Discrete Structures: Sample Questions, Exam 2, SOLUTIONS

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1 Discrete Structures: Sample Questions, Exam 2, SOLUTIONS (This is longer than the actual test.) 1. Show that any postage of 8 cents or more can be achieved by using only -cent and 5-cent stamps. We proceed by induction. There are basis steps. This corresponds to the fact that the smaller stamp is for cents. Basis case n = 8: 8 = 5 + one 5-cent stamp plus one -cent stamp. Basis case n = 9: 9 = + + three -cent stamps. Basis case n = 10: 10 = two 5-cent stamps. Inductive case. We will add -cent stamps to prove the result. If we can make n cents out of - and 5-cent stamps, then we can add another -cent stamp to the total to make n + cents. This shows P (n) P (n + ). It remains to note that we can reach any integer n 8 by starting at one of the base cases and adding nonnegative multiple of. Starting at 8, we get 8, 11, 14, 17, 20,...; starting at 9, we get 9, 12, 15, 18, 21,...; starting at 10, we get 10, 1, 16, 19, 22,.... This covers all the integers 8. (It s possible to make a more rigorous argument by considering n mod for n 8.) 2. Consider the function f(x) = 5x + 4, where f : R R. Find the inverse function. If y = 5x + 4, solve for x y = 5x + 4, y 4 = 5x, 1 (y 4) = 5 x, (y 4) = x. 1 5

2 . Let a n be the sequence recursively defined by a 1 = 2, a 2 =, a n = a n 1 a n 2 for n. Compute the first five terms a 1,..., a 5. a 1 = 2, a 2 =, a = a 1 a 2 = 6, a 4 = a a 2 = 18, a 5 = a 4 a = List all the strings over X = {a, b, c} of length 2 or less. λ, a, b, c, aa, ab, ac, ba, bb, bc, ca, cb, cc. 5. Short answer. No partial credit. (a) A relation R on X = {1, 2, } has matrix digraph.. Draw its 1 2 (b) Write the relation in part (a) as a subset of X X. R = {(1, 1), (1, ), (2, 2), (2, )}. (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, 2,, 4, 5, 6} and the equivalence relation on Y given by xry if and only if x y is a multiple of. Write down the elements of the equivalence class [1]. [1] = {1, 4}, since 1R1 and 4R1, but 1 is not related to 2,,5,6.

3 (e) Write down the digraph of the equivalence relation from part (d) (f) Write down the matrix of the equivalence relation from part (d) (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 +. 2,. 2 doesn t divide and doesn t divide 2. (h) For the partial order from part (g), find an element k of Z + so that n(krn). Write down k divides n for all n Z +. 2 (i) For a i = i, compute a i. i=0 2 a i = a 0 + a 1 + a 2 = = = 1. i=0 (j) On X = {1, 2,, 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 = {(2, 1), (, 1), (, 2), (4, 1), (4, 2), (4, )}, while S = {(1, 1), (1, 2), (1, ), (1, 4), (2, 2), (2, 4), (, ), (4, 4)}. So R S =

4 and there are no elements in R S. (k) List the elements of R S as ordered pairs, where R, S are from part (o). R S = {(1, 1), (1, 2), (1, ), (1, 4), (2, 1), (2, 2), (2, 4), (, 1), (, 2), (, ), (4, 1), (4, 2), (4, ), (4, 4)}. 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 occurrence(s, n){ smallest = s 1 last = 1 for i = 2 to n if (s i smallest){ smallest = s i last = i } return last }

5 7. True/False. Circle T or F. No explanation needed. (a) T F Let f : R R be given by f(x) = x 5. Then f is one-to-one. T. If y = x 5, then x = 1 (y + 5). So for each y, there is only one x so that y = x 5. (b) T F f from (a) is onto. T. See (a) above. (c) T F For f from (a), f 1 (x) = x + 5. F. f 1 (x) = 1 (x + 5), as we compute in (a) above. (d) T F Consider A = {0, 1, 2, }. The function g : A A defined by g(x) = 2x mod 4 is a bijection. F. Compute g(0) = 0, g(1) = 2, g(2) = 0, and g() = 2. This shows g is not onto (since 1 and are not in the range), and also g is not one-to-one, since g(0) = g(2). (e) T F For A as in (d), the function h : A A defined by h(x) = x mod 4 is a bijection. T. Compute h(0) = 0, h(1) =, h(2) = 2, h() = 1. This is clearly one-to-one and onto, and thus is a bijection. (f) T F If V = {1, 2, } and W = {a, b, c, d}, then there exists a one-to-one function f : V W. T. For example, define f by f(1) = a, f(2) = b, f() = c. (g) T F For V, W as in (f), there exists an onto function g : V W. F. If f : V W, then the range {f(1), f(2), f()} can have at most elements. This cannot cover all of W, whose cardinality is 4. (h) 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 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 b((a = b b = a) a = b). This is obviously true.

6 (i) T F On Z +, consider the relation R defined by mrn if there is an integer k so that m = 2 k n. Then R is an equivalence relation. T. It s reflexive, since m = 2 0 m and so mrm for all m. It s symmetric since if mrn, then m = 2 k n and so n = 2 k m, which means that nrm. It s transitive since if mrn and nrp, then m = 2 k n and n = 2 l p, which implies m = 2 k (2 l p) = 2 k+l p and so mrp. (j) T F R from part (i) is a partial order. F. R is not antisymmetric, since for example 1R2 and 2R1, but 1 2. (k) T F The relation from problem 5(a) is reflexive. F. R. (l) T F The relation from problem 5(a) is symmetric. F. For example, 1R but R1. (m) T F The relation from problem 5(a) is antisymmetric. T. There is no pair of 1 s across the diagonal. (n) T F The relation from problem 5(a) 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. (o) T F ( If R is a relation ) from {1, 2} to {a, b, c} with matrix, and S is a relation from {a, b, c} to {1, 2} with matrix 1 1, then the relation R S has ma- 0 1 ( ) 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. (p) 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 a one-to-one correspondence from R to R 1. Therefore, the cardinalities must be the same. (q) T F 5! = 720. F. 5! = = 120.