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Thread: Big O - Algorithm Analysis

  1. #1
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    Default Big O - Algorithm Analysis

    How do I analyze a for loop that looks like

    for (j = 0; j < 2*n; j += 2)

    I know that if it was just j++ it would loop 2n times but the += is throwing me off. This loop is nesting in a normal loop of

    for(i=0; i < n; i++)

    I know this will make it n^2 but I don't know how the j += 2 affects the time.


    Thanks for the help,
    PeskyToaster


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    Default Re: Big O - Algorithm Analysis

    j += 2
    just adds 2 to j, it is the same as
     j = j + 2
    , and because j is incrementing twice every loop, it is going to reach it's end twice as fast, so it is going to be looping
    1/2 * 2n
    or
    n
    times.

  3. The Following User Says Thank You to Tjstretch For This Useful Post:

    PeskyToaster (March 15th, 2012)

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    Default Re: Big O - Algorithm Analysis

    That loop is the same as for (int index = 0; index < N; index++) in terms on time complexity.
    Please use [highlight=Java]//code goes here...[/highlight] tags when posting your code

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    Default Re: Big O - Algorithm Analysis

    And, does the factor of 2x or 0.5x or whatever was worrying you, really matter?

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    Default Re: Big O - Algorithm Analysis

    I have another problem. First here's the code snippet. It's just general code for analysis

     
    for (j=0; j<2n; j++) {
         i = 1;
        Generic Statement;
        while (j<n) {
              Generic statement;
              Generic statement;
              i = i *2;
       }
    }

    How does the i = 2n and the j < 2n affect the Big O and execution time. The execution time is just the equation with each statement being one time unit.

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