我将如何为该不变式编写循环? [英] How would I write a loop for this invariant?

问题描述

这些是算法的断言，用于找到数组b [h.k]的最小值:

These are assertions for an algorithm to find the minimum of an array b[h.k]:

Precondition: h <= k < b.length
Postcondition: b[x] is the minimum of b[h...k]

这是该不变式的正确循环吗?

Is this the correct loop for this invariant?

不变式:b [x]是b [h ... t]的最小值

invariant: b[x] is the minimum of b[h...t]

int x = t;    int t = h;
// {inv: b[x] is the minimum of b[h...t]}
while (t != k) {
   t = t+1;
   if (b[t] < b[x])
      { x = t;}
}

推荐答案

您可以通过以下方式找到数组的最小值(伪代码):

You can find the minimum of an array this way (pseudocode):

// assume b.length > 0
min = b[0]
for i=1 to b.length
  if b[i] < min
    min = b[i]

将其限制为b[h, ..., k]:

min = b[h]
for i=h+1 to k
  if b[i] < min
    min = b[i]

因此，您基本上只需更改循环的上限和下限

So you basically just change the upper and lower bound of the loop

由于h<=k<b.length，b[h]有效，因此从下一个元素开始执行循环，直到k遍历所需的元素为止(如果h==k，则该循环为空)

Since h<=k<b.length, b[h] is valid and executing the loop from the next element until k iterates over the reqiured elements (if h==k, the loop is empty)

更新:由于您始终无法将伪代码实现到Java中，因此我将为您翻译它:

UPDATE: as you are consistently failing with the implementation of the pseudocode into java, I'll translate it for you:

// assume: int b[]; int h; int k; h<=k<=b.length and b.length>0
// find min == b[i] such that b[i]<=b[j] for all h<=j<=k
int min = b[h];
for (int i=h+1; i<k; i=i+1) {
  if (b[i] < min) {
    min = b[i];
  }
}
// here: min contains the (first) minimum element within b[h, ..., k]

注意:您也可以将i=i+1写为++i

Note: you could write i=i+1 as ++i as well

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