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Consider this recurrence relation: T(1) = 1 T(2) = 1 T(n) = 4 T(n-2) + 2n2 for n>2 The Master Theorem tells us Group of answer choices

Question

Consider this recurrence relation:

T(1) = 1
T(2) = 1
T(n) = 4 T(n-2) + 2n^2   for n > 2

The Master Theorem tells us
Group of answer choices

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Solution

The Master Theorem provides a method for analyzing the time complexity of recurrence relations of the form:

T(n)=aT(nb)+f(n) T(n) = aT\left(\frac{n}{b}\right) + f(n)

where a1a \geq 1 and b>1b > 1. However, the recurrence relation you've given does not fit this form directly since the recursive call is based on n2n-2 rather than dividing nn.

To analyze this recurrence, we can use substitution or expansion to find a pattern. Here are the steps for doing so:

  1. Break Down the Problem
    We have the recurrence: T(n)=4T(n2)+2n2for n>2 T(n) = 4T(n-2) + 2n^2 \quad \text{for } n > 2 with initial conditions T(1)=1T(1) = 1 and T(2)=1T(2) = 1.

  2. Relevant Concepts
    We will expand the recurrence to express T(n)T(n) in terms of previous values: T(n)=4T(n2)+2n2 T(n) = 4T(n-2) + 2n^2 We can substitute T(n2)T(n-2) based on the same form: T(n2)=4T(n4)+2(n2)2 T(n-2) = 4T(n-4) + 2(n-2)^2

  3. Analysis and Detail
    Substituting back into the original equation iteratively:

    • First substitute T(n2)T(n-2): T(n)=4(4T(n4)+2(n2)2)+2n2=16T(n4)+8(n2)2+2n2 T(n) = 4(4T(n-4) + 2(n-2)^2) + 2n^2 = 16T(n-4) + 8(n-2)^2 + 2n^2
    • Now substitute further: T(n)=16(4T(n6)+2(n4)2)+8(n2)2+2n2 T(n) = 16(4T(n-6) + 2(n-4)^2) + 8(n-2)^2 + 2n^2

    Continuing this substitution until reaching the base cases leads to identifying how many terms will contribute to T(n)T(n).

  4. Verify and Summarize
    We can explore the pattern. Each recursive step effectively adds terms of the form 2k22k^2 with a multiplying factor increasing by 4 for each step back. It will be complex without exact forms so setting bounds directly might be easier.

    Ultimately, recognizing that this generates a quadratic sum leads to overall polynomial growth.

Final Answer

The recurrence grows as we recursively apply the relation, indicating a non-linear polynomial time complexity, roughly of the order O(n3)O(n^3) when approximated through expansion and verifying dominating terms in calculations. However, without a straightforward fit to the Master Theorem or an explicit coefficient identification, it may require deeper methods (like generating functions) for a concrete exact closed form. Thus, identify that growth resembles cubic behavior due to iterated summing of quadratic contributions.

This problem has been solved

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