big-theta

问题 I am stuck at this problem. I think it is equivalent to show 2m choose m is big theta of 4 to the power n, but still find difficult to prove it. Thanks of @LutzL's suggestion. I thought of stirling's approximation before. 回答1: The O -part should be easy. Choosing exactly n /2 elements out of n is a special case of choosing arbitrary combinations out of n elements, i.e. deciding for each of these n elements whether to choose it or not. The Ω -part is harder. In fact, plotting 4n / binomial(2 n

问题 I am stuck at this problem. I think it is equivalent to show 2m choose m is big theta of 4 to the power n, but still find difficult to prove it. Thanks of @LutzL's suggestion. I thought of stirling's approximation before. 回答1: The O -part should be easy. Choosing exactly n /2 elements out of n is a special case of choosing arbitrary combinations out of n elements, i.e. deciding for each of these n elements whether to choose it or not. The Ω -part is harder. In fact, plotting 4n / binomial(2 n

问题 In What exactly does big Ө notation represent?, the most upvoted answer contains the following statement: For example, merge sort worst case is both O(n*log(n)) and Omega(n*log(n)) - and thus is also Ө(n*log(n)) , but it is also O(n^2) , since n^2 is asymptotically "bigger" than it. However, it is not Ө(n^2) , Since the algorithm is not Omega(n^2) . I have two questions: How do you determine that the worst case is O(n*log(n)) and Omega(n*log(n)) . In "Introduction to Algorithms" you determine

问题 Let's suppose that a recursive formula is a big-o(n^2), and at the same time a big-omega(n^2). Does this imply that the recursion is a big-Theta(n^2)? 回答1: To make the long story short: the answer is Yes, it does . See proof below. Though everybody have heard about big-o notation lets recall what exactly does these notations mean with a help of Introduction to Algorithms. For a general case it is said Ο(g(n)) , Ω(g(n)) , Θ(g(n)) , but we will consider yours. Ο(n 2 ) Ο(n 2 ) notation defines a

问题 Let's suppose that a recursive formula is a big-o(n^2), and at the same time a big-omega(n^2). Does this imply that the recursion is a big-Theta(n^2)? 回答1: To make the long story short: the answer is Yes, it does . See proof below. Though everybody have heard about big-o notation lets recall what exactly does these notations mean with a help of Introduction to Algorithms. For a general case it is said Ο(g(n)) , Ω(g(n)) , Θ(g(n)) , but we will consider yours. Ο(n 2 ) Ο(n 2 ) notation defines a

问题 How to decide on expressing time complexity of algorithm ? Should we choose to express time complexity in terms of O(n) or theta(n) ? Because a function f(n) can be expressed as either Big-Oh(g(n)) or theta (g(n)) . When do we choose big-oh over theta ? 回答1: Use Big Theta notation when you want to also specify a lower bound. f(n) = O(g(n)) says that f is bounded above by g , whereas f(n) = Theta(g(n)) says that f is bounded both above and below by g . In other words, there are constants k1

问题 I'm trying to analyze the worst case order of growth as a function of N for this algorithm: for (int i = N*N; i > 1; i = i/2) for (int j = 0; j < i; j++) { total++; } What I'm trying is to analyze how many times the line total++ will run by looking at the inner and outer loops. The inner loop should run (N^2)/2 times. The outer loop I don't know. Could anyone point me in the right direction? 回答1: The statement total++; shall run following number of times: = N^2 + N^2 / 2 + N^2 / 4 ... N^2 / 2

问题 I have the homework question: Find a theta notation for the number of times the statement x = x + 1 is executed. (10 points). i = n while (i >= 1) { for j = 1 to n { x = x + 1 } i = i/2 } This is what I have done: Ok first let’s make it easier. We will fist find the order of growth of: while (i >= 1) { x = x + 1 i = i/2 } that has order of growth O(log(n)) actually log base 2 the other inner for loop will execute n times therefore the algorithm should be of order: O(log(n)*n) The part where I

问题 Is 2^n = Θ(4^n)? I'm pretty sure that 2^n is not in Ω(4^n) thus not in Θ(4^n), but my university tutor says it is. This confused me a lot and I couldn't find a clear answer per Google. 回答1: 2^n is NOT big-theta (Θ) of 4^n , this is because 2^n is NOT big-omega (Ω) of 4^n . By definition, we have f(x) = Θ(g(x)) if and only if f(x) = O(g(x)) and f(x) = Ω(g(x)) . Claim 2^n is not Ω(4^n) Proof Suppose 2^n = Ω(4^n) , then by definition of big-omega there exists constants c > 0 and n0 such that: 2

问题 For the following code fragment, what is the order of growth in terms of N? int sum = 0; for (int i = 1; i <= N; i = i*2) for (int j = 1; j <= N; j = j*2) for (int k = 1; k <= i; k++) sum++; I have figured that there is lgN term, but I am stuck on evaluating this part : lgN(1 + 4 + 8 + 16 + ....). What will the last term of the sequence be? I need the last term to calculate the sum. 回答1: You have a geometric progression in your outer loops, so there is a closed form for the sum of which you