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  • Dynamic Programming: Fibonacci

    Recently I watched an interesting video in youtube, the vbloger use calculating Fibonacci number to explain dynamic programming

    after watch this video, I decide to write it down in English, also for practice my written English

    ok, in this article, we will assume you already know what's Finabonacci number

    commonly, we always use recursion to get the number, it's pretty easy to implement it

    Recursion:

    int Fib(int n)
    {
        if (n <= 2) return 1;
        return Fib(n - 1) + Fib(n - 2);
    }

    This is also the example when we learn recursion

    the time complexity is O(X=2^n), it's calcualted like this

    Fib(n) once

    Fib(n -1) once

    Fib(n-2) twice

    Fib(n-3) Third

    the total time is equal = 1+2+3...+(n - 1) = 2^n

    this approach works for most of cases, but it's no effective and will cause stack over exception if the number is big, because there's call stacks

    it cost really long time when I set n = 100

    so we need to improve the recursion

    we can see some numbers are calculated multiple times

    for instance, Fib(5) = Fib(4) + Fib(3), Fib(4) = Fib(3) + Fib(2), Fib(3) will be calculated twice

    Let's think about an approach to avoid it

    Recursion and Memoize

    in this appraoch, we will store the number when it's calculated

    int Fib(int n, int[] memoized)
    {
        if (memoized[n] != 0) return memoized[n];
        if (n <= 2) return 1;
        int f = Fib(n - 1) + Fib(n - 2);
        memoized[n] = f;
        return f;
    }

    ok, we will only calculate once for one number, and the time complexity is O(n)

    however there are still lots of call stack while calculating

    the above 2 approaches are calculated from top to bottom, from n, n-1,...,1

    How about calculate from bottom, just like the exmaple number see, 1,1,2,3,5,6...

    Bottom Up

    int Fib(int n)
    {
        if (n <= 2) return 1;
        var memoized = new int[n + 1];
        memoized[1] = 1;
        memoized[2] = 1;
        for (int i = 3; i <= n; i++)
        {
            memoized[i] = memoized[i - 1] + memoized[i - 2];
        }
        return memoized[n];
    }

    in this approach, we calcuate from bottom to up, altthough we add extra space for new array, but there are not so many call stacks, it's effective

    The time complexity is also O(n) 

    ok, this is the summary of the video, I also found a video which explain dynamic programming by MIT

    Please also find this video for reference

    Dynamic Programming I: Fibonacci, Shortest Paths

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  • 原文地址:https://www.cnblogs.com/zhaohuayang/p/10584629.html
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