yichao chen, jonathan l. gross and toufik mansour- genus distributions of mobius ladders: a website...

8/3/2019 Yichao Chen, Jonathan L. Gross and Toufik Mansour- Genus Distributions of Mobius Ladders: A Website Supplement

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GENUS DISTRIBUTIONS OF MOBIUS LADDERS

A WEBSITE SUPPLEMENT

YICHAO CHEN, JONATHAN L. GROSS, AND TOUFIK MANSOUR

Abstract. The genus distribution of the Mobius ladderMLn is re-derived here, using overlap

matrices and Chebyshev polynomials. This website document in intended as a supplementto [4].

1. Some sets of overlap matrices for ladders

The genus distribution, as well as the total embedding distribution, of a graph can be calculatedfrom the rank distribution of its overlap matrices. We define for the Mobius ladder M Ln+1some sets of overlap matrices:

(1) Mn+2 as the set of all matrices over Z2 of the form Mc,x,y,X,Y,Zn+2 ;

(2) Mn+2(z) =n+2

j=0 Mn+2(j)zj as the rank-distribution polynomial of the set Mn+2, that

is, of the overlap matrices for general rotation systems for the Mobius ladder M Ln+1;

(3) Nn+2 as the set of all matrices of the form Mc,x,y,0,Y,Zn+2 ; and

(4) Nn+2(z) =n+2

j=0 Nn+2(j)zj as the rank-distribution polynomial of the set Nn+2, that

is, of the overlap matrices for pure rotation systems for the Mobius ladder M Ln+1.

In a matrix of the form Mc,x,y,X,Y,Zn+2 as in display (1) of [4], suppose that we first add the second

row to the first row and next add the second column to the first column. Without changing therank of the matrix, these operations produce a matrix of the following form:

(1) Mc,x+z1,y+zn,X,Y,Zn+2 =

xe c + xf x + z1 0 0 0 y + znc + xf xf z1 z2 zn2 zn1 yx + z1 z1 x1 y1

0 z2 y1 x2. . . 0

......

. . .. . .

0 zn2 0 yn20 zn1 0 yn2 xn1 yn1

y + zn y yn1 xn

.

For xy {00, 01, 10, 11}, we employ the notations(1) Mxyn+2 as the set of all matrices over Z2 of the form M

c,x,y,X,Y,Zn+2 ;

(2) Mxy

n+2(z) =n+2

j=0 M

xy

n+2(j)zj

as the rank-distribution polynomial of the set Mxy

n+2;

2000 Mathematics Subject Classification. Primary: 05C10; Secondary: 30B70, 42C05.Key words and phrases. graph embedding; total embedding distribution; Mobius ladders; overlap matrix;

Chebyshev polynomials.

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2 YICHAO CHEN, JONATHAN L. GROSS, AND TOUFIK MANSOUR

(3) Nxyn+2 as the set of all matrices of the form Mc,x,y,0,Y,Zn+2 ; and

(4) Nxyn+2(z) =n+2

j=0 Nxyn+2(j)z

j as the rank-distribution polynomial of the set Nxyn+2.

Clearly, we have the following property.

Property 1.1. For all n 1,Mn+2(z) =

xy=00,01,10,11

Mxyn+2(z) and Nn+2(z) =

xy=00,01,10,11

Nxyn+2(z).

For the vectors X = (x0, x1, . . . , xn), Y = (y1, y2, . . . , yn1), and Z = (z1, z2, . . . , zn), withxi, yj , zk Z2, we define the matrices

(2) MX,Yn =

x1 y1y1 x2 y2 0

y2 x3 y3. . .

. . .. . .

0 yn2 xn1 yn1yn1 xn

and

(3) MX,Y,Zn+1 =

x0 z1 z2 z3 . . . zn1 znz1 x1 y1z2 y1 x2 y2 0

z3 y2 x3. . .

.... . .

. . . yn2zn1 0 yn2 xn1 yn1

zn yn1 xn

.

As described by [1] and [3], every overlap matrix of the closed-end ladder Ln1 has the form

MX,Yn+1 , and every overlap matrix of the Ringel ladder Rn1 has the form M

X,Y,Zn+1 . (Note that

the subscripts of Rn1 and MX,Y,Zn+1 differ by two.) Now we further define

(1) Pn as the set of all matrices over Z2 of the form M0,Y,Zn ;

(2) Cn(j) as the number of overlap matrices for Rn that are of rank j;

(3) Pn(z) =n+1

j=0 Cn(j)zj as the rank-distribution polynomialof the set Rn; and

(4) On(z) as the rank-distribution polynomial of the overlap matrices of the closed-endladder Ln1 over the set of matrices of the form M

0,Yn .

Theorem 1.2. (see [2]) The polynomialOn(z) satisfies the recurrence relation

On(z) = On1(z) + 2z2On2(z)

with the initial conditions O1(z) = 1 andO2(z) = z2 + 1. Moreover, the generating function

O(t; z) =

n1 On(z)tn is given by

O(t; z) =t + z2t2

1 t 2z2t2 .


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GENUS DISTRIBUTIONS OF MOBIUS LADDERS A WEBSITE SUPPLEMENT 3

Theorem 1.3. [3] LetOn1(z) be the rank-distribution polynomial of the closed-end ladderLn2. Then the polynomialPn(z) (n 3) satisfies the recurrence relation

Pn+1(z) = Pn(z) + 8z2Pn1(z) + 2

n1z2On1(z).(4)

with the initial conditions P2(z) = z2

+ 1, P3(z) = 7z2

+ 1, and P4(z) = 12z4

+ 19z2

+ 1.Moreover, the generating functionP(t; z) =

n2 Pn(z)tn is given by

P(t; z) =t2(1 + z2 2(1 2z2)t 4z2(4 + z2)t2 32z4t3)

(1 2t 8z2t2)(1 t 8z2t2) .

2. Rank-distribution polynomial for Mobius ladders

Proposition 2.1. [6] For all n 3,

PEn (z) =

1

2(Pn(z) + 1 z2).

Moreover, the generating functionPE(t; z) =

n3 PEn (z)tn is given by

t3(1 + 3z2 3(1 z2)(1 + 2z2)t + 2(1 10z2 13z4)t2 + 8z2(3 2z2 4z4)t3 + 64z4t4)(1 2t 8z2t2)(1 t 8z2t2)(1 t) .

Lemma 2.2. The polynomialN00n (z) (n 4) equals

N00n+2(z) = P

On+1(z) + 2

n1z2On(z) =1

2(Pn+1(z) 1 + z2) + 2n1z2On(z).(5)

where On1(z) is the rank-distribution polynomial of the closed-end ladder Ln2, and wherePOn (z) is the rank distribution polynomial over the set M

O,Y,Zoddn .

Proof. Note that this case must have x = z1 and y = zn, so the matrix has the following form.

Mc,x+z1,y+zn,0,Y,Zn+2 =

0 c 0 0 0 0 0c 0 z1 z2 zn2 zn1 zn0 z1 0 y1

0 z2 y1 0. . . 0

......

. . .. . .

0 zn2 0 yn20 zn1 0 yn2 0 yn10 zn yn1 0

.

If c = 0, then it contributes POn+1(z) to N00n+2(z). Otherwise c = 1, since there are 2

n1 choices

of z1, z2, , zn such that the number of variables in {z1, z2, , zn} equals to 1 is even, itcontributes 2n1z2On(z) to N

00n+2(z). Note that P

En (z) + P

On (z) = Pn(z), and the rest follows

immediately by Proposition 2.1.


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Lemma 2.3. The polynomialN10n (z) (n 4) equals

N10n+2(z) = 4z

2Pn(z),

wherePn(z) is the rank-distribution polynomial of the Ringel ladder Rn2.

Proof. Note that this case must have x = 1 + z1 and y = zn, and the matrix has the followingform.

Mc,x+z1,y+zn,0,Y,Zn+2 =

0 c 1 0 0 0 0c 0 z1 z2 zn2 zn1 zn1 z1 0 y1

0 z2 y1 0. . . 0

......

. . .. . .

0 zn2 0 yn20 zn1 0 yn2 0 yn10 zn yn1 0

.

Multiply row 3 by the constant c and add it to the second row. Similarly, multiply column 3 bythe constant c and add it to the second row. If z1 = 1, then we add the first row to the thirdrow and then add the first column to the third column. A similar discussion follows for y1. Weinterchange rows 2 and 3 and then columns 2 and 3, whereby we transform Mc,1,0,0,Y,Zn+2 into thefollowing form:

0 11 0 0

0 z2 z3 zn1 znz2 0 y2

0 z3 y2 0. . . 0

.... . .

. . .

zn1 0 0 yn1zn yn1 0

.

Note that the lower-right n n submatrix has the form M0,Y,Zn . There are 22 different assign-ments of the variables y1 and z1 in this case, and it contributes 4z

2Pn(z) to the polynomialN10n+2(z).

Lemma 2.4. The polynomialN01n (z) (n

4) equals

N01n+2(z) = 4z

2Pn(z),

where Pn(z) is the rank-distribution polynomial of the Ringel ladder Rn2.


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Proof. Note that this case must have x = z1 and y = 1 + zn, and the matrix has the followingform.

Mc,0,1,0,Y,Zn+2 =

0 c 0 0 0 0 1c 0 z1 z2 zn2 zn1 1 zn0 z1 0 y1

0 z2 y1 0. . . 0

......

. . .. . .

0 zn2 0 yn20 zn1 0 yn2 0 yn11 1 zn yn1 0

.

Multiply the last row by the constant c and add it to the second row. Similarly, multiply the lastcolumn by the constant c and add it to the second row. The resulting matrix has the followingform.

Mc,0,1,0,Y,Zn+2 =

0 0 0 0 0 0 10 0 z1 z2

zn2 zn1 + cyn1 1 + zn

0 z1 0 y1

0 z2 y1 0. . . 0

......

. . .. . .

0 zn2 0 yn20 zn1 + cyn1 0 yn2 0 yn11 1 + zn yn1 0

.

If zn = 0, We first add the first row to the second row then add the first column to the secondcolumn. A similar discussion to yn1. we can transform M

c,0,1,0,Y,Zn+2 to the following form:

Mc,0,1,0,Y,Zn+2 =

0 0 0 0 0 0 10 0 z1 z2

zn2 zn1 + cyn1 0

0 z1 0 y1

0 z2 y1 0. . . 0

......

. . .. . .

0 zn2 0 yn20 zn1 + cyn1 0 yn2 0 01 0 0 0

.

There are 22 different assignments of the variables yn1 and zn, and in this case, it contributes4z2Pn(z) to the polynomial N

01n+2(z).

Lemma 2.5. The polynomialN11n (z) (n 4) is given by the equationN11n+2(z) = 2z2Pn(z) + 4z2QEn (z),

where Pn(z) is the rank-distribution polynomial of the Ringel ladder Rn2, and where QEn (z) is

the rank-distribution polynomial over the set MO,Y,Zeven,1n .


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Proof. Note that this case must have x = 1+z1 and y = 1+ zn, and the matrix has the followingform:

Mc,1,1,0,Y,Zn+2 =

0 c 1 0 1c 0 z1 z2 zn2 zn1 1 + zn1 z1 0 y1

z2 y1 0. . . 0

0...

. . .. . .

zn2 0 yn2zn1 0 yn2 0 yn1

1 1 + zn yn1 0

.

Multiply the last row constant c and add to the second row. Similarly, multiply the last columnby the constant c, and add it to the second column. A similar discussion continues for the row(column) 3. The above matrix is thereby transform into

0 0 0 0 0 0 10 0 1 + z1 + zn z2 zn2 zn1 + cyn1 1 + zn0 1 + z1 + zn 0 y1 0 yn1 00 z2 y1 0

. . . 0...

......

. . .. . .

0 zn2 0 0 yn20 zn1 + cyn1 yn1 0 yn2 0 yn11 1 + zn 0 yn1 0

.

(1) Case 1: c = 0. There are four subcases. Subcase 1: zn = 1, yn1 = 0. Note that the lower-right submatrix has the form

MO,Y,Zevenn . In this case, it contributes z2PEn (z) to the polynomial N

11n+2(z).

Subcase 2: zn = 0, yn1 = 0. We add the first row to the second row, and then addthe first column to the second column. Note that the lower-right submatrix has the

form MO,Y,Zeven

n . In this case, it contributes z2

P

E

n (z) to the polynomialN11

n+2(z). Subcase 3: zn = 1, yn1 = 1. We first delete the first row and the last row, thendelete the first column and the last column. At last we obtain a matrix of the formM

O,Y,Zeven,1n1 . In this case, it contributes z

2QEn1(z) to the polynomial N

11n+2(z).

0 0 0 0 0 0 10 0 z1 z2 zn2 zn1 00 z1 0 y1 0 1 00 z2 y1 0

. . . 0...

......

. . .. . .

0 zn2 0 0 yn20 zn1 1 0 yn2 0 1

1 0 0 1 0

.

Subcase 4: zn = 0, yn1 = 1. We delete the first row and the last row, andthen delete the first column and the last column. We obtain a matrix of the formM


2QEn1(z) to the polynomial N

11n+2(z).


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(2) Case 2: c = 1. In this case, there are again four subcases:

0 0 0 0 0 0 10 0 1 + z1 + zn z2 zn2 zn1 + yn1 1 + zn0 1 + z1 + zn 0 y1

0 yn1 0

0 z2 y1 0. . . 0

......

.... . .

. . .

0 zn2 0 0 yn20 zn1 + yn1 yn1 0 yn2 0 yn11 1 + zn 0 yn1 0

.

Subcase 1: zn = 1, yn1 = 0. Note that the lower-right submatrix has the formMO,Y,Zoddn . In this case, it contributes z

2POn (z) to the polynomial N11n+2(z) .

Subcase 2: zn = 0, yn1 = 0. We add the first row to the second row, and then addthe first column to the second column. Note that the lower-right submatrix has theform MO,Y,Zoddn . In this case, it contributes z

2POn (z) to the polynomial N11n+2(z).



2QEn1(z) to the polynomial N11n+2(z):

0 0 0 0 0 0 10 0 z1 z2 zn2 zn1 + 1 00 z1 0 y1 0 1 00 z2 y1 0

. . . 0...

......

. . .. . .

0 zn2 0 0 yn20 zn1 + 1 1 0 yn2 0 11 0 0 1 0

.



2QEn1(z) to the polynomial N11n+2(z):

Theorem 2.6. [6] For all n 4,

gCLn(x) = 1 x +1 3x 2x

4x(2x)n + 1 3x + 2

x

4x(2

x)n

+ 2nx(i

2x)n

Un

1

2i

2x

Un2

1

2i

2x

+ (1 x)(2i2x)n

Un

1

4i

2x

Un2

1

4i

2x

,

where Us is the sth Chebyshev polynomial of the second kind and i2 = 1


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Theorem 2.7. For all n 4,

gMLn(x) = x 1 +1 3x 2x

4x(2x)n + 1 3x + 2

x

4x(2

x)n

+ 2n

x(i2x)n

Un 1

2i2x Un2

1

2i2x

+ (1 x)(2i

2x)n

Un

1

4i

2x

Un2

1

4i

2x

,

where Us is the sth Chebyshev polynomial of the second kind and i2 = 1

Proof. By Lemmas 2.2, 2.3, 2.4, 2.5, we have gMLn(x) = gCLn(x) 2 + 2x. The theoremfollows.

Theorem 2.8. The generating functionN(t; z) =

n4 Nn(z)tn is given by

N(t; z) =4z2t4f(t; z)

(1 2t 8z2

t2

)(1 t 8z2

t2

)(1 4t2

z2

)(1 t),

where

f(t; z) = 5 + 3z2 + 13(z2 1)t + (20z4 106z2 + 6)t2 + (3 + 109z2 160z4)t3+ (2 + 2z2 + 448z4 + 32z6)t4 + 24z2(1 7z2 + 16z4)t5 64z4(1 + 7z2)t6.

Proof. By Lemmas 2.2, 2.3, 2.4 and 2.5, we obtain

Nn+2(z) =1

2(Pn+1(z) 1 + z2) + 10z2Pn(z) + 2n1z2On(z) + 4z2QEn (z).

Multiplying the above equation by tn and summing over n 2, we obtain

N(t; z) =t

2P(t; z) P2(z)t2

t3(1

z2)

1 t

+ 10z2t2P(t; z) +z2t2

2(O(2t; z) 2t) + 4z2t2Q(t; z).

By Proposition 2.1 and [2, Proposition 2.4], we now complete the proof.

References

[1] J. Chen, J. L. Gross, and R. G. Rieper, Overlap matrices and total embeddings, Discrete Math. 128 (1994)7394.

[2] Y. Chen, T. Mansour, and Q. Zou, Embedding distributions and Chebyshev polynomial, Graphs and Com-bin., DOI: 10.1007/s00373-011-1075-5.

[3] Y. Chen, L. Ou, and Q. Zou, Total embedding distributions of Ringel ladders, Discrete Math. 311 (2011)

24632474.[4] Y. Chen, J. L. Gross and T. Mansour, Total embedding distributions of Mobius ladders, 2011.

[5] L. A. McGeoch, Genus distribution for circular and Mobius ladders, Technical report extracted from PhDthesis, Carnegie-Mellon University, 1987.

[6] http://www.cs.columbia.edu/ gross/supplementary.html


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College of Mathematics and Econometrics, Hunan University, 410082 Changsha, China

E-mail address: [email protected]

Department of Computer Science, Columbia University, New York, NY 10027 USA


Department of Mathematics, University of Haifa, 31905 Haifa, Israel


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