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Register a new userPolynomial-Time Algorithms for Sliding Tokens on Cactus Graphs and Block Graphs
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Added by
Duc A. Hoang
Publication date
2017
fields
Informatics Engineering
and research's language is
English
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Given two independent sets $I, J$ of a graph $G$, and imagine that a token (coin) is placed at each vertex of $I$. The Sliding Token problem asks if one could transform $I$ to $J$ via a sequence of elementary steps, where each step requires sliding a token from one vertex to one of its neighbors so that the resulting set of vertices where tokens are placed remains independent. This problem is $mathsf{PSPACE}$-complete even for planar graphs of maximum degree $3$ and bounded-treewidth. In this paper, we show that Sliding Token can be solved efficiently for cactus graphs and block graphs, and give upper bounds on the length of a transformation sequence between any two independent sets of these graph classes. Our algorithms are designed based on two main observations. First, all structures that forbid the existence of a sequence of token slidings between $I$ and $J$, if exist, can be found in polynomial time. A sufficient condition for determining no-instances can be easily derived using this characterization. Second, without such forbidden structures, a sequence of token slidings between $I$ and $J$ does exist. In this case, one can indeed transform $I$ to $J$ (and vice versa) using a polynomial number of token-slides.
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Suppose that we are given two independent sets $I_b$ and $I_r$ of a graph such that $|I_b|=|I_r|$, and imagine that a token is placed on each vertex in $I_b$. Then, the sliding token problem is to determine whether there exists a sequence of independent sets which transforms $I_b$ into $I_r$ so that each independent set in the sequence results from the previous one by sliding exactly one token along an edge in the graph. This problem is known to be PSPACE-complete even for planar graphs, and also for bounded treewidth graphs. In this paper, we thus study the problem restricted to trees, and give the following three results: (1) the decision problem is solvable in linear time; (2) for a yes-instance, we can find in quadratic time an actual sequence of independent sets between $I_b$ and $I_r$ whose length (i.e., the number of token-slides) is quadratic; and (3) there exists an infinite family of instances on paths for which any sequence requires quadratic length.
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