我在读CLRS的《算法导论》。在第二章中,作者提到了“循环不变量”。什么是循环不变量?


当前回答

值得注意的是,循环不变量可以帮助迭代算法的设计,因为它被认为是一个断言,表示变量之间的重要关系,在每次迭代开始时和循环结束时,这些关系必须为真。如果这是成立的,计算是在有效的道路上。如果为false,则算法失败。

其他回答

对不起,我没有评论权限。

正如你提到的@Tomas Petricek

另一个较弱的不变式也是成立的,即i >= 0 && i < 10(因为这是连续条件!)”

为什么它是循环不变量?

我希望我没有错,据我理解[1],循环不变将在循环开始时为真(初始化),它将在每次迭代(维护)之前和之后为真,它也将在循环结束后为真(终止)。但是在最后一次迭代之后,i变成了10。因此,条件i >= 0 && i < 10变为假值并终止循环。它违反了循环不变量的第三个性质(终止)。

[1] http://www.win.tue.nl/~kbuchin/teaching/JBP030/notebooks/loop-invariants.html

《如何思考算法》的定义,Jeff Edmonds著

循环不变式是放置在循环和循环顶部的断言 每次计算返回到循环的顶部时,这必须成立。

循环不变量是在循环执行前后为真的断言。

除了这些不错的答案,我想Jeff Edmonds在《如何思考算法》(How to Think About Algorithms)中举的一个很好的例子可以很好地说明这个概念:

EXAMPLE 1.2.1 "The Find-Max Two-Finger Algorithm" 1) Specifications: An input instance consists of a list L(1..n) of elements. The output consists of an index i such that L(i) has maximum value. If there are multiple entries with this same value, then any one of them is returned. 2) Basic Steps: You decide on the two-finger method. Your right finger runs down the list. 3) Measure of Progress: The measure of progress is how far along the list your right finger is. 4) The Loop Invariant: The loop invariant states that your left finger points to one of the largest entries encountered so far by your right finger. 5) Main Steps: Each iteration, you move your right finger down one entry in the list. If your right finger is now pointing at an entry that is larger then the left finger’s entry, then move your left finger to be with your right finger. 6) Make Progress: You make progress because your right finger moves one entry. 7) Maintain Loop Invariant: You know that the loop invariant has been maintained as follows. For each step, the new left finger element is Max(old left finger element, new element). By the loop invariant, this is Max(Max(shorter list), new element). Mathe- matically, this is Max(longer list). 8) Establishing the Loop Invariant: You initially establish the loop invariant by point- ing both fingers to the first element. 9) Exit Condition: You are done when your right finger has finished traversing the list. 10) Ending: In the end, we know the problem is solved as follows. By the exit condi- tion, your right finger has encountered all of the entries. By the loop invariant, your left finger points at the maximum of these. Return this entry. 11) Termination and Running Time: The time required is some constant times the length of the list. 12) Special Cases: Check what happens when there are multiple entries with the same value or when n = 0 or n = 1. 13) Coding and Implementation Details: ... 14) Formal Proof: The correctness of the algorithm follows from the above steps.

简单地说,它是一个循环条件,在每次循环迭代中都为真:

for(int i=0; i<10; i++)
{ }

在这里,我们可以说i的状态是i<10并且i>=0