Counting domino trains

Let G be a directed graph with set of vertices V G and edge set E G . A path of length m on G is a sequence of vertices v 0 , v 1 , . . . , v m such that (v i-1 , v i ) ∈ E G for all 1 ≤ i ≤ m. A path is a cycle if v 0 = v m and (v i-1 , v i ) = (v j-1 , v j ) for all 1 ≤ i < j ≤ m. An eulerian path or cycle is a path or cycle of length |E G |. These concepts can be extended to the case when G is an undirected graph in a natural way (see [4] for these and other elementary concepts about Graph Theory).

Two eulerian cycles are called equivalent if one is a cyclic permutation of the other. Let Eul(G) denote the number of equivalence classes of eulerian circuits. If G is a directed graph there is a well-known theorem, the so called BEST Theorem (see [2,3]), which computes Eul(G) but if G is an undirected graph the situation is much more difficult. Nevertheless, in the case of complete graphs interesting results exist. In [1] for instance, an asymptotic value for Eul(K n ) (as well as the exact number for n ≤ 21) is given, with K n being the complete graph with n vertices. Now let us denote by Eul j i (G) the number of eulerian paths in G starting in vertex v i and ending in vertex v j (with no equivalence relation taken into account). In this paper we introduce a new approach and present a new method to compute Eul j i (G) for any undirected graph G. In particular we will count the number of trains that we can construct with a given set of domino pieces, where a train is a chain constructed following the rules of domino and using all the pieces from the given set. Since any graph gives rise to a set of domino pieces where trains correspond to eulerian paths, our method applies to any graph.

The paper is organized as follows. In the first section we present some elementary definitions and fix the notation. The second section is devoted to prove the main result in the paper, namely to present a method to compute the number of trains constructible from a given set of domino pieces. Finally, in the third section we translate this result to a graph theory setting.

Let us denote by {e ij | 1 ≤ i, j ≤ n} the set of matrix units in M n (R); i.e., e ij is the square matrix of size n having a 1 in position (i, j) and 0 elsewhere. Now, we define e ij = e ij + e ji for all i = j and e ii = e ii . Clearly the set B = {e ij | 1 ≤ i ≤ j ≤ n} is a basis for the vector space S n (R) of real symmetric matrices of size n. If we define a new product over M n (R) by A • B = AB + BA our basis multiply in the following way:

(1)

. With the previous convention and choosing e t = e itjt for t = 1, . . . , m it makes sense to compute S m (e 1 , . . . , e t ) =

In what follows we will denote by (i, j) the domino piece marked with numbers i and j. We recall that in the game of domino two pieces can be placed together if they share at least one of their numbers. Now let us suppose that we are given certain set of dominoes that we shall denote by D = {(i 1 , j 1 ), . . . , (i m , j m )}, we define a train as a sequence (i k1 , j k1 ) . . . (i km , j km ) admissible by the rules of domino; i.e., such that j kr = i kr+1 for all 1 ≤ r ≤ m -1.

Given a set of domino pieces D = {(i 1 , j 1 ), . . . , (i m , j m )} we are interested in counting the number of trains that we can construct using all the pieces of D. If an element of D appears more than once, we will assume that we can distinguish them.

For any domino piece (i, j) we will identify (i, j) ↔ e ij . Clearly, two pieces (i, j) and (k, l) can be placed together following the rules of domino if and only if e ij • e kl = 0. Consequently we have the following:

Proof. By induction on n. Cases n = 1, 2 are obvious due to (1). Now let us suppose that (i 1 , j 1 )(i 2 , j 2 ) . . .

Then, there must exist 2 ≤ k < n such that e i1j1 •• • ••e i k j k = 0 and by our induction hypothesis (i 1 , j 1 ) . . . (i k , j k ) is not a train which is a contradiction. Conversely if e i1j1 • e i2j2 • • • • • e injn = 0, then e i1j1 • e i2j2 = 0 and e i2j2 • • • • • e injn = 0 so, by induction, both (i 1 , j 1 )(i 2 , j 2 ) and (i 2 , j 2 ) . . . (i n , j n ) are trains and consequently so is (i 1 , j 1 )(i 2 , j 2 ) . . . (i n , j n ). Now, given the set D and since the set B is a basis for S n (R), we have that S m (e i1j1 , . . . , e imjm ) = α ij e ij . In fact, the following lemma holds. Lemma 2. In the previous situation, α i k j k = 0 if and only if we can construct a train starting with i k and ending with j k (or viceversa) using the pieces of the set D.

Proof. It is an easy consequence of (1) and Lemma 1.

Observe that, given a train (i 1 , j 1 )(i 2 , j 2 )(i 3 , j 3 ) we could have placed its pieces in different ways according to the order. Namely we could have firstly placed the piece (i 1 , j 1 ) then (i 2 , j 2 ) on its right and finally the piece (i 3 , j 3 ) on the right of the latter. We could also have started by piece (i 2 , j 2 ), then (i 1 , j 1 ) on its left and finally (i 3 , j

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