graph pegging by jason counihan. the rules of pegging we start with a graph, such as this graph...

Graph Pegging

By Jason Counihan

The Rules of PeggingThe 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.


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


• 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.


• Notice that we can move a peg to any vertex we want. This means our distribution is solvable.

The Goals of PeggingThe 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

The Goals of PeggingThe Goals of Pegging

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

gopt(G) = 3

Pegging on PathsPegging 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.


• 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.


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


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.


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.


From the results of these two steps, we have

n-2 < g(Pn) n-1

Thus, g(Pn) = n-1 for n > 3.


• 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.


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


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.


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.


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.


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.


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.


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

Other ResultsOther 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.

Where to go from hereWhere 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)

Thanks to:Thanks to:

• Dr. Wyels for the project idea and assistance throughout

• Dr. Fogel for keeping the project on schedule

graph pegging by jason counihan. the rules of pegging we start with a graph, such as this graph...

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