The Master Theorem can be applied to solve recurrence relations of the form T(n) = aT(n/b) + f(n), where a = 1, b 1, and f(n) is an asymptotically positive function. In this case, a = 3, b = 2, and f(n) = 1. The Master Theorem has three cases: 1. If f(n) = O(n^c), where c = 0, then T(n) = Θ(n^c * log^(k+1)(n)). 3. If f(n) = Ω(n^c), where c log_b(a), if a*f(n/b)

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The Master Theorem can be applied to solve recurrence relations of the form T(n) = aT(n/b) + f(n), where a >= 1, b > 1, and f(n) is an asymptotically positive function. In this case, a = 3, b = 2, and f(n) = 1. The Master Theorem has three cases: 1. If f(n) = O(n^c), where c < log_b(a), then T(n) = Θ(n^log_b(a)). 2. If f(n) = Θ(n^c * log^k(n)), where c = log_b(a) and k >= 0, then T(n) = Θ(n^c * log^(k+1)(n)). 3. If f(n) = Ω(n^c), where c > log_b(a), if a*f(n/b) <= k*f(n) for some constant k < 1 and sufficiently large n, then T(n) = Θ(f(n)). In this case, f(n) = 1, which is Θ(n^0). The value of c is 0, which is less than log_b(a) = log_2(3). Therefore, we are in case 1 of the Master Theorem. So, T(n) = Θ(n^log_b(a)) = Θ(n^log_2(3)).

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The Master Theorem can be applied to solve recurrence relations of the form T(n) = aT(n/b) + f(n), where a >= 1, b > 1, and f(n) is an asymptotically positive function. In this case, a = 3, b = 2, and f(n) = 1. The Master Theorem has three cases: 1. If f(n) = O(n^c), where c < log_b(a), then T(n) = Θ(n^log_b(a)). 2. If f(n) = Θ(n^c * log^k(n)), where c = log_b(a) and k >= 0, then T(n) = Θ(n^c * log^(k+1)(n)). 3. If f(n) = Ω(n^c), where c > log_b(a), if a*f(n/b) <= k*f(n) for some constant k < 1 and sufficiently large n, then T(n) = Θ(f(n)). In this case, f(n) = 1, which is Θ(n^0). The value of c is 0, which is less than log_b(a) = log_2(3). Therefore, we are in case 1 of the Master Theorem. So, T(n) = Θ(n^log_b(a)) = Θ(n^log_2(3)).

What is the tight bound using Big-Theta notation for the time complexity of the following recurrence relation: T(n) = 3T(n/2) + 1 when the Master Theorem is used?group of answers: Θ(n^(log2(3))) Θ(n^(log3(2))) Θ(nlogn) Θ(log2(n))+O(1)

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