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A minimum spanning tree based heuristic for the travelling salesman tour

机译:基于跨越的旅游推销局跨越式启发式

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Abstract This paper presents a heuristic to find the travelling salesman tour (TST) in a connected network. The approach first identifies a node and two associated arcs that are desirable for inclusion in the required TST. If we let this node be denoted by $$p$$ p and two selected arcs emanating from this node be denoted by $$left( {p,q} ight),{ext{and}},left( {p,k} ight),$$ p , q and p , k , then we find a path joining the two nodes $$q,{ext{and}}, k$$ q and k passing through all the remaining nodes of the given network. A sum of these lengths, i.e. length of the links $$left( {p,q} ight),{ext{and}},left( {p,k} ight)$$ p , q and p , k along with the length of the path that joins the nodes $$q,{ext{and}}, k$$ q and k passing through all the remaining nodes will result in a feasible TST, hence gives an upper bound on the TST. A simple procedure is outlined to identify: (1) the node $$p$$ p , (2) the two corresponding links $$left( {p,q} ight),{ext{and}},left( {p,k} ight),$$ p , q and p , k , and (3) the path joining the nodes $$q,{ext{and}}, k$$ q and k passing through all the remaining nodes. The approach is based on the minimum spanning tree; hence the complexity of the TST is reduced. The network in the present context has been assumed to be a connected network with at least two arcs emanating from each node.
机译:摘要本文提出了一个启发式,可以在连接的网络中找到旅行的推销员巡回赛(TST)。该方法首先识别一种节点和两个相关的电弧,其希望包含在所需的TST中。如果我们让这个节点用$$ p $$ p,并且两个选定的弧从这个节点发出的弧度由$$ left({p,q} 右),{ text {}}}表示。 left({p,k}右),$$ p,q和p,k,然后我们找到了一个路径加入两个节点$$ q ,{ text {}} ,k $$ q和k通过给定网络的所有剩余节点。这些长度的总和,即链接的长度$$ left({p,q}右),{ text {and}} , left({p,k} 右)$$ p, q和p,k以及加入节点$$ q ,{ text {}} ,k $$ q和k通过所有剩余节点的路径的长度将导致可行的Tst,因此在TST上给出一个上限。概述了一个简单的程序来识别:(1)节点$$ p $$ p,(2)两个相应的链接$$ left({p,q} 右),{ text {和}} ,左({p,k}右),$$ p,q和p,k和(3)路径加入节点$$ q ,{ text {}} ,k $$ q k通过所有剩余的节点。该方法基于最小的生成树;因此降低了TST的复杂性。已经假设本文中的网络是连接网络,其具有从每个节点发出的至少两个弧。

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