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Time Complexity of Algorithm

  1. Sep 1, 2009 #1
    Ok, I am brand new at this so I am kind of confused how to figure this out.

    Code (Text):

     i [tex]\leftarrow[/tex] n
     while i >= 1    
      j [tex]\leftarrow[/tex] i
      while j <= n
       <body of the j loop>, needs (1)
       j [tex]\leftarrow[/tex]j*2
      end j while
      i [tex]\leftarrow[/tex]i-1
     end i while

     
    I know that with nested loops that n goes to a power of 2, but it states that in my problem. What exactly am I missing and how do I show each step?
     
  2. jcsd
  3. Sep 2, 2009 #2

    berkeman

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    Staff: Mentor

    n is an initializer, it would appear. So it does not change in value, does it? And the middle loop doesn't seem right. j gets initialized to n the first time, is left as n^2 after the firsts pass, and would appear to fail the comparison the 2nd time through? I must be missing something as well...


    EDIT -- Oh, I see that j gets re-initialized before the 2nd comparison, so will not fail. In fact, it just keeps getting decremented until the outer loop comparison faiils, I guess.
     
  4. Sep 2, 2009 #3
    The body of the inner j loop is executed [tex]1 + \lfloor \log_2 (n / i) \rfloor[/tex] times for a given pass of the outer i loop. Now sum this over the different values of i from the outer loop.
     
  5. Sep 11, 2009 #4
    Alright, I didn't get this one quite right on the homework, but I am trying to figure out exactly how to solve this thing. The answer was n log n , but how do you total this up when going through the loop


    Code (Text):

    i [tex]\leftarrow[/tex] n                                                                         1
       while i >= 1                                                                                       n
          j [tex]\leftarrow[/tex] i                                                                    1
             while j <= n                                                                                 n?
                  <body of the j loop>, needs [tex]\Theta[/tex](1)                  [tex]\Theta[/tex](1)
                  j [tex]\leftarrow[/tex]j*2                                                         1
             end j while
         i [tex]\leftarrow[/tex]i-1                                                                   1
    end i while
     
    Now when I total up I disregard so I left with n. Now I assume that the inner loop is log n, but I don't fully understand. Thanks for the reply justsam and berkeman, but further direction would be greatly appreciated
     
  6. Sep 11, 2009 #5
    While I get what JustSam is saying, that's really not what O(N) is all about (keep the math out of it.).
    "Summing up" the parts of an algorithm is, as far as I'm concerned, not how I solve these.

    I rewrote what you put down so I could visualize it:

    Code (Text):
    for(i = n; i>= 1; i--){    -LOOP B
       for(j = i; j <= n; j*=2){   -LOOP A
          <constant time steps>
       }
    }
    Alright, so we have linear-over-n many logarithmic-over-n loops. Multiply to get n*log(n).

    Don't overanalyze O(N). These problems will introduce things (like the variable i above) just to waste your time. The goal is not to know exactly how many times a line will be executed given a certain N, the problem is to know how changing N will change the number of times the inner step is executed (O(N) tells you how Scalable your algorithm is)

    So, think this: If I increase N, there will be linearly more outer loops, and logarithmically more inner runnings of the middle statement.

    Consider these cases. If I gave you,

    Code (Text):
    for(i = 0; i < n; i++){
       for(j = 0; j < n; j++){
          <statement>
       }
    }
    the runtime is obviously O(N^2)

    Now try
    Code (Text):
    for(i = 0; i < n; i++){
       for(j = i; j < n; j++){
          <statement>
       }
    }
    It's STILL O(N^2) man. we have linearly-over-n many linear-over-n inner loops. Multiply to get n*n.

    Another example:
    Code (Text):

    for(i = 1; i < n; i*=2){
      for(j = i; j <= n; j*=2){
        <statement>
      }
    }
     
    Go through it with me: We have logarithmic-in-n many logarithmic-in-n loops. Multiply to get O(N) = (log2(n))^2
     
  7. Sep 11, 2009 #6

    Hurkyl

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    Staff Emeritus
    Science Advisor
    Gold Member

    In all fairness, that doesn't quite work: the inner loop varies with respect to i, and it is easy to construct examples where a simplistic multiplication give the wrong value, such as:
    Code (Text):
    for(int i = 1; i < N; i *= 2) {
      for(int j = 0; j < i; ++j) {
        // Something
      }
    }
    which executes "Something" O(N) times, not O(N^2) times.


    How can you keep the math out of it? :confused: The question we're trying to answer is already math!

    In fact (assuming your intuition isn't simply flat out wrong), the way you are approaching this problem is most likely just appling the basic ideas of calculus! You intuitively recognize that most of the N times through the outer loop take nearly log(N)/log(2) steps and everything else is negligible, which heuristically justifies just multiplying the two.

    But, for the record, it is very easy to turn it into a fully rigorous argument. For simplicity, let me define two functions:
    A(N) = the time to run the whole algorithm
    B(i,N) = the time to run the inner loop

    First, we just replace the inner loop time with something that's always bigger, but still logarithmic:
    We know B(i,N) < 1 + log(N) / log(2), and therefore
    A(N) < N * (1 + log(N) / log(2)) = O(N log N)

    Now, we isolate "most" of the iterations and find a logarithmic lower bound for those:
    If i < N/2, then B(i,N) = log(N) / log(2), and therefore
    A(N) > (N/2) log(N) / log(2) = O(N log N)

    And we put it together to get A(N) = O(N log N)
     
  8. Sep 11, 2009 #7

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    It would be log(n), base 2, if j were initialized to 1 instead of i each time. Are you sure that is not the case?

    EDIT: just read Hurkyl's last post more carefully ... looks like a good argument to me.
     
  9. Sep 11, 2009 #8
    @hurkyl: Ah, that makes sense.

    Minor correction, do you mean

    Well, I was curious why flipping the inner with the outer loop mattered so much, and I was really surprised. Here are some graphs:

    http://taif09.exofire.net/report/ONdata/

    So yes, it looks like every situation with nested loops is a little different. I still don't think the inclusion of i / a "middle" variable is really the important thing, but I guess I don't have any basis for this assumption. Neat, though.
     
  10. Sep 12, 2009 #9
    Thanks for all the help.

    Wow, this seems to be a very interesting problem. So with looking at the arguments I am gathering the inner loop is a log n while the outer loop remains n and you multiply together.
     
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