Answer:
Follows are the explanation of the choices:
Explanation:
Following are the Pseudocode for selection sort:
for j = 0 to k-1 do:
SS = i
For l = i + 1 to k-1 do:
If X(l) < X(SS)
SS= l
End-If
End-For
T = X(j)
X(j) = X(SS)
X(SS) = T
End-For
Following are the description of Loop invariants:
The subarray A[1..j−1] includes the lowest of the j−1 components, ordered into a non-decreasing order, only at beginning of the iteration of its outer for loop.
A[min] is the least amount in subarray A[j.. l−1] only at beginning of the each loop-inner iterations.
Following are the explanation for third question:
Throughout the final step, two elements were left to evaluate their algorithm. Its smaller in A[k-1] would be placed as well as the larger in A[k]. One last is the large and medium component of its sequence because most and the last two components an outer loop invariant has been filtered by the previous version. When we do this n times, its end is a repetitive, one element-sorting phase.
Following is the description of choosing best-case and worst-case in run- time:
The body the if has never been activated whenever the best case time is the list is resolved. This number of transactions are especially in comparison also as a procedure, that will be (n-1)(((n+2)/2)+4).
A structure iterator at every point in the worst case that array is reversed, that doubles its sequence of iterations in the inner loop, that is:(n−1)(n+6) Since both of them take timeΘ(n2).
<span>Pinhole cameras were one of the most sophisticated devices of the period, it made tasks much easier it basically worked exactly like the human eye and is something just like tracing.
Hope it helps! :)</span>
Mr. Jones put 1.2 bags of potting soul around each tree. To solve this problem divide the number of bags by the number of plants.
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Where is the attached file?
Answer:
Option d) B is 1.33 times faster than A
Given:
Clock time, 
No. of cycles per instructions, 
Solution:
Let I be the no. of instructions for the program.
CPU clock cycle, 
= 2.0 I
CPU clock cycle, 
= 1.0 I
Now,
CPU time for each can be calculated as:
CPU time, T = 
Thus B is faster than A
Now,
Performance of B is 1.33 times that of A
