Majority Consensus and the Local Majority Rule

we only show one further result that emphasizes our point that understanding LMP is fundamental to understanding any generalization of this process. A simple generalization of the local majority process would allow vertex v to have some resistivity towards color switch. Formally, for a nonnegative integer kv, we de ne a kv-local majority rule for vertex v: ct+1 v = ( ctv if jfw 2 Nv : ctw = ctvgj jN(v)j 2 + kv 1 ctv if jfw 2 Nv : ctw 6= ctvgj > jN(v)j 2 + kv (8) The value kv is called the resistivity value of vertex v and we call the graph G = (V;E) together with the set of vertex resistivities fkv : v 2 V g a varied-resistivity graph. Similarly, the process de ned by (8) is called the local majority process with resistivities. Note that the local majority process with resistivities where kv = 0, v 2 V , is exactly the local majority process. As the next theorem shows, introducing vertex resistivities does not bring additional di culties: the local majority process with resistivities can be simulated by the (standard) local majority process. Theorem 7. Let G(V;E) be a varied-resistivity graph, with the vertex set V = fv1; : : : ; vng, and the corresponding resistities R = fkv1 ; : : : ; kvng. The local majority process with resistivities on the varied-resistivity graph G can be simulated by the local majority process on some graph G0(V 0; E0). Proof. We will derive the graph G0(V 0; E0) iteratively. Let the graph G0(V0; E0) = G(V;E), where the vertices of G0 are labelled di erently, i.e. V0 = fv1 1 = v1; : : : ; vn n = vng. The edges are the same. Also, let c0 be the coloring de ned on G, and let the corresponding coloring for G0 be c00. We now show how to derive Gi from Gi 1. The vertices of the graph Gi 1(Vi 1; Ei 1) are Vi 1 = fv1 1 ; : : : ; vm 1 ; : : : ; v1 n; : : : ; vm n g, where kvj l = kvj0 l for all 1 j; j0 m and for all l. If kvj i = 0, then we set Gi = Gi 1, i.e. Vi = Vi 1 and Ei = Ei 1. Otherwise, we do the following. For ease of notation, we refer to kvj i as ki, since all vj i vertices have the same resistivity by assumption. For each vertex vj l 2 Vi 1, we add ki new vertices. We shall denote these ki new vertices for each vertex vj l 2 Vi 1 as vd l , where d = (m + (j 1)kvj l + 1); : : : ; (m + jkvj l ). We shall call vj l the parent vertex of all such ki new vd l vertices. More formally, Vi = Vi 1 [ki u=1 fvm+(j 1)ki+u l : vj l 2 Vi 1g. The resistivity values of all the vertices are reassigned as kvj l = ( 0 if l = i, or vj0 i is parent of vj l , for any j0 kvj l if l 6= i, or 8j0, vj0 i is not parent of vj l (9) That explains all the vertices we add to the graph. We add the edges to the new graph as follows. First, for each vertex vd i , (m+ (j 1)ki) d (m+ jki), we add the edge (vd i ; vj i ), where vj i is the parent vertex of the vertex vd i . For each pair of vertices vd i ; ve i , for all (m + (j 1)ki) d < e (m + jki), we add the edges (vd i ; ve i ) to Ei. Then, for every pair of vertices vd l ; vd l0 such that the parent of vd l is vj l and the parent of vd l0 is vj l0 , we add the edge (vd l ; vd l0) to Ei whenever (vj l ; vj l0 ) 2 Ei 1. More formally, Ei is equal to Ei 1 [ (f(vd i ; vj i ) : (m+ (j 1)ki) d (m+ jki)g) [ ([mj=1f(vd i ; ve i ) : (m+ (j 1)ki) d < e (m+ jki)g) [ ([ki u=1f(vm+(j 1)ki+u l ; vm+(j0 1)ki+u l0 ) : (vj l ; vj0 l0 ) 2 Ei 1): This completes the construction of the new graph. Note that jVij = kijVi 1j. Also, by the assignment of resistivities, kvj i = 0 for all j. We now extend the coloring c0i 1 for the coloring c0i of the new graph simply by making the color of each new vertex as the color of its parent vertex. More formally, c0i = fc0i (vm+(j 1)ki+u l ) = c0i 1(vj l ) : 1 u ki; 1 j mg. Note that now c1 according to the local update procedure can be found on Gi by extending c1 from Vi 1to Vi, or by applying the local update on c0 in Gi directly.Finally, note that Gn has djV j vertices, where d =Qnu=1 ku. Also, since for the graph Gi, kj = 0 for allj i, Gn has the resistivity values of all vertices as zero. At each step, Gi simulates Gi 1, and thus Gnsimulates G0 = G.ut6 Conclusions and DirectionsThe main result of this paper is that failure-free computation of majority consensus by iterative applicationsof the local majority rule is possible only in the networks that are nowhere truly local (Theorem 4). Inother words, the idea of solving a truly global task (reaching consensus on majority) by means of trulylocal computation only (local majority rule) is doomed for failure. However, even well connected networksof agents that are nowhere truly local might fail to reach majority consensus when iteratively applying thelocal majority rule. We have investigated the properties of majority consensus computers, i.e., the networksin which iterative application of the local majority rule always yields consensus in the initial majority state.There are several directions that might be of potential interest. One such direction that was not of ourinterest here is to determine the complexity of the decision problem:MCC. Input is a nite graph G. Is G a majority consensus computer?Clearly,MCC is in co-NP because of Theorem 3 and it is very likely thatMCC is co-NP complete. 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Suppose that there exists t such that maj(ct ) 6= maj(c0). De ne another initial coloring d0 = ct ,and observe that dt = ct +t.Since G is a m.c.c., the local majority process reaches consensus on the maj(c0). In other words, thereexists a t0 such that ct = maj(c0) for every t t0. Thus, for t t0 t , maj(dt) = maj(ct+t ) = maj(c0) 6=maj(ct ) = maj(d0). Therefore G is not a m.c.c. since it does not admit majority consensus for d0. Acontradiction.utLemma 8 (1). LetVi(ct) = fv 2 V :ctv = ig, i = 0; 1, where ct is a coloring of G = (V;E). If there existsi 2 f0; 1g and colorings ct and dt0 such thatVi(ct) Vi(dt0) thenVi(ct+k) Vi(dt0+k) for k = 0; 1; 2; : : : .Proof. By induction on k. If k = 0 there is nothing to prove. SupposeVi(ct+k) Vi(dt0+k). We have to showthat, for every v 2 V , ct+k+1v= i ) dt0+k+1v= i. It follows from the assumption that N(v) \Vi(ct+k)N(v) \Vi(dt0+k) and, in particular, ct+kv = i ) dt0+kv = i. Hence, if ct+k+1v= i because jN(v) \Vi(ct+k)j >jN(v)j=2, then jN(v) \Vi(dt0+k)j > jN(v)j=2 also, and dt0+k+1v= i. If ct+k+1v= i because ct+kv = i andjN(v)\Vi(ct+k)j = jN(v)j=2, then dt0+kv = i andjN(v)\Vi(dt0+k)j jN(v)j=2 which shows that dt0+k+1v= i.utTheorem 8 (2). Suppose G admits majority consensus for any c0 such that sum(c0) = n=2+1. Then G isa majority consensus computer.Proof. By symmetry, if G admits majority consensus for all c0 with maj(c0) = 1, then G admits majorityconsensus for all c0 with maj(c0) = 0 also. (If G does not admit majority consensus for c0 with maj(c0) = 0,the n G does not admit majority consensus for 1 c0 also. Note that maj(1 c0) = 1 maj(c0) = 1.)Suppose d0 such that maj(d0) = 1, i.e., sum(d0) (n + 1)=2. Thus, there exists a c0 with sum(c0) =(n + 1)=2 such that c0 and d0 satisfy conditions of Lemma 1 with i = 1 (e.g., construct c0 from d0 bychanging color of any sum(d0) sum(c0) vertices w such thatd0w = 1). By assumption, G admits majorityconsensus for c0. Thus, using terminology of Lemma 1, there exists t such thatV1(ct) = V and, by thelemma,V1(ct) V1(dt). Therefore, dt is a consensus with maj(dt) = 1 which shows that G admits majorityconsensus for d.utProposition 6 (2). If v is a master in G, then ct+1v = maj(ct). More generally, if v is a k-master in Gand jsum(ct) n=2j (k + 1)=2, then ct+1v = maj(ct).Proof. Recall that a vertex v is a k-master if deg(v) = n (k+1). Note that jfw 2 V :ctw = 1 maj(ct)gjdeg(v)=2 implies ct+1(v) = maj(ct) (at time t, at most half of v's neighbors have color 1 maj(ct); if each ofthe two colors is the color of exactly half of v's neighbors, then no other vertex has color 1 maj(ct) and, inparticular,ctv = maj(ct) and the local majority process ensures ct+1v =ctv = maj(ct)). In order to completethe proof note that jsum(ct) n=2j (k+1)=2 is equivalent to jfw 2 V :ctw = 1 maj(ct)gj (n (k+1))=2.utProposition 7 (4). If (G) n 3, then G admits majority consensus for every c0 such that sum(c0)(n+ 3)=2.Proof. Note that every v 2 V is either a master, a 1-master, or a 2-master. Thus, by Proposition 2,c1v =maj(c0) for every v 2 V .utLemma 9 (4). Let ct be a coloring of G, (G) n 3, such that sum(ct) = (n+ 1)=2. Let H = (VH ; EH)be a connected component of Gc with jVH j 2. Suppose thatctv =ctw for every v; w 2 VH . Then ct+1v = 1ctvfor every v 2 VH . Proof. Let jVH j = l. Ifctv = 1 for all v 2 VH , then result follows from Lemma 4 with jSj = l. If Ifctv = 0 forall v 2 VH , then result follows from Lemma 4 with jSj = 0.utLemma 10 (5). Let ct be a coloring of G, (G) n 3, such that sum(ct) = (n+1)=2. Let C2k Gc be aconnected component in Gc. Suppose the colors assigned by Ct alternate along the cycle: if u is adjacent tov in C2k thenctu = 1ctv. Then ct+1v =ctv for every v 2 C2k.Proof. Every v 2 C2k is a 2-master and, by (c) of Lemma 2, ct+1v = 1ctuctw = 1 (1ctv), because in C2k,v is adjacent to both u and w.ut