Computer Science Section 3.2

Transcription

1 Computer Science 180 Solutions for Recommended Exercises Section 3.. (a) If x > 4, then x > 16 x x. Also, if x > 17, then x > 17x x 17x 17x x. Seventeen is greater than four, so if x > 17, we have 0 < 17x x + x x + x = x 17x + 11 x. So, we let C = and k = 17. Accordingly, 17x + 11 is O ( x ). (b) If x > 3, then x > 104 x x x x + x 0 < x x x x. So, C = and k = 3 are our witnesses to the fact that x is O ( x ). (c) Using the Principle of Mathematical Induction (discussed in this course), we can prove that n < n for all positive integers n. Using calculus, we can prove that x < x for all real x. We can also use calculus to prove that logx (we assume that the base of this logarithm is ) is an increasing function, which essentially means that it preserves order. For now, we will assume that these results are true. Accordingly, for all x 1, 1 x < x log1 logx < log x 0 logx < x 0 xlogx < x xlogx < x xlogx 1 x. So, xlogx is O ( x ), with our witnesses being C = k = 1. 1

2 (d) Assume that x 4 / is O ( x ). Then, there exist real constants C and k such that for all x > k, x 4 / C x x C. Clearly, C must be positive. Now, let x 0 be the maximum of k +1 and C + 1. Then, x 0 > k, so by the statement above, x 0 C. On the other hand, x 0 C x0 C x0 > C 0 x 0 > C, which is a contradiction. Thus, x 4 / is not O ( x ). (e) Recall that x < x for all real x. Like in (e), we assume that x is O ( x ), so that we can find real constants C and k such that for all x > k, x C x x Cx. Of course, C must be positive. Now, let x 0 be the maximum of k + 1 and 7(C + 1). Then, x 0 > k, so by the statement above, x 0 Cx0. However, x 0 = ( x 0/3 ) 3 > (x0 /3) 3 = x 0 7 x 0 7(C + 1) 7 x 0 = (C + 1)x 0 > Cx 0 i.e. x 0 > Cx 0, which is a contradiction. Thus, x is not O ( x ). (f) x, known as the floor function, returns the largest integer n that is less than or equal to x i.e. n is the unique integer satisfying x 1 < n x n x x x. x, known as the ceiling function, returns the smallest integer n that is greater than or equal to x i.e. n is the unique integer satisfying Then, for all x 1, x n < x + 1 n < x + 1 x < x x x x (x + 1) = x + x x + x x = x, so x x x. Ergo, x x is O ( x ), with witnesses C = and k = 1.

3 6. For all x 1, 0 < x3 + x x + 1 so x 3 + x x + 1 < x3 + x x = x3 x + x x = 1 x x + x < x, x. Choosing C = and k = 1, we obtain the desired result. 8. (a) For all x 1, 0 < x + x 3 logx < x + x 3 (x) x x + x 4 = 3x 4 x + x 3 logx 3 x 4, so n = 4. (b) For all x 1, 0 < 3x 5 + (logx) 4 3x 5 + (x) 4 3x 5 + x 4 x = 4x 5 3x 5 + (logx) 4 4 x 5, so n = 5. (c) For all x 1, 0 < x4 + x + 1 x 3 x4 + x x + 1 x x x4 + x + 1 x x, so n = 1. = 3x4 x < 3x4 x 3 = 3x (d) For all x 1, 0 < x3 + 5logx x 4 < x3 + 5(x) + 1 x x3 + 5logx x , so n = 0. < x3 + 5x x 4 = x3 x 4 + 5x x 4 = 1 x + 5 x = x 1 0 < 1 x 0 < 1 x 3 x x 3 0 < x 3 x 4, so for all x > 1, x 3 1 x 4. Thus, x 3 is O ( x 4) with witnesses C = k = 1. Now assume that x 4 is O ( x 3), so that we can find real constants C and k such that for all x > k, x 4 C x 3 x C. Picking any x that is larger than both k and C yields a contradiction, so x 4 is not O ( x 3). 3

4 1. For all x 1, 0 logx < x 0 x logx x x 0 xlogx < x xlogx 1 x so xlogx is O ( x ) with witnesses C = k = 1. Now assume that x is O(xlogx), so that we can find real constants C and k such that for all x > k, x C xlogx x C logx. Again, C must be positive. Next, let x 0 be the maximum of k + 1 and 4C. Then, x 0 > k, so by our assumption, x 0 C logx 0. However, x 0 = > logx 0 logx 0 = = logx 0 4 (logx 0)/ logx 0 > log 4C 4 logx 0 = 4C 4 logx 0 = C logx 0 i.e. x 0 > C logx 0, which is a contradiction. Therefore, x is not O(xlogx). 14. (a) Assume that x 3 is O ( x ), so that we can find real constants C and k such that for all x > k, x 3 C x x C. Choosing any x that is greater than both k and C produces a contradiction, so the answer is no. (b) For all real x, x 3 1 x 3, so the answer is yes. (c) For all real x, x 3 x + x 3, and for all nonnegative x, x 3 x + x 3 = 1x + x 3, so the answer is yes. (d) For all x 1, 0 < x 3 x 4 x + x 4 x 3 1 x + x 4, so the answer is yes. (e) 3 x = log3 x = xlog3 = (xlog3)/3 3 xlog3 > 3 x 3 (f) For all real x, ( ) x, so the answer is yes. log3 x 3 1 x 3, so the answer is yes. 3 = ( ) log3 3 x 3 3 4

5 ( 0. (a) We use Theorems and 3: n 3 + n logn ) (logn + 1) + (17logn + 19) ( n 3 + ) = O (( n 3 + n logn ) (logn) + (17logn + 19) ( n 3 + )) = O (( n 3 + n logn ) (logn) + (17logn) ( n 3 + )) = O (( n 3 + n logn ) (logn) + (17logn) ( n 3)) = O ( n (n + logn)(logn) + (17logn) ( n 3)) = O ( n (n)(logn) + (17logn) ( n 3)) = O ( n 3 logn + 17n 3 logn ) = O ( n 3 logn ). (b) ( n + n )( n n) = O ( ( n ) ( n n)) = O(( n )(3 n )) = O(( 3) n ) = O(6 n ). (c) (n n + n n + 5 n )(n! + 5 n ) = O((n n + n n + 5 n )(n!)) = O((n n )(n!)) = O(n! n n ). 4. (a) We first show that 3x + 7 is O(x): for all x 7, 0 < 3x + 7 3x + x = 4x 3x x, so our witnesses are C = 4 and k = 7. Next, we show that 3x + 7 is Ω(x): for all x 1, 3x + 7 > 3x > 0 3x x, so our witnesses are C = 3 and k = 1. Since 3x + 7 is both O(x) and Ω(x), it is Θ(x), as required. (b) x + x 7 = ( x ) + (x 7) x + x 7 (Triangle Inequality) = x + (x) + ( 7) x + x + 7 (Triangle Inequality again) = x + x + 7 = x + x + 7 (provided that x is nonnegative) 5 (continued)

6 (continued) x + x + 7 (provided that x 1) x + x + x (provided that x 3) = 4x = 4 x. Accordingly, x + x 7 is O ( x ) with witnesses C = 4 and k = 3. Next, x + x 7 = x + (x 7) x (provided that x 7) 0 x + x 7 x, so x + x 7 is Ω ( x ) with witnesses C = and k = 7. Consequently, x + x 7 is Θ ( x ), as required. (c) x + 1 is the unique integer n satisfying ( x + 1 ) 1 < n x + 1 x 1 < x + 1 x + 1. Then, for all x 1, 0 < 1 x < 1 x + 1 (x 1) = x 1 < x + 1 x x, so x + 1 is Ω(x) with witnesses C = 1/ and k = 1. Also, for all x 1, 0 < x + 1 x + 1 < x + x = x x + 1 x, so x + 1 is O(x) with witnesses C = and k = 1. Ergo, x + 1 is Θ(x), as required. (d) For all x, x = x < x + 1 < x + x = x = ( ) x < (x)x = x. In other words, for all x, 1 < x < ( x + 1 ) 1/ < x log1 < logx < log ( x + 1 ) 1/ < logx 0 < logx < 1 log( x + 1 ) < logx (continued) 6

7 (continued) 0 < logx < log ( x + 1 ) < 4logx logx log ( x + 1 ) 4 logx. To show that log ( x + 1 ) is O(logx), we choose C = 4 and k =, and to show that it is Ω(logx), we make C = and k = ; both of these show that it is Θ(logx), as required. (e) For all positive x, x = log x log 10 x = log 10 log x log 10 x = (log x)log 10 log 10 x = (log 10 )log x log 10 x = (log 10 ) log x. Note that if a = b, then a b and a b are both true! Thus, log 10 x is Θ(log x) with witnesses C = log 10 and k = To prove this biconditional statement, we must prove each implication. First, assume that f (x) is O(g(x)). Then, there exist real constants C and k such that for all x > k, f (x) C g(x). Clearly, C must be posi...waitaminute! It could be zero! Precisely, C must be nonnegative. Say that C = 0. Then, for all real x, f (x) = 0, so g(x) 0 g(x) f (x) g(x) 1 f (x). Hence, g(x) is Ω( f (x)) with witnesses C = k = 1. Now say that C > 0. Then, for all x > k, f (x) C g(x) C g(x) f (x) g(x) 1 C f (x). Thus, g(x) is Ω( f (x)) with witnesses C = 1 C and k = k + 1 (this ensures that k > 0 and k > k). Either way, g(x) is Ω( f (x)). Next, assume that g(x) is Ω( f (x)). Then, there are positive numbers C and k such that for all x > k, g(x) C f (x) C f (x) g(x) f (x) 1 C g(x). Ergo, f (x) is O( f (x)) with witnesses C = 1 C and k = k. 7