an algebraic analogue of a formula of Knuth

Transcription

1 An algebraic analogue of a formula of Knuth (MIT) FPSAC, San Francisco, August 6, 2010

2 Talk Outline Knuth s formula: generalizing n n with weights: generalizing (x x n ) n with group structure: generalizing (Z/nZ) n 1. Recent developments!

3 Starting Point: Cayley s Theorem The number of rooted trees on n labeled vertices is n n 1. Refinement: The number of trees with degree sequence (d 1,...,d n ) is the coefficient of x d x d n n in nx 1...x n (x x n ) n 2. We can break this out by root: n r=1 x i x r (x x n ) n 2 i r outdegrees indegrees

4 Oriented Spanning Trees An oriented spanning tree of K 3,3. Let G = (V,E) be a finite directed graph. An oriented spanning tree of G is a subgraph T = (V,E ) such that one vertex, the root, has outdegree 0; all other vertices have outdegree 1; T has no oriented cycles v 1 v 2 v k v 1.

5 Complexity of A Directed Graph The number κ(g) = # of oriented spanning trees of G is sometimes called the complexity of G. Examples: κ(k n ) = n n 1 κ(k m,n ) = (m + n)m n 1 n m 1 κ(db n ) = 2 2n 1

6 The De Bruijn Graph DB n DB 0 DB 1 DB 2 vertices {0,1} n, edges {0,1} n+1. The endpoints of the edge e = b 1...b n+1 are its prefix and suffix: e b 1...b n b 2...b n+1.

7 Directed Line Graphs G = (V,E) : finite directed graph s,t : E V Edge e is directed like this: s(e) t(e) The directed line graph LG = (E,E 2 ) of G has Vertex set E, the edge set of G. Edge set E 2 = {(e,f ) E E s(f ) = t(e)}. e f e f e e (e,f ) E 2 (e,f ) / E 2 f (e,f ) / E 2

8 A Graph G and Its Directed Line Graph LG a b c a b c

9 Examples of Directed Line Graphs Kn = L(one vertex with n loops). Km,n = L(two vertices {a,b} with m edges a b and n edges b a). DB n = L(DB n 1 ). Iterated line graphs: L n G = (E n,e n+1 ), where E n = {directed paths of n edges in G}.

10 Spanning Tree Enumerators Let (x v ) v V and (x e ) e E be indeterminates, and let κ edge (G,x) = T κ vertex (G,x) = T x e e T x t(e) e T The sums are over all oriented spanning trees T of G. Example: κ vertex (K n,x) = (x x n ) n 1.

11 Knuth s Formula G = (V,E) : finite directed graph with no sources outdegrees a 1,...,a n indegrees b 1,...,b n 1 LG : the directed line graph of G Theorem (Knuth, 1967). For any edge e of G, where κ(lg,e ) = α(g,e ) n i=1 a b i 1 i α(g,e ) = κ(g,t(e )) 1 a and a is the outdegree of t(e ). κ(g,s(e)). t(e)=t(e ) e e

12 Weighted Knuth s Formula G : finite directed graph with no sources LG : its directed line graph b 1,...,b n 1 : the indegrees of G. Theorem (L.) ( ) bi 1 κ vertex (LG,x) = κ edge (G,x) x e. i V s(e)=i Both sides are polynomials in the edge variables x e.

13 Specializing x e = 1 Complexity of a line graph: Examples: κ(lg) = κ(g) n i=1 a b i 1 i. G = one vertex with n loops, LG = K n, get n n 1. G = two vertices, LG = K m,n, get (m + n)m n 1 n m 1. G = DB n 1, LG = DB n : κ(db n ) = κ(db n 1 ) 2 2n 1 = κ(db n 2 ) 2 2n 1 2 2n 2 =... = 2 2n 1.

14 Rooted Version Fix an edge e = (w,v ) of G. Let b be the indegree of v. Theorem (L.) If b i 1 for all i, and b 2, then κ vertex (LG,e,x) = x e κ edge (G,w,x) ( ) bi 1 i V s(e)=i x e. s(e)=v x e

15 Matrix-Tree Theorem κ edge (G,x) = [t]det(t Id edge ). κ vertex (G,x) = [t]det(t Id vertex ). Goal: relate edge G with vertex LG.

16 The Missing Link: Directed Incidence Matrices Consider the K-linear maps Then A : K E K V, B : K V K E e t(e) v s(e)=v x e e. edge G vertex LG = AB D = BA D L where D and D L are the diagonal matrices ( ( D(v) = x e )v, D L (e) = x f )e. s(e)=v s(f )=t(e)

17 Intertwining edge G and vertex LG A vertex LG = A(BA D L ) = ABA DA = (AB D)A = edge G A In particular vertex LG (ker A) ker A. Writing K E = ker A Im(A T ) puts vertex LG in block triangular form.

18 The Proof Falls Into Place Since G has no sources, A : K E K V is onto. So AA T has full rank. So A : Im(A T ) K V is an isomorphism. So det( vertex LG Im(A T ) ) = det edge G = κ(g,x). Eigenvalues of vertex LG ker A are s(e)=i x e, each with multiplicity b i 1.

19 Comparison with Knuth Knuth s formula involved the strange quantity α(g,e ) = κ(g,t(e )) 1 a Why is it missing from our formulas? κ(g,s(e)). t(e)=t(e ) e e

20 The Unicycle Lemma A unicycle of G is an oriented spanning tree together with an outgoing edge from the root. By counting unicycles through v in two ways, we get: Lemma. κ edge (G,v,x) x e = κ edge (G,s(e),x)x e. s(e)=v t(e)=v So Knuth s formula simplifies to κ(lg,e ) = 1 a κ(g,s(e )) n i=1 a b i 1 i.

21 The Sandpile Group of a Graph K(G,v ) Z n 1 / Z n 1, where = D A is the reduced Laplacian of G. Lorenzini 89/ 91 ( group of components ), Dhar 90, Biggs 99 ( critical group ), Baker-Norine 07 ( Jacobian ). Directed graphs: Holroyd et al. 08 Matrix-tree theorem: #K(G,v ) = det = #{spanning trees of G rooted at v }. Choice of sink: K(G,v ) K(G,v ) if G is Eulerian.

22 Maps Between Sandpile Groups Theorem (L.) If G is Eulerian, then the map Z E Z V e t(e) descends to a surjective group homomorphism K(LG,e ) K(G,t(e )).

23 Maps Between Sandpile Groups Theorem (L.) If G is balanced k-regular, then the map Z V Z E v s(e)=v descends to an isomorphism of groups e K(G) k K(LG). Analogous to results of Berget, Manion, Maxwell, Potechin and Reiner on undirected line graphs. arxiv:

24 The Sandpile Group of DB n De Bruijn Graph DB n = L n (a single vertex with 2 loops). Theorem (L.) n 1 K(DB n ) = (Z/2 j Z) 2n 1 j. j=1 Generalized by Bidkhori and Kishore to k-ary De Bruijn graphs for any k.

25 Equating Exponents By counting spanning trees, we know that Now write #K(DB n ) = κ(db n,v ) = 2 2n n 1. K(DB n ) = Z a 1 2 Za 2 4 Za Za m 2 m for some nonnegative integers m and a 1,...,a m satisfying m ja j = 2 n n 1. (1) j=1 By the previous theorem and inductive hypothesis K(DB n 1 ) 2K(DB n ) Z 2n 3 2 Z 2n 4 4 Z 2 n 2 Z a 2 2 Za Za m 2 m 1. So m = n 1 and a j = 2 n j 1.

26 A (Formerly) Open Problem From EC1 In EC1, Stanley asks for a bijection {pairs of binary De Bruijn sequences of order n} {all binary sequences of length 2 n } Both sets have cardinality 2 2n.

27 Recent Developments In arxiv: , Bidkhori and Kishore give a bijective proof of the weighted Knuth formula ( ) bi 1 κ vertex (LG,x) = κ edge (G,x) x e i V s(e)=i and use it to solve Stanley s problem! Perkinson, Salter and Xu give a surjective map even when G is not Eulerian. K(LG,e ) K(G,s(e ))

28 Now What? unweighted weighted enumeration κ(g) κ(g, x) algebra K(G)?

29 Thank You! a b c a b c References: D. E. Knuth, Oriented subtrees of an arc digraph, J. Comb. Theory 3 (1967), L., Sandpile groups and spanning trees of directed line graphs, J. Comb. Theory A, to appear. arxiv: