PageRank问题

发布于 2024-10-17 16:37:39 字数 746 浏览 2 评论 0原文

我不好意思问这样的问题；但我已经很长时间没有使用数学了，我记不起很多年前学过的许多概念。

在网址 http://www.javadev.org/files/Ranking.pdf 中，用一个例子来说明页面排名机制。页面 A、B 和 C 之间的关系是 A 链接到 B 和 C，B 链接到 C，C 链接到 A。因此 PageRank 方程如下

方程 A)

    PR(A) = 0.5 + 0.5 PR(C) 
    PR(B) = 0.5 + 0.5 (PR(A) / 2)
    PR(C) = 0.5 + 0.5 (PR(A) / 2 + PR(B))

得出结果结果

    PR(A) = 14/13 = 1.07692308 
    PR(B) = 10/13 = 0.76923077 
    PR(C) = 15/13 = 1.15384615

My问题是如何从方程 A 得出结果 B？

我尝试例如替换方程 PR(A) 中的 PR(C)

    PR(A) = 0.5 + 0.5 (0.5 + 0.5 (PR(A) / 2 + PR(B)))

这似乎以无限循环结束。所以我很困惑它如何得出结果，例如 PR(A) 值为 1.07692308？

为这样愚蠢的问题道歉。

我很感激任何建议。

原文

I am embarrassed to ask such question; but I haven't use math for a long time I can not recall many concepts learned many years ago.

In the url http://www.javadev.org/files/Ranking.pdf, an example is used for illustrate the page rank mechanism. The relation between page A, B, and C is A links to B and C, B links to C, and C links to A. So the PageRank equation is as below

Equation A)

    PR(A) = 0.5 + 0.5 PR(C) 
    PR(B) = 0.5 + 0.5 (PR(A) / 2)
    PR(C) = 0.5 + 0.5 (PR(A) / 2 + PR(B))

and it comes up with the result

Result B)

    PR(A) = 14/13 = 1.07692308 
    PR(B) = 10/13 = 0.76923077 
    PR(C) = 15/13 = 1.15384615

My question is how Result B is derived from Equation A?

I try e.g. replacing PR(C) in equation PR(A)

    PR(A) = 0.5 + 0.5 (0.5 + 0.5 (PR(A) / 2 + PR(B)))

this seems to end up with an infinite loop. So I am confused how it can derive the result e.g. PR(A) value is 1.07692308?

Appologize for such stupid question.

I appreciate any advice.

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梦太阳 2024-10-24 16:37:45

这不是一个愚蠢的问题，你只是生锈了。

将你的方程乘以 16（不是真的必要，但它会让事情看起来更好）：

16 PR(A) = 12 + 2 PR(A) + 4 PR(B)

现在从两边减去 2 PR(A)：

14 PR(A) = 12 + 4 PR(B)

现在替换 PR(B)，使用“方程A”的第二部分：

14 PR(A) = 12 + 2 + PR(A)
13 PR(A) = 14
   PR(A) = 14/13

其他遵循相同的方式。如果您发现方程两边都相同 (X = X)，则可能意味着您进行了两次相同的替换；备份并重试。经过一些练习，您就会掌握它的窍门。

Not a stupid question, you're just rusty.

Take your equation and multiply by 16 (not really necessary, but it makes things look nicer):

16 PR(A) = 12 + 2 PR(A) + 4 PR(B)

Now subtract 2 PR(A) from both sides:

14 PR(A) = 12 + 4 PR(B)

Now replace PR(B), using the second part of "equation A":

14 PR(A) = 12 + 2 + PR(A)
13 PR(A) = 14
   PR(A) = 14/13

And the others follow the same way. If you find that an equation winds up being the same on both sides (X = X), it probably means that you did the same substitution twice; just back up and try again. With a little practice you'll get the hang of it.

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