1. Graph Pegging By Jason Counihan

2. The Rules of Pegging We start with a graph, such as this graph representation a cube. –Graphs are made of vertices (dots) connected by edges; two vertices are “adjacent” if there is an edge connecting them.

3. The Rules of Pegging Now we lay out some pegs on the graph. The layout is called a distribution of pegs.

4. The Rules of Pegging We can move a peg by “jumping” it over another peg, and landing on an empty vertex. The peg it jumped over is then removed.

5. The Rules of Pegging Notice that we can move a peg to any vertex we want. This means our distribution is solvable.

6. The Goals of Pegging The pegging number of a graph G, written g(G), is the minimum number of pegs we need to guarantee a solvable distribution. g(G) = 5

7. The Goals of Pegging The optimal pegging number of G, written g opt (G), is the minimum number of pegs needed for some solvable distribution. g opt (G) = 3

8. Pegging on Paths One family of graphs that I have worked with is paths. These are just a string of vertices connected only to the vertices immediately before and after. P n represents a path of n vertices. On a path, k unpegged vertices in a row is called a k-gap.

9. Pegging on Paths Lemma 1:On a path or cycle, no peg can be moved beyond the first hole of a gap without jumping a peg from the other side.

10. Pegging on Paths Theorem:g(P n ) = n-1 for n > 3 Proof:To prove this, we break this into two inequalities. That is, we want to show 1)g(P n ) > n-2, and 2)g(P n )  n-1

11. Pegging on Paths Step 1: g(P n ) > n-2 Proof:Observe the distribution of n-2 pegs shown above. By Lemma 1, the first vertex cannot be pegged, and thus the distribution is unsolvable and g(P n ) > n-2.

12. Pegging on Paths Step 2: g(P n )  n-1 Proof:Observe that in a distribution of n-1 pegs on n vertices (where n > 3), only one vertex is not pegged. No matter where this unpegged vertex is, we can peg it.

13. Pegging on Paths From the results of these two steps, we have n-2 < g(P n )  n-1 Thus, g(P n ) = n-1 for n > 3.

14. Pegging on Paths Lemma 1:On a path or cycle, no peg can be moved beyond the first hole of a gap without jumping a peg from the other side. Lemma 2:A solvable distribution on a cycle or path cannot contain a 3-gap.

15. Pegging on Paths Theorem:g opt (P 2k+r ) = k+r for k > 2 Proof: To prove this, we break this into two inequalities. That is, we want to show 1) g opt (P 2k+r )  k+r, and 2) g opt (P 2k+r )  k+r

16. Pegging on Paths Step 1: g opt (P 2k+r )  k+r Proof:Observe the above distribution of pegs. Since it is solvable and contains k+r pegs, we know that the optimal pegging number is, at most, k+r.

17. Pegging on Paths Step 2: g opt (P 2k+r )  k+r Proof:We will break this into two cases: r = 0 and r = 1. In each case, let us assume the theorem is false, and that k is the smallest integer for which the theorem fails.

18. Pegging on Paths Step 2 (Case 1): Suppose g opt (P 2k ) < k. Also, g opt (P 2k )  g opt (P 2k-1 ) = g opt (P 2(k-1)+1 ) = (k-1)+1 = k Putting these inequalities together gives: k > g opt (P 2k )  g opt (P 2k-1 ) = k This yields a contradiction, and we must conclude that no such k can exist.

19. Pegging on Paths Step 2 (Case 2):If the theorem is false for P k+1, then some solvable distribution of k pegs must exist. If no 2-gaps exist in a distribution of k pegs, then the distribution looks like above and is clearly not solvable. Also, by Lemma 2, no 3-gaps can exist. Thus we conclude that some 2-gap exists.

20. Pegging on Paths Step 2 (Case 2):Since a 2-gap exists, some portion of the solvable distribution must appear as above. Now, two vertices and a peg can be removed and the distribution will remain solvable, contradicting k+1 being the smallest integer for which the theorem fails. We must again conclude that no such k can exist.

21. Pegging on Paths From these two cases, we can conclude that no k can exist where g opt (P 2k+r ) < k+r, and thus g opt (P 2k+r )  k+r. Combining the results of the two steps, we see that k+r  g opt (P 2k+r )  k+r, and thus, g opt (P 2k+r ) = k+r for k > 2

22. Other Results g(C n ) = n-2 g opt (C 2k+r ) = k+r g(K n ) = g opt (K n ) = 2 If G is bipartite, then g(G) is greater than the cardinality of its larger subset of vertices. If d is the diameter of a graph G, then g(G)  d-1.

23. Where to go from here Examine other families of graphs (such as trees) Examine Graham’s Conjecture (I have found a few cases like C 5  C 5 for which the conjecture fails, but are there more?) Examine a weight function, for which a solvable distribution exists when for each vertex, the total weight sums to one (I have found a function that works for simple graphs involving the golden ratio)

24. Thanks to: Dr. Wyels for the project idea and assistance throughout Dr. Fogel for keeping the project on schedule