Let G be a connected graph with vertex set V ( G ) and edge set E ( G ) . A set X ⊂ V ( G ) is a cut-set of G, if G − X is disconnected. For a cut-set X of G, we by ω ( G − X ) and τ ( G − X ) , respectively, denote the number of components and the order of the largest component in G − X . The score of X is defined as S c ( X ) = ω ( G − X ) − | X | − τ ( G − X ) . The rupture degree of a noncomplete- connected graph G is defined by

r ( G ) = max { S c ( X ) : X ⊂ V ( G ) , ω ( G − X ) > 1 } .

We call X a r-set of G, if S c ( X ) = r ( G ) .

The rupture degree is well used to measure the vulnerability of graphs, for it can measure not only the amount of work done to damage the network, but also how badly the network is damaged. The references about this parameter see [1] [2] [3].

For a vertex set S ⊆ V ( G ) , we by G [ S ] denote the subgraph of G that is induced by S. And by N ( S ) denote neighbor set of S that contains vertex, not in S, but has neighbor in S.

Let G 1 , G 2 , ⋯ , G k be connected graphs. The Cartesian product G 1 × G 2 × ⋯ × G k is a graph that has vertex set V ( G 1 ) × V ( G 2 ) × ⋯ × V ( G k ) with two vertices u = ( u 1 , u 2 , ⋯ , u k ) and v = ( v 1 , v 2 , ⋯ , v k ) adjacent if for exactly one i, u i ≠ v i and ( u i , v i ) is an edge in G i . As usual, we by P n and C n denote the path and the cycle on n vertices, respectively. It is well known that Cartesian products are highly recommended for the design of interconnection networks [4]. In this paper, we first determine the rupture degree for some Cartesian products such as P m × C n and C m × C n . Then, discuss the rupture degree of grids P n 1 × P n 2 × ⋯ × P n k , and tori C n 1 × C n 2 × ⋯ × C n k .

For terminology and notations not defined here, we refer to the book [5].

In this section, we determine the rupture degree of Cartesian product P m × C n and C m × C n . First, give some useful lemmas, which have been proved in [2].

Lemma 2.1. If H is a spanning subgraph of G, then r ( H ) ≥ r ( G ) .

Lemma 2.2. The rupture degree of path P n and cycle C n are

r ( P n ) = { 0 ,     if n isodd; − 1 ,   if n iseven . r ( C n ) = { − 2 , if n isodd ; − 1 , if n iseven .

Lemma 2.3. Let X be a cut-set of G ( = P m × C n ) . If n is odd, then ω ( G − X ) ≤ m n − m 2 .

Proof. Suppose S is a cut set of C n , then ω ( C n − S ) ≤ n − 1 2 . Notice that m C n is a spanning subgraph of G, we have that ω ( G − X ) ≤ m ω ( C n − S ) ≤ m n − m 2 for any cut-set X of G.

Lemma 2.4. Let X be a r-set of G ( = C m × C n ) with m , n are odd, then ω ( G − X ) ≤ ( m − 1 ) ( n − 1 ) 2 .

Proof. Suppose that V ( C m ) = { u 1 , u 2 , ⋯ , u m } and V ( C n ) = { v 1 , v 2 , ⋯ , v n } , then V ( G ) = { w i j = ( u i , v j ) | 1 ≤ i ≤ m ; 1 ≤ j ≤ n } . Let X be a r-set of G and W i = { w i 1 , w i 2 , ⋯ , w i n } for 1 ≤ i ≤ n . Clearly, the induced subgraphs G [ W i ] is cycle with order n, named as C n i . And for any cut set S i of C n i , we have

ω ( C n i − S ) = n − 1 2 . And call S i the optimal cut-set if equality holds. Clearly, the

optimal cut-set of C n i is either { w i 1 , w i 3 , ⋯ , w i n } or { w i 2 , w i 4 , ⋯ , w i ( n − 1 ) } for

1 ≤ i ≤ n . Consider X be a r-set of G and m is odd, there exist C n i and C n i + 1 such

that ω ( G [ V ( C n i ∪ C n i + 1 ) ] − X ) ≤ n − 1 2 for some i. Now, let

G 1 = G [ V ( C n i ∪ C n i + 1 ) ] and G 2 = G − G 1 = P m − 2 × C n . Consider ω ( G − X ) ≤ ω ( G 1 − X ) + ω ( G 2 − X ) , by Lemma 2.3, we get

ω ( G − X ) ≤ n − 1 2 + ( m − 2 ) ( n − 1 ) 2 = ( m − 1 ) ( n − 1 ) 2 .

Lemma 2.5. Let m ≥ 2 and n ≥ 3 be positive integers. Then r ( P m × C n ) ≥ r ( P m + 1 × C n ) .

Proof. Let G = P m + 1 × C n , G ′ = P m × C n . Then G − G ′ is a cycle. Support that X is a r-set of G, then X 1 = X ∩ V ( G ′ ) and X 2 = X ∩ V ( G − G ′ ) are vertex cut set of G ′ and G − G ′ , respectively. Denote ω 1 = ω ( G ′ − X 1 ) , ω 2 = ω ( G − G ′ − X 2 ) . Since r ( G ′ ) ≥ ω 1 − | X 1 | − τ ( G ′ − X 1 ) and τ ( G − X ) ≥ τ ( G ′ − X 1 ) , we have r ( G ′ ) ≥ ω 1 − | X 1 | − τ ( G − X ) . So r ( G ) = ω ( G − X ) − | X | − τ ( G − X ) ≤ ω 1 + ω 2 − | X 1 | − | X 2 | − τ ( G − X ) ≤ r ( G ′ ) + ω 2 − | X 2 | .

Notice that G − G ′ is a cycle and thus ω 2 ≤ | X 2 | . Thus r ( G ) ≤ r ( G ′ ) . This means r ( P m × C n ) ≥ r ( P m + 1 × C n ) .

Theorem 2.6. Let m ≥ 2 and n ≥ 3 be positive integers. Then the rupture degree of P m × C n is

r ( P m × C n ) = { − 1 , if   n   is   even; { − 4 + m 2 , if   m   is   even; − 3 + m 2 , if   m   is   odd . if   n   is   odd .

Proof. Suppose V ( P m ) = { u 1 , u 2 , ⋯ , u m } and V ( C n ) = { v 1 , v 2 , ⋯ , v n } , then V ( P m × C n ) = { ( u i , v j ) | 1 ≤ i ≤ m ; 1 ≤ j ≤ n } . For narrative purposes, we let P m × C n = G and distinguish three cases to complete the proof.

Case 1. n is even.

Notice that G contains a Hamilton cycle C m n , we by Lemmas 2.1 and 2.2 get r ( G ) ≤ − 1 . On the other hand, let X * = { ( u i , v j ) | i ≡ 0 ( mod 2 ) , j = 1 , 3 , ⋯ , n − 1 ; i ≡ 1 ( mod 2 ) , j = 2 , 4 , ⋯ , n } for

1 ≤ i ≤ m . Clearly, ω ( G − X * ) = m n 2 , | X * | = m n 2 and τ ( G − X * ) = 1 . By the

definition of rupture degree, we have r ( G ) ≥ ω ( G − X * ) − | X * | − τ ( G − X * ) = − 1 . Thus r ( P m × C n ) = − 1 while n is even.

Case 2. n is odd, m is even.

First, let X * = { ( u i , v j ) | i ≡ 0 ( mod 2 ) , j = 1 , 3 , ⋯ , n ; i ≡ 1 ( mod 2 ) , j = 2 , 4 , ⋯ , n − 1 } for

1 ≤ i ≤ m . Since ω ( G − X * ) = m n − m 2 , | X * | = m n 2 and τ ( G − X * ) = 2 . Then S c ( X * ) = ω ( G − X * ) − | X * | − τ ( G − X * ) = − m + 4 2 . Now, we by showing S c ( X ) ≤ − m + 4 2 for any cut-set X of G to get r ( G ) = − m + 4 2 . Now, distinguish some cases to discuss.

Subcase 2.1. | X | ≥ | X * | = m n 2 .

If τ ( G − X ) = 1 , by Lemma 2.3, then | X | = m n − ω ( G − X ) ≥ m n + m 2 . Consider m ≥ 2 , thus S c ( X ) = ω ( G − X ) − | X | − τ ( G − X ) ≤ m n − m 2 − m n + m 2 − 1 = − 2 m + 2 2 ≤ − m + 4 2 .

If τ ( G − X ) ≥ 2 , consider | X | ≥ m n 2 , we have S c ( X ) = ω ( G − X ) − | X | − τ ( G − X ) ≤ m n − m 2 − m n 2 − 2 = − m + 4 2 .

Subcase 2.2. | X | < | X * | = m n 2 .

Let A i = { ( u i , v j ) | 1 ≤ j ≤ n } , X i = A i ∩ X , X i * = A i ∩ X * for i = 1 , 2 , ⋯ , m and B j = { ( u i , v j ) | 1 ≤ i ≤ m } , Y j = B j ∩ X , Y j * = B j ∩ X * for j = 1 , 2 , ⋯ , n . Clearly, G [ A i ] and G [ A i ] are cycles with order n and path with order m,

respectively. We discuss by n ≥ m 2 and n ≤ m 2 − 1 .

Subcase 2.2.1. n ≥ m 2 .

Notice that G [ A i ] ∪ G [ A i + 1 ] is a spanning subgraph of G [ A i ∪ A i + 1 ] with 1 ≤ i ≤ m − 1 , we can get ω ( G [ A i ∪ A i + 1 ] − X i ∪ X i + 1 ) ≤ | X i ∪ X i + 1 | − 1 . In fact, if 2 ≤ | X i ∪ X i + 1 | < n , then ω ( G [ A i ∪ A i + 1 ] − X i ∪ X i + 1 ) ≤ ω ( G [ A i ] − X i ) + ω ( G [ A i + 1 ] − X i + 1 ) − 1 . Thus, ω ( G [ A i ∪ A i + 1 ] − X i ∪ X i + 1 ) ≤ | X i | + | X i + 1 | − 1 = | X i ∪ X i + 1 | − 1 .

If | X i ∪ X i + 1 | ≥ n , then ω ( G [ A i ∪ A i + 1 ] − X i ∪ X i + 1 ) ≤ n − 1 . Therefore, ω ( G [ A i ∪ A i + 1 ] − X i ∪ X i + 1 ) ≤ | X i ∪ X i + 1 | − 1 . Combine this with the fact ω ( G [ A i ∪ A i + 1 ] − X i * ∪ X i + 1 * ) = n − 1 , it is clear that ω ( G [ A i ∪ A i + 1 ] − X i * ∪ X i + 1 * ) − ω ( G [ A i ∪ A i + 1 ] − X i ∪ X i + 1 ) ≥ n − | X i ∪ X i + 1 | for | X i ∪ X i + 1 | ≥ 2 with 1 ≤ i ≤ m − 1 .

Now let a = ∑ | X i ∪ X i + 1 | ≥ 2 ( n − | X i ∪ X i + 1 | ) , M = { X i ∪ X i + 1 : | X i ∪ X i + 1 | = 0 } and N = { X i ∪ X i + 1 : | X i ∪ X i + 1 | = 1 } for i = 1 , 3 , ⋯ , m − 1 . Consider X is a vertex cut set of G, then | M | + | N | ≤ m 2 − 1 . Furthermore, since

G [ A 1 ∪ A 2 ] ∪ G [ A 3 ∪ A 4 ] ∪ ⋯ ∪ G [ A m − 1 ∪ A m ] is a spanning subgraph of G with ω ( G [ A i ∪ A i + 1 ] − X i * ∪ X i + 1 * ) − ω ( G [ A i ∪ A i + 1 ] − X i ∪ X i + 1 ) ≥ n − | X i ∪ X i + 1 | for | X i ∪ X i + 1 | ≥ 2 , we have | X * | = m n 2 = | X | + | M | n + | N | ( n − 1 ) + a .

And thus

ω ( G − X * ) = m ( n − 1 ) 2 = ∑ i     ω ( G [ A i ∪ A i + 1 ] − X i * ∪ X i + 1 * ) ≥ ∑ i     ω ( G [ A i ∪ A i + 1 ] − X i ∪ X i + 1 ) + | M | ( n − 2 ) + | N | ( n − 2 ) + a ≥ ω ( G − X ) + ( | M | + | N | ) ( n − 2 ) + a .

Now, we estimate the value S c ( X ) in details. If | M | ≥ 1 , then τ ( G − X ) ≥ 2 n . We have

If | M | = 0 , | N | ≥ 1 , then τ ( G − X ) ≥ 2 n − 1 . Consider | N | ≤ m 2 − 1 and m ≥ 4 . Thus we get

S c ( X ) = ω ( G − X ) − | X | − τ ( G − X ) ≤ m ( n − 1 ) 2 − | N | ( n − 2 ) − a + | N | ( n − 1 ) + a − m n 2 − 2 n + 1 = − m 2 + | N | − 2 n + 1 ≤ − 2 n ≤ − m ≤ − m + 4 2 .

If | M | = 0 , | N | = 0 , then | X | = m n 2 − a . Combine τ ( G − X ) ≥ 2 . We have S c ( X ) = ω ( G − X ) − | X | − τ ( G − X ) ≤ m ( n − 1 ) 2 − a − m n 2 + a − 2 = − m + 4 2 .

Subcase 2.2.2. n ≤ m 2 − 1 .

Notice that G [ B j ∪ B j + 1 ] is a ladder with order 2m and G [ B 1 ∪ B 2 ] ∪ G [ B 3 ∪ B 4 ] ∪ ⋯ ∪ G [ B n − 2 ∪ B n − 1 ] ∪ G [ B n ] is a spanning subgraph of G. We similarly get

ω ( G [ B j ∪ B j + 1 ] − Y j * ∪ Y j + 1 * ) − ω ( G [ B j ∪ B j + 1 ] − Y j ∪ Y j + 1 ) ≥ m − | Y j ∪ Y j + 1 | while | Y j ∪ Y j + 1 | ≥ 1 for j = 1 , 3 , ⋯ , n − 2 . Now, let S = { Y j ∪ Y j + 1 : | Y j ∪ Y j + 1 | = 0 } and b = ∑ | Y j ∪ Y j + 1 | ≥ 1 ( m − | Y j ∪ Y j + 1 | ) with j = 1 , 3 , ⋯ , n − 2 and discuss S c ( X ) by | S | ≥ 1 and | S | = 0 .

If | S | ≥ 1 , then τ ( G − X ) ≥ 2 m . Similarly, we get | X * | = m n 2 = | X | + | S | m + b + m 2 − | Y n | and

ω ( G − X * ) = m ( n − 1 ) 2 = ∑ j     ω ( G [ B j ∪ B j + 1 ] − Y j * ∪ Y j + 1 * ) ≥ ∑ j     ω ( G [ B j ∪ B j + 1 ] − Y j ∪ Y j + 1 ) + | S | ( m − 1 ) + b ≥ ω ( G − X ) + | S | ( m − 1 ) + b .

Therefore, we get

S c ( X ) = ω ( G − X ) − | X | − τ ( G − X ) ≤ m ( n − 1 ) 2 − [ | S | ( m − 1 ) + b ] − [ m n 2 − ( | S | m + b + m 2 − | Y n | ) ] − 2 m = | S | − | Y n | − 2 m ≤ n − 3 2 − 2 m ≤ − 7 m + 8 4 < − m + 4 2 .

If | S | = 0 , we discuss by the value of | Y n | . While | Y n | ≥ m 2 , consider | X | < m n 2 , then τ ( G − X ) ≥ 2 . Combine | X * | = m n 2 = | X | + m 2 − | Y n | + b and ω ( G − X * ) = m ( n − 1 ) 2 ≥ ω ( G − X ) + b , we have

S c ( X ) = ω ( G − X ) − | X | − τ ( G − X ) ≤ m ( n − 1 ) 2 − b − ( m n 2 − m 2 + | Y n | − b ) − 2 = − | Y n | − 2 ≤ − m + 4 2 .

while | Y n | ≤ m 2 − 1 , we would get ∑ j = 1 , 3 , ⋯ , n − 2 ω ( G [ B j ∪ B j + 1 ] − Y j ∪ Y j + 1 ) ≥ ω ( G − X ) − | Y n | + m 2 . In fact, since the optimal cut set of G [ B n ] always contains m 2 vertices, each vertex of Y n * ∖ Y n would connect at least two components in G − X . Thus, we get ∑ j = 1 , 3 , ⋯ , n − 2 ω ( G [ B j ∪ B j + 1 ] − Y j ∪ Y j + 1 ) ≥ ω ( G − X ) + m 2 − | Y n | . Combine this with | X * | = m n 2 = | X | + m 2 − | Y n | + b , τ ( G − X ) ≥ 2 and m ( n − 1 ) 2 ≥ ∑ j ω ( G [ B j ∪ B j + 1 ] − Y j ∪ Y j + 1 ) + b , we have

S c ( X ) = ω ( G − X ) − | X | − τ ( G − X ) ≤ ∑ j = 1 , 3 , ⋯ , n − 2     ω ( G [ B j ∪ B j + 1 ] − Y j ∪ Y j + 1 ) + | Y n | − m 2 + m 2     − | Y n | + b − m n 2 − 2 ≤ m ( n − 1 ) 2 − b + b − m n 2 − 2 = − m + 4 2 .

Case 3. m , n are both odd.

By Lemma 2.5, we get r ( P m − 1 × C n ) ≥ r ( P m × C n ) . Notice that n − 1 is even, we get r ( G ) ≤ r ( P m − 1 × C n ) = − 3 + m 2 .

On the other hand, let

X * = { ( u i , v j ) | i ≡ 0 ( mod 2 ) , j = 1 , 3 , ⋯ , n ; i ≡ 1 ( mod 2 ) , j = 2 , 4 , ⋯ , n − 1 } for 1 ≤ i ≤ m . Clearly, X * is a cut set of P m × C n with ω ( G − X * ) = m n − m 2 , | X * | = m n − 1 2 and τ ( G − X * ) = 2 . This implies that r ( G ) ≥ ω ( G − X * ) − | X * | − τ ( G − X * ) = − 3 + m 2 .

Therefore, r ( P m × C n ) = − 3 + m 2 while m , n are both odd. This completes the proof.

Theorem 2.7. Let m , n be positive integers with n ≥ m ≥ 3 . Then the rupture degree of C m × C n is

Proof. Suppose V ( C m ) = { u 1 , u 2 , ⋯ , u m } and V ( C n ) = { v 1 , v 2 , ⋯ , v n } , then V ( C m × C n ) = { w i j = ( u i , v j ) | 1 ≤ i ≤ m ; 1 ≤ j ≤ n } . Similarly, we let C m × C n = G and W i = { w i j | 1 ≤ j ≤ n } , W j = { w i j | 1 ≤ i ≤ m } . Clearly, both ∪ i G [ W i ∪ W i + 1 ] and ∪ j G [ W j ∪ W j + 1 ] are spanning subgraph of G. Now, we distinguish three cases to complete the proof.

Case 1. Both m and n are even.

Notice that P m × C n is a spanning subgraph of G, by Lemma 2.1 and Theorem 2.6, we have r ( G ) ≤ − 1 . On the other hand, let

X * = { ( u i , v j ) | i ≡ 0 ( mod 2 ) , j = 1 , 3 , ⋯ , n − 1 ; i ≡ 1 ( mod 2 ) , j = 2 , 4 , ⋯ , n } for 1 ≤ i ≤ m . Since ω ( G − X * ) = m n 2 , | X * | = m n 2 and τ ( G − X * ) = 1 . Thus r ( G ) ≥ ω ( G − X * ) − | X * | − τ ( G − X * ) = − 1 . Therefore, r ( G ) = − 1 .

Case 2. One of n and m is even.

Without loss generality, suppose m is even, then n is odd. We first let

X * = { ( u i , v j ) | i ≡ 0 ( mod 2 ) , j = 1 , 3 , ⋯ , n ; i ≡ 1 ( mod 2 ) , j = 2 , 4 , ⋯ , n − 1 } for 1 ≤ i ≤ m . Clearly, ω ( G − X * ) = m n − m 2 , | X * | = m n 2 and τ ( G − X * ) = 2 . Thus r ( G ) ≥ ω ( G − X * ) − | X * | − τ ( G − X * ) = − m + 4 2 . Consider P m × C n is a spanning subgraph of G, by Lemmas 2.1 and 2.5, we have r ( G ) ≤ − m + 4 2 . So, we get r ( G ) = − m + 4 2 while m is even and n is odd. Similar to the case for n is even and m is odd, here omitted.

Case 3. Both m and n are odd.

First, let

X * = { ( u i , v j ) | i ≡ 0 ( mod 2 ) , j = 2 , 4 , ⋯ , n − 1 ; i ≡ 1 ( mod 2 ) , j = 1 , 3 , ⋯ , n } for 1 ≤ i ≤ m . Clearly, X * is a cut set of G with ω ( G − X * ) = ( m − 1 ) ( n − 1 ) 2 , | X * | = m n + 1 2 and τ ( G − X * ) = 2 . Thus r ( G ) ≥ ω ( G − X * ) − | X * | − τ ( G − X * ) = − 4 + m + n 2 . The following we by proving claim to show r ( G ) ≤ − 4 + m + n 2 .

Claim. Let m , n be two odd numbers with 3 ≤ m ≤ n and S be a r-set of G ( = C m × C n ) . Then G − S = s K 2 ∪ t K 1 with s ≤ m + n − 2 2 .

Proof. Let S be a r-set of G with τ ( G − S ) as small as possible. Suppose G 1 , G 2 , ⋯ , G k are components of G − S . We first show | G i | ≤ 2 for 1 ≤ i ≤ k . If not, assume that G 1 , G 2 , ⋯ , G k 1 with | G j | ≥ 3 for 1 ≤ j ≤ k 1 ≤ k , then each G j has at least one cut vertex (unless G j = K 2 or K 1 ). In fact, assume G j ( ≠ K 2 , K 1 ) has no cut vertex, we exchange vertices in N ( G j ) with vertices in V ( G j ) to keep | S | constant and find that either ω ( G − S ) − | S | − τ ( G − S ) would be greater or τ ( G − S ) would be smaller, which contradicts to the choose of S. So, each G j ( 1 ≤ j ≤ k 1 ) has cut vertex and suppose w 1 , w 2 , ⋯ , w k 1 are cut vertices of G 1 , G 2 , ⋯ , G k 1 , respectively. Let S ′ = S ∪ { w 1 , w 2 , ⋯ , w k 1 } . Then τ ( G − S ′ ) ≤ τ ( G − S ) − 2 . Thus we get

ω ( G − S ′ ) − | S ′ | − τ ( G − S ′ ) ≥ ω ( G − S ) + k 1 − ( | S | + k 1 ) − ( τ ( G − S ) − 2 ) > ω ( G − S ) − | S | − τ ( G − S ) .

This contradicts to the choice of S. So | G i | ≤ 2 for 1 ≤ i ≤ k and then denote G − S = s K 2 ∪ t K 1 . Further, it finds that there are at most one component as K 2 in G [ C i ∪ C i + 1 ] − S for 1 ≤ i ≤ m − 1 and G [ C j ∪ C j + 1 ] − S for 1 ≤ j ≤ n − 1 . Otherwise, if G [ C i ∪ C i + 1 ] − S or G [ C j ∪ C j + 1 ] − S has at least two components as K 2 , then τ ( G − S ) ≥ 3 , contradiction. This implies that

s ≤ m − 1 2 + n − 1 2 = m + n − 2 2 .

By Lemma 2.4 and the above Claim, we get

| S | ≥ m n − 2 m + n − 2 2 − ( ( m − 1 ) ( n − 1 ) 2 − ( m + n − 2 ) 2 ) = m n − m − n + 2 − ( m n − 2 m − 2 n + 3 ) 2 = m n + 1 2 .

Thus, we get

r ( G ) = ω ( G − S ) − | S | − τ ( G − S ) ≤ ( m − 1 ) ( n − 1 ) 2 − m n + 1 2 − 2 = − 4 + m + n 2 .

This completes the proof.

Let n 1 , n 2 , ⋯ , n k be positive integers. We discuss the rupture degree of grids P n 1 × P n 2 × ⋯ × P n k with n i ≥ 2 and tori C n 1 × C n 2 × ⋯ × C n k with n i ≥ 3 .

Lemma 3.1. [2] Let m , n be integers with n ≤ m . Then r ( K m , n ) = m − n − 1 .

Lemma 3.2. [1] Let m , n be integers with 1 ≤ m ≤ n . Then r ( K m × K n ) = m + n − m n − ⌈ n m ⌉ .

Theorem 3.3. For all positive integers n 1 , n 2 , ⋯ , n k , the rupture degree of grids is

r ( P n 1 × P n 2 × ⋯ × P n k ) = { 0 , if   all   n i   are   odd , − 1 , if   some   n i   is   even .

Proof. Notices that P n 1 × P n 2 × ⋯ × P n k contains a Hamilton path P n 1 n 2 ⋯ n k . So by Lemmas 2.1 and 2.2, we have

r ( P n 1 × P n 2 × ⋯ × P n k ) ≤ r ( P n 1 n 2 ⋯ n k ) = { 0 , if   all   n i   are   odd , − 1 , if   some   n i   is   even .

On the other hand, it is well known that P n 1 × P n 1 × P n 2 × ⋯ × P n k is a bipartite graph and thus

Then by Lemmas 2.1 and 3.1, we get

r ( P n 1 × P n 2 × ⋯ × P n k ) ≥ { 0 , if   all   n i   are   odd , − 1 , if   some   n i   is   even .

This completes the proof.

Lemma 3.4. Let N = { n 1 , n 2 , ⋯ , n k } be integer number set with k ≥ 3 and | n i | ≥ 3 . If S and T and disjoint sets of N with S ∪ T = N , such that

x = ∏ n i ∈ S n i ≤ y = ∏ n i ∈ T n i , then x + y − x y − ⌈ y x ⌉ meets maximum while x = min { n 1 , n 2 , ⋯ , n k } .

Proof. Without lose generality, suppose min { n 1 , n 2 , ⋯ , n k } = n 1 and

n 1 n 2 ⋯ n k = x y = a . Consider k ≥ 3 and y ≥ x , we have a ≥ x 2 and y ≥ n 1 2 . Thus a ≥ n 1 2 x . Now, let I = n 1 + a n 1 − a − ⌈ a n 1 2 ⌉ , estimate the deference of x + y − x y − ⌈ y x ⌉ and I for x = ∏ n i ∈ S n i > n 1 .

Clearly, if n 1 ≥ 4 , we have

I − ( x + y − x y − ⌈ y x ⌉ ) = n 1 + a n 1 − ⌈ a n 1 2 ⌉ − ( x + a x − ⌈ a x 2 ⌉ ) = ( n 1 − x ) + ( x − n 1 ) a n 1 x − ⌈ a n 1 2 ⌉ + ⌈ a x 2 ⌉ ≥ ( x − n 1 ) ( a n 1 x − 1 ) − a n 1 2 − 1 + a x 2 = ( x − n 1 ) ( a n 1 x − x − n 1 n 1 2 x 2 − 1 ) − 1 ≥ ( x − n 1 ) ( a n 1 x − 2 x n 1 2 x 2 − 1 ) − 1 ≥ ( x − n 1 ) ( n 1 − 3 ) − 1 ≥ 0.

If n 1 = 3 and x ≥ 9 , we similarly have

I − ( x + y − x y − ⌈ y x ⌉ ) = 3 + a 3 − ⌈ a 3 2 ⌉ − ( x + a x − ⌈ a x 2 ⌉ )

= ( 3 − x ) + ( x − 3 ) a 3 x − ⌈ a 3 2 ⌉ + ⌈ a x 2 ⌉ ≥ ( x − 3 ) ( a 3 x − 1 ) − a 3 2 − 1 + a x 2 ≥ ( x − 3 ) ( a − 3 x 3 x ) − a 3 2

= 1 3 [ ( x − 3 ) ( y − 3 ) − a 3 ] ≥ 1 3 ( 2 3 a − 6 y + 9 ) = 1 3 ( 2 x − 18 3 y + 9 ) ≥ 0.

If n 1 = 3 and 4 ≤ x ≤ 8 , this means that | S | = 1 . Suppose x = n s and let a 3 n s = b ≥ 3 . By simple checking, we find that the value x + y − x y − ⌈ y x ⌉ meet maximum while x = n 1 for N = { 3 , 3 , 4 } , { 3 , 4 , 4 } or { 3 , 3 , 5 } . Thus, we get

I − ( x + y − x y − ⌈ y x ⌉ ) = 3 + a 3 − ⌈ a 3 2 ⌉ − ( n s + a n s − ⌈ a n s 2 ⌉ ) = ( 3 − n s ) + ( n s − 3 ) b − ⌈ a 3 2 ⌉ + ⌈ a n s 2 ⌉ ≥ ( n s − 3 ) ( b − 1 ) − a 3 2 − 1 + a n s 2 = ( n s − 3 ) ( b − 1 ) − b n s 3 + 3 b n s − 1

= ( n s − 3 ) ( b − 1 ) − ( n s − 3 ) ( n s + 3 ) 1 3 n s b − 1 = ( n s − 3 ) [ ( b − 1 ) − n s + 3 3 n s b ] − 1 = ( n s − 3 ) [ b ( 1 − n s + 3 3 n s ) − 1 ] − 1.

Clearly, If n s ≥ 6 , then ( n s − 3 ) ( b ( 1 − n s + 3 3 n 3 ) − 1 ) − 1 ≥ 3 ( b × 1 2 − 1 ) − 1 ≥ 0 . If n s = 5 , consider b ≥ 4 , then we have ( n s − 3 ) ( b ( 1 − n s + 3 3 n 3 ) − 1 ) − 1 ≥ 2 ( 4 × 7 15 − 1 ) − 1 ≥ 0 . If n s = 4 , consider b ≥ 5 , then ( n s − 3 ) ( b ( 1 − n s + 3 3 n 3 ) − 1 ) − 1 ≥ 5 × 5 12 − 1 − 1 ≥ 0 .

This completes the proof.

By the above argument, we discuss the rupture degree of tori C n 1 × C n 2 × ⋯ × C n k . Notice that cycle C n 1 n 2 ⋯ n k is a spanning subgraph of C n 1 × C n 2 × ⋯ × C n k . Firstly, by Lemmas 2.1 and 2.2, we directly get the upper bound.

Theorem 3.5. For all integers n 1 , n 2 , ⋯ , n k ≥ 3 , the rupture degree of tori is

r ( C n 1 × C n 2 × ⋯ × C n k ) ≤ { − 2 , if   all   n i   are   odd , − 1 , if   some   n i   is   even .

Theorem 3.6. Let n 1 , ⋯ , n k be integers with 3 ≤ n 1 ≤ n 2 ≤ ⋯ ≤ n k . If some n i is odd, then the rupture degree of tori is

r ( C n 1 × C n 2 × ⋯ × C n k ) ≥ n 1 + ( n 2 ⋯ n k ) ( 1 − n 1 ) − ⌈ n 2 ⋯ n k n 1 ⌉ .

Proof. Notice that C n 1 × C n 2 × ⋯ × C n k is spanning subgraph of K x × K y such that x y = ∏ i = 1 k n i with x = ∏ n i ∈ S n i ≤ y = ∏ n i ∈ T n i . By Lemmas 2.1, 3.2 and Theorem 2.8, we have

r ( C n 1 × C n 2 × ⋯ × C n k ) ≥ r ( K n 1 × K n 2 × ⋯ × n k ) = n 1 + ( n 2 ⋯ n k ) ( 1 − n 1 ) − ⌈ n 2 ⋯ n k n 1 ⌉ .

In particular, if all n i are even, C n 1 × C n 2 × ⋯ × C n k is a spanning subgraph of K n 1 n 2 ⋯ n k 2 , n 1 n 2 ⋯ n k 2 , so by Lemmas 2.1, 3.1 and Theorem 3.5, we get

Theorem 3.7. If all n i are evens, the rupture degree of tori is r ( C n 1 × C n 2 × ⋯ × C n k ) = − 1 .

The authors would like to thank anonymous reviewers for their valuable comments and suggestions to improve the quality of the article. This work was supported by NSFC QHAFFC No. 2022-ZJ-753.

The authors declare no conflicts of interest regarding the publication of this paper.

Li, Y.K. and Zhu, T.Y. (2023) Rupture Degree of Some Cartesian Product Graphs. Open Journal of Discrete Mathematics, 13, 16-26. https://doi.org/10.4236/ojdm.2023.131002