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.

在线性搜索(根据书中给出的练习)中，我们需要在给定的数组中找到值V。

它很简单，从0 <= k < length开始扫描数组并比较每个元素。如果找到V，或者扫描到数组的长度，就终止循环。

根据我对上述问题的理解-

循环不变量(初始化): 在k - 1迭代中找不到V。第一次迭代，这是-1因此我们可以说V不在-1位置

保养: 在下一次迭代中，V不在k-1中成立

Terminatation: 如果V位于k个位置，或者k达到数组的长度，则终止循环。

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。

Loop invariant is a mathematical formula such as (x=y+1). In that example, x and y represent two variables in a loop. Considering the changing behavior of those variables throughout the execution of the code, it is almost impossible to test all possible to x and y values and see if they produce any bug. Lets say x is an integer. Integer can hold 32 bit space in the memory. If that number exceeds, buffer overflow occurs. So we need to be sure that throughout the execution of the code, it never exceeds that space. for that, we need to understand a general formula that shows the relationship between variables. After all, we just try to understand the behavior of the program.

之前的回答已经很好地定义了循环不变量。

以下是CLRS的作者如何使用循环不变量来证明插入排序的正确性。

插入排序算法(见书):

INSERTION-SORT(A)
    for j ← 2 to length[A]
        do key ← A[j]
        // Insert A[j] into the sorted sequence A[1..j-1].
        i ← j - 1
        while i > 0 and A[i] > key
            do A[i + 1] ← A[i]
            i ← i - 1
        A[i + 1] ← key

循环不变量在这种情况下: 子数组[1到j-1]始终被排序。

现在让我们检查一下，证明这个算法是正确的。

初始化:在第一次迭代j=2之前。所以子数组[1:1]就是要测试的数组。因为它只有一个元素，所以它是有序的。这样不变性就被满足了。

维护:这可以通过在每次迭代后检查不变量来轻松验证。在这种情况下，它被满足了。

终止:这是我们将证明算法正确性的步骤。

当循环结束时，j=n+1。循环不变量再次被满足。这意味着子数组[1到n]应该排序。

这就是我们想用算法做的。因此，我们的算法是正确的。

《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。