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我正在研究管道网络优化,我将染色体表示为一串数字,如下所示

例子

chromosome [1] = 3 4 7 2 8 9 6 5

其中,每个数字指的是井,并定义了井之间的距离。因为,一个染色体的孔不能重复。例如

chromosome [1]' = 3 4 7 2 7 9 6 5 (not acceptable) 

可以处理这样的表示的最佳突变是什么?提前致谢。

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1 回答 1

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不能说“最好”,但我用于类似图形问题的一个模型是:对于每个节点(井号),计算整个人口中相邻节点/井的集合。例如,

population = [[1,2,3,4], [1,2,3,5], [1,2,3,6], [1,2,6,5], [1,2,6,7]]
adjacencies = { 
  1 : [2]         ,   #In the entire population, 1 is always only near 2
  2 : [1, 3, 6]   ,   #2 is adjacent to 1, 3, and 6 in various individuals
  3 : [2, 4, 5, 6],   #...etc...
  4 : [3]         ,
  5 : [3, 6]      , 
  6 : [3, 2, 5, 7],
  7 : [6]         
}
choose_from_subset = [1,2,3,4,5,6,7] #At first, entire population

然后通过以下方式创建一个新的个人/网络:

 choose_next_individual(adjacencies, choose_from_subset) : 
   Sort adjacencies by the size of their associated sets
   From the choices in choose_from_subset, choose the well with the highest number of adjacent possibilities (e.g., either 3 or 6, both of which have 4 possibilities)
   If there is a tie (as there is with 3 and 6), choose among them randomly (let's say "3")
   Place the chosen well as the next element of the individual / network ([3])
   fewerAdjacencies = Remove the chosen well from the set of adjacencies (see below)
   new_choose_from_subset = adjacencies to your just-chosen well (i.e., 3 : [2,4,5,6])
   Recurse -- choose_next_individual(fewerAdjacencies, new_choose_from_subset)

这个想法是具有大量邻接的节点适合重组(因为总体尚未收敛,例如 1->2),较低的“邻接计数”(但非零)意味着收敛,而零邻接计数(基本上)是一种突变。

只是为了展示一个示例运行..

#Recurse: After removing "3" from the population
new_graph = [3]
new_choose_from_subset = [2,4,5,6] #from 3 : [2,4,5,6] 
adjacencies = { 
  1: [2]             
  2: [1, 6]      ,  
  4: []          ,
  5: [6]         , 
  6: [2, 5, 7]   ,
  7: [6]         
}


#Recurse: "6" has most adjacencies in new_choose_from_subset, so choose and remove
new_graph = [3, 6]
new_choose_from_subset = [2, 5,7]    
adjacencies = { 
  1: [2]             
  2: [1]         ,  
  4: []          ,
  5: []          , 
  7: []          
}


#Recurse: Amongst [2,5,7], 2 has the most adjacencies
new_graph = [3, 6, 2]
new_choose_from_subset = [1]
adjacencies = { 
  1: []              
  4: []          ,
  5: []          , 
  7: []          
]

#new_choose_from_subset contains only 1, so that's your next...
new_graph = [3,6,2,1]
new_choose_from_subset = []
adjacencies = {
  4: []          ,
  5: []          , 
  7: []          
]

#From here on out, you'd be choosing randomly between the rest, so you might end up with:
new_graph = [3, 6, 2, 1, 5, 7, 4] 

完整性检查?3->6在原始中出现 1x,6->2出现 2x,2->1出现 5x,1->5出现 0,5->7出现 0,7->4出现 0。因此,您保留了最常见的邻接 (2->1) 和其他两个“可能很重要”的邻接。否则,您将在解决方案空间中尝试新的邻接。

更新:最初我忘记了递归时的关键点,您选择与刚刚选择的节点连接最多的节点。这对于保留高适应性链至关重要!我已经更新了描述。

于 2012-05-29T23:24:40.390 回答