递归算法的复杂性
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【中文标题】递归算法的复杂性【英文标题】:Complexity in a recursion algorithm 【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】:
【参考方案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?【参考方案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】:
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