# Reduction from 3SAT [closed]

You are given a directed acyclic graph G = (V, E) in which each node has one “left” out-arc and one “right” out-arc, with a distinguished source node s and sink node t. You are also given a list of “ties” (u, v) which say that if you take the left [right] edge out of u, then you must also take the left [right] edge out of v. Is there a path from s to t subject to the ties? Show that this decision problem is NP-complete.

For this I know we can reduce a 3SAT to a problem of checking whether a path exists or not between 's' and 't'. But I am finding difficult to add the constraints of "ties" into it

## closed as unclear what you're asking by Evil, David Richerby, Luke Mathieson, Juho, vonbrandDec 19 '15 at 3:34

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• So you know there is a reduction, but you don't know how it works? – G. Bach Dec 4 '15 at 14:09
• I know the reduction for 3SAT to Check whether there exists a path between 's' and 't' but this problem is a variant of it. As, there are constraints on the nodes in terms of "ties". – letsBeePolite Dec 4 '15 at 14:14
• Hint: The ties represent pairs (or, by symmetry and transitivity, sets) of decisions for which you are required to make the same choice. Similarly, in 3SAT, for all occurrences of a given variable $x$ in a CNF you have to assume the same value. – Klaus Draeger Dec 4 '15 at 15:33
• @Check I rolled back your edit. If you want to ask a completely different question, ask it as a new question, not an edit of an existing one. – David Richerby Dec 4 '15 at 21:12
• Assuming that P$\,\neq\,$NP, you cannot reduce 3SAT to $s$-$t$-connectivity (at least, not with deterministic polynomial-time many-one reductions), since the former is NP-complete and the latter is in P. Do you have the reduction the wrong way round? – David Richerby Dec 4 '15 at 21:14

Suppose you are given a CNF $C_1\wedge\dots\wedge C_k$, with clauses of the form $C_i = L_i^1\vee L_i^2\vee L_i^3$, each literal $L_i^j$ being either $x_i^j$ or $\neg x_i^j$ for some variable $x_i^j\in V$. Construct a graph as follows:
• Set of nodes is $N=\{n_i^j\ |\ 1\le i\le k, 1\le j\le 3\}\cup\{\top,\bot\}$.
• For all $i,j$: let the "sat" successor be $n_{i+1}^1$ for $i<k$, and $\top$ otherwise; let the "unsat" successor be $n_i^{j+1}$ for $j<3$, and $\bot$ otherwise. If $L_i^j=x_i^j$, the left and right edge from $n_i^j$ go to the sat and unsat successor, respectively; if $L_i^j=\neg x_i^j$, it is the other way around.
• Add ties between all pairs of nodes $(n_i^j,n_p^q)$ with $x_i^j=x_p^q$.
Then a valuation of the variables in $V$ corresponds exactly to a choice of left or right such that for all nodes belonging to the same variable, the same direction is chosen; a satisfying valuation corresponds to a choice of directions which leads from $s=n_1^1$ to $t=\top$.