用任意翻转切片进行煎饼分选的推广？

Question

在pancake sort中，主要操作是：将所有煎饼翻转到给定位置。如何在两个给定位置之间翻转所有煎饼呢？有谁知道这是否已被研究过？

为了说明这个问题，这里有一个快速的暴力，贪婪的Python 2.7实现（不太确定它总是收敛）：

def sort(a):
    entropy = 0
    for i in range(len(a) - 1):
        entropy += abs(a[i] - a[i+1])
    while True:
        max_improvement = 0
        for i in range(len(a) - 1):
            for j in range(i + 1, len(a)):
                improvement = 0
                if i > 0:
                    improvement += abs(a[i] - a[i-1]) - abs(a[j] - a[i-1])
                if j < len(a) - 1:
                    improvement += abs(a[j] - a[j+1]) - abs(a[i] - a[j+1])
                if improvement > max_improvement:
                    max_improvement = improvement
                    (next_i, next_j) = (i, j)
        if max_improvement == 0:
            if a and a[0] > a[-1]:
                a[:] = a[::-1]
                print a
            return
        entropy -= max_improvement
        a[next_i:next_j+1] = a[next_i:next_j+1][::-1]
        print a

a = [7, 1, 3, 8, 6, 0, 4, 9, 2, 5]
print a
sort(a)

输出：

[7, 1, 3, 8, 6, 0, 4, 9, 2, 5]
[7, 6, 8, 3, 1, 0, 4, 9, 2, 5]
[7, 6, 8, 9, 4, 0, 1, 3, 2, 5]
[7, 6, 8, 9, 4, 5, 2, 3, 1, 0]
[9, 8, 6, 7, 4, 5, 2, 3, 1, 0]
[9, 8, 7, 6, 4, 5, 2, 3, 1, 0]
[9, 8, 7, 6, 5, 4, 2, 3, 1, 0]
[9, 8, 7, 6, 5, 4, 3, 2, 1, 0]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Answer 1

根据评论者的建议，我在CS StackExchange Beta上交叉发布了这个问题，很快就收到了我正在寻找的答案：这个问题叫做逆转排序。

这是一个有效的Python implementation。

注意：自there are no guarantees that a current SE site will live past beta以来，我想在这里打开我的问题，恕我直言，它同样是主题。如果它与StackExchange礼仪相反，请随意关闭它。