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Steve the presenter. Cayley Graphs & Expanders. Some history. Arthur Cayley. Son of a British merchant in Russia who wedded a Russian girl. English Super Genius Did everything Very smart and knew more than 6 languages including yours and math.

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Cayley Graphs & Expanders

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Steve the presenter

Steve the presenter

Cayley Graphs & Expanders

Some history

Some history

Arthur Cayley

Cayley graphs expanders

  • Son of a British merchant in Russia who wedded a Russian girl.

  • English Super Genius

  • Did everything

  • Very smart and knew more than 6 languages including yours and math.

  • Occupied Lucasian Chair and fathered British pure math

  • Died at 74

The Real Arthur Cayley

For more demoralizing facts, visit his homepage on Wikipedia

Some definitions

Some Definitions

Let G be a graph (finite and infinite), and let S be a non-empty, finite subset of G. Here we assume S to be symmetric, S= S −1

Definition: The Cayley graph G(G, S) is the graph with the vertex set V = G and edge set

E = {{x,y} : x,y in G, there exist s in S : y = xs}

Cayley graphs expanders

  • Hence, two vertices are adjacent if one is obtained from the other by right multiplication by some element of S.

  • Note, since S is symmetric, this adjacency relation is also symmetric, i.e, a ~ b implies b ~ a, so that the resulting graphs are undirected.

Cayley graphs expanders

  • Cayley graphs are motivated by the Cayley Theorem, which states hat every group G is isomorphic to a subgroup of the symmetric group on G.

Let s look at some examples

Let's look at some examples:

If G = Zn is the finite cyclic group of order n and the set S consists of two elements, the standard generator of G and its inverse, then the Cayley graph is the cycle Cn.

G z 6 z s 1 1

G = Z/6Z, S = {1, -1}

G z 6z s 2 2

G = Z/6Z, S = {2, -2}

G z 6z s 2 21

G = Z/6Z, S = {2, -2}

G z 6z s 3

G = Z/6Z, S = {3}

G z 6z s 31

G = Z/6Z, S = {3}

G z 6z s 2 2 3

G = Z/6Z, S = {2, -2, 3}

G z 6z s 2 2 31

G = Z/6Z, S = {2, -2, 3}

G sym 3 s 123 132 12

G = Sym(3), S = {(123), (132), (12)}

Non isomorphic groups isomorphic cayley graphs

non-isomorphic groups, isomorphic Cayley Graphs

More examples

MOre Examples

Similarly, suppose now that G= Z is the infinite cyclic group and the set S consists of the standard generator 1 and its inverse (−1 in the additive notation) then the Cayley graph is an infinite chain.

G z s 1 1

G = Z, S = {1, -1}

G z s 1 11

G = Z, S = {1, -1}

This of course simply the number line.

G z s 2 2 3 3

G = Z, S = {2, -2, 3, -3}

G z 2 s 0 1 1 0 0 1 0 1

G = Z2, S = {(0,1), (-1,0), (0,1), (0,-1)}

G l2 the free group on 2 generators a b s a a 1 b b 1

G = L2, the free group on 2 generators a, b; S = {a, a-1, b, b-1}

Cayley graphs expanders

Alternatively, if we draw each new edge horizontal and vertical to half the length, this gives rise to fractal images. The Cayley graphs of infinite groups provide interesting geometries!



Now what is an expander?

Cayley graphs expanders

The investigation of an expander centres around the following question:

Given a sparse set of points, how do we systematically construct a highly connected yet efficient network between the points?

Conceivably, the answer to such a question is of great interest to electric grid constructors and designers of network between computers among many others.

Cayley graphs groups and expanders

Cayley Graphs, Groups and expanders

In the previous slides, we have seen that Cayley graphs give very nice encoding of the structure of discrete groups.

Conversely, it also helps us sometimes to translate a graph problem back into its group presentation. This way, we can I will not tell you how now because I don't know yet. But do remember to ask me about it if you see me next semester!



We consider graphs X = (V, E), where V again is the set of vertices and E the set of edges of X. Here we shall limit ourselves to the case that X is undirected and finite.

Cayley graphs expanders

  • A path in X is a sequence v1, v2,...., vk of vertices, where vi is adjacent to vi+1 (i.e. and {vi, vi+1} is an edge).

  • A graph X is connected if every two vertices can be joined by a path. (One can trace from one vertice to another one through a series of edges.)‏

  • Finally a graph is k-regular if every vertex is also connected to k other vertices through k-edges.

Cayley graphs expanders

  • This is the celebrated Petersen Graph.

  • It is clearly connected.

  • In fact, it is a 3-regular expander.

Cayley graphs expanders

  • Let F be a subset of V, The boundary ∂F is the the set of edges connecting F to V – F.

  • In our case here, F = 3, V – F = 7, ∂F are the seven black “fat” edges.

Cayley graphs expanders

  • The expanding constant, or the isoperimetric constant of X, is:

  • h(X) = inf {abs(∂F)/(min{abs(F),abs(V – F)}) : F is subset of V, 0 < abs(F) < +∞}

Cayley graphs expanders

  • h(X) = inf {abs(∂F)/(min{abs(F),abs(V – F)}) : F is subset of V, 0 < abs(F) < +∞}

  • If we view X as a network transmitting information (where retained by some vertex propagates, say in one unit of time, to neighboring vertices), then h(X) measures the quality of X as a network:

  • h(X) is large, information propagates well.

Let s consider two extreme examples

Let's consider two extreme examples:

  • The complete graph Km on m vertices is defined s.t. Every vertex is connected to one another, e.g. m = 5.

  • It's clear in this case, if F = i, ∂F = i(m-i) s.t. h(Km) = m/2

Cayley graphs expanders

  • The cycle Cn on n vertices:

  • When n = 6, if F is a half-cycle, ∂F = 2

  • h(Cn) is smaller than 2/(n/2) = 4/n

  • In particular, h(Cn) goes to zero as n goes to infinity

Cayley graphs expanders

  • Thus we see that highly connected complete graph has a large expanding constant that grows proportionately with the number of vertices.

  • On the other hand, the minimally connected graph has a small expanding constant that goes to zero as n increases.

  • In this sense, h(X) does indeed give us a measure of the “quality” of the network.

Cayley graphs expanders

That's all I know about expanders at this point.

Thank you!

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