Download Exam 2 CMSC 203 Spring 2011: Mathematics Problems and Solutions and more Exams Discrete Structures and Graph Theory in PDF only on Docsity!
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- (16 points) Circle T if the corresponding statement is True or F if it is False.
T F GCD(10, 0) = 0.
T F If a prime divides the square of a Natural number, then it divides the number.
T F If a ≡ b MOD 5, then 5 divides both a and b.
T F Recursive algorithms generally use memory more efficiently than their equivalent Iterative version.
T F 1 + 2 + 3 + 4 + ... + 1000 = 500(1001).
T F Algorithms with O( n^2 ) are less efficient than those with O(2 n ).
T F The two Principles of Mathematical Induction are logically equivalent.
T F GCD(1000, 678) = GCD(678, 322).
- (8 points) Find the GCD and LCM of a = 2^45873111176 and b = 2^354112132172192
ab = 24+3 5 8+4 7 3+0 11 1+2 13 0+2 17 6+2 19 0+ so GCD( a,b ) = 2^35470111130172190 = 2^354111172 and LCM( a,b ) = 2^45873112132176192.
- (8 points) List out the search intervals of the Binary Search algorithm to find 6 in the list: 3 4 6 9 13 18 21 34 55 72 83 85 92 104 111 133
Pass 1: {3, 4, 6, 9, 13, 18, 21, 34} and {55, 72, 83, 85, 92, 104, 111, 133}
Pass 2: {3, 4, 6, 9} and {13, 18, 21, 34}
Pass 3: {3, 4} and {6, 9}
Pass 4: {6} and {9}
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- (10 points) Find a numeric expression for.
- (12 points) Trace the Division Algorithm below to find (52 MOD 6).
PROCEDURE MOD(A,B: integers) WHILE (A > B) A = A − B ENDWHILE OUTPUT (A)
Step 0 1 2 3 4 5 6 7 8 A 52 46 40 34 28 22 16 10 4 B 6 6 6 6 6 6 6 6 6 (A > B)? 1 1 1 1 1 1 1 1 0 OUTPUT 4.
- (8 points) Give a Recursive Definition for the set S = { n ∈ N | n ≡ 3 MOD 7}:
Basis: 3 ∈ S
Induction: If n ∈ S, then ( n + 7) ∈ S.
- (8 points) Show n^19 is the Big-Oh of the algorithm with complexity:
(12 n^4 + 3 n^3 log^3 n )(3 n^7 + 4 n^3 )( n^6 + 5 n^2 + 4).
(12 n^4 + 3 n^3 log^3 n )(3 n^7 + 4 n^3 )( n^6 + 5 n^2 + 4) < (12 n^4 + 3 n^3 n^3 )(3 n^7 + 4 n^7 )( n^6 + 5 n^6 + 4 n^6 )
< (12 n^6 + 3 n^6 )(7 n^7 )(10 n^6 ) = (15 n^6 )(7 n^7 )(10 n^6 ) = 1050 n^19 which is O( n^19 ).
4 i +5 7 ( i ) i = 0
10
4 i +5 7 ( i ) i = 0
10
∑ i
i = 0
10
i
i = 0
10
10 11( ) 2
----------------- 57
10 + (^1) Ð 1 7 Ð 1
- ------------------------- 220 5 7 ( 11 Ð 1 ) 6 +--------------------------
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- (10 points) Prove ONE of the TWO Theorems below:
Theorem 1: For all Integers, n, if n is odd, then n^2 ≡ 1 MOD 8. (Hint: If an Integer is odd, then its successor is even.)
Theorem 2: Between any two distinct Real Numbers is another Real Number.
Theorem 1:
Proof: Let n be an odd Integer, so n = 2 k + 1, for some Integer k. We want to show
n^2 ≡ 1 MOD 8; that is ( n^2 − 1) = 8 p , for some Integer p.
Now, ( n^2 − 1) = (2 k + 1)^2 − 1 = (4 k^2 + 4 k + 1) − 1 = 4 k^2 + 4 k = 4 k ( k + 1). However, since k is an odd Integer, we see that ( k + 1), the successor of k , is even. This lets us assert that ( k + 1) = 2 m , for some Integer m.
Combining all this, we see that ( n^2 − 1) = 4 k ( k + 1) = 4 k (2 m ) = 8 km. Moreover, since k and m are
Integers, we conclude that p = km is an Integer, thus ( n^2 − 1) = 8 p , for some Integer p.
Therefore n^2 ≡ 1 MOD 8 for any odd Integer n. QED
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Theorem 2:
Proof: Let X and Y be distince Real Numbers and, without loss of generality, assume X < Y.
Now, X = (2X) / 2 = (X + X) / 2 < (X + Y) / 2 < (Y + Y) / 2 = (2Y) / 2 = Y. Moreover, since X and Y are Real, we see that (X + Y) is Real, hence [ (X + Y) / 2 ] is Real. Since X < (X + Y) / 2 < Y, we conclude, therefore, that there exists a Real Number between distinct Real Numbers. QED
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- (10 points) Prove ONE of the TWO Theorems below by Contradiction or Contraposition.
Theorem 1: The set of Prime Numbers is infinite.
Theorem 2: For all Integers, n > 2, if n is prime, then n ≡ 1 MOD 2.
Theorem 1: Proof: (Contradiction) Assume the set of Prime Numbers is finite. Denote the finite set of the Primes as { p 1 , p 2 , p 3 , ... , pn } for some Natural Number n.
Now, construct the Natural Number, M = [ p 1 ( p 2 )( p 3 )( ... )( pn )] + 1. Since M is a Natural, it has a
prime factor, but since any prime number is also a factor of the product [ p 1 ( p 2 )( p 3 )( ... )( pn )], we
conclude that this prime factor must also divide 1. However this is a contradiction since no primes divide 1.
Therefore, the set of Primes is infinite. QED
Theorem 2: Proof: (Contraposition) We shall show for any Integer n > 2, if n ≡ 0 MOD 2, then n is composite.
Now, let n ≡ 0 MOD 2, for any Integer n > 2. This means that 2 divides ( n − 0) = n , hence n /2 is an Integer. Moreover, since n > 2, we see that n /2 > 1, thus n /2 = m for some Integer m. Consequently, n = 2 m with m > 1, hence n is composite.
Therefore, for all Integers, n > 2, if n is prime, then n ≡ 1 MOD 2. QED
