在一个生成随机学生组的程序中,我给用户提供了强制将特定学生组合在一起并阻止学生配对的选项.我已经尝试了两天来制作我自己的算法来实现这一点,但是我在所有的递归中都迷失了
students = {
Student1 = {name=Student1, force={"Student2"}, deny={}};
Student2 = {name=Student2, force={"Student1","Student3"}, deny={}};
Student3 = {name=Student3, force={"Student2"}, deny={}};
}-- A second name property is given in the case that the student table needs to be accessed by students[num] to retrieve a name
然后我创建临时表：
forced = {}--Every student who has at least 1 entry in their force table is placed here, even if they have 1 or more in the deny table
denied = {}--Every student with 1 entry for the deny table and none in the force table is placed here
leftovers = {}--Every student that doesn't have any entries in the force nor deny tables is placed here
unsortable = {}--None are placed here yet -- this is to store students that are unable to be placed according to set rules(i.e. a student being forced to be paired with someone in a group that also contains a student that they can't be paired with
SortStudentsIntoGroups()--predefined; sorts students into above groups
在每个学生都被安排在这些小组之后(注意他们也仍留在学生桌上),我首先插入被迫在小组中配对的学生(好吧,我已经尝试过),插入有一个学生的学生或者将deny表中的更多条目放入可以放置它们的组中,然后用剩余的组填充剩余的组.
有几件事情会有所帮助：
numGroups--predefined number of groups
maxGroupSize--automatically calculated; quick reference to largest amount of students allowed in a group
groups = {}--number of entries is equivalent to numGroups(i.e. groups = {{},{},{}} in the case of three groups). This is for storing students in so that the groups may be displayed to the end user after the algorithm is complete.
sortGroups()--predefined function that sorts the groups from smallest to largest; will sort largest to smallest if supplied a true boolean as a parameter)
正如我之前所说,我不知道如何为此设置递归算法.每当我尝试将强迫学生插入到一起时,我最终会将同一个学生分成多个组,强制对没有配对,等等.还要注意格式.在每个学生的强制/拒绝表中,给出了目标学生的姓名 – 而不是直接参考学生.我应该使用什么样的算法(如果这种情况存在的话)？任何帮助是极大的赞赏.
对我来说就像你在这里遇到 NP-Hard问题一样.这相当于k001颜色的graph-coloring问题,其中edge是拒绝列表.
图形着色：
Given a graph G=(V,E), and an integer `k`, create coloring function f:V->{1,2,..,k} such that:
f(v) = f(u) -> (v,u) is NOT in E
从图形着色到您的问题的减少：
给定图形着色问题(G,k),其中G =(V,E),用以下方法创建问题的实例：
students = V
for each student: student.denial = { student' | for each edge (student,student')}
#groups = k
直观地,每个顶点由学生表示,并且学生否认所有学生在表示它们的顶点之间存在边缘.
组的数量是给定的颜色数.
现在,给你一个问题的解决方案 – 我们得到k组,如果学生你否认学生v – 他们不在同一组,但这与用不同颜色着色u和v相同,所以对于每个边缘(u, v)在原始图中,u和v是不同的颜色.
另一种方式是类似的
所以,我们从图形着色问题得到了一个polynomial reduction,因此找到问题的最佳解决方案是NP-Hard,并且没有已知的有效解决方案,并且大多数人认为不存在.
一些替代方案使用诸如Genetic Algorithms之类的启发式方法,其不提供最佳解决方案,或使用耗时的brute force方法(对于大量学生来说不可行).蛮力只会产生对k组的所有可能的分裂,并检查它是否是可行的解决方案,最后 – 将选择找到的最佳解决方案.