Graph Pegging

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. g(G) = 5 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.

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

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. • Pn 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(Pn) = n-1 for n > 3 Proof: To prove this, we break this into two inequalities. That is, we want to show 1) g(Pn) > n-2, and 2) g(Pn) n-1

11. Pegging on Paths Step 1: g(Pn) > 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(Pn) > n-2.

12. Pegging on Paths Step 2: g(Pn) 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(Pn) n-1 Thus, g(Pn) = 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:gopt(P2k+r) = k+r for k > 2 Proof: To prove this, we break this into two inequalities. That is, we want to show 1) gopt(P2k+r) k+r, and 2) gopt(P2k+r) k+r

16. Pegging on Paths Step 1:gopt(P2k+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:gopt(P2k+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 gopt(P2k) <k. Also, gopt(P2k) gopt(P2k-1) = gopt(P2(k-1)+1) = (k-1)+1 = k Putting these inequalities together gives: k > gopt(P2k) gopt(P2k-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 Pk+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 gopt(P2k+r) <k+r, and thus gopt(P2k+r) k+r. Combining the results of the two steps, we see that k+rgopt(P2k+r) k+r, and thus, gopt(P2k+r) = k+r for k > 2

22. Other Results • g(Cn) = n-2 • gopt(C2k+r) =k+r • g(Kn) = gopt(Kn) = 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 C5 C5 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