Question

重写以下解决的问题以供参考：

原始问题：

T(n) = 2T(n - 1) - 1, if n > 0
1,                    otherwise

第一次迭代

T(n) = 2T(n - 1) - 1
T(n - 1) = 2T(n - 1 - 1) - 1 
T(n - 1) = 2T(n - 2) - 1

第二次迭代

T(n) = 2(2T(n - 2) - 1) - 1
T(n) = 4T(n - 2) - 2 - 1
T(n) = 4T(n - 2) - 3
T(n - 2) = 2T(n - 2 - 1) - 1
T(n - 2) = 2T(n - 3) - 1

第3次迭代

T(n) = 4(2T(n - 3) - 1) - 3
T(n) = 8T(n - 3) - 4 - 3
T(n) = 8T(n - 3) - 7
T(n - 3) = 2T(n - 3 - 1) - 1
T(n - 3) = 2T(n - 4) - 1

第4次迭代

T(n) = 8T(n - 3) - 7
T(n) = 8(2T(n - 4) - 1) - 7
T(n) = 16T(n - 4) - 8 - 7
T(n) = 16T(n - 4) - 15

决赛桌

At k=1, T(n) = 2T(n - 1) - 1
At k=2, T(n) = 4T(n - 2) - 3
At k=3, T(n) = 8T(n - 3) - 7
At k=4, T(n) = 16T(n - 4) - 15
        T(n) = 2ᴷT(n - k) - (2ᴷ - 1)
At k=n  T(n) = 2ᴺT(n - n) - (2ᴺ - 1)
        T(n) = 2ᴺT(0) - (2ᴺ - 1)
        T(n) = 2ᴺ - (2ᴺ - 1)
        T(n) = 2ᴺ - 2ᴺ + 1
        T(n) = 1

Answer 1

T（n）= 1的归纳证明如下：

感应开始。

什么是T（0）？ 1。

归纳步骤。

假设任何n> = 0表示T（n）= 1.现在让我们看看声明是否适用于T（n + 1）。

T（n + 1）= 2 * T（n） - 1 = 2 * 1 - 1 = 1

<强>结论

对于所有自然n，

T（n）= 1。 1是在O（1）。

您的错误。

我在计算中发现的第一个错误是：

<VirtualHost *:443>
    ServerName website.com

    RewriteEngine On

    # When Upgrade:websocket header is present, redirect to ws
    # Using NC flag (case-insensitive) as some browsers will pass Websocket
    RewriteCond %{HTTP:Upgrade} =websocket [NC]
    RewriteRule ^/ws/(.*)    wss://localhost:8888/ws/$1 [P,L]

    # All other requests go to http
    ProxyPass "/" "http://localhost:8888/"

Answer 2

T(n) = 2T(n-1) -1
     = 2(2T(n-2)-1) - 1 = 4T(n-2)-3 = (2^2)T(n-2)-(2^2-1)
     = 2(4T(n-3)-3) - 1 = 8T(n-3)-7 = (2^3)T(n-3)-(2^3-1)
     ...
     = (2^n)T(n-n)-(2^n-1) = (2^n)1 - 2^n + 1 = 1

为什么这个递归关系O（1）？

2 个答案: