To compute the average-case number of $\geq$ comparisons using Algorithm 5.8 for making change, let us proceed step by step, incorporating Definition 5.6 (likely related to the concept of average-case complexity from probability theory):

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Key Steps

1. Understand Algorithm 5.8:

   • Typically, such an algorithm iteratively determines how many coins are needed for a given amount $N$.
   • The comparisons ($\geq$) happen as the algorithm checks if $N$ is greater than or equal to each coin denomination in the list.

2. Denominations:

   • Assume common denominations like $\{25, 10, 5, 1\}$ (U.S. currency), unless explicitly stated otherwise.
   • For this algorithm, the comparisons depend on the structure of $N$ and the denomination set.

3. Definition 5.6 (Average-Case Complexity):

   • The average-case complexity is calculated as the expected number of comparisons, considering all possible inputs and their probabilities.

4. Inputs $N \in \{10, 15, 20, 25\}$:

   • All these values are equally likely.
   • Probability of each $N = \frac{1}{4}$.

5. Analyzing the Algorithm for $N = 10, 15, 20, 25$:

   • For each $N$, count the number of comparisons needed to make change.

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Detailed Computation

Case 1: $N = 10$

• Denominations: $25, 10, 5, 1$.
• Comparisons:
  1. $N \geq 25$: False (1 comparison).
  2. $N \geq 10$: True (2nd comparison).
  • $N = 0$ (no further checks needed).

Total comparisons for $N = 10$: 2.

Case 2: $N = 15$

• Denominations: $25, 10, 5, 1$.
• Comparisons:
  1. $N \geq 25$: False (1 comparison).
  2. $N \geq 10$: True (2nd comparison).
  • Subtract 10, $N = 5$.
  3. $N \geq 5$: True (3rd comparison).
  • $N = 0$.

Total comparisons for $N = 15$: 3.

Case 3: $N = 20$

• Denominations: $25, 10, 5, 1$.
• Comparisons:
  1. $N \geq 25$: False (1 comparison).
  2. $N \geq 10$: True (2nd comparison).
  • Subtract 10, $N = 10$.
  3. $N \geq 10$: True (3rd comparison).
  • $N = 0$.

Total comparisons for $N = 20$: 3.

Case 4: $N = 25$

• Denominations: $25, 10, 5, 1$.
• Comparisons:
  1. $N \geq 25$: True (1st comparison).
  • $N = 0$.

Total comparisons for $N = 25$: 1.

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Average-Case Computation

The average number of comparisons is the expected value:

$\text{Average Comparisons} = \frac{1}{4} \sum_{N \in \{10, 15, 20, 25\}} (\text{Comparisons for } N)$

Substitute the values:

$\text{Average Comparisons} = \frac{1}{4} (2 + 3 + 3 + 1) = \frac{9}{4} = 2.25$

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Final Answer

The average-case number of $\geq$ comparisons is:

$\boxed{2.25}$

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Would you like further clarifications on the methodology or another example?

Related Questions

1. How does the choice of denominations affect the average-case complexity?
2. What happens if $N$ follows a different probability distribution (e.g., non-uniform)?
3. How does the worst-case number of comparisons differ from the average-case number?
4. Can the number of comparisons be minimized using a different coin set?
5. How does the algorithm scale for larger $N$ values?

Tip

For problems involving average-case complexity, always consider the probability distribution of inputs and focus on expected values rather than worst