Counting the spanning trees of the 3-cube using edge slides Christopher Tuffley Institute of Fundamental Sciences Massey University, Manawatu 2009 New Zealand Mathematics Colloquium Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 1/15 Outline 1 Introduction Cubes and spanning trees Counting spanning trees: ways and means 2 The 3-cube Edge slides Counting the trees 3 Higher dimensions Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 2/15 Introduction Cubes and spanning trees Cubes {1, 2, 3} Definition The n-cube is the graph Q with n {1, 2} {2, 3} vertices the subsets of {1, 3} [n]= {1, 2,..., n}; an edge between S and R if they {2} {1} {3} differ by adding or deleting a single element. ∅ Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 3/15 Introduction Cubes and spanning trees Spanning trees Definition A spanning tree of a connected graph G is a maximal subset of the edges that contains no cycle; equivalently, a minimal subset of the edges that connects all the vertices. Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 4/15 Introduction Counting spanning trees: ways and means Counting trees — the matrix way Theorem (Kirchoff’s Matrix-Tree Theorem) The number of spanning trees of a simple connected graph G is given by the determinant of a matrix associated with G — the Laplacian of G, with row i, column i deleted. 1 4 3 − 1 − 1 − 1 − 1 2 −1 0 − 1 −1 2 0 − 1 0 0 1 2 3 Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 5/15 Introduction Counting spanning trees: ways and means Counting trees — the matrix way Theorem (Kirchoff’s Matrix-Tree Theorem) The number of spanning trees of a simple connected graph G is given by the determinant of a matrix associated with G — the Laplacian of G, with row i, column i deleted. 1 4 2 −1 0 det = 3 −1 2 0 0 0 1 2 3 Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 5/15 Introduction Counting spanning trees: ways and means Counting trees — the combinatorial way Model: Prüfer code for spanning trees of Kn (Prüfer, 1918) The Prüfer code is a bijection n−2 spanning trees of Kn ←→ {1,..., n} n−2 — recovering Cayley’s Theorem that Kn has n spanning trees. 1 2 6 Prüfer code 3411 5 3 4 Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 6/15 Introduction Counting spanning trees: ways and means Spanning trees of the n-cube Known result The n-cube has n 2n−n−1 n 2|S| = 2 k(k) Y Y S⊆[n] k=1 |S|≥2 spanning trees. For n = 3 this gives 24 · 23 · 3 = 384 spanning trees. Proof. The Matrix-Tree Theorem + clever determination of eigenvalues. See e.g. Stanley, Enumerative Combinatorics, Vol II. Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 7/15 Introduction Counting spanning trees: ways and means Spanning trees of the n-cube Known result The n-cube has n 2n−n−1 n 2|S| = 2 k(k) Y Y S⊆[n] k=1 |S|≥2 spanning trees. For n = 3 this gives 24 · 23 · 3 = 384 spanning trees. Proof. The Matrix-Tree Theorem + clever determination of eigenvalues. See e.g. Stanley, Enumerative Combinatorics, Vol II. Problem Stanley: “A direct combinatorial proof of this formula is not known.” Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 7/15 Introduction Counting spanning trees: ways and means A weighted count Theorem (Martin and Reiner, 2003) With respect to certain weights q1,..., qn, x1,..., xn we have − qdir(T )x dd(T ) = q · · · q q (x 1 + x ). X 1 n Y X i i i s. trees S⊆[n] i∈S |S|≥2 of Qn dir(T ) degree of qi in q is the number of edges in direction i dd(T ) degree of xi in x is the number of edges in the “upper” i-face minus the number in the “lower”. Suggests that a spanning tree of Qn l a choice of element and sign at each vertex of cardinality 2. Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 8/15 Introduction Counting spanning trees: ways and means A weighted count Theorem (Martin and Reiner, 2003) With respect to certain weights q1,..., qn, x1,..., xn we have − qdir(T )x dd(T ) = q · · · q q (x 1 + x ). X 1 n Y X i i i s. trees S⊆[n] i∈S |S|≥2 of Qn dir(T ) degree of qi in q is the number of edges in direction i dd(T ) degree of xi in x is the number of edges in the “upper” i-face minus the number in the “lower”. Suggests that a spanning tree of Qn l a choice of element and sign at each vertex of cardinality 2. Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 8/15 The 3-cube Edge slides Edge slides {1, 2, 3} Definition An edge of a spanning tree is slidable if it can be “slid” across a face of the {1, 2}{1, 3} {2, 3} cube to give a second spanning tree. Observation {1} {3} {2} An edge that may be slid in direction i must lie on the path joining two i-edges. ∅ Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 9/15 The 3-cube Edge slides Edge slides {1, 2, 3} Definition An edge of a spanning tree is slidable if it can be “slid” across a face of the {1, 2}{1, 3} {2, 3} cube to give a second spanning tree. Observation {1} {3} {2} An edge that may be slid in direction i must lie on the path joining two i-edges. ∅ Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 9/15 The 3-cube Edge slides Edge slides {1, 2, 3} Definition An edge of a spanning tree is slidable if it can be “slid” across a face of the {1, 2}{1, 3} {2, 3} cube to give a second spanning tree. Observation {1} {3} {2} An edge that may be slid in direction i must lie on the path joining two i-edges. ∅ Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 9/15 The 3-cube Edge slides Edge slides {1, 2, 3} Definition An edge of a spanning tree is slidable if it can be “slid” across a face of the {1, 2}{1, 3} {2, 3} cube to give a second spanning tree. Observation {1} {3} {2} An edge that may be slid in direction i must lie on the path joining two i-edges. ∅ Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 9/15 The 3-cube Edge slides Edge slides {1, 2, 3} Definition An edge of a spanning tree is slidable if it can be “slid” across a face of the {1, 2}{1, 3} {2, 3} cube to give a second spanning tree. Observation {1} {3} {2} An edge that may be slid in direction i must lie on the path joining two i-edges. ∅ Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 9/15 The 3-cube Edge slides Existence Lemma A minimal path joining two i-edges contains a unique edge that may be slid in direction i. Proof (length three case only). u v Vertices u and v must meet edges of the tree. There are three possibilities. Corollary A tree with k edges in direction i has k − 1 edges that may be slid in direction i, for a total of exactly four possible slides. Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 10/15 The 3-cube Edge slides Existence Lemma A minimal path joining two i-edges contains a unique edge that may be slid in direction i. Proof (length three case only). u v Vertices u and v must meet edges of the tree. There are three possibilities. Corollary A tree with k edges in direction i has k − 1 edges that may be slid in direction i, for a total of exactly four possible slides. Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 10/15 The 3-cube Edge slides Existence Lemma A minimal path joining two i-edges contains a unique edge that may be slid in direction i. Proof (length three case only). u v Vertices u and v must meet edges of the tree. There are three possibilities. Corollary A tree with k edges in direction i has k − 1 edges that may be slid in direction i, for a total of exactly four possible slides. Christopher Tuffley (Massey University) Counting the spanning trees of the 3-cube NZMC2009 10/15 The 3-cube Edge slides Existence Lemma A minimal path joining two i-edges contains a unique edge that may be slid in direction i. Proof (length three case only). u v Vertices u and v must meet edges of the tree. There are three possibilities. Corollary A tree with k edges in direction i has k − 1 edges that may be slid in direction i, for a total of exactly four possible slides.