The travel time function of an algorithm has the form: f(n) = 3n^2 + 4n + 8 Prove that Big Oh is n^2 which is denoted by f(n) = O(n^2);
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![The travel time function of an algorithm has the form: f(n) = 3n^2 + 4n + 8
Prove that Big Oh is n^2 which is denoted by f(n) = O(n^2);](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ef2496e-5225-493d-ac65-b5bbb9faabe8%2F55e5067b-8d9f-44d3-8cae-b9d146a83c9b%2Fl55vvcj_processed.png&w=3840&q=75)
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- A computer science student designed two candidate algorithms for a problem while working on his part-time job The time complexity of these two algorithms are T,(n) = 3 n logn and T2(n) = n6/5 a) Which algorithm is better? Why? b) If we run both algorithms at the same time with an input size of 105, which algorithm produces results faster than the other one? Why?An algorithm is made up of two modules M1 and M2. If order of M1 is f(n) and M2 is g(n) then the order of algorithm is 1. max (f(n), g(n)) 2. min (f(n), g(n)) 3. f(n) + g(n) 4. f(n) x g(n) XQuestion: If f(n) is the number of executing steps of an algorithm, find the values of Big O, Theta and Omega of the following functions: 1. F(n)=n^2/2 - 3n 2. F(n)=n−n^2+7n^5 3. F(n)=sqrt(n) 4. F(n)=n.log(n) Note: g(n) can be any value
- AlgorithmThis function uses a curious mix of iteration and recursion: function F(n) if n < 1 t <- O return 1 for i <- 0 to n for j <- i to n t<- t+j return t + F(n-1) The number of basic operations (additions and subtractions) performed is: ○ O(n) Ⓒ (n²) (n² log n) Ⓒ (n³) Ө Ⓒ (n4)2) A computer science student designed two candidate algorithms for a problem while working on his part-time job The time complexity of these two algorithms are T1(n) = 3 n logn and T2(n) = nº/5 . a) Which algorithm is better? Why? b) If we run both algorithms at the same time with an input size of 10°, which algorithm produces results faster than the other one? Why?
- Simple Computer Science 3 question: Determining the big-O runtimes of a loop? void X(int n){ for(int i=0; i<n; i++) for(int j=0; j<n; j++) for(int k=0; k<n; k++) for(int w=1; w<n; w=2*w) time++;} My answer: n x logn x log n = O(n log n^2) I am not sure if that is correct though.Function f grows no slower than function g. O f (n) = O(g(n) O f (n) = 2(g(n)) O f (n) = o(g(n)) O f(n) = O(g(n))Computer Science Use the non-restoring division algorithm to compute the division of a dividend X = 14 = (1110) and a divisor D = 6 = (0110).
- Computer science2) A computer science student designed two candidate algorithms for a problem while working on his part-time job The time complexity of these two algorithms are T1(n) = 3 n log n and T2(n) = nº/5 . a) Which algorithm is better? Why? b) If we run both algorithms at the same time with an input size of 105, which algorithm produces results faster than the other one? Why?Find the time complexity of the given loop.
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