递归算法的复杂性答案

【问题标题】：Complexity in a recursion algorithm递归算法的复杂性
【发布时间】：2015-11-05 13:43:46
【问题描述】：

我目前正在大学学习数据结构，偶然发现了一个关于递归复杂性的问题。

鉴于此代码：

Unsigned func (unsigned n)
{
 if (n ==0) return 1;
 if(n==1) return 2;

 \\do somthing(in O(1) )

 return func(n-1) + func(n-1);
}

我知道代码的作用。我知道现在的形式，时间复杂度是O(2^n)。

然而，我的问题是：如果不是我将编写的代码的最后一个返回调用：return 2*func(n-1)，时间复杂度会改变吗？

我知道，就内存复杂度而言，我们说的是递归占用空间的显着减少，但就时间复杂度而言，会有什么变化吗？

我使用递归函数进行了数学运算，并了解到时间复杂度不会发生变化，对吗？

【问题讨论】：

标签： algorithm recursion big-o time-complexity complexity-theory

【解决方案1】：

这个方法只有O(n)，因为如果你用 5 运行它，它会用 4 递归，然后用 3 等等。

Unsigned func (unsigned n)
{
 if (n ==0) return 1;
 if(n==1) return 2;

 \\do somthing(in O(1) )

 return 2*func(n-1);
}

但是这个呢：

Unsigned func (unsigned n)
{
 if (n ==0) return 1;
 if(n==1) return 2;

 \\do somthing(in O(1) )

 return func(n-1) + func(n-1);
}

例如 func(5) 它将首先执行如下

5 -> 4 -> 3 -> 2 -> 1

然后它返回到2，但在那里它会执行“第二”部分，所以整个过程看起来像

5 -> 4 -> 3 -> 2-> 1; 2-> 1; 3->2->1; etc.

因此，它确实将复杂性从 O(n) 大幅更改为 O(2^n)

让我们试试这个代码的实际例子：

public class Complexity {
    private static int counter;
    public static void main(String[] args) {
        for (int i = 0; i < 20; i++) {
            counter = 0;
            func(i);
            System.out.println("For n="+i+" of 2*func the number of cycles is " + counter);
            counter = 0;
            func2(i);
            System.out.println("For n="+i+" of func + func the number of cycles is " + counter);
        }
    }

    public static int func(int n) {
        counter++;
        if (n == 0) {
            return 1;
        }
        if (n == 1) {
            return 2;
        }
        return 2 * func(n - 1);
    }

    public static int func2(int n) {
        counter++;
        if (n == 0) {
            return 1;
        }
        if (n == 1) {
            return 2;
        }
        return func2(n - 1) + func2(n - 1);
    }    
}

有这个输出：

For n=0 of 2*func the number of cycles is 1
For n=0 of func + func the number of cycles is 1
For n=1 of 2*func the number of cycles is 1
For n=1 of func + func the number of cycles is 1
For n=2 of 2*func the number of cycles is 2
For n=2 of func + func the number of cycles is 3
For n=3 of 2*func the number of cycles is 3
For n=3 of func + func the number of cycles is 7
For n=4 of 2*func the number of cycles is 4
For n=4 of func + func the number of cycles is 15
For n=5 of 2*func the number of cycles is 5
For n=5 of func + func the number of cycles is 31
For n=6 of 2*func the number of cycles is 6
For n=6 of func + func the number of cycles is 63
For n=7 of 2*func the number of cycles is 7
For n=7 of func + func the number of cycles is 127
For n=8 of 2*func the number of cycles is 8
For n=8 of func + func the number of cycles is 255
For n=9 of 2*func the number of cycles is 9
For n=9 of func + func the number of cycles is 511
For n=10 of 2*func the number of cycles is 10
For n=10 of func + func the number of cycles is 1023
For n=11 of 2*func the number of cycles is 11
For n=11 of func + func the number of cycles is 2047
For n=12 of 2*func the number of cycles is 12
For n=12 of func + func the number of cycles is 4095
For n=13 of 2*func the number of cycles is 13
For n=13 of func + func the number of cycles is 8191
For n=14 of 2*func the number of cycles is 14
For n=14 of func + func the number of cycles is 16383
For n=15 of 2*func the number of cycles is 15
For n=15 of func + func the number of cycles is 32767
For n=16 of 2*func the number of cycles is 16
For n=16 of func + func the number of cycles is 65535
For n=17 of 2*func the number of cycles is 17
For n=17 of func + func the number of cycles is 131071
For n=18 of 2*func the number of cycles is 18
For n=18 of func + func the number of cycles is 262143
For n=19 of 2*func the number of cycles is 19
For n=19 of func + func the number of cycles is 524287

但如果你记得已经计算过的结果，即使使用第二种方法，复杂度仍然是O(n)：

public class Complexity {
    private static int counter;
    private static int[] results;

    public static void main(String[] args) {
        for (int i = 0; i < 20; i++) {
            counter = 0;
            func(i);
            System.out.println("For n="+i+" of 2*func the number of cycles is " + counter);
            counter = 0;
            func2(i);
            System.out.println("For n="+i+" of func + func the number of cycles is " + counter);
            counter = 0;
            results = new int[i+1];
            func3(i);
            System.out.println("For n="+i+" of func + func with remembering the number of cycles is " + counter);
        }
    }

    public static int func(int n) {
        counter++;
        if (n == 0) {
            return 1;
        }
        if (n == 1) {
            return 2;
        }
        return 2 * func(n - 1);
    }

    public static int func2(int n) {
        counter++;
        if (n == 0) {
            return 1;
        }
        if (n == 1) {
            return 2;
        }        
        return func2(n - 1) + func2(n - 1);
    }

    public static int func3(int n) {
        counter++;
        if (n == 0) {
            return 1;
        }
        if (n == 1) {
            return 2;
        }

        if (results[n] == 0){
            results[n] = func3(n - 1) + func3(n - 1);
        }

        return results[n];
    }        
}

有这个输出：

For n=0 of 2*func the number of cycles is 1
For n=0 of func + func the number of cycles is 1
For n=0 of func + func with remembering the number of cycles is 1
For n=1 of 2*func the number of cycles is 1
For n=1 of func + func the number of cycles is 1
For n=1 of func + func with remembering the number of cycles is 1
For n=2 of 2*func the number of cycles is 2
For n=2 of func + func the number of cycles is 3
For n=2 of func + func with remembering the number of cycles is 3
For n=3 of 2*func the number of cycles is 3
For n=3 of func + func the number of cycles is 7
For n=3 of func + func with remembering the number of cycles is 5
For n=4 of 2*func the number of cycles is 4
For n=4 of func + func the number of cycles is 15
For n=4 of func + func with remembering the number of cycles is 7
For n=5 of 2*func the number of cycles is 5
For n=5 of func + func the number of cycles is 31
For n=5 of func + func with remembering the number of cycles is 9
For n=6 of 2*func the number of cycles is 6
For n=6 of func + func the number of cycles is 63
For n=6 of func + func with remembering the number of cycles is 11
For n=7 of 2*func the number of cycles is 7
For n=7 of func + func the number of cycles is 127
For n=7 of func + func with remembering the number of cycles is 13
For n=8 of 2*func the number of cycles is 8
For n=8 of func + func the number of cycles is 255
For n=8 of func + func with remembering the number of cycles is 15
For n=9 of 2*func the number of cycles is 9
For n=9 of func + func the number of cycles is 511
For n=9 of func + func with remembering the number of cycles is 17
For n=10 of 2*func the number of cycles is 10
For n=10 of func + func the number of cycles is 1023
For n=10 of func + func with remembering the number of cycles is 19
For n=11 of 2*func the number of cycles is 11
For n=11 of func + func the number of cycles is 2047
For n=11 of func + func with remembering the number of cycles is 21
For n=12 of 2*func the number of cycles is 12
For n=12 of func + func the number of cycles is 4095
For n=12 of func + func with remembering the number of cycles is 23
For n=13 of 2*func the number of cycles is 13
For n=13 of func + func the number of cycles is 8191
For n=13 of func + func with remembering the number of cycles is 25
For n=14 of 2*func the number of cycles is 14
For n=14 of func + func the number of cycles is 16383
For n=14 of func + func with remembering the number of cycles is 27
For n=15 of 2*func the number of cycles is 15
For n=15 of func + func the number of cycles is 32767
For n=15 of func + func with remembering the number of cycles is 29
For n=16 of 2*func the number of cycles is 16
For n=16 of func + func the number of cycles is 65535
For n=16 of func + func with remembering the number of cycles is 31
For n=17 of 2*func the number of cycles is 17
For n=17 of func + func the number of cycles is 131071
For n=17 of func + func with remembering the number of cycles is 33
For n=18 of 2*func the number of cycles is 18
For n=18 of func + func the number of cycles is 262143
For n=18 of func + func with remembering the number of cycles is 35
For n=19 of 2*func the number of cycles is 19
For n=19 of func + func the number of cycles is 524287
For n=19 of func + func with remembering the number of cycles is 37

【讨论】：

那么我对 func(n-1) + func(n-1) 的递归调用以 O(2^n) 的复杂度运行是否正确？另外，为什么我的递归公式，当我把它写在纸上并计算时，我发现它们都有相同的时间复杂度？
@Tom - 是的，它以 O(2^n) 的复杂度运行...关于您的公式 - 您使用错误或公式错误。将该公式发布到您的问题中，也许我们可以告诉您出了什么问题。
我的公式是：T(n) = 2T(n-1) while T(0)=1,T(1)= 2。我使用了“发布方法”，得到了一般函数：T(n) = 2^i *T(n-i)。在定位我的停止条件（如 i = n-1 或 i = n-2）后，我达到了 T(n) = 2^(n-1) *T(1) = (2^n)/2 的复杂性 - ------> O(2^n)。其他停止条件的结果相同。我做错了什么，因为我为两种类型的代码得到了相同的公式。
T(n) = 2T(n-1) 在2*func(n-1) 中是不正确的，因为T(n) 不做T(n-1) 两次，它只做一次然后将结果乘以2，这是不同的事情。
你能帮我写出适合 2*func(n-1) 场景的 T(n) 吗？我如何模拟它只调用一个然后将结果乘以 2 的事实？

【解决方案2】：

这取决于您的编程/算法语言的语义。

如果f(n) 的意思是“无论之前是否使用相同的参数调用该函数”（大多数编程语言都是如此），那么您的更改会将复杂性显着降低到 O(n ）。每减少一个参数，您就有一个 O(1) 函数调用。

否则（如果您谈论的是纯函数并记住已知结果），这两种算法的复杂度都已经为 O(n)。

【讨论】：