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Finding matchings in sequences
Sarah Holliday
Southern Polytechnic State University
Finding matchings that occur in sequences
Sarah Holliday
Southern Polytechnic State University
Sequences in matchings
Sarah Holliday
Southern Polytechnic State University
Story time!
Sarah Holliday
Southern Polytechnic State University
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
Southern Polytechnic State University
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
OEIS
• Fn+2 = number of binary sequences of length n that have no consecutive 0s.
Forbidden Substrings
Sarah Holliday
Southern Polytechnic State University
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
Southern Polytechnic State University
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)|
Lemon
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
Southern Polytechnic State University
Generalized Fibonacci Numbers
• Fn+2 = aFn+1 + bFn, F0 = c, F1 = d.
Generalized Fibonacci Numbers
Generalized Fibonacci Numbers
• 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
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