树的恰当$\boldsymbol{\{a,b\}}$边赋权
Proper $\{a,b\}$-edge-weightings of trees
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中文总结 AI 辅助
本文刻画了不存在恰当{a,b}边赋权的树的结构，针对不同的a、b取值给出具体判定条件，并提出线性时间算法来判断该赋权是否存在且可构造。
中文摘要 AI 辅助
设$a$和$b$为不同的实权重，树的$\boldsymbol{\{a,b\}}$边赋权是指为每条边分配其中一个权重，若相邻顶点的关联边权重和不同，则称该赋权是恰当的。对于每对这样的权重，本文明确刻画了不存在恰当$\boldsymbol{\{a,b\}}$边赋权的树的结构：当$ab(a+b)\neq0$时，唯一不存在此类赋权的树是$K_2$；当$a+b=0$时，树不存在恰当$\boldsymbol{\{a,b\}}$边赋权当且仅当每个顶点的度数为1或3，且由度数为3的顶点诱导的子图存在完美匹配；对于剩余的$ab=0$的情况，生成由删除后会产生两个奇数阶分量的边构成的生成森林，树$T$不存在恰当$\boldsymbol{\{a,b\}}$边赋权当且仅当二分图的两个划分类均为奇数阶，且该森林的每个分量满足两个条件：一是满足前述的度数与匹配条件，二是每个分量中，顶点$v$在森林中的度数，加上两倍的、满足“$T-e$中不含$v$的分量包含来自每个二分划分类的奇数个顶点”的关联边$e$的数量，该值与$v$无关。对于每对固定的不同实权重，证明过程可得到一个线性时间算法，用于判断恰当$\boldsymbol{\{a,b\}}$边赋权是否存在，若存在则构造出该赋权。
英文摘要
Let $a$ and $b$ be distinct real weights. An $\{a,b\}$-edge-weighting of a tree assigns one of these weights to each edge and is proper if adjacent vertices have different sums of incident edge weights. For every such pair, we give an explicit structural characterization of the trees that do not admit a proper $\{a,b\}$-edge-weighting. If $ab(a+b)\neq0$, then $K_2$ is the only tree without such a weighting. If $a+b=0$, then a tree has no proper $\{a,b\}$-edge-weighting exactly when every vertex has degree $1$ or $3$ and the subgraph induced by the degree-$3$ vertices has a perfect matching. For the remaining case $ab=0$, form the spanning forest consisting of the edges whose deletion leaves two odd-order components. A tree $T$ has no proper $\{a,b\}$-edge-weighting exactly when both bipartition classes have odd order and every component of this forest satisfies two conditions. First, every component satisfies the preceding degree-and-matching condition. Second, within each component, the degree of a vertex $v$ in the forest plus twice the number of incident edges $e$ outside the forest for which the component of $T-e$ not containing $v$ has an odd number of vertices from each bipartition class is independent of $v$. For every fixed pair of distinct real weights, the proofs yield a linear-time algorithm that decides whether a proper $\{a,b\}$-edge-weighting exists and constructs one when it does.
