Discrete Structures: Sample Questions, Exam 2, SOLUTIONS

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1 Discrete Structures: Sample Questions, Exam, SOLUTIONS This is longer than the actual test.) 1. List all the strings over X = {a, b, c} of length or less. λ, a, b, c, aa, ab, ac, ba, bb, bc, ca, cb, cc.. Consider the Fibonacci sequence defined recursively by f 1 = f = 1, f n = f n 1 + f n for n 3. Use induction to prove that f n = φn ψ n, where φ = 1 +, ψ = 1. For the basis cases, compute φ 1 ψ 1 = ) = 1 = 1 = f 1, φ = = 1 + ) = ψ = = 1 ) = φ ψ = ) = 1 = 1 = f, = 3 +, = 3, Now we prove the inductive case. The form of the Fibonacci sequence means that we should show P n 1) P n ) P n) for all n 3 where P n) is the formula we are trying to prove for all n 1.) So we assume P n 1) and P n ). For P n), compute f n = f n 1 + f n

2 = 1 φ n 1 ψ n 1 ) + 1 φ n ψ n ) by P n 1), P n ) = 1 φ n 1 + φ n ) 1 ψ n 1 + ψ n ) = 1 φ n φ + 1) 1 ψ n ψ + 1) ) = φn 1 + = φn ) = φn φ ) ψn ψ ) = φn ψ n ψn 1 ψn 3 ) + 1 ) The second to last line follows from the formulas for φ, ψ computed above.) This proves the inductive case, and thus the formula holds for all n Show that n! = On n ). Compute n! = n and n n = n n n n n. Note they have the same number of terms in the product, and that all the terms in the product for n! are positive and are all the corresponding terms in n n i.e., 1 n, n,... n n.) Thus the product n! n n, for all n 1. Thus for constant C = 1, we have n! 1 n n for all n 1, and this shows n! = On n ). 4. Short answer. No partial credit. a) A relation R on X = {1,, 3} has matrix digraph Draw its

3 1 3 b) Write the relation in part a) as a subset of X X. R = {1, 1), 1, 3),, ),, 3)}. c) Compute the matrix of the relation R R. The matrix of a composition of two relations is computed using the product of their matrices in the opposite order. In this case, the two matrices are the same, and the product is = Typically we would then have to replace all nonzero numbers by 1, but in this case, all the nonzero entries are 1, and so this is not necessary. d) Consider Y = {1,, 3, 4,, 6} and the equivalence relation on Y given by xry if and only if x y is a multiple of 3. Write down the elements of the equivalence class [1]. [1] = {1, 4}, since 1R1 and 4R1, but 1 is not related to,3,,6. e) Write down the digraph of the equivalence relation from part d) f) Write down the matrix of the equivalence relation from part d).

4 g) On the set Z + of positive integers, consider the partial order mrn if m divides n. Write down an incomparable pair of elements of Z +., 3. doesn t divide 3 and 3 doesn t divide. h) For the partial order from part g), find an element k of Z + so that nkrn). Write down k divides n for all n Z +. i) For a i = 3 i, compute a i. i=0. a i = a 0 + a 1 + a = = = 13. i=0 j) For parts j)-n), choose a theta class for the function or number of times the line x = x+1 executes in the algorithm. The possible answers are Θ1), Θlg n), Θn), Θn lg n), Θn ), Θ n ), Θn!). What is the theta class of fn) = n ) 1 + lg n)? n fn) = Θn lg n), since fn) = 1 + n lg n n lg n, and the highest order term is n lg n. k) gn) = n gn) = Θn ), since 1++ +n = 1 n +n) = Θn ). l) j = n while j 1){ for i = 1 to n x = x + 1 j = j/ } Θn lg n). The inner loop always takes n steps note it does not depend on the outer loop index j). As we saw in class, the outer loop takes Θlg n). So all together, it takes Θn lg n) steps.

5 m) j = n while j 1){ for i = 1 to j x = x + 1 j = j/ } Θn), as we saw in class, and in Example in the book. n) for j = 1 to n for i = j to n x = x + 1 Θn ). The inner loop goes n j + 1 times which depends on j). So adding up this for all j from 1 to n, we have n + n 1) + n ) = 1 n + n) = Θn ). o) On X = {1,, 3, 4}, let the relation R be given by xry if x > y, and let the relation S be given by xsy if x divides y. List the elements, as ordered pairs, of the relation R S. R = {, 1), 3, 1), 3, ), 4, 1), 4, ), 4, 3)}, while S = {1, 1), 1, ), 1, 3), 1, 4),, ),, 4), 3, 3), 4, 4)}. So R S = and there are no elements in R S. p) List the elements of R S as ordered pairs, where R, S are from part o). R S = {1, 1), 1, ), 1, 3), 1, 4),, 1),, ),, 4), 3, 1), 3, ), 3, 3), 4, 1), 4, ), 4, 3), 4, 4)}.. Refer to tables in the book for this problem. Write a sequence of operations select, project, join) to answer the query. Also, provide the answer to the query. Find all buyers who bought a part from Yu s department. TEMP1 = DEPARTMENT [Dept=Dept] SUPPLIER TEMP = TEMP1 [Part No = Part No] BUYER TEMP3 = TEMP [Manager = Yu] selection) ANSWER = TEMP3 [Name] projection) join) join)

6 The answer is Name United Supplies ABC Unlimited Danny s 6. Write an algorithm which returns the index of the last occurrence of the smallest element in the sequence s 1,..., s n. So if the sequence is , the output would be 4. last occurrences, n){ smallest = s 1 last = 1 for i = to n if s i smallest){ smallest = s i last = i } return last }

7 7. True/False. Circle T or F. No explanation needed. a) T F There is a relation on the set {α, β, γ} which is both an equivalence relation and a partial order. T. The equality relation given by arb if a = b is both an equivalence relation and a partial order. To show this, check it s reflexive, symmetric, and transitive these are just the standard 3 properties of equality). To show it s also a partial order, we need to show it s also antisymmetric. In other words, we need to show a ba = b b = a) a = b). This is obviously true. b) T F On Z +, consider the relation R defined by mrn if there is an integer k so that m = k n. Then R is an equivalence relation. T. It s reflexive, since m = 0 m and so mrm for all m. It s symmetric since if mrn, then m = k n and so n = k m, which means that nrm. It s transitive since if mrn and nrp, then m = k n and n = l p, which implies m = k l p) = k+l p and so mrp. c) T F R from part b) is a partial order. F. R is not antisymmetric, since for example 1R and R1, but 1. d) T F The relation from problem 4a) is reflexive. F. 3 R3. e) T F The relation from problem 4a) is symmetric. F. For example, 1R3 but 3 R1. f) T F The relation from problem 4a) is antisymmetric. T. There is no pair of 1 s across the diagonal. g) T F The relation from problem 4a) is transitive. T. Compute the square of the matrix to be again. Since all of the 1 s in the square are also 1 s in the original matrix, the relation is transitive. h) T F n 3 + lg n = Θn 3 ). T. This is true since lg n is lower order of growth than n 3.

8 i) T F n n = Θn 3 ). F. 3 n exponential growth) has much higher order of growth than n 3, and so n n = Θ3 n ). j) T F If R is a relation ) from {1, } to {a, b, c} with matrix, and S is a relation from {a, b, c} to {1, } with matrix , then the relation R S has ma- ) 0 1 trix. 1 1 F. The matrix of R S is determined by the matrix product in the opposite order. So compute 0 0 ) 1 1 = is the matrix of R S. k) T F n = Θn 4 ). F. It is Θn 3 ). l) T F n = Θ n ). T. This geometric series has sum 1 n+1 1 = n+1 1 = n 1 has highest order term n, which is = Θ n ). m) T F Let X, Y be finite sets, and let R be a relation from X to Y. Then R = R 1. Here recall R is the cardinality of R as a subset of X Y.) T. If R 1 is defined by {y, x) : x, y) R}, and so clearly x, y) y, x) is an one-to-one correspondence from R to R 1. Therefore, the cardinalities must be the same. n) T F! = 70. F.! = = 10. o) T F For the Fibonacci sequence f n as in problem above, f =. T. f 1 = f = 1, f 3 = f + f 1 =, f 4 = f 3 + f = 3, f = f 4 + f 3 =.