Math Problem Statement

“Derivation and Optimization of Pivot Selection i. Assume the pivot is at position p (fraction of the array length n, p = k * n, where k ∈ (0,1)). ii. Derive the total number of comparisons C(n, k) as a continuous function of k. iii. Use calculus to determine the value of k that minimizes C(n, k). iv. Verify your result using the second derivative test.” i don’t know two cents about what’s being discussed here. But I’m good at calculus. Just give me the function & how to easily derive it. I’ll find out how to find minimum value of C and apply the second derivative test.

Solution

Let me break this down into the essential calculus framework and assumptions. You need a function C(n,k)C(n, k) that represents the total number of comparisons made in some process (likely related to quicksort or similar algorithms where pivots are involved).

Here’s the derived framework and how you can approach it step-by-step:


1. General Expression for C(n,k)C(n, k):

The total number of comparisons, C(n,k)C(n, k), depends on how the pivot splits the array at each step. The pivot divides the array into two subarrays, proportional to kk and 1k1 - k. The number of comparisons can be expressed as:

C(n,k)=n1+C(kn,k)+C((1k)n,k),C(n, k) = n - 1 + C(k \cdot n, k) + C((1 - k) \cdot n, k),

where:

  • n1n - 1: comparisons made to partition the array around the pivot.
  • C(kn,k)C(k \cdot n, k): recursive call for the left partition.
  • C((1k)n,k)C((1 - k) \cdot n, k): recursive call for the right partition.

For simplicity, we assume a continuous approximation: C(n,k)nlnn+ϕ(k),C(n, k) \approx n \ln n + \phi(k), where ϕ(k)\phi(k) represents the pivot-dependent term (logarithmic complexity from partitioning into proportions).

In terms of ϕ(k)\phi(k), a heuristic form is: ϕ(k)=klnk+(1k)ln(1k).\phi(k) = k \ln k + (1 - k) \ln(1 - k).


2. The Function for Optimization:

The simplified continuous approximation for the number of comparisons is: C(n,k)=nlnn+n[klnk+(1k)ln(1k)].C(n, k) = n \ln n + n [k \ln k + (1 - k) \ln(1 - k)].

3. Steps to Minimize C(n,k)C(n, k):

You need to minimize the term dependent on kk: g(k)=klnk+(1k)ln(1k).g(k) = k \ln k + (1 - k) \ln(1 - k).


4. Derivatives for Optimization:

Take the derivative g(k)g'(k): g(k)=lnk+1ln(1k)1.g'(k) = \ln k + 1 - \ln(1 - k) - 1.

Simplify: g(k)=lnkln(1k).g'(k) = \ln k - \ln(1 - k).

Set g(k)=0g'(k) = 0 to find critical points: lnk=ln(1k).\ln k = \ln(1 - k).

Exponentiate both sides: k=1k.k = 1 - k.

Solve: k=12.k = \frac{1}{2}.


5. Second Derivative Test:

Find g(k)g''(k): g(k)=1k+11k.g''(k) = \frac{1}{k} + \frac{1}{1 - k}.

At k=12k = \frac{1}{2}: g(12)=112+1112=2+2=4.g''\left(\frac{1}{2}\right) = \frac{1}{\frac{1}{2}} + \frac{1}{1 - \frac{1}{2}} = 2 + 2 = 4.

Since g(k)>0g''(k) > 0, k=12k = \frac{1}{2} is a local minimum.


Final Answer:

  • k=12k = \frac{1}{2} minimizes g(k)g(k), and hence C(n,k)C(n, k).
  • Second derivative test confirms k=12k = \frac{1}{2} is a minimum.

Key Takeaway:

You primarily need to derive and minimize g(k)=klnk+(1k)ln(1k)g(k) = k \ln k + (1 - k) \ln(1 - k). Let me know if you'd like more details or clarification on any step.


5 Expanding Questions:

  1. Why is g(k)g(k) a good approximation for optimization of pivot selection?
  2. Can the function g(k)g(k) have multiple critical points in different cases?
  3. How does this model change when pivot selection is randomized?
  4. Can the method handle edge cases like k=0k = 0 or k=1k = 1?
  5. What is the real-world significance of minimizing C(n,k)C(n, k)?

Tip: Focus on interpreting g(k)g(k)'s behavior near k=0.5k = 0.5 to understand its symmetry and why it optimizes comparisons.

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Math Problem Analysis

Mathematical Concepts

Calculus
Optimization
Logarithmic Functions

Formulas

C(n, k) = n - 1 + C(k * n, k) + C((1 - k) * n, k)
C(n, k) ≈ n ln n + n [k ln k + (1 - k) ln(1 - k)]
g(k) = k ln k + (1 - k) ln(1 - k)

Theorems

Critical Points of Differentiable Functions
Second Derivative Test

Suitable Grade Level

Undergraduate