Answer:
Explanation:
For the first iteration of i for loop 1 to n, the j for loop will run from 2 to n times. i.e. n-1 times.
For the second iteration of i for loop, the j for loop will run from 3 to n times. i.e. n-2 times.
From the third to the last iteration of i for loop, the j for loop will run n-1 to n times. i.e. 2 times.
From the second to the last iteration of i for loop, the j for loop will run from n to n times. i.e. 1 time.
For the last iteration of i for loop, the j for loop will run 0 times because i+1 >n.
Hence the equation looks like below:
1 + 2 + 3 + ...... + (n-2) + (n-1) = n(n-1)/2
So the number of total iterations is n(n-1)/2.
There are two operations per loop, i.e. Comparison and Multiplication, so the iteration is 2 * n(n-1)/2 = n ^2 - n
So f(n) = n ^ 2 - n
f(n) <= n ^ 2 for n > 1
Hence, The algorithm is O(n^2) with C = 1 and k = 1.
Answer:
The value of k is 4
Explanation:
Solution
Given that:
k = integer
Input size = 500
The algorithm takes a run of = 16 seconds
Input size = 750
The algorithm takes a run of = 81 seconds
Now,
We have to determine the value of k
The equation is shown below:
(500)^k /16 = (750) ^k /81
Thus
(750/500)^ k = 81/16
= (3/2)^k
=(3/2)^ 4
k is = 4
A <u>sequence</u> number is a 32-bit number that's used to indicate where you are in a sequence of TCP segments.
<h3>What is TCP?</h3>
TCP is an acronym for Transmission Control Protocol and it is an essential and important standard protocol of the Internet protocol network.
In Computer technology, TCP is an essential part of the transmission control protocol and internet protocol (TCP/IP) network which has a wide range of applications in the following areas:
In TCP segments, a <u>sequence</u> number is a 32-bit number that's typically used to indicate where an end user is in a sequence.
Read more on TCP here: brainly.com/question/17387945
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