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Consider the following two problems:

A. Given a directed graph and a parameter $k$, determine if it contains a path (not necessarily simple) of length $k$.

B. Given a directed graph, two vertices $s,t$ and a parameter $k$, determine if the graph contains a path from $s$ to $t$ (not necessarily simple) of length $k$.

How can I reduce problem B to problem A?

I know that I can make a DFS tree at height $k$ with repeating vertices, however it solves the problem directly rather than by reduction.

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  • $\begingroup$ What kind of reduction are you looking for? $\endgroup$ – Steven Oct 22 at 16:22
  • $\begingroup$ k is a constant, and I need to convert graph G to G' s.t by giving a solution to A I can find a solution to B by run the algorithm on G' $\endgroup$ – daniTo Oct 22 at 16:37
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    $\begingroup$ Are you sure you're not interested in a reduction in the other direction? $\endgroup$ – Yuval Filmus Oct 22 at 17:22
  • $\begingroup$ I'm sure. The opposite it's much easier. $\endgroup$ – daniTo Oct 22 at 18:31
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Since $k$ is a constant you can solve $B$ in polynomial-time by exploring all $O(n^{k-1})$ paths of length $k$ starting from $s$ and ending in $t$, as you point out.

At this point the reduction is trivial: If the answer to an instance $\langle G, s, t\rangle$ of $B$ is "yes" then the corresponding instance of $A$ is $G$. If the answer to the instance $\langle G, s, t \rangle$ is "no", then the instance of $A$ is the empty graph.

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  • $\begingroup$ The problem that is directe solution and I need reduction I starting to believe that there's is typo at the assignment. thank you anyway $\endgroup$ – daniTo Oct 22 at 19:53
  • $\begingroup$ Formally, this is a reduction... $\endgroup$ – Steven Oct 22 at 19:54
  • $\begingroup$ This "solve-it-up-front" reduction is legitimate because $k$ is not part of the input, so $O(n^{k-1})$ is polynomial in $n$ (and you're allowed to spend that much time converting the problem instance from $B$ to $A$). If you want a reduction that allows $k$ to be part of the input, I can't see one. Problem $A$ gives you very little flexibility: For a digraph containing a cycle, the answer is YES for every $k$. If you restrict inputs to DAGs, then assuming $k \le n$, a reduction that attaches an $n$-path before $s$, and another $n$-path after $t$, and increases $k$ by $2n$, would work. $\endgroup$ – j_random_hacker Oct 24 at 17:26
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    $\begingroup$ If $k$ is part of the input and is polynomially bounded w.r.t. the size of $G$ then you can reduce $B$ to $A$. A rough sketch of the idea is to create $k$ "layers" where the first layer contains only $s$, the last layer contains only $t$, and the intermediate $k-2$ layers are copies of $G$. The edge edges of $G$ are rewirted to only go from one layer to the next. $\endgroup$ – Steven Oct 24 at 17:35
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    $\begingroup$ Ah of course, making a nice DAG out of copies! I was thinking too narrowly. I think this is probably the solution the question was looking for -- and perhaps OP wrote "k is a constant" without that actually being in the question, and without realising the implications. $\endgroup$ – j_random_hacker Oct 26 at 23:38

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