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DISCRETE
MATHEMATICS
Assignment – 3
Name: SHIKHAR GUPTA
Enrollment Number: IIT
Q1. (a) Show that Subtraction Rule and Division Rule follows from the Sum Rule. Ans. Since Subtraction rule says, for finite set S1 and S |S1 U S2| = |S1| + |S2| - |S1 ^ S2| -------------(1) Let’s assume there exist a set(finite) S3 such that, |S3^S1| = Φ ----------(2) Therefore, |S1 U S3| = |S1| + |S3| which is the addition rule. Division rule states that, |S2/S1| = |S2| - |S1| where S1⊆ S Now, from subtraction rule, |S1’ U S2| = |S1’| + |S2| - |S1’ ^ S2| 1=|S1’| + |S2| - |S1’ ^ S2| => |S1’^S2| = |S2| - |S1| = |S2/S1| Thus, proved.
(b) Show the expansion of inclusion-exclusion principle for n=4. Ans. Inclusion exclusion principle states Now for n= |A1 ∪ A2 ∪ A3 ∪ A4 | = |A1| + |A2| + |A3| + |A4| − (|A1 ∩ A2| + |A1 ∩ A3| + |A2 ∩ A3| + |A1 ∩ A4|+|A2 ∩ A4|+|A4 ∩ A3|) + ( |A1 ∩ A2 ∩ A3| + |A1 ∩ A2 ∩ A4| + |A1 ∩ A4 ∩ A3| + |A4 ∩ A2 ∩ A3|) − ( |A1 ∩A2 ∩ A3 ∩ A4|) Q2. For matrices A, B, C ∈ M (i.e. set of integers), proof: (a) Associative Law: A+(B+C) = (A+B)+C Ans. Proof: Let A = [aij]m × n, B = [bij]m × n and C = [cij]m × n Let B + C = D = [dij]m × n, A + B = E = [eij]m × n, A + D = P = [pij]m × n, E + C = Q = [qij]m × n Then, dij = bij + cij , eij = aij + bij , pij = aij + dij and qij = eij + cij Now, A + (B + C) = A + D = P = [pij]m × n and (A + B) + C = E + C = Q = [qij]m × n Therefore, P and Q are the matrices of the same order and pij = aij + dij = aij + (bij + cij) = (aij + bij) + cij , (by the definition of addition of matrices) = eij + cij
= [aij] = A Hence, proved. Q3. (a) Proof Pigeon-Hole Principle: If a finite set S is partitioned into k sets, then at least one of the sets has |S|/k or more elements. Ans. The claim is that at least one of the k boxes contain at least [N/k] objects. The proof goes by contradiction: Assumption: - Suppose the claim is false, then each box must have strictly less than [N/k] objects, i.e., at most [N/k]−1 objects (the greatest integer strictly less than n is n−1). Solution: - Now, then since there are k boxes and each box has objects ≤[N/k]−1, the total number of objects from all the boxes is ≤k([N/k]−1)<k⋅N/k=N (we use [x]−1<x) which gives us a contradiction since the total number of objects from all the boxes cannot be strictly less than N (since N is the total number of objects from all the boxes). Notice the one single < in the chain of inequalities in the penultimate step which makes the overall inequality strict, i.e, gives us N<N. Which is just a false statement. This contradiction arose due to inappropriate assumption. Hence our assumption was wrong and at least one of the k boxes contain at least [N/k] objects. (b) How many cards must be selected from a standard deck of 52 cards to guarantee that at least three cards of the same suit are chosen? Ans. Suppose that for each suit, we have a box that contains cards of that suit. The number of boxes is 4, by the generalized pigeonhole principle, to have at least 3 (= N/4) cards in the same box, the total number of the cards must be at least N = (3-1). 4 + 1 = 2. 4 + 1 = 9.
(c) How many cards must be selected to guarantee that at least three hearts are selected? Ans. The worst case, we may select all the clubs, diamonds, and spades (39 cards) before any Hearts. So, to guarantee that at least three hearts are selected, 39+3=42 cards should be selected. Q4. (a) Proof: If c is an edge in a simple closed path in G, then e belongs to some cycle. Ans. (x1)=================➔(x ) In the above figure, the path ee with vector sequence x1, x2, x3 in the graphs is a closed path but not a cycle as it is not simple (as all the vertices are not distinct) and same goes for the path ee with vertices sequence x2, x1, x3. So, the graph is acyclic. |====================== | V | (x1)=================➔(x ) The path ef in the vector sequence x1, x2, x1 in the above graph is simple as well as closed and the path keeps on traversing forming a cycle the vector sequence x2, x1, x2 with the path fe also forms the cycle in the same way. (b) The sum of the degrees of the vertices of a graph is twice the number of edges. That is, 2|E|=Σv ∈ Vdeg(v) Ans. We want to proof 2|E|=Σv∈Vdeg(v) for a simple graph (no loops). For our proof we
x<=z. Ans. Given that x<=y. So, y-x>= ----------------(1) And, y<=z. So, z-y>= ------------------------(2) Now z-y=(z-y) + (y-x) -------------------------(3) From 1,2,3 we have z-y>= or z>=x Thus, proved.
