کد کردن معادلات ریاضی و تولید شبیه سازی

---

شرح پروژه

با سلام 

یه تئوری ریاضی دارم که می خوام چندتا شبیه سازی به وسیله کدینگ ازش بیرون بیارم 

Main result

Let G be a finite abelian group of order n and S be a subset of G, and Irr(G) =

{χ1, . . . , χ

n

} be the set of all irreducible characters of G. Then, by Theorem 1, the eigenvalues of BCay(G, S) are ±|χi

(S )|, where i = 1, . . . , n and χi

(S) =

∑

s∈S

χi

(s). Hence

BCay(G, S) is integral if and only if |χi

(S )| ∈ Z for all i = 1, . . . , n. To characterize

bi-Cayley graphs over abelian groups having PST, we need the following result:

Proposition 1. ([2, Corollary 4.8]) Let A be an abelian group, Irr(A) = {1 = χ1, . . . , χ

n

},

G = Dih(A, x) or G = Dic(A, y, x), and Γ = Cay(G, T ), where T = T

1 ∪ xT

2

for some

T

1

, T

2 ⊆ A. if 1 ̸ = g

−1

1

g

2 ∈ A then Γ has a PST between g

1

and g

2

if and only if the

following conditions hold

(i) the order of g

−1

1

g

2

is two,

(ii) Γ is integral, and for each i, χi

(T

1

) and |χi

(T

2

)| are integers,

(iii) ν

2

(|T | − λ

+

i

) = ν

2

(|T | − λ

−

i

) is the same integer, say k, for all i that χi

(g

−1

1

g

2) = −1

and for all i with χi

(g

−1

1

g

2

) = 1, ν

2

(|T | − λ

+

i

) > k and ν

2

(|R| + |S| − λ

−

i

) > k,

where λ

+

i

= χi

(T

1) + |χi

(T

2

)| and λ

−

i

= χi

(T

1

) − |χi

(T

2

)|, i = 1, . . . , n.

Also, if g

−1

1

g

2

/ ∈ A, then Γ has a PST between g

1

and g

2

at time t if and only if the

following conditions hold

(i) χj

(T

2

) ̸ = 0 for each j ,

(ii) if g

1 ∈ A and g

2

/ ∈ A then χj

(g

−1

1

g

2) =

|χj (T2)|

χj (T2)

exp(−i(λ

+

1

− λ

+

j

)t), and if g

1

/ ∈ A

and g

2 ∈ A then χj

(g

−1

1

g

2) =

|χj (T2)|

χj (T2)

exp(−i(λ

+

1

− λ

+

j

)t), where in the both cases

exp(−i(λ

+

1

− λ

+

j

)t) = ±1.

(ii) χj

(T

1

), |χj

(T

2

)| ∈ Z for each j , in particular Γ is integral,

(vi) v

2

(|T

2

|) = v

2

(|χj

(T )|) for all j .

Also Γ is periodic if and only if Γ is integral. Furthermore, the minimum period of the

vertices is

2π

M

, where M = gcd{λ − λ

+

1

| λ ∈ Spec(Γ) \ {λ

+

1

}}.

Theorem 2. Let Γ = HT G(m, 2n, s). Then Γ has no PST between two distinct vertices.

Also Γ is periodic if and only if one of the following holds:

5

(1) m = n = 2 and s ∈ {2(2k − 1) | k ≥ 1}.

(2) m = 1, n = 3 and s ∈ {3(2k + 1) | k ≥ 0}.

(3) m = 1, n = 4 and s ∈ {8k + 3, 8k + 5, 4k, 4k + 2 | k ≥ 0}.

(4) m = 1, n = 6 and s ∈ {6k, 12k + 3,

12k+3

5

, 12k + 9,

6k+3

2

| k ≥ 0}.

(5) m = 2, n = 3 and s ∈ {6k | k ≥ 1}.

(6) m = 2, n = 6 and s ∈ {12k +6,

12k−6

5

,

12k+6

5

, | k ≥ 0}, s ∈ {6k | k ≥ 1}∩{4k −2 | k ≥

1}, s ∈ {4k − 2 | k ≥ 1} ∩ {

12k−6

5

| k ≥ 1}, s ∈ {4k − 2 | k ≥ 1} ∩ {12k + 6 | k ≥ 0}

or s ∈ {4k − 2 | k ≥ 1} ∩ {3k | k ≥ 1}.

Furthermore, the minimum period of Γ is

2π

M

, where M = gcd{3 − |χi

(S )|, 3 + |χi

(S )|, 6 |

i = 2, . . . , mn}.

Proof. By [15, Theorem 5.2] Γ is periodic if and only if it is integral. Also, by [2,

Theorem 1], if Γ has a PST between two distinct vertices u = (g, i) and h, j ), then Γ is

integral, i = j , g

−1

h has order 2 and v

2

(3 − |χi

(S )|) = v

2(3 + |χi

(S )|) for each i with

χi

(g

−1

h) = −1. Thus we first determine the integrality of Γ. Also to the prove that

there is no PST between distinct vertices (g, i) and (h, j ), we show that in each case there

exists an irreducible character χi

with χi

(g

−1

h) = −1 and |χi

(S )| = 1. Then we get the

contradiction 1 = v

2(2) = v

2

(4) = 2.

Let Γ be integral. Then, Lemma 2.1 implies that the abelian group A = ⟨x, y |

x

n

= 1, y

m

= x

(m+s)/2

, xy = yx⟩ admits a connected cubic integral bi-Cayley graph and

|1 + χ(x) + χ(y)| ∈ Z for all χ ∈ Irr(A). Hence, by [5, Theorem A], A is isomorphic to

one of the groups Z

2

2

, Z3

, Z4

, Z6

, or Z2 × Z6

. On the other hand, A

∼

= Zd × Z mn

d

, where

d = gcd(m, n,

m+s

2

), see [1, p. 707].

If A = ⟨a⟩ × ⟨b⟩

∼

= Z

2

2

then mn = 4, which implies that m = n = d = 2 and 2 |

m+s

2

.

Hence x

2

= y

2

= 1. By a tedious computing, by using GAP, we see that Γ is integral for

all possibilities of S . In fact, S = {1, a, b}, {1, ab, b} or {1, a, ab}. If Γ has a PST between

two distinct vertices (g, i) and (h, i), by the above discussion, i = j , g

−1

h has order 2 and

v

2

(3 − |χi

(S )|) = v

2(3 + |χi

(S )|) for each i with χi

(g

−1

h) = −1. Then g

−1

h = a, b or

ab. By considering the charater table of Z

2

2

, see Table 1, in the first and second cases,

6

Table 1: Character Table of A = ⟨a⟩ × ⟨b⟩

∼

= Z

2

1 b a ab

χ1

1 1 1 1

χ2

1 −1 −1 1

χ3

1 −1 1 −1

χ4

1 1 −1 −1

since |χ2

(S )| = 1, we get 1 = v

2(2) = v

2

(4) = 2, a contradiction. In the later case, since

|χ3

(S )| = 1, again we get 1 = 2, a contradiction.

If A

∼

= Z3

then mn = 3 which implies that m = 1, n = 3, y = x

1+s

2 = x

2

, 6 | s − 3

and A = ⟨x⟩. By a tedious computing, by using GAP, we see that Γ is integral. Also [2,

Corollary 4.6] implies that Γ has no PST between distinct vertices.

If A

∼

= Z4

then mn = 4, which implies that m = d = 1, n = 4, y = x

1+s

2 , A = ⟨a⟩

and x = a or a

2

. In the first case 8 | s − 3 or 8 | s − 5 and in the later case 4 | s or

4 | s − 2. Furthermore, in both cases, a tedious computing, by using GAP, shows that Γ is

integral for all possibilities of S . In fact, S = {1, a, a

2

}, {1, a, a

3

} or {1, a

2

, a

3

}. Consider

the charater table of Z4

, see Table 2. Let Γ has a PST between two distinct vertices

Table 2: Character Table of A = ⟨a⟩

∼

= Z4

1 a a

2

a

3

χ1

1 1 1 1

χ2

1 −1 1 −1

χ3

1 i −1 −i

χ4

1 −i −1 i

u = (g, i) and v = (h, j ). Then by [2, Theorem 1], i = j = 0 or i = j = 1 and g

−1

h = a

2

.

If S = {1, a, a

2

}, S = {1, a, a

3

} or S = {1, a

2

, a

3

}, then |χ3

(S )| = 1, and similar to the

above, by [2, Theorem 1], we should have 1 = 2, a contradiction.

If A

∼

= Z6 = ⟨a⟩ then mn = 6, which implies that d = 1 and (m, n) = (1, 6), (2, 3) or

(3, 2). In the first case, y = x

1+s

2 and in the second case y

2

= x

2+s

2 , x = a

2

and y

2

= x.a

s

.

Furthermore, if y = a or a

4

then 6 | s, if y = a

2

or a

5

then 6 | s − 2 and if y = a

3

then

6 | s + 2. In the later case, x = a

3

, y

3

= x

3+s

2 , if y = a, a

3

or a

5

then 4 | s − 1 and if

7

y = a

2

or a

4

then 4 | s + 3. One can see, by using GAP, that Γ is integral if and only if

S = {1, a, a

2

}, {1, a, a

5

} or {1, a

4

, a

5

}. This implies that the later case can not happen,

in the first case we have 6 | s, 12 | s − 3, 12 | 5s − 3, 12 | s − 9 or 6 | 2s − 3, and in the

second case we have 6 | s. Suppose, towards a contradiction, that Γ has a PST between

distinct vertices (g, i) and (h, j ). Then by [2, Theorem 1], i = j = 0 or i = j = 1 and

g

−1

h = a

3

. Clearly,

χ : G → C, a

k

7→ (−1)

k

, k = 0, . . . , 5

مهارت ها و تخصص های مورد نیاز

---

---