finding matchings in sequences - clemson universitygoddard/mini/2010/holliday.pdf · finding...

Finding matchings in sequences

Sarah Holliday

Southern Polytechnic State University

Finding matchings that occur in sequences

Sarah Holliday


Sequences in matchings

Sarah Holliday


Story time!

Sarah Holliday


Once upon a time,

• in 12th century Italy

there was a

businessman named

Gugliemo Bonacci of

Pisa. He traveled

around the

Mediterranean

region, buying and

selling.

Story time…

• In 1170, his son

Leonardo was born.

He took Leonardo

with him on his trips.

During longer

voyages, he taught

Leonardo some

mathematics

(arithmetic).

Story time…

• Leonardo made

contact with some

Greek math (Euclid’s

elements) and Arab

Algebra (Al-

Khowarizmi).

Story time…

• Europe was still

using Roman

Numerals, but

Leonardo of Pisa

thought the Hindu-

Arabic numerals with

their fancy place-

system were soooo

cool!

Story time…

• Leo was so excited,

he wrote a book in

1202, called Liber

abaci. He wrote it

primarily to teach

Europeans about the

Hindu-Arabic

numeral system.

Story time…

• He was so proud of

himself (and clearly,

rightly so), that in

1228, he released a

second edition that

included the

following problem:

Story problem:

• A certain man had one pair of rabbits together in a certain enclosed place, and one wishes to know how many are created from the pair in one year when it is the nature of them in a single month to bear another pair, and in the second month those born to bear also.

Fibonacci sequences

Sarah Holliday


Fibonacci sequence

• 0

• 1

• 1

• 2

• 3

• 5

• 8

• 13

OEIS

• Online Encyclopedia of Integer Sequences

• Neil Sloane (Pres., OEIS Foundation) : oeis.org

OEIS

• Calls the Fibonacci sequence by A000045, and uses 0,1,1,2,3,5,8,13,…

• In other words, F0=0, F1=1, Fn+2=Fn+1+Fn.

OEIS

• Fn+2 = number of binary sequences of length n that have no consecutive 0s.

Forbidden Substrings

Sarah Holliday


Forbidden substrings

• From Schilling, the number of binary sequences without a run of k zeros or ones is given by an = an-1 + an-2 + …+ an-k+1

• In the case of k = 2, we repeat the Fibonacci sequence.

Forbidden substrings

True Fact

Let |B| = r ≤ q = |A|, and S be generated by the k-action of P={σ1, σ2, ... σp} on B, where each σi is a permutation of B and P defines a latin rectangle, then for n ≥ k ≥ 3 the number of S-free strings of length n over A, an, satisfies an=(q-1)an-1 + (q-1)an-2

+ ... + (q-1)an-k+2 + (q-p)an-k+1 + p(q-r)an-k, ai=qi, i=0,1,...,k-1.

OEIS

• Fn+2 = number of binary sequences of length n that have no consecutive 0's.

• Fn+2 = number of subsets of {1,2,...,n} that contain no consecutive integers.

• Fn+1 = number of tilings of a 2 X n rectangle by 2 X 1 dominoes.

OEIS

• Fn+1 = number of matchings in a path graph on n vertices

(Emeric Deutsch)

Getting Fibonacci sequences out of graphs

Sarah Holliday


Definition: Path

• Path on n vertices (not n edges.)

• Our notation is Pn

Definition: Matching

• A matching is a set of disjoint edges.

• The set of all matchings of a graph G is M(G)

OEIS

• Fn+1 = number of matchings in a path graph on n vertices

(Emeric Deutsch)

Number of matchings in a path graph on n vertices

A B C D E

M(P5)={Ø,{AB},{BC},{CD},{DE},{AB,CD},{AB,DE},{BC,DE}}

Five vertices, eight matchings. F5 = 8

Definition: Product

• The cartesian product of graphs G and H is denoted G □ H.

• V(G □ H) = V(G) X V(H)

• E(G □ H) = E(G) X V(H) U V(G) X E(H)

G=

H=G □ H=

OEIS

• Fn+1 is the number of perfect matchings in the ladder graph Ln = P2 □ Pn

(Sharon Sela)

Defintion: Perfect matching

• A matching that uses all of the vertices of a graph.

• The set of all perfect matchings of a graph is PM(G)

OEIS

• Fn+1 is the number of perfect matchings in the ladder graph Ln = P2 □ Pn

(Sharon Sela)

Fn+1 is the number of perfect matchings in the ladder graph

Ln = P2 □ Pn

A B C D E

1

2

PM(L5)={{(A1,A2), (B1,B2), (C1,C2), (D1,D2), (E1,E2)},{(A1,B1), (A2,B2), (C1,C2), (D1,D2), (E1,E2)},{(A1,A2), (B1,C1), (B2,C2), (D1,D2), (E1,E2)},{(A1,A2), (B1,B2), (C1,D1), (C2,D2), (E1,E2)},{(A1,A2), (B1,B2), (C1,C2), (D1,E1), (D2,E2)},{(A1,B1), (A2,B2), (C1,D1), (C2,D2), (E1,E2)},{(A1,B1), (A2,B2), (C1,C2), (D1,E1), (D2,E2)},{(A1,A2), (B1,C1), (B2,C2), (D1,E1), (D2,E2)}}

Lemma

• Pn has the same number of total matchings as Ln=Pn □ P2 has perfect matchings.

• |M(Pn)| = |PM(Ln)|

Lemma

• The argument is constructive; each matching of the M(Pn) corresponds to a perfect matching of the PM(Ln) as follows: For each edge of m of M(Pn), add the two corresponding edges of Ln, and for each unsaturated vertex in m, add the edges of the K2 that use those vertices.

Lemon

• For any graph G, G □ P2 has the same number of perfect matchings as G has total matchings.

• |M(G)| = |PM(G □ P2)|

True Fact:

• |PM(G □ P2)| =

|M(G)| + |PM(G)|(|PM(G)| - 1)

|PM(G □ P2)| = |M(G)| + |PM(G)|(|PM(G)| - 1)

• Each matching m in M(G) becomes a perfect matching p of PM(G □ P2) by duplicating the edges of m on both copies of G, and using P2 edges of G □ P2 to match the pairs of vertices unsaturated by m. Each pi of PM(G) can be placed on one of the G of G □ P2 and a distinct pj of PM(G) on t'other G of G □ P2.

Generalized Fibonacci Numbers

Sarah Holliday



• Fn+2 = aFn+1 + bFn, F0 = c, F1 = d.


• Fn+2 = aFn+1 + bFn, F0 = c, F1 = d.

• If c = 0, with a = 1, b = 1, d = 1, these are the traditional Fn Fibonacci numbers.

• If c=1, with a = 1, b = 1, d = 1, these are the Fn+1 Fibonacci numbers.

• If c = 2, with a = 1, b = 1, d = 1, these are the Ln Lucas numbers.

H, Komatsu

• For the generalized Fn+2 = Fn+1 + Fn with F0 = c and F1 = 1, we can count the number of matchings in a “bloated” cycle.

H, Komatsu

• For the generalized Fn+2 = Fn+1 + Fn with F0 = 6 and F1 = 1, we use the cycle with (6-1) edges in the “bloat”.

A B C D E

{Ø, AB1, AB2 , AB3, AB4, AB5, BC,CD,DE,AE, AB1CD, AB2CD, AB3CD, AB4CD, AB5CD, AB1DE, AB2DE, AB3DE, AB4DE, AB5DE, BCDE, BCAE, CDEA}

True Fact

• For the generalized Fn+2 = Fn+1 + Fn with F0 = 6 and F1 = 1, we can use the path with (6) edges in the “bloat”.

A B C D E

{Ø, AB1, AB2 , AB3, AB4, AB5, AB6, BC,CD,DE, AB1CD, AB2CD, AB3CD, AB4CD, AB5CD, AB6CD, AB1DE, AB2DE, AB3DE, AB4DE, AB5DE, AB6DE, BCDE}

False fact

• |M(L’n)| = |PM(Pn □ K2)|, using bloated Pn

|M(L’n)| = |PM(P’n □ K2)|,

using bloated Pn

{vwxyz, AB1ab1xyz, AB2 ab2xyz, AB3ab3xyz, AB4ab4xyz, AB5ab5xyz, AB6ab6xyz, BCbcyz,CDcdwz,DEdevwx, AB1ab1CDcdz, AB2ab2CDcdz, AB3ab3CDcdz, AB4ab4CDcdz, AB5ab5CDcdz, AB6ab6CDcdz, AB1ab1DEdex, AB2ab2DEdex, AB3DEab3dex, AB4ab4DEdex, AB5ab5DEdex, AB6ab6DEdex, BCbcDEdev, AND AB1ab2xyz, AB1ab3xyz, AB1ab4xyz,...}

New Fact

• L’n = P’n □’ K2, using bloated Pn

• We need to define a new type of product operation, □’ in which the first copy P’n is bloated and the second copy is Pn not bloated.

Definition

• L’n = P’n □’ K2

True Fact

• For the generalized Fn+2 = Fn+1 + Fn with F0 = c and F1 = 1, we can count the number of perfect matchings in a “bloated” path “multiplied” against a path.

• |PM(L’n)| = |PM(P’n □’ K2)| = |M(P’n)|

True Fact

• For Fn+2 = Fn+1 + bFn, F0 = 1 F1 = d, we construct a graph as follows: start with a path on n vertices, with each edge inflated b times, but the first edge bloated to b-d+1 edges.

• The total number of matchings of this graph is Fn

True Fact

• Each edge inflated b times, and the first edge bloated to b-d+1 edges.

A different generalization

• Gn = Gn-1 + Gn-2 + Gn-3 + … + Gn-k+1

Definition: Matching

• A matching is a set of disjoint edges.

• A subgraph of Pn that does not contain a P3.

Gn=Gn-1 + Gn-2 + Gn-3 +…+ Gn-k+1

• The number of subgraphs of Pn+1 that do not contain Pk.

Goals

• To identify the family of graphs whose perfect matchings generate the sequences given by:

Gn = a1Gn-1 + a2Gn-2 + a3Gn-3 + … + an-1G1

finding matchings in sequences - clemson universitygoddard/mini/2010/holliday.pdf · finding...

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