hw2solutions

F n = n 2 on all other real nonnegative arguments n

Info iconThis preview shows pages 4–6. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: f ( n ) = n 2 , on all other real nonnegative arguments n we have f ( n ) = n 2 . Note Most solutions assumed that f is only defined on natural numbers and we accepted such solutions, of course. A. Prove that f ( n ) is O ( n 2 ). Answer Let N = 0 . 5 and c = 1. We will show that ∀ n ≥ N we have f ( n ) ≤ cn 2 . 4 Indeed, fix an arbitrary nonnegative real number n that is bigger than N . Either n is an odd natural number and then f ( n ) = n 2 ≤ n 2 = cn 2 . Otherwise f ( n ) = n/ 2 ≤ n 2 = cn 2 because n ≥ N = 1 / 2. Note Some of the submitted solutions proved first that n/ 2 is O ( n 2 ) and then that n 2 is O ( n 2 ) and then jumped to the conclusion that f ( n ) is O ( n 2 ). This is a big jump that needs proof! Some people stated that they used the same N,c in both proofs of n/ 2 is O ( n 2 ) and n 2 is O ( n 2 ). That’s a key observation in bridging the jump we are talking about so we have some partial credit for that. But it’s still not a complete proof of the required statement. Just for fun try to prove the following. Let f,g,h be three functions such that f is O ( h ) and g is O ( h ). Let A be an infinite set of nonnegative real numbers and define a function k such that k ( n ) = f ( n ) when n ∈ A and k ( n ) = g ( n ) otherwise. Prove that k is O ( h ). B. Prove that f ( n ) is not O ( n ) Answer We must show that, for any c,N > 0, there exists an n ≥ N such that f ( n ) > cn . Let c,N > 0 arbitrary. Note that if n > c then n 2 > cn . So it suffices to use an odd n , so that f ( n ) = n 2 such that n > c and n ≥ N . Such n ’s exist, for example n = 2 d max( c,N ) e + 1 Extra Credit 2: 15 points Analyze the following code fragment and give a Big-Oh characterization of the running time of doMoreWork . public static void doMoreWork (int barr) { for (int i=1; i < barr.length; i *= 2) { for (int j=0; j < i; j++) { barr[j]=j; } foo(barr); } } public static void foo(int a){ for (int x : a) System.out.print("foo!"); } } Answer 5 Let us first analyze foo . Let m = a . length be the size of its input. The for loop always iterates m times. The body of the loop takes constant time. Hence foo runs in time O ( m ). Now let n = barr . length () be the size of the input of doMoreWork . For simplicity we assume that n is a power of 2 (just like we did in the lecture notes when we analyzed the running time of binary search). At every iteration of the outer loop in doMoreWork i doubles so after k iterations i is 2 k . The iterations stop when 2 k = n so the outer loop iterates log n times. Now we analyze the running time of the body of the outer loop. It has two components: the inner loop and the statement foo(barr); . We have analyzed the latter and it runs in time O ( n ). The body of the inner loop runs in time O (1) and for each i the inner loop iterates i times. So the inner loop runs in time O ( i ). But i < n so it runs in time O ( n ). From both components the body of the outer loop runs in time O ( n + n ) which is O ( n ). We conclude that the method doMoreWork runs in time O ( n log n ) 6...
View Full Document

{[ snackBarMessage ]}

Page4 / 6

f n = n 2 on all other real nonnegative arguments n we have...

This preview shows document pages 4 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online