1. DIAM About the number of vines on n nodes

2. TU Delft

3. Outline of the talk  Introduction  The Prüfer code  Trees  Vines  Line Graph & Regular Vines  Some Results  Final Comments

4. TU Delft Enumeration Problems  How much free beer have you received since the age of 12?  Jedediah Buxton (1702-1772) British  5,116 pints (5 ounces a day in 56 years)  2,906 lt (142 ml a day in 56 years)  How many hairs are there in a cow’s tail?  Thomas Fuller (1710-1790) African shipped to US as Slave  2,872  How many labeled trees are there on n nodes?  How many Vines are there on n nodes?

5. TU Delft Labeled Trees  A tree is a connected graph with no cycles  Cayley in 1889  # of isomers of hydrocarbons C n H 2n+2  The number of labeled trees on n nodes is n n−2  Ernst Paul Heinz Prüfer (1896 - 1934)  Prüfer code (1918)

6. TU Delft The Prüfer code  Every sequence R = ( a 1, a 2,..., a n−2 ) where each a i is an integer not greater than n is a Prüfer Code for some tree on n nodes.  Remove the endpoint with the smallest label and let a 1 be the label of the unique node which was adjacent to it. Delete endpoint & edge. Repeat this process to find a 2, a 3 and so on until a tree on two nodes has been found.  R = (2, 2, 4, 5, 5)  The process may be reversed. a 1 = 2 a 2 = 2 a 3 = 4 a 4 = 5 a 5 = 5

7. TU Delft  R = (2, 2, 4, 5, 5) 7 2 2 4 5 5 7 1 2 2 4 5 5 7 1 3 2 2 4 5 5 7 1 3 2 2 2 4 5 5 7 1 3 2 4 2 2 4 5 5 7 1 3 2 4 6 2 2 4 5 5 7 1 3 2 4 6 5 The Extended Prüfer code  R = (a 1, a 2,..., a n−2 )  Write the root in the right most position of R.  Write another row of integers on the bottom of R from left to right.  Each entry b i in this new row is the smallest integer that has not been already written in this new row (the row of b’is) nor in the first row (the row of a’is) in the position exactly above it or every other position to the right

8. TU Delft Vines & Prüfer codes

9. TU Delft Vines & Prüfer codes

10. TU Delft Vines & Prüfer codes

11. TU Delft Trees and Vines  Number of Vines  With 7 nodes there are:  16,807 labeled trees 5.85 * # hairs in a cow’s tail  2,580,480 regular vines 898.5 * # hairs in a cow’s tail 153.5 * # labeled trees  130,691,232,000 vines 45,505,303.6 * # hairs in a cow’s tail 7,776,000 * # labeled trees 50,646.1 * # regular vines

12. TU Delft Catalogue

13. TU Delft 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

14. TU Delft  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 Regular Vines & the Line Graph

15. TU Delft 1 2 3 4 5 6 2 2 4 5 5 7 1 3 2 4 6 5 1 2 3 4 5 3 3 4 4 6 1 2 3 5 4 Regular Vines & the Line Graph

16. TU Delft 1 2 3 4 5 6 2 2 4 5 5 7 1 3 2 4 6 5 1 2 3 4 5 3 3 4 4 6 1 2 3 5 4 1 2 3 4 1 3 3 5 2 1 4 3 Regular Vines & the Line Graph

17. TU Delft Regular Vines & the Line Graph

18. TU Delft More Counts

19. TU Delft Catalogue

20. TU Delft Catalogue

21. TU Delft Some results  If the first tree of a vine on n nodes has one node with maximal degree, then the number of regular vines possible with this tree equals the number of regular vines on (n-1) nodes.  Since every edge is adjacent in the tree is adjacent to each other then the line graph of this tree is a complete graph on (n-1) that has (n-1) (n-1)-2 possible spanning trees with all regular vines on each tree being possible

22. TU Delft Some results  If the first tree of a vine on n nodes has (n-2) nodes with degree 2, then the number of regular vines possible with this tree equals 1.  The line graph of this tree will also be a tree and hence it has only one possible spanning tree.

23. TU Delft Final comments  Finding all possible spanning trees of a graph is not easy except if this graph is the complete graph or a tree.  If vines on less than 6 nodes are required using only the Prüfer codes might be better idea.  The number of non-regular vines is much larger than the number of regular vines for n > 6 then using the line graph might become a better idea.  It is important to know not only how many spanning trees are there in the line graph of a tree on n nodes but how many of each non-isomorphic tree on (n-1) nodes to get the recursion.  There are a lot more regular vines on 6 or more nodes than hairs in a cow’s tail.

24. TU Delft

25.  Construct all Prufer codes for the first tree in the vine.  The edges of each one of the n n−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 vines on n nodes.

26. TU Delft  Construct all Prufer codes for the first tree in the vine.  The edges of each one of the n n−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. Constructing all possible vines on n nodes.