About the number of regular vines on n nodes

Outline of the talk Previous Approach The Prüfer code Trees (Cayley’s theorem) Vines Line Graph & Regular Vines Some Results About the number of regular vines Unlabeled regular vines Final Comments

Regular & Non-Regular vines

The Prüfer code Every sequence R = (a1, a2, ..., an−2) where each ai is an integer not greater than n is a Prüfer Code for some tree on n nodes. a1 = 2 a2 = 2 a3 = 4 a4 = 5 a5 = 5

Vines & Prüfer codes

Vines & Prüfer codes

Vines & Prüfer codes

Catalogue

Regular Vines & the Line Graph The line graph LG(G) of a graph G has as its nodes the edges of G, with two nodes being adjacent if the corresponding edges are adjacent in G 1 2 3 4 5 6 2 2 4 5 5 7 1 3 2 4 6 5

Regular Vines & the Line Graph A spanning subgraph T of a graph G is a subgraph with the same set of nodes as G. If T is a tree, it is called a spanning tree of G 1 2 3 4 5 6 2 2 4 5 5 7 1 3 2 4 6 5

Catalogue

Catalogue

Lesson learned It is a good idea to store a regular vine on an array such as V2 Information regarding nodes adjacent to each other is lost in lower order trees

Solution Build an array that keeps the information required. Start by defining a Natural Order similarly to a Prüfer code A natural order of the elements of a regular vine on n elements is a sequence of numbers NO(n) = (An,An-1,…, A1) where each Ai is an integer not greater than n obtained as follows: (3,6,1,5,2,4)

With NO(n) construct a lower triangular array as follows: Write NO(6) = (3,6,1,5,2,4) in the main diagonal Aj-1,j equals element j-1 in the NO Element Ai,j codes node (Ai,j , Aj,j | Ai-1,j , …, A1,j )

Example Ai-1,h is a child of Ai,j if {Ah,h, Ai-1,h, Ai-2,h,…, A1,h} ⊂ {Aj,j , Ai,j , Ai-1,j, Ai-2,j,…, A1,j} |{Ah,h, Ai-1,h, Ai-2,h,…, A1,h}| = |{Aj,j , Ai,j , Ai-1,j, Ai-2,j,…, A1,j}| -1 |{Ah,h, Ai-1,h} ∩ {Aj,j, Ai,j}| = 1 Example {6, 1, 2, 4, 5} ⊂ {3, 6, 5, 1, 4, 2} |{6, 1, 2, 4, 5}| = |{3, 6, 5, 1, 4, 2}| -1 |{6, 1} ∩ {3, 6}| = 1

When does TA(n) code a regular vine? TA(n) represents a regular vine⇔for all i ≥ 2, element Ai,j = Ah,h or Ai,j = Ai-1,h for some h such that i ≤ h < j and {Aj,j ,…, Ai+1,j} ∩ {Ai-1,h,…, A1,h} = ∅ ⇒Note that if TA(n) represents a regular vine then every Ai,j will have two children in i-1 one is in j the other in h and verify the previous definition ⇐ By induction taking care to show that Ak,n is a child of Ak+1,n it has a second child Ak,h which together with Ak,n have a common child.

How many ways of extending TA(n-1)? The top two elements are fixed by the construction of TA(7). The last element is fixed by the choice of the previous. Node Ak,n satisfies regularity if it has a sibling which is a child of Ak+1,n and has a child which is also a child of its sibling. A5,7 = 6 or 1

Node Ak,n satisfies regularity if it has a sibling which is a child of Ak+1,n and has a child in TA(n-1) which is also a child of its sibling.

Node Ak,n satisfies regularity if it has a sibling which is a child of Ak+1,n and has a child in TA(n-1) which is also a child of its sibling. 6 ⇒ 5 appears in both children of 1,6|2,4,5 7 , {1,2,6,5} ⇒ 4 appears in both children of 1,6|2,4,5 7 , {1,4,6,5} ⇒ 2 appears in both children of 1,6|2,4,5

Node Ak,n satisfies regularity if it has a sibling which is a child of Ak+1,n and has a child in TA(n-1) which is also a child of its sibling.

Extending an n-1 reg. vine 2 possibilities for each Ai,n 1 < i < n-1 For any regular vine on n-1 elements, the number of regular n vines which extend this vine, preserving the natural ordering is 2n-3

With a fixed NO The number of regular vines possible with a fixed natural order NO(n) = (An,An-1,…, A1)

Finally How many NO(n) = (An,An-1,…, A1) are there? There are regular vines in total

Previous results verify the calculations. 2520 + 9 * 2520 + 19 * 5040 + 840 * 33 + 630 * 80 + 2520 * 168 + 840 * 168 + 1260 * 342 + 420 * 1452 + 210 * 2928 + 7 * 23040 = 2,580,480 = 7C2 * 5! * 2^(5C2)

Unlabeled vines check for tree isomorphism in each level of the vine Equivalence classes (Joe) Very large number of regular vines for representing multivariate distributions

Constructing all possible vines on n nodes. Construct all Prufer codes for the first tree in the vine. The edges of each one of the nn−2 trees in the previous step become nodes in the next tree. Hence, for each tree in the previous step: Label the edges of each tree giving the label 1 to the edge appearing in the first column in its extended Prufer code, 2 to the edge in the second column and so on until all edges have been labeled Construct all Prufer codes possible for this new tree and connect the new labeled edges as nodes according to these new Prufer codes. Repeat this process until two edges must be connected in the last tree. At this point there is only one way to connect them and no Prufer code is required.

Constructing all possible regular vines on n nodes. Construct all Prufer codes for the first tree in the vine. The edges of each one of the nn−2 trees in the previous step become nodes in the next tree. Hence, for each tree in the previous step: Label the edges of each tree giving the label 1 to the edge appearing in the first column in its extended Prufer code, 2 to the edge in the second column and so on until all edges have been labeled Construct the line graph of each one of the trees from step 2. For each line graphs from step 3 find all possible spanning trees. Connect the edges of each tree in step one according to all spanning trees from its line graph. Repeat this process until two edges must be connected in the last tree. At this point there is only one way to connect them and no operation is required. There are more regular vines on six nodes than hairs in a cow’s tail.