It is known that SAT is hard even when restricted to, e.g., formulas with exactly 3 literals per clause and at most 4 occurrences per variable. On the other hand it is easy if there are exactly 3 literals per clause and at most 3 occurrences per variable. I am interested in other restrictions on the number of occurrences per variable.

For a set $S$ of positive integers, define $(k,S)$-SAT to be the satisfiability problem restricted to boolean formulas with exactly $k$ literals per clause, and the number of occurrences of each variable is in the set $S$. For example $(3,\{1,2,3,4\})$-SAT is hard, $(3,\{1,2,3\})$-SAT is easy.


If $k=3$ and $S$ is the even numbers, then you can easily pad an instance of $3$-SAT to get an equivalent instance of $(3,S)$-SAT. Is there anything known more generally? Is it always hard if $S$ is infinite? Are there any easy cases for $k=3$ besides $S\subseteq \{1,2,3\}$?

Main Question

I am really interested in 1-in-3-SAT restricted to the case where each variable appears an even number of times - the standard reduction from 3-SAT introduces new variables some of which occur an odd number of times. My feeling is that the problem should be NP-hard. Is there a suitable reduction from SAT? Or from normal 1-in-3-SAT? Or perhaps some sort of dichotomy theorem I don't know about that can be applied?


The answer to my question as stated is that the problem is indeed hard, because you can simply copy all clauses of a 1-in-3-SAT instance. I now insist that the clauses should not be repeated (we can efficiently preprocess an instance of SAT so that there are no repeat clauses after all)

  • $\begingroup$ Take a hard instance $I$ of 1-in-3-SAT, and add a copy of every individual clause. If a variable occurs $t$-times in $I$, then it occurs $2t$-times in the new instance with the copied clauses. $\endgroup$ – Gamow Sep 11 '19 at 11:05
  • $\begingroup$ @Gamow :facepalm: yes of course.. unfortunately I'm not sure that helps me in my research... perhaps I should insist that clauses aren't repeated? The clauses should be a set, not a multiset... I will edit the question I think. $\endgroup$ – Christopher Purcell Sep 11 '19 at 11:50

Theorem: The special case of 1-in-3-SAT where each variable appears an even number of times is NP-hard.

Proof: Consider an instance $I$ of 1-in-3-SAT, and let $a_1,\ldots,a_n$ be an enumeration of the variables in $I$. Assume that variables $a_1,\ldots,a_m$ occur an odd number of times, whereas $a_{m+1},\ldots,a_n$ occur an even number of times. Without loss of generality $m$ is an even number (and otherwise, add a new dummy clause with three newly created dummy variables).

From instance $I$, we create a new instance $I'$ of 1-in-3-SAT in which every variable occurs an even number of times. Furthermore, $I$ is satisfiable if and only if $I'$ is satisfiable:

  • Instance $I'$ contains all variables and all clauses of $I$.
  • For every variable $a_i$, instance $I'$ contains a (new) corresponding variable $b_i$.
  • Furthermore, $I'$ contains a (new) variable $c$.
  • For every clause $a_i,a_j,a_k$ in $I$, instance $I'$ contains a new clause $\lnot b_i, \lnot b_j,\lnot b_k$.
  • For $i=1,\ldots,m$, instance $I'$ contains a new clause $a_i,b_i,c$.

It is easy to check that every variable in $I'$ occurs in an even number of clauses. If $I'$ is satisfiable, then this induces a satisfying assignment for $I$.

Finally consider a satisfying assignment for $I$. Use truth values $b_i:=\lnot a_i$ for $1\le i\le n$ and $c:=$FALSE. The resulting assignment satisfies all clauses in $I'$.

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  • $\begingroup$ This absolutely answers my question and I think it would be unfair to move the goalposts now, so I am happy to accept (I am interested in this question for more general reasons) However, the problem I am really interested in is the monotone (actually, positive) version of 1-in-3-SAT. If you have any thoughts please let me know. $\endgroup$ – Christopher Purcell Sep 11 '19 at 14:36

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