by 1. According to this, we have that the required dual expressions are following: (1) (x · y) + (z · y), (2) ((0 · x) + y) · (x + y + z), (3) ((x + y) · 0) + (1 · x) + z. Exercise 2. 1) Let the natural number N be our universe. We start by defining the property P (q): q is even. We define the relation R on N as follows: 8n;m (n; m) 2 R iff P (n + m). a) Show that R is an equivalence relation. b) Can you determine how many equivalence classes the relation R has? 2) Given the set F = fa; b; c; d; eg with an ordering described by the diagram d e b c a Write down all subsets of F in which the element c is a minimal element. Solution. 1) a) We will show that R is reflexive, symmetric and transitive. Take m 2 N. Note that (m; m) 2 R since P (m + m) is true (m + m = 2m is an even number). Hence R is reflexive. For the symmetric property, assume that (m; n) 2 R. This means that P (m + n) is true. Since m + n = n + m, then P (n + m) is true. Hence (n; m) 2 R and so R is symmetric. Finally, for transitivity, assume that (m; n); (n; p) 2 R. This implies that P (m + n) and P (n + p) are both Date: April 24, 2020. 1 2 MA0301 ELEMENTARY DISCRETE MATHEMATICS NTNU, SPRING 2020 true, i.e. m + n and n + p are both even integers. Their sum is also an even integer, let us say, (m + n) + (n + p) = 2r, for some r 2 N. Note that 2r = m + n + n + p = m + p + 2n ) m + p = 2r − 2n = 2(r − n): Then m + p is also an even integer. We have that P (m + p) is true and hence (m; p) 2 R. Then R is transitive. We conclude that R is an equivalence relation. 1) b) We show that R has two equivalence classes: the even natural numbers and the odd natural numbers. First note that (1; 2) 62 R since 1 + 2 = 3 is not an even number. Now, let m be an odd natural number. This means that there is a non-negative natural number k 2 N0 such that m = 2k + 1. Then we have that (1; m) 2 R since 1 + m = 1 + 2k + 1 = 2k + 2 = 2(k + 1) is even. Then the odd natural numbers belong to the same equivalence class. In order to see that they indeed form an equivalence class, notice that if n = 2r is even, then 1 + n = 2r + 1 is odd, and so (1; n) 62 R. Hence f1; 3; 5; 7;:::g is an equivalence class. Finally, if n = 2r is even, then (2; n) 2 R since 2 + n = 2 + 2r = 2(r + 1) is even. Hence the even natural numbers form an equivalence class. We conclude that R has the two following equivalence classes: f1; 3; 5; 7;:::g; f2; 4; 6; 8;:::g: 2) Recall the definition of minimal element: x 2 P is minimal if there is no y 2 P such that x 6= y and y ≤ x. Now, let G be a subset of F such that c is minimal in G. Obviously c 2 G. According to the diagram, we have that a ≤ c, c ≤ d, c ≤ e and b and c are not comparable. Then a 62 G since otherwise it would contradict that c is minimal in G. Since there is no problem with the remaining three elements of F , we can freely choose them to belong to G. This corresponds to the following eight subsets: fcg; fc; bg; fc; dg; fc; eg; fc; b; dg; fc; b; eg; fc; d; eg; fc; b; d; eg: Exercise 3. a) Let G be a planar graph with eight vertices. What is the maximal number of edges possible in G? b) Let G be a finite graph. Can G have a subgraph H (H is supposed to be different from G) so that G and H are isomorphic? c) Consider the graph G with V (G) := fa; b; c; d; e; fg and E(G) := ffa; cg; fa; dg; fa; eg; fb; eg; fb; fgg. i) Determine G − a ii) Find the connected components of G − a. Solution. a) We know that a maximal planar graph with jV j > 2 vertices has precisely 3jV j − 6 edges. Then the maximum number of edges possible in G is 3 · 8 − 6 = 18. b) No. We will show it by contradiction. Let G = (V; E) be the initial graph and assume that there is a subgraph H = (V1;E1) such that H 6= G and G and H are isomorphic. The definition of graph isomorphism establishes that there exists a function f : V ! V1 such that f is a bijection and (1) fa; bg 2 E , ff(a); f(b)g 2 E1 Since H is subgraph of G, we have that V1 ⊆ V and E2 ⊆ E. However, since V is finite and f is a bijection, then V1 has the same size that of V and hence V = V1. Note that f also induces a bijective correspondence between E and E1 given by (1). Since E is also finite and E1 ⊆ E, we MA0301 ELEMENTARY DISCRETE MATHEMATICS NTNU, SPRING 2020 3 have that E = E1. Then we conclude H = (V; E) = G, which is a contradiction. Hence, a finite graph cannot have a subgraph isomorphic to itself. c) i) The set of vertices of G−a is given by V (G−a) = fb; c; d; e; fg. For the case of set of edges, we have to delete all the edges with contain a. The resulting set of edges is E(G − a) = ffb; eg; fb; fgg. c) ii) Note that there is no edge in G − a such that contains the vertex c. The same happens with d. Hence these vertices are isolated. On the other hand, the remaining three vertices belong to the same connected component since we have the path e ! b ! f. Hence the connected components of G − a are given by (fag; ;); (fcg; ;) and (fb; e; fg; ffb; eg; fb; fgg). Exercise 4. Let G be a planar graph with fewer than 12 vertices. Show that G has a vertex of degree at most 4. (Hint: Recall the formula jEj ≤ 3jV j − 6.) Solution. If jV j ≤ 2, the result follows immediately. Assume now that 2 < jV j < 12. Recall the formula jEj ≤ 3jV j − 6 for a planar graph G = (V; E) such that jV j > 2. By the formula of the sum of the degrees we have X (2) deg(v) = 2jEj ≤ 2(3jV j − 6) = 6jV j − 12 < 6jV j − jV j = 5jV j; v2V where in the last inequality we used that jV j < 12. If every vertex has degree at least 5, we would have that X X deg(v) ≥ 5 = 5jV j; v2V v2V and according to (2), we would get that 5jV j < 5jV j, which is clearly a contradiction. Then our assumption is false and we conclude that G has a vertex of degree at most 4. Exercise 5. Show that the following graph is not planar. (Hint: Notice that, in the case that the graph is planar, it contains no triangles. Then, any face would be bordered by at least 4 edges) Solution. We proceed by contradiction. Assume that the graph is planar. We see that jV j = 11 and jEj = 20. Since we are assuming that the graph is planar, we can apply Euler’s formula in order to get that jF j = 2 − 11 + 20 = 11. The main observation is that the graph does not contain any 4 MA0301 ELEMENTARY DISCRETE MATHEMATICS NTNU, SPRING 2020 triangle and then any face is bordered by at least 4 edges. On the other hand, each edge borders two faces since there are no cut-edges in the graph. Then X X 2jEj = (no. of edges bordering f) ≥ 4 = 4jF j: f2F f2F Since jEj = 20 and jF j = 11, the above inequality tell us that 40 = 2 · 20 ≥ 4 · 11 = 44, which is a contradiction. We conclude then that the graph is not planar as we wanted to show. Pn 2 2 Exercise 6. Use induction to show that for all natural numbers: 4 k=1(k + 2k)(k + 4) = (n + n)(n + 4)(n + 5): Solution. We proceed by mathematical induction on n. For the base case n = 1, note that 1 X 4 (k2 + 2k)(k + 4) = 4(1 + 2)(1 + 4) = 60 = (12 + 1)(1 + 4)(1 + 5); k=1 and then the base case holds. For the induction hypothesis, assume that the formula holds for a natural number m. We will show that the formula holds for m + 1. Indeed, by splitting the sum and using the induction hypothesis, we have m+1 X 4 (k2 + 2k)(k + 4) = (m2 + m)(m + 4)(m + 5) + 4((m + 1)2 + 2(m + 1))((m + 1) + 4) k=1 = (m2 + m)(m + 4)(m + 5) + 4((m + 1)2 + 2(m + 1))(m + 5) = (m + 5)[m(m + 1)(m + 4) + 4(m + 1)(m + 1 + 2)] = (m + 1)(m + 5)[m(m + 4) + 4(m + 3)] = (m + 1)(m + 5)(m2 + 8m + 12) = (m + 1)(m + 5)(m + 2)(m + 6) = (m + 1)(m + 2)(m + 5)(m + 6) = (m + 1)((m + 1) + 1)((m + 1) + 4)((m + 1) + 5) = ((m + 1)2 + (m + 1))((m + 1) + 4)((m + 1) + 5): Hence the formula holds for m + 1.