Question: The running time of algorithm A is described by the recurrence TA(n) = 7 * TA(n/2) + n2. The running time of algorithm B is

The running time of algorithm A is described by the recurrence TA(n) = 7 * TA(n/2) + n2. The running time of algorithm B is given by TB(n) = a * TB(n/4) + n2. What is the largest value of a for which algorithm B is asymptotically faster than algorithm A, and what does this tell us about how algorithm B should be designed? Hint: Consider the Master Theorem/Method.

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