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In graph theory, it is well-known to be NP-complete the problem of given a set of $k$ pairs of source-sink, deciding whether there exists $k$ vertex-disjoint paths connecting these pairs.

Our problem is a variant in which the requirement of being vertex-disjoint is replaced by being induced.

A set of $k$ paths is said to be induced if:

  1. They are vertex-disjoint.
  2. Each one is itself an induced path.
  3. No edge connects two vertices of two different paths.

Input: A graph $G(V,E)$ and $k$ pairs of source-sink $\{(s_1,t_1),\dots,(s_k,t_k)\}$

Output: YES if there exists (a set of) $k$ induced paths connecting $s_i$ to $t_i$ for every $1\leq i\leq k$, NO otherwise.

Is this problem NP-complete?

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  • $\begingroup$ "(a set of) $k$ induced paths" as in simply $k$ different (not necessarily vertex-disjoint) paths? $\endgroup$ – dkaeae May 28 at 7:35
  • $\begingroup$ No, a set of $k$ paths who induce a graph with $k$ components, each being a path, I would guess. $\endgroup$ – Pål GD May 28 at 7:43
  • $\begingroup$ @dkaeae they have to be induced aka chordless. $\endgroup$ – Juho May 28 at 7:44
  • $\begingroup$ What is an induced path? $\endgroup$ – xskxzr May 29 at 17:59
  • $\begingroup$ @xskxzr A set of vertices that induce a subgraph that is a path. So specifically, not every path is an induced path. $\endgroup$ – Juho May 29 at 20:11
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Your problem is NP-hard (and so NP-complete), by reduction from 3SAT.

Consider a 3SAT instances with variables $x_1,\ldots,x_n$ and clauses $C_1,\ldots,C_m$.

For each variable $x_i$, there are four vertices $a_i,T_i,F_i,b_i$, connected as follows: $a_i-T_i-b_i$ and $a_i-F_i-b_i$.

For each clause $C_j = \ell_{j,1} \lor \ell_{j,2} \lor \ell_{j,3}$ (where $\ell_{j,1},\ell_{j,2},\ell_{j,3}$ are literals), there are five vertices $c_j,L_{j,1},L_{j,2},L_{j,3},d_j$, connected as follows: $c_j-L_{j,k}-d_j$ (for $k=1,2,3$). Additionally, if $\ell_{j,k} = x_i$ then we connect $L_{j,k}$ with $F_i$, and if $\ell_{j,k} = \bar{x}_i$ then we connect $L_{j,k}$ with $T_i$.

The source-sink pairs are $(a_i,b_i)$ (for $i=1,\ldots,n$) and $(c_j,d_j)$ (for $j=1,\ldots,m$).

Each $a_i-b_i$ path corresponds to a truth assignment of $x_i$, and each $c_j-d_j$ path corresponds to a choice of a satisfied literal in $C_j$.

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  • $\begingroup$ as a side question, can your reduction be adapted to work when the set of sources and sinks does not come in pair, i.e. a feasible solution is allowed to pair them arbitrarily. $\endgroup$ – Thinh D. Nguyen May 31 at 15:14
  • $\begingroup$ This seems like a completely different question. $\endgroup$ – Yuval Filmus May 31 at 15:59
  • $\begingroup$ So, I will ask another question. $\endgroup$ – Thinh D. Nguyen Jun 1 at 9:49

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