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    If algorithm A and another algorithm B take log2(n)log_{2} (n)log2​(n)​ and n\sqrt{n}n​​  microseconds, respectively to solve a problem
    Question



    If algorithm A and another algorithm B take log2(n)log_{2} (n)​ and n\sqrt{n}​  microseconds, respectively to solve a problem, then the largest size n of a problem these algorithms can solve, respectively, in one second are _____________ and _________.

    A.

    2106and 106\mathrm{2^{10^6} and \ 10^6}​​

    B.

    2106and 1012\mathrm{2^{10^6 } and \ 10^{12}}​​

    C.

    2106and 6.106\mathrm{2^{10^6 } and \ 6.10^6}​​

    D.

    2106and 6.1012\mathrm{2^{10^6 } and \ 6.10^{12}}​​

    Correct option is B

    Let's break down and solve the question step by step:
    Problem Given:
    • Algorithm A takes log2(n)\log_2(n)​ microseconds.
    • Algorithm B takes n\sqrt{n}​  microseconds.
    • You need to find the largest size n that these algorithms can solve in one second.
    Step 1: Convert time into microseconds

    • We know that 1 second = 10610^6​ microseconds.
    Step 2: Solve for the maximum n for each algorithm
    For Algorithm A:

    • The time taken by Algorithm A is log2(n)\log_2(n)​ microseconds.
    • We are asked to find the maximum size n such that it can be solved within 1 second, i.e., within 10610^6​ microseconds.
    log2(n)=106\log_2(n) = 10^6​​
    • Solve for n:
    n=2106n = 2^{10^6}​​
    • So, n=2106n = 2^{10^6}​​
    For Algorithm B:
    • The time taken by Algorithm B is n\sqrt{n}​  microseconds.
    • We are asked to find the maximum size n such that it can be solved within 1 second, i.e., within 10610^6​ microseconds.
    n=106\sqrt{n} = 10^6​​
    • Solve for n:
    n=(106)2=1012n = (10^6)^2 = 10^{12}​​
    So, n=1012n = 10^{12}​.
    Final Answer:
    The largest sizes n that each algorithm can solve in 1 second are:
    • For Algorithm A: 21062^{10^6}​​
    • For Algorithm B: 101210^{12}​​

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