What is the count for the instruction CountMe as a function of n for the fragment below?
Line 1: j = 1
Line 2: while (j<= n/2) {
Line 3: i = 1
Line 4: while (i<=j) {
Line 5: CountMe
Line 6: i++
Line 7: }
Line 8: j++
Line 9: }
Well, let's count it through a clownish lens! 🎪
First, we start with Line 1, where j is assigned to 1. 🎈
So, that's one CountMe already!
Next, we move to Line 2, the while loop where j is checked against n/2. Let's say the loop runs m times. 🎢
Inside the while loop, we have Line 3 where i is assigned to 1. That's another CountMe! 🤡
Then, we encounter Line 4, another while loop, where i is checked against j. Let's suppose this inner loop runs k times.
Inside this inner loop, there's Line 5 calling for a CountMe. Get ready for some laughter! 🤹♀️
After Line 6, we increment i. And we repeat Lines 5 and 6 until i exceeds j.
Once that inner loop is finished, we move to Line 8, where we increment j. 🎪
Now, we go back to Line 2, where we check if j is still less than or equal to n/2. If it is, we go through the whole process again.
So, in total, we'll have m iterations of the outer loop. And for each of those m iterations, we'll have k iterations of the inner loop.
Therefore, the count for CountMe as a function of n is m * k. 🎉
Now, the exact values for m and k depend on the specific value of n, which is not mentioned here. But boy, will those clowns keep you entertained with all those CountMes! 🎪🤹♂️🤡
The count for the instruction "CountMe" as a function of n can be determined by analyzing the nested loops in the given fragment.
The outer loop (Lines 2-9) is executed n/2 times, since j starts at 1 and increments by 1 until it is greater than n/2.
The inner loop (Lines 4-7) is executed j times for each iteration of the outer loop. In the first iteration of the outer loop, j = 1, in the second iteration j = 2, and so on, until j reaches n/2.
Therefore, the total count for the instruction "CountMe" can be calculated by summing up the counts of the inner loop for each iteration of the outer loop.
The count for the inner loop can be calculated as the sum of integers from 1 to j using the formula n(n+1)/2.
So, the count for the instruction "CountMe" is given by the sum of the counts for the inner loop for each iteration of the outer loop:
Count = 1 + 2 + 3 + ... + n/2
This is an arithmetic sum and can be calculated using the formula n(n+1)/2.
Therefore, the count for the instruction "CountMe" as a function of n is n(n+1)/4.
To find the count for the instruction "CountMe" as a function of n for the given fragment, we need to analyze the nested loops and determine how many times the "CountMe" instruction will be executed.
Let's break down the fragment:
Line 1 initializes the variable 'j' to 1.
Line 2 starts a while loop that will continue as long as 'j' is less than or equal to n/2.
Lines 3 to 7 contain another nested while loop. Each time the outer loop iterates, the inner loop will execute 'j' times.
Line 5 is the instruction "CountMe" that we are interested in.
Line 8 increments the variable 'j' by 1.
Therefore, the number of times the "CountMe" instruction will be executed can be calculated as follows:
- The outer loop runs (n/2) - 1 times. This can be determined by the initial value of 'j' (1) and the condition (j <= n/2). For example, if n=10, the outer loop will run 4 times.
- The inner loop runs 'j' times for each iteration of the outer loop. So, the total number of times the inner loop will run is the sum of 1 to n/2.
- This can be calculated using the arithmetic series formula: (n/2) * (n/2 + 1) / 2
Hence, the count for the instruction "CountMe" as a function of n for the given fragment is (n/2) * (n/2 + 1) / 2.