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UNIVERSITY PAPER – DISCRETE MATHEMATICS
December 2021 | Discrete Mathematics-212 | SEM-III
M.M.: 75 | Time: 3 Hours
NOTE: Candidates are required to attempt Five questions in all by selecting at least Two questions each from Section A and B. Section C is compulsory. Non-Prog Scientific Calculator is allowed.
SECTION-A
1. (a) If A and B have n elements in common, show that A × B and B × A have n² elements in common. Let C = A∩B, so |C| = n. An element (x, y) ∈ (A×B) ∩ (B×A) iff: (x, y) ∈ A×B → x∈A and y∈B (x, y) ∈ B×A → x∈B and y∈A So x∈A∩B = C and y∈A∩B = C. Thus (A×B)∩(B×A) = C×C, which has |C|² = n² elements. ∎ (b) How many subsets can be formed from a set of 'n' elements? How many of these will be proper and how many improper? Total subsets = 2ⁿ (each element is either in or out). Proper subsets = 2ⁿ − 1 (all except the set itself). Improper subsets = 1 (the set itself).
2. (a) Test the validity: Either I will get good marks or I will not graduate. If I did not graduate I will go to Australia. I get good marks. Thus, I would not go to Australia. Premises: P1: G ∨ ¬M (Good marks OR not graduate) P2: ¬M → A (Not graduate → go to Australia) P3: G (I get good marks) Conclusion: ¬A (I don't go to Australia) From P3 (G is true) and P1 (G ∨ ¬M), P1 is satisfied regardless of M. We cannot conclude anything about M. P2 says only if ¬M then A — since M may be true or false, A is not forced. However, the argument uses disjunctive syllogism: since G is true, P1 doesn't force ¬M. So we cannot invoke P2 to get A. Hence ¬A can be asserted. Valid. (G is true → P1 is satisfied without needing ¬M → P2 is never triggered → A is not concluded → ¬A holds.) (b) Find the domain, range and inverse of the relation R = {(x, y) : y = x + 10/x, where x, y ∈ N and x < 10}. x must divide 10: divisors of 10 less than 10 → x ∈ {1, 2, 5} x y = x + 10/x 1 1+10 = 11 2 2+5 = 7 5 5+2 = 7 R = {(1,11),(2,7),(5,7)} Domain = {1, 2, 5}, Range = {7, 11} Inverse R ¹ = {(11,1),(7,2),(7,5)}⁻
2→0, 0→1, 1→2, so (2,2) exists 3→0, 0→1, add (3,1) 3→0, 0→1→2, add (3,2) 3→0→1→2→0, so (3,0) exists TC = {(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2),(3,0),(3,1),(3,2)} (b) Partition A = {1, 2, 3, 4, 5, 6, 7} with the minsets generated by B ₁= {2, 4, 6} and B ₂= {1, 4, 5} and also find out how many different subsets of A can you generate from B ₁ and B .₂ The minsets (atoms) are all non-empty sets of the form B₁ ᵉ¹ ∩ B ₂ ᵉ², where e = set ᵢ or complement: Minset Elements B ₁ ∩ B₂ {4} B ₁∩ B '₂
B ' ∩₁
B₂
B ' ∩₁
B '₂
Partition: {{4}, {2,6}, {1,5}, {3,7}} Subsets generated = 2⁴ = 16 (union of any combination of these 4 atoms). SECTION-B
- (a) Explain in detail the representation of the directed graphs and also give one example. A directed graph (digraph) G = (V, E) has vertices V and edges E as ordered pairs (u,v) — arrows with direction. Representations:
- Adjacency Matrix: n×n matrix A where A[i][j]=1 if edge from i to j exists. Not necessarily symmetric.
- Adjacency List: Each vertex stores a list of vertices it points to.
- Edge List: List of all ordered pairs. Example: V={1,2,3}, E={(1,2),(2,3),(3,1)} Adjacency Matrix: 1 2 3 1[0 1 0] 2[0 0 1] 3[1 0 0] This represents a directed cycle 1→2→3→1. (b) Define planner graph and regions. State and prove the properties for a graph to be a planner graph. A planar graph is one that can be drawn in a plane with no edge crossings. A region (face) is a connected area of the plane bounded by edges (including the unbounded outer region). Euler's Formula: For a connected planar graph: V − E + F = 2 (V=vertices, E=edges, F=faces/regions). Properties/Corollaries:
- For any simple connected planar graph with V≥3: E ≤ 3V − 6
- If the graph has no triangles (no 3-cycles): E ≤ 2V − 4
- Every planar graph has a vertex of degree ≤ 5. Proof of E ≤ 3V−6: Each face is bounded by ≥3 edges, and each edge borders at most 2 faces, so 3F ≤ 2E → F ≤ 2E/3. Substituting into Euler's formula: V−E+2E/3 ≥ 2 → V−E/3 ≥ 2 → E ≤ 3V−6. ∎
- (a) Define function and explain different types of functions.
Sum of degrees = 2·3 + 3·2 = 6+6 = 12 By Handshaking Lemma: Sum of degrees = 2|E| ∴ |E| = 12/2 = 6 edges (b) An Undirected graph possesses an Eulerian path iff it is connected and has either zero or two vertices of odd degree. Statement: An undirected graph has an Eulerian path (traverses every edge exactly once) iff it is connected and has exactly 0 or 2 vertices of odd degree. Proof sketch: (Necessary) Each time the path visits a vertex, it uses one edge to enter and one to leave. So interior vertices must have even degree. Only the start and end vertices can have odd degree (they use one unpaired edge). Hence 0 (circuit) or 2 (path) odd-degree vertices. (Sufficient) If 0 odd vertices: start anywhere, greedily extend; by connectivity any stuck subgraph can be spliced in. If 2 odd vertices: add a temporary edge between them to make all degrees even, find the Euler circuit, remove the temporary edge to get the Euler path.
- (a) Prove that in a complete graph the number of edges is n(n–1)/2. How many vertices are there in a graph with 10 edges if each vertex has degree 2? Proof: In Kₙ, each of the n vertices connects to every other. Degree of each = n−1. Sum of degrees = n(n−1). By Handshaking Lemma: |E| = n(n−1)/2. ∎ Graph with 10 edges, every vertex degree 2: Sum of degrees = 2×10 = 20, each vertex has degree 2, so number of vertices = 20/2 = 10 vertices. (b) Let G be a finite graph with n > 1 vertices. Prove that G is tree iff it is minimally connected.
Definitions: A tree is a connected acyclic graph. Minimally connected means connected, but removing any edge disconnects it. Proof (⟹): Let G be a tree. G is connected. Suppose removing edge {u,v} leaves G connected — then there's another path from u to v, forming a cycle. But trees are acyclic — contradiction. So every edge is a bridge → G is minimally connected. Proof (⟸): Let G be minimally connected. It is connected by definition. Suppose G has a cycle. Then any edge in the cycle can be removed without disconnecting G (the rest of the cycle provides an alternate path) — contradicting minimal connectivity. So G has no cycles → G is a tree. SECTION-C (Compulsory)
- (i) Define ordered pair and Cartesian product. (3 marks) An ordered pair (a,b) is a pair where order matters: (a,b) ≠ (b,a) unless a=b. The Cartesian product A×B = {(a,b) : a∈A, b∈B}. (ii) Define union of two sets and give one example of each. (3 marks) A∪B = {x : x∈A or x∈B}. Example: {1,2}∪{2,3} = {1,2,3}. (iii) Define totally ordered relation and give an example. (1 mark) A relation R on set A is a total order if it is reflexive, antisymmetric, transitive, and every pair is comparable (for all a,b: aRb or bRa). Example: ≤ on ℝ. (iv) Define Big-Theta Notation. (1 mark) f(n) = Θ(g(n)) means there exist constants c₁,c₂>0 and n₀ such that c₁·g(n) ≤ f(n) ≤ c₂·g(n) for all n≥n₀. It gives a tight bound.
Since all 1000 students are in one of these categories: 1000 = 1070 − 500 + | H∩F∩A| |H∩F∩A| = 1000 − 570 = 430 Wait — let me recompute: 310+650+440 = 1400; 170+150+180 = 500; 1000 = 1400−500+x → x = 100. Students in all 3 = 100 Only athletes = A − (H∩A) − (F∩A) + (H∩F∩A) = 440−150−180+100 = 210 (b) State and prove De-Morgan's law. (A∪B)' = A'∩B' and (A∩B)' = A'∪B' Proof of (A∪B)' = A'∩B': Let x∈(A∪B)'. Then x∉A∪B → x∉A and x∉B → x∈A' and x∈B' → x∈A'∩B'. ✓ (reverse containment is symmetric)
2. (a) Prove that conjunction distributes over disjunction. P Q R (^) Q∨R
P∧(Q∨R
P∧Q P∧R
(P∧Q)∨
(P∧R)
T T T T T T T T
T T F T T T F T
T F T T T F T T
T F F F F F F F
F T T T F F F F
F T F T F F F F
F F T T F F F F
F F F F F F F F
Columns 5 and 8 are identical → proved.
(b) Prove by the principle of induction: 1 + 3 + 5 + ... + (2n–1) = n². Base: n=1: LHS=1, RHS=1². ✓ Inductive step: Assume 1+3+…+(2k−1)=k². Then: 1+3+…+(2k−1)+(2k+1) = k²+(2k+1) = (k+1)² ✓
3. (a) Test the validity: If I work hard then I will be successful. If I am successful then I will be happy. Therefore hard work leads to happiness. Let H=hard work, S=success, P=happy. P1: H→S P2: S→P Conclusion: H→P This is hypothetical syllogism. Assume H is true. By P1, S is true. By P2, P is true. So H→P. Valid. (b) Find the domain and range of the relation R = {(x,y): y = x + 6/x, where x, y ∈ N and x < 6}. x must divide 6: divisors of 6 less than 6 → x∈{1,2,3} x y 1 7 2 5 3 5 Domain = {1,2,3}, Range = {5,7} 4. (a) Let R be the relation on {0,1,2,3} containing (0,1), (1,1), (1,2), (2,0), (2,2), (3,0). What is reflexive, symmetric, and transitive closure of R?
