Question

PART B- The Efficiency of Algorithms, 10 points 1. Show how you count the number of operations (not only . Consider the following two loops, 4 points: basic operations) required by the algorithm, 4 points: //Loop λ sumList (aList, n) ( thǐssum = 0; for (j = 1; j <= 10000; j++) sum - sum + j; lastSum = 5000; for (int i = 0; i <= n; i++) { // Loop B for (j = 1; j <= n; j++) sumsum + j return thisSum lastSum; Although Loop A is O(n) and Loop B is O(n2), Loop B can be faster than Loop A for small values of n. Design and implement an experiment to find a value of n for which Loop B is faster.

Answer 1

1.

sumList(aList, n) {

thisSum = 0; //runs O(1) time

lastSum = 5000; //runs O(1) time

for(int i=0; i<=n; i++) { //runs O(n+1) times

thisSum = thisSum + aList[i] * 2; //runs O(n+1) times

}

return thisSum >= lastSum; //runs O(1) time

}

In total => O(1 + 1 + (n+1) + (n+1) + 1) time

Remove the constant times, which gives the overall time complexity of O(n) time complexity.

2.

Loop A runs n*10000 times. If n = 100, then it will execute 100*10000 times, which in total 1000000 times.

On the other hand, Loop B runs n*n times. If n = 100, then it will execute 100*100 times, which in total 10000 times.

As you can see, Loop B is faster than Loop A if and only if, n is less than 10000. If n is greater than 10000 then loop A runs faster.