To determine the worst-case number of comparisons performed by the pseudocode, we need to carefully analyze the logic behind the given code. The pseudocode performs a while loop, with the condition being that the loop runs while $n < 18$. Inside the loop, $n$ is incremented by 5 on each iteration.

Pseudocode breakdown:

while n < 18 do
    n ← n + 5

• Preconditions: $n$ is an integer in the set {5, 10, 15, 20, 25}.
• The loop executes as long as $n < 18$, and on each iteration, $n$ is increased by 5.

Step-by-step analysis:

1. If $n = 5$:

   • The condition $5 < 18$ is true, so the loop runs.
   • After the first iteration, $n$ becomes 10. The comparison is made again.
   • After the second iteration, $n$ becomes 15. The comparison is made again.
   • After the third iteration, $n$ becomes 20, and $20 < 18$ is false, so the loop exits.
   • Total comparisons: 4 (one for each iteration before the loop exits).

2. If $n = 10$:

   • The condition $10 < 18$ is true, so the loop runs.
   • After the first iteration, $n$ becomes 15. The comparison is made again.
   • After the second iteration, $n$ becomes 20, and $20 < 18$ is false, so the loop exits.
   • Total comparisons: 3 (one for each iteration before the loop exits).

3. If $n = 15$:

   • The condition $15 < 18$ is true, so the loop runs.
   • After the first iteration, $n$ becomes 20, and $20 < 18$ is false, so the loop exits.
   • Total comparisons: 2 (one for the iteration before the loop exits).

4. If $n = 20$:

   • The condition $20 < 18$ is false, so the loop doesn't run at all.
   • Total comparisons: 1 (the first comparison is made to check if the loop condition is satisfied).

5. If $n = 25$:

   • The condition $25 < 18$ is false, so the loop doesn't run at all.
   • Total comparisons: 1 (the first comparison is made to check if the loop condition is satisfied).

Summary of results:

• When $n = 5$, 4 comparisons are made.
• When $n = 10$, 3 comparisons are made.
• When $n = 15$, 2 comparisons are made.
• When $n = 20$, 1 comparison is made.
• When $n = 25$, 1 comparison is made.

Worst-case number of comparisons:

• The worst-case number of comparisons occurs when $n = 5$, where 4 comparisons are made.

Answer:

• Worst-case number of comparisons: 4
• Values of $n$ that produce the worst case: 5

Let me know if you would like further details or clarification!

Here are some related questions for you:

1. How does the number of comparisons change if we change the increment in the pseudocode?
2. What would happen if the loop condition were $n \leq 18$ instead of $n < 18$?
3. What would the number of comparisons be if $n$ started with a value greater than 18?
4. How can you generalize this problem to determine the number of comparisons for other sets of values for $n$?
5. What other types of loops can have similar analysis for determining the worst-case scenario?

Tip: When analyzing loops, focus on how the loop conditions interact with the input values, as this is key to determining the number of iterations (and comparisons) performed.