Graphs & Algorithms: Advanced Topics Nowhere-Zero Flows
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1 Graphs & Algorithms: Advanced Topics Nowhere-Zero Flows Uli Wagner ETH Zürich
2 Flows Definition Let G = (V, E) be a multigraph (allow loops and parallel edges). An (integer-valued) flow on G (also called a circulation) is a pair (D, f ), where D = (V, E) is an orientation of G; f : E Z is a function such that every v V satisfies f (uv) = f (vw) (Flow Conservation) u N + D (v) w N D (v) A flow is nowhere zero (a NFZ, for short) if f ( e) 0 for all e E. It is positive if f ( e) > 0 for all e E. A k-flow is a flow such that f ( e) < k for all e E. We write k-nfz as a shorthand for nowhere zero k-flow. The definitions of flows and nowhere zero flows make sense for functions that take values in an arbitrary abelian group Γ, e.g. Γ = Z k. We then speak of a Γ-flow or Γ-NFZ.
3 First Observations Proposition The following conditions are equivalent for a graph G: (i) G admits a k-nfz. (ii) G admits a positive k-flow. (iii) For every orientation of G there is a k-nfz. Corollary Admitting a k-nfz is a property of the underlying, undirected graph. (The same argument works for Γ-NFZ s.) Observation If G has a k-nzf then it has a Z k -NZF. (Later: Converse also true.) Proposition If (D, f 1 ) and (D, f r ) are flows (or Γ-flows) on G (same orientation) then so is (D, f 1 + f 2 ).
4 Which Graphs Admit a NFZ? If (D, f ) is a flow on G = (V, E) and if V = X Y is a partition of V then let E(X, Y ) be the set of directed edges e = xy with x X and y Y, and analogously for E(Y, X ). Lemma The flow along any (directed) cut is zero: For any cut V = X Y, e E(X,Y ) f ( e) }{{} =:f ( E(X,Y )) e E(Y,X ) f ( e) }{{} =:f ( E(Y,X )) Proposition A graph G admits a NFZ iff G is bridgeless. = 0 (Proof: Exercise) Question Is there a fixed k such that every bridgeless graph admits a k-nfz? (We will answer this question later.) (Exercise)
5 Small Values of k Observation No graph admits a 1-NZF. A graph admits a 2-NFZ iff it is even (all vertices have even degree). Theorem (Tutte, 1949) A cubic graph admits a 3-NZF iff it is bipartite. Proof. First assume that G = (A B, E) is bipartite. By Hall s theorem, the regular bipartite graph G has a perfect matching M. Orient the edges of M from A to B and give them flow value f ( e) = 2. Orient the remaining edges from B to A and give them flow value f ( e) = 1. Conversely, let G be a cubic graph with a (w.l.o.g.) positive k-flow. Each vertex must be incident to one edge with flow value 2 and two edges with flow value 1. Thus, the 2-edges form a perfect matching. Let A be the set of tails of the 2-edges, and let B = V \ A. Then each 2-edge points from A to B, and so each 1-edge must point from B to A, so V = A B is a bipartition.
6 Flow-Coloring Duality Observation A graph G = (V, E) admits a proper k-coloring (for some k that might depend on G) iff it is loopless. A coloring f : V {0,..., k 1} induces, for any orientation, a function g : E { (k 1),..., 0,..., k 1} by setting g( uv) = f (v) f (u). Note that g is nowhere zero iff f is proper. Moreover, g vanishes along directed cycles. Definition Let G = (V, E) be a plane graph. The dual of G is a graph G = (V, E ), where V is the set of faces of G, and (f 1, f 2 ) E iff they share an edge in G. Note that there is a natural 1-1 correspondence e e between E and E. If e is oriented, orient e by requiring that it cross e from left to right. Observation If e E is a bridge, then e E is a loop. If e 1,..., e k E are the edges of a cycle of G, then e1,..., e k E are the edges of a cut of G.
7 Flows and Colorings of Plane Graphs Theorem (Tutte 1954) A plane bridgeless graph G = (V, E) is k-face-colorable (its dual is k-colorable) iff it admits a k-nzf. Proof. Let c be a proper k-coloring of the faces of G. Let D be an arbitrary orientation of G. Define a function f : E Z by setting f ( e) = c(fl e ) c(f R e e ), where fl and f R e are the faces to the left of e, respectively. Then 0 < f ( e) < k for all e E and f is a flow. Conversely, let (D, f ) be a Z k -NZF on G. Define a k-coloring c of the faces of G as follows. Color the unbounded face with color 0. For any other face F, traverse faces (in an arbitrary route) to the outer face. Define c(f ) to be the (directed) sum of flows through edges you cross: Add f ( e) if e points to your right and subtract otherwise. All calculations are modulo k. Note that c is a well-defined proper k-coloring.
8 Group-Valued Flows Theorem (Tutte, 1954) Let Γ be an abelian group of order Γ = k, and let G be a graph. Then G admits a k-nzf iff it admits a Γ-NZF. (Exercise) Proposition G has a 4-NZF iff it is the union G = G 1 G 2 of two even subgraphs (use the group Z 2 Z 2 ). Theorem A cubic graph is 3-edge colorable iff it admits a 4-NZF. Proof. Use the group Z 2 Z 2. A Z 2 Z 2 -NZF is equivalent to a an edge coloring with the nonzero elements of Z 2 Z 2 (these sum up to zero). Corollary (Tait, 1878) A simple bridgeless cubic plane graph is 3-edge colorable iff it is 4-face colorable.
9 Tutte s Flow Conjectures Conjecture (3-NZF) Every 4-edge connected graph admits a 3-NZF. Conjecture (4-NZF) Every bridgeless graph without a Petersen graph minor admits a 4-NZF. Conjecture (5-NZF) Every bridgeless graph has a 5-NZF. Remark All three conjectures are true for planar graphs, by duality. For these, the first is dual to Grötzsch s Theorem 1. The second is dual to the 4-color theorem (since the Petersen graph is nonplanar). The third is dual to the 5-color theorem. 1 Every triangle-free planar graph is 3-colorable. (Not true for list colorings!)
10 Towards the NZF Conjectures (off by 1) Theorem Every 4-edge connected graph admits a 4-NZF. (Proof omitted.) Theorem (Kilpatrick 1975, Jaeger 1979) Every bridgeless graph admits an 8-NZF. Theorem (Seymour 1981) Every bridgeless graph admits a 6-NZF. Basic Idea of the Proof: Find a good even subgraph H G. Prove that G has a Z 3 -flow f that is (possibly) zero only on H. Let g be a Z 2 -NZF on H. Then (f, g) is a Z 3 Z 2 -NZF on G.
11 Construction of a good H W.l.o.g. G is connected, hence 2-edge connected. Construct J 0... J r G. J 0 is an arbitrary vertex of V, and J i = J i 1 H i F i. Stop with first J r that spans G. Definition of H i and F i : Let X i V \ V (J i 1 ) be minimal such that X i and e(x i, (V \ V (H i 1 ) \ X i ) 1 (exists since V \ V (H i 1 ) is a candidate). G is 2-edge connected, so by definition of X i, e(x i, V (J i 1 )) 1. For F i choose two arbitrary edges connecting X i and V (J i 1 ) if possible, else one. By minimality, G[X i ] is a single vertex or 2-edge connected. Let H i be any even connected subgraph of G[X i ] that contains the X i -endpoints of the edges in F i (possible by Menger s theorem for edge connectivity). Let G = (V, E ) := G E(J r ) and let H = J 0 r i=1 H i. Note that H is even and that J r is spanning and connected.
12 Construction of a Z 3 -Flow Define Z 3 -flows f r, f r 1,..., f 0 in reverse induction, such that f i is nowhere zero on E r j=i+1 F j, for every 0 i r. Induction base: For every e E, let C e denote an arbitrary circuit in J r {e} that contains e, and let f e be a flow with values ±1 on the edges of C e and 0 elsewhere. Let f r = e E f e. Induction step: f i f i 1 Case 1: F i = 1, say F i = {e}. Let f i 1 = f i. We claim f i (e) 0. It suffices to prove f i (e ) 0, where e is the unique edge in E(X i, (V \ V (J i 1 ) \ X i ). By induction, it suffices to show that e E r j=i+1 F j. Clearly, e i j=1 F j. Moreover, e is a bridge of G[V \ V (J i 1 )] and for every i k r, H k is a bridgeless subgraph of G[V \ V (J i 1 )], hence e r k=i H k. Case 2: F i = 2, say F i = {e 1, e 2 }. Since J i 1 and H i are both connected, there exists a cycle C in (J i 1 H i ) + F i that contains e 1 and e 2. If f i (e 1 ) 0 and f i (e 2 ) 0, then let f i 1 = f i. Otherwise w.l.o.g. f i (e 1 ) = 0 and f i (e 1 ) {0, 1}. Circulate a flow of 1 on C in the direction agreeing with e 2 ; denote this flow by g. Let f i 1 = f i + g. Note that f i 1 (e 1 ) 0, f i 1 (e 2 ) 0 and f i 1 (e) = f i (e) for e E r j=i+1 F j.
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