Consider the following pseudocode for a sorting algorithm, for 0 < α < 1 and n > 1.
badsort( - 1])
if (n = 2) and (a[0] > a[1])
swap a[0] and a[1]
else if (n > 2)
m = [α . n]
badsort( – 1])
badsort(a[n – – 1])
badsort( – 1])
1. implement badstoogesort from pseudocode to sort an array of integers. the value of α should be an input parameter to your program.
2. implement the algorithm in c/c++. your program should be able to read inputs from a file called "data. txt", where the first value of each line is the number of integers that need to be sorted, followed by the integers. the output will be written to a file called "bad. out". then, modify the code to collect running time data. call the new timing program badstoogesort.
3. instead of reading arrays from the file data. txt and sorting, you will now generate arrays of size n containing random integer values from 0 to 10,000 to sort. use the system clock to record the running times of each algorithm for n = 5000, 1, 15000, 20,000, for two values of α = 2/3 and α = 3/4. you may need to modify the values of n if an algorithm runs too fast or too slow.

Answervirtual memory

Middleware

The answer is c. 36======analyzing the for-loop, we see that the control variable, k, starts at 0 and increments by after every loop iteration. the condition  tells us that the loop will stop when k is above 7; in this case, when k is 8.0 1 2 3 4 5 6 7 are the values for k that will be used. each number is for each iteration; therefore, the for-loop will have 8 iterations in total.the code inside the loop itself goes like this: 1st iteration (k=0)starting with sum=0, val=1    means that add  to  .since sum is 0 and val is 1, 0 + 1 = 1. the new value of sum is 1.    means to increment  by 1.since val is 1, the new value of val of 2.end of iteration with  k=0; k is increased by one to become  k=12nd iteration (k=1)starting with sum=1, val=2    ======let us establish a pattern. this is moreso logic than programming: we can  rewrite so that we have the elements from the previous iteration. the 1 came from 0 + 1 before. so we can write it as      .and if we did the third iteration, by the end, sum would look like      notice how the number of plus signs is equal to the iteration you are on (which is just k+1), and what is being added is increasing by one each time.  for example, the third iteration (k=2)  ends up with  with 3 plus signs.the second iteration (k=1) ends up with  with two plus signs.the first iteration (k=0) ends up with  with one plus sign.this is a pattern. the nth iteration will have n plus signs.since we established before that there are 8 iterations of this loop in total, the value of sum at k=7 (the 8th and last iteration) will have sum with 8 plus signs.    so the value of sum at the end of the loop is 36.

Is this a joke i can’t tell