A graph G has an S-factor if there exists a spanning subgraph F of G such that for all v ∈ V : deg F (v) ∈ S. The simplest example of such factor is a 1-factor, which corresponds to a perfect matching in a graph. In this paper we study the computational complexity of finding S-factors in regular graphs. Our techniques combine some classical as well as recent tools from graph theory. ACM Subject Classification Mathematics of computing → Matchings and factors; Theory of computation → Problems,

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... uctions and completeness ϕ(0, 1) = ϕ(1, 0) = 1. In this case: ϕ(x i , y i ) = NE(x i , y i ). ϕ(0, 1) = 0, ϕ(1, 0) = 1. In this case: ∃y i ϕ(x i , y i ) = x i and ∃x i ϕ(x i , y i ) = ¬y i . ϕ(0, 1) = 1, ϕ(1, 0) = 0. In this case: ∃y i ϕ(x i , y i ) = ¬x i and ∃x i ϕ(x i , y i ) = y i . Next, we show that for symmetric relations we can derive further closure properties under some technical conditions. Lemma 29. Let R be a symmetric 2 -ary relation such that: ∈ Spec(R) and { − 1, + 1} ⊆ Spec(R). Then {R, NE} read-twice-implements EQ 3 or {x, ¬x}. Proof. We define the following two sets: Furthermore, let a − = min S − and a + = min S + . We define a − or a + to be infinity if S − or S + is empty, respectively. We consider three cases: Case 1: a + = a − . Observe that a − ≥ 2. Using NE and Lemma 25, we plug − a − pairs z i , ¬z i into the relation R. Formally, consider, Consequently, ∃zR(z,ȳ) = EQ k (ȳ), where k = 2a − ≥ 4. Case 2: a + > a − . Observe that a − ≥ 1 and consider R(z,ȳ) as above. Now, however, since a + > a − : Hence, we obtain ¬y i . Using NE, we can obtain y i . Case 3: a − > a + . Observe that a + ≥ 1. We repeat the argument of Case 2 for the dual relation R * of R. As R * read-twice-implements {x, ¬x}, so does R. We use a similar argument for relations of odd arity. Lemma 30. Let R be a symmetric 2 + 1-ary relation such that: { , + 1} ⊆ Spec(R). Then {R, NE} read-twice-implements EQ 3 or {x, ¬x}. Proof. We define the following two sets: S − = {a | R( − a) = 1 } and S + = {a | R( + 1 + a) = 1 }. Furthermore, let a − = min S − and a + = min S + . We define a − or a + to be infinity if S − or S + is empty, respectively. We consider three cases: Case 1: a + = a − . Observe that a − ≥ 1. Consider, Consequently, ∃zR(z,ȳ) = EQ k (ȳ), where k = 2a − + 1 ≥ 3. Case 2: a + > a − . Observe that a − ≥ 0 and consider R(z,ȳ) as above. Now, however, since a + > a − : R(z,ȳ) = 1 ⇐⇒ w H (ȳ) = 0. Hence, we obtain ¬y i . Using NE, we can obtain y i . Case 3: a − > a + . Observe that a + ≥ 0. We repeat the argument of Case 2 for the dual relation R * of R. As R * read-twice-implements {x, ¬x}, so does R.