Interlace polynomials of 4 n -snowflake graphs

In this paper, we study the interlace polynomial of a special graph with n vertices, called 4 n - snowflake graph. It is similar as the friendship graph F n of n vertices, which is made of n 3-cycles sharing one center vertex. In stead of 3-cycles, the 4 n -snowflake graph Q n is constructed by gluing n 4-cycles to one center vertex. We describe certain properties of such graphs, provide recursive and explicit formulas for the interlace polynomials


History and Background
Graph polynomials have been studied and applied to characterize certain properties of the considered graphs. Two well-known polynomials are Tutte and Martin polynomials [6], [7]. The concept of interlace polynomial first appeared along with a study of Euler circuits of undirected 4-regular graphs and the transformations [4]. Arratia, Bollóbas, Coppersmith, and Sorkin [3] introduced the interlace polynomial of a graph, a polynomial that represents the information gained from doing a toggling process on the graph.
The interlace polynomial of a graph gives important information about the graph, for example, the number of k-component circuit partitions, for any k ∈ N [3]. The interlace polynomials of certain special graphs, such as paths, cycles, stars, and complete graphs, were studied ( [1,3,4].) Recently, ladder graphs, n-claw graphs, forest graphs, Friendship graphs, and butterfly graphs were studied ( [2,9,10,12].) Work has been done on multi-variate interlace polynomials as well, see [2,8].
In this paper, we focus on snowflake graphs, which is similar as the Friendship graphs. In [12], Friendship graphs were studied. In the Friendship graph F n with n vertices, n 3-cycles share exactly one center vertex v. For the 4n-snowflake graphs, instead of 3-cycles, we glue n 4cycles to one center vertex. We name it the 4n-snowflake graph. We describe certain properties of such graphs, provide recursive and explicit formulas for the interlace polynomials, and give some properties of the polynomials such as special values and patterns for certain coefficients.

Defining the pnterlace Polynomial
The interlace polynomial of a given simple graph G is recursively defined by Arratia, Bollobás, and Sorkin in [3]. First, the interlace polynomial of the empty graph E n with n vertices is defined to be x n . For any non-empty undirected graph G, we first select one edge ab from G, where a and b are two adjacent vertices of G. We consider three special neighborhoods related to a and b: . A toggling process on the edge ab is applied to G to create a "pivot" of G based on the neighborhoods for the two end vertices of a and b. This process creates a new graph G ab such that G and G ab have the same vertex set and for every pair of vertices u, v belonging to different neighborhoods V a , V b , V ab shown above, uv is an edge of G ab if and only if uv is not an edge of G. The resulting graph G ab is called the pivot of G at ab. The interlace polynomial of a graph G is defined recursively by applying the pivot process.
Definition 1 (Interlace Polynomial). [3] Let G be any undirected graph with n vertices and ab be an edge of G. The interlace polynomial q(G, x) of G is defined by Arratia, Bollobás, and Sorkin in [3] showed that the interlace polynomial is well-defined, that is, it is independent on the choice of the edge. The graph Q n is similar to the Friendship Graph F n ( [12]), which is the graph resulted by gluing n 3-cycles, instead of n 4-cycles. Obviously, the graph Q 1 = C 4 . In this paper, we study the properties of the graph Q n and its interlace polynomial q(Q n , x). We give a useful recursive formula for q(Q n , x) and use it to investigate certain special values, coefficients, the properties of this polynomial. The adjacency matrix of Q n and related matrices are studied. The interlace polynomial helps to describe the ranks of the matrices and some properties of the ground graph Q n .

Interlace Polynomial q(Q n , x) and Its Properties
First we list some basic graph theory properties about the graph Q n . We skip the proof because it is straightforward.
3. Q n has exactly one cut vertex, the center vertex of degree 2n. 4. The independence number of Q n is α(Q n ) = 2n. 5. The size of a maximal matching of Q n is µ(Q n ) = n + 1.
Since Q 1 = C 4 , the interlace polynomial q(Q 1 , x) = q(C 4 , x) = 3x 2 + 2x. For n = 2, we decompose the graph Q 2 by two toggling processes. The graph Q 2 and its pivot Q ab 2 at the edge ab are shown in Figure 3. By toggling at the edge ab, we decompose Q 2 into the two graphs Q 2 − a and Q ab 2 − b shown in Figure 4. Thus, we have We further toggle the graph Q 2 − a at the edge bu and then at the edge uc (see Figure 4). It results in q(Q 2 − a, x) = x 2 q(P 2 , x) + q(C 4 , x) + xq(P 2 , x). By Theorem 2(1)(2), q(P 1 , x) = 2x, q(P 2 , x) = x 2 + 2x, and q(C 4 , x) = 3x 2 + 2x. Combining these formulas, Next, we give a recursive formula for the interlace polynomial of Q n for any positive integer n > 1. The process is very similar as the pivot process for Q 2 .
Proof. We fix an edge ab of Q n . The graph Q n and the pivot Q ab n are show in Figure 5. We perform a similar toggling process on ab as shown for the graph Q 2 : 1. Toggling Q n at ab to obtain two graphs Q n − a and Q ab n − b (refer to Figure 6). Obviously, Q ab n − b = P 1∪ Q n−1 , the disjoint union of the path P 1 and Q n−1 . Thus q(Q ab n − b, x) = 2xq(Q n−1 , x). 2. Toggling the graph Q n − a at the edge bu which decomposes the graph into (Q n − a) − b and the disjoint union of the empty graph E 2 with n − 1 copies of P 2 ( Figure 7). Thus, 3. Similarly, toggling the graph (Q n − a) − b at the edge cu, we obtain the graph Q n−1 and the disjoint union of E 2 with n − 1 copies of P 2 .

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Interlace polynomials of 4n-snowflake graphs | J. Champanerkar and A. Li The above procedure produces the recursive relation: Since P 2 = x 2 + 2x by Theorem 2, we obtain Theorem 6 enables us to recursively describe the interlace polynomials q(Q n , x) for all positive integers n. Below we show those for n = 1, 2, 3, 4, 5, 6, 7, 8. One can see some interesting patterns on the coefficients. They will be discussed later.
Using the iterative formula given in Theorem 6 we develop an explicit formula for q(Q n , x) at the value x = 1.

Explicit Formula for q(Q n , x)
Theorem 7 gives an explicit formula for q(Q n , x) at x = 1. In this section, we develop an explicit formula for q(Q n , x) for every x ̸ = 1.
Theorem 8. For x ̸ = 1 and n ≥ 1, Proof. It is by mathematical induction again. For n = 1, the formula is true because by Example 1: Apply the induction hypothesis and the recursive formula given in Theorem 6 again, Many properties of q(Q n , x) can be developed from the above recursive formula and the explicit formula. It is known that some special values of the interlace polynomial can tell something about the underlying graph. The following proposition gives special values of q(Q n , x) at x = 2, −1, and 2. Formulas for the degree of the polynomial Q n and some coefficients are given as well.
Proof. From Theorem 6, for n > 1 we have From Theorems 7 and 8, we have We use equations (1) and (2) to prove the assertions stated in Proposition 9.
1. We apply mathematical induction on the number k > 0 which is the number of copies of C 4 in Q k . Since Q 1 = C 4 , q(Q 1 , x) = 3x 2 + 2x. Hence deg(q(Q 1 , x)) = 2. Suppose deg(q(Q k , x)) = 2k. We will prove the claim for Q k+1 . From Equation (1), Obviously, deg(x k+1 (x + 1)(x + 2) k ) = 2k + 2. By the induction hypothesis, deg((2x + 1)q(Q k , x)) = 2k + 1 and the leading term of q(Q k+1 , x) is x 2k+2 . Hence, the degree of the polynomial q(Q k+1 , x) is 2k + 2 = 2(k + 1) with leading coefficient 1. Thus by mathematical induction, the leading term of q(Q n , x) is x 2n and deg(q(Q n , x)) = 2n for every positive integer n; 2. Since q(Q 1 , x) = 3x 2 + 2x, we see that the coefficient of the x-term is 2. Suppose that the coefficient of the x-term in q(Q k , x) is 2. We will prove the claim for q(Q k+1 , x). Refer to Equation (1) again. Since the constant terms of q(Q k , x) is 0, the lowest degree term is 2x in q(Q k+1 , x). Thus, by the principle of mathematical induction, the x-term in q(Q n , x) has coefficient 2 for all n ≥ 1. 3-5. The statements 3, 4, 5 can be shown by substituting 2, −1, −2 in the formula given in Theorem 8. 6. By Theorem 7, it is true for n = 1. For n ̸ = 1, note that in Equation (1), x n (x + 1)(x + 2) n−1 is even for all integers x. Then the parity of q(Q n+1 , x) is the same as that of q(Q n , x). However, q(Q 1 , x) is even if and only if x is even.
Next, we examine the coefficients of q(Q n , x). Since the degree of q(Q n , x) is 2n and the constant is 0, we write q(Q n , x) = a n,(2n) x 2n + a n,(2n−1) x 2n−1 + · · · + a n,2 x 2 + a n,1 x = 2n k=1 a n,k x k .
From the previous results, a n,1 = 2, a n,2n = 1. To obtain the other coefficients of q(Q n , x), we introduce another function f n (x): By Theorem 8, for x ̸ = 1, f n (x) = (x − 1)q(Q n , x). Assume the coefficient of x k for f n (x) is b n,k , that is, We now describe the coefficients of f n (x).

It proves (3). For (4), applying (1) and (3).
The above lemma and the Theorem 10 can help in developing the explicit formula for each coefficient a n,k of the polynomial q(Q n , x). Now we discuss these coefficients.
Theorem 12. For n ≥ 1, q(Q n , x) can be expressed as Proof.
By the binomial expansion formula, the second part of q(Q n−1 , x) can be written as Combining all the above, we obtain the desired expression for q(Q n , x).
Applying Theorems 12 and 10, we can describe the coefficients a n, k as follows: Theorem 13. For any integer n ≥ 4,
3. Similarly as in the above proof, a n, n = a n, n+1 + b n, n+1 . Applying the result from the above part 4 and Theorem 10, we have a n, n = 3 n + (n − 2)2 n−1 . 4. (4) and (5) are straightforward. We skip the proof.
We claim that the coefficient sequence is one mode, but the maximum does not necessarily occur in the middle. Also, the leading coefficient is the only even coefficient. Consider the sequence (a n,k ) 2n k=1 , representing the coefficients of the polynomial q(Q n , x).
For 3 ≤ k ≤ n, since a n, k = a n, k−1 − b n,k , a n, k ≥ a n, k−1 if and only if b n,k ≤ 0. While, by Theorem 10, Thus, a n, 3 ≤ a n, 4 ≤ · · · ≤ a n, rn and a n, rn ≥ a n, k+1 · · · ≥ a n, n .

The Rank of a Related Matrix
In graph theory, the adjacency matrix of a given graph is used to describe the adjacency of the vertices and some connectivity properties. In particular, certain operations on the adjacency matrix help in counting the numbers of paths and cycles in the graph. In [5], Balister, Bollóbas, Cutler, and Peabody gave an explicit formula for the interlace polynomial of a graph at x = −1 involving the rank of the adjacency matrix plus the identity matrix modulo 2.
Theorem 16. [5] Let A be the adjacency matrix of a graph G with m vertices and let r = rank(A + I) modulo 2, where I is the m × m identity matrix. Then Recall that the graph Q n has 3n + 1 vertices. Denote the adjacency matrix of Q n by A 3n+1 and let B 3n+1 = A 3n+1 + I 3n+1 , where I 3n+1 is the (3n + 1) × (3n + 1) identity matrix. In this section, we examine the adjacency matrix A 3n+1 and B 3n+1 . We apply the above result to determine the rank of the adjacency matrix B 3n+1 . Refer to the graph of Q n with vertices labeled as follows: . . Assume the i th row of the adjacency matrix A 3n+1 represents the vertex v i (i = 0, 1, . . . , 3n + 1). The 1 × 3 matrix C = [1, 0, 1] occurs repeatedly as a sub-matrix of A 3n+1 . Then the first row of A 3n+1 , representing v 0 , is [1, C, C, . . . , C] with n copies of C. For 1 ≤ k ≤ n, the adjacency matrix for the induced subgraph by v 3k−2 , v 3k−2 , v 3k is given by where I 3 is the 3 × 3 identity matrix. We now give the structures of the matrices A 3n+1 and B 3n+1 : Lemma 17. Let n be any positive integer and A 3n+1 , B 3n+1 , C, M A , M B be as above. Then the two (3n + 1) × (3n + 1) matrices A 3n+1 and B 3n+1 are structured below: where the empty places are occupied by 0's.
As an application of the result for interlace polynomials stated in Theorem 16, we claim that for every positive integer n, the matrix B 3n+1 is of full rank modulo 2.
Of course, the above result can be achieved by the traditional linear algebra method through elementary row operations. But the process is more complicated. We sketch the process briefly.
Using the number 1 in the (1,1) position of B ′ 3n+1 , we can perform elementary row operations to make the resulting matrix having 0 in the (i, 1) positions for all i > 1. Similarly, by performing corresponding elementary column operations, we can reduce the matrix into one with 0 in all of the (1, j) positions. That is, by multiplying an invertible matrix A from left and some invertible matrix A ′ from right, we obtain 1 0 0 · · · · · · 0 0 I 3 * · · · · · · * 0 0 M B * · · · * . . . Since M B a 3 × 3 matrix of full tank, the rank of B ′′ 3n+1 is 3n + 1 and so does that of B 3n+1 . Certainly the method using the interlace polynomial of Q n to determine the rank of B 3n+1 is easier (Theorem 18). It is nice to see an application of a graph theory result in solving a linear algebra problem.