Suppose we are comparing implementations of insertion sort...

Name: suppose we are comparing implementations of insertion sort and merge sort on the same machine for inputs of size n insertion sort runs in 05n2 steps while merge sort runs in 256nlogn steps f 01846
Uploaded: 2023-09-22T03:00:52-08:00
Duration: 5 min 22 s
Channel: Akash M
Description: suppose we are comparing implementations of insertion sort and merge sort on the same machine for inputs of size n insertion sort runs in 05n2 steps while merge sort runs in 256nlogn steps f 01846

Suppose we are comparing implementations of insertion sort and merge sort on the same machine. For inputs of size n, insertion sort runs in 0.5n2 steps, while merge sort runs in 256nlogn steps. For which values of n does insertion sort take more steps than merge sort?

Transcript

00:01 This is a c+ + in this question we have to implement a c+ + code to compare the insertion sort and the merge sort for worst case input so code for this is hashtag include iostream, hashtag include vector, hashtag include ctype, hashtag include algorithm using namespace std then we have to put function to perform the insertion sort the code for this is void insertion sort vector int and are for int i is equal to 1 i less than are size i plus plus so int key is equal to are i int j is equal to i minus 1 while j greater than equal to 0 and and are j greater than key so are j plus 1 is equal to are j j minus minus and are j plus is equal to key then we have to make a function to perform merge sort so void merge sort vector int and are if are size less than equal to 1 return int middle is equal to are size divided by 2 vector int left are begin are begin plus middle and vector int right are begin plus middle are end then merge sort left merge sort right merge left begin and left end right end right end are begin now this now coming to the main program so int main int n cout enter the value of n c in n vector int are n then we have to initialize the array which worst case input reverse order you using for loop for it i is equal to 0 i less than n i plus plus array i is equal to n minus 1 clock t start end then measure the time for insertion sort so a start is equal to clock insertion sort are and is equal to clock double insertion time is equal to double and minus start divided by clocks per second then reset the array so using for loop for i is equal int is it int i is equal to 0 i less than n i plus plus array i in n minus 1 after that, we have to measure the time for the merge sort so code for this is start is equal to clock merge sort are end clock and double merge time is equal to double time minus end divided by clock clocks per second now c out insertion sort time for n n insertion time and second slash n now in next line, we have to write c out much sort time for n n merge time second slash n then return 0 basically this code achieve allows you to input the value of n and measure the running time of insertion sort and merge sort for the worst case input reverse order...

Question

Suppose we are comparing implementations of insertion sort and merge sort on the same machine. For inputs of size n, insertion sort runs in 0.5n2 steps, while merge sort runs in 256nlogn steps. For which values of n does insertion sort take more steps than merge sort?

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