This question asks you to perform competitive analysis of transpose (TR) and frequency count (FC).
(i) [5 marks] Suppose that you are maintaining a list of n elements under access operation only. The cost of access to the i-th element in the list is i. Let S be a
request sequence of m access operations over this list. For any su?ciently large m, construct a request sequence S such that for this request sequence, the total cost of TR divided by the total cost of MTF is ?(1).
(ii) [5 marks] Use the result of (i) to argue that TR is not competitive.
Hint: You just have to use the de?nition of competitiveness. Therefore, You can
solve this subquestion without having to solve (i) ?rst.
(iii) [5 marks] Further prove that FC is not competitive, either.
The Correct Answer and Explanation is :
Let’s analyze the problem and provide the answers systematically.
(i) Request Sequence Construction for TR and MTF
We are tasked to construct a request sequence ( S ) where the ratio of the total cost of the Transpose (TR) strategy to the Move-To-Front (MTF) strategy is ( \Theta(1) ).
Key Observations:
- Cost of Access (TR): In the TR strategy, when an element is accessed at position ( i ), it is swapped with the element in position ( i-1 ). This changes the order of elements, but it does not move the accessed element to the front of the list. Over time, frequently accessed elements move closer to the front, but this process is gradual.
- Cost of Access (MTF): In the MTF strategy, an accessed element at position ( i ) is moved directly to the front of the list after access. This ensures that frequently accessed elements are always at the front, minimizing their access cost in future operations.
Construction of Sequence ( S ):
- Consider a list of ( n ) elements indexed as ( 1, 2, \ldots, n ).
- Let ( S = [1, 2, \ldots, n] ) repeated ( m/n ) times, ensuring each element is accessed sequentially.
Cost Analysis:
- TR Cost:
- For each access, TR incurs a cost equal to the current position ( i ) of the accessed element.
- For ( m ) accesses, the total cost is approximately:
[
C_{\text{TR}} = \frac{m}{n} \cdot \left(1 + 2 + \cdots + n\right) = \frac{m}{n} \cdot \frac{n(n+1)}{2} = \frac{m(n+1)}{2}.
] - MTF Cost:
- For each access, MTF incurs a cost equal to ( i ) for the first access, but subsequent accesses to the same element have a cost of ( 1 ).
- Total cost ( C_{\text{MTF}} ) is asymptotically smaller than ( C_{\text{TR}} ), as MTF moves frequently accessed elements to the front.
Thus, the ratio ( C_{\text{TR}} / C_{\text{MTF}} = \Theta(1) ) for this sequence.
(ii) TR is Not Competitive
The competitive ratio for a strategy ( A ) is defined as:
[
\text{Competitive Ratio} = \sup_S \frac{\text{Cost}A(S)}{\text{Cost}{\text{OPT}}(S)},
]
where ( \text{OPT} ) is the optimal offline algorithm.
- For the sequence ( S ) constructed in (i), MTF achieves a near-optimal cost since it quickly adapts to the access pattern by moving frequently accessed elements to the front.
- In contrast, TR adapts more slowly and incurs significantly higher costs for ( S ).
- Hence, TR has a competitive ratio greater than a constant for certain sequences ( S ), implying TR is not competitive.
(iii) FC is Not Competitive
The Frequency Count (FC) strategy maintains the list in decreasing order of access frequency. While this seems optimal, it has limitations:
- The cost of access for FC depends on the frequency of elements but does not directly minimize the position ( i ) of recently accessed elements.
- For certain sequences (e.g., adversarial ones where the most recently accessed element is always far in the list), FC incurs significantly higher costs compared to MTF or the optimal strategy.
Thus, FC fails to adapt optimally to access patterns and has a competitive ratio greater than a constant, proving that it is not competitive.
Explanation (300 words)
This question explores the competitiveness of list maintenance strategies. A strategy is competitive if its cost is within a constant factor of the optimal offline strategy for any access sequence ( S ).
(i) Construction of ( S ): We constructed ( S ) as a sequence of sequential accesses over ( n ) elements. For this sequence, TR incurs a higher cost than MTF. TR swaps the accessed element with its predecessor, resulting in slow adaptation to the access pattern. In contrast, MTF moves frequently accessed elements directly to the front, minimizing future costs. The cost ratio ( C_{\text{TR}} / C_{\text{MTF}} = \Theta(1) ) highlights TR’s inefficiency for ( S ).
(ii) TR’s Non-Competitiveness: Competitive ratio evaluates a strategy’s performance relative to the optimal. TR’s slower adaptation makes it suboptimal for sequences like ( S ), where frequent accesses do not lead to immediate cost reduction. Therefore, TR has a competitive ratio greater than a constant, failing the competitiveness criterion.
(iii) FC’s Non-Competitiveness: FC orders elements by frequency but neglects recency. Adversarial sequences can exploit this, leading to high costs. For example, alternating accesses between distant elements results in frequent reordering but no cost minimization. As a result, FC’s competitive ratio exceeds a constant, proving it is not competitive.
In conclusion, both TR and FC strategies lack the adaptability of MTF, resulting in suboptimal performance for certain access sequences and rendering them non-competitive.