Let \(G\) be a simple, finite and connected graph. A graph is said to be decomposed into subgraphs \(H_1\) and \(H_2\) which is denoted by \(G= H_1 \oplus H_2\), if \(G\) is the edge disjoint union of \(H_1\) and \(H_2\). Assume that \(G= H_1 \oplus H_2 \oplus \cdots \oplus H_k\) and if each (H_i\), \(1 \leq i \leq k\), is a path or cycle in \(G\), then the collection of edge-disjoint subgraphs of \(G\) denoted by \(\psi\) is called a path decomposition of \(G\). If each \(H_i\) is a path in \(G\) then \(\psi\) is called an acyclic path decomposition of \(G\). The minimum cardinality of a path decomposition of \(G\), denoted by \(\pi (G)\), is called the path decomposition number and the minimum cardinality of an acyclic path decomposition of \(G\), denoted by \(\pi_a(G)\), is called the acyclic path decomposition number of \(G\). In this paper, we determine path decomposition number for a number of graphs in particular, the Cartesian product of graphs. We also provided bounds for \(\pi(G)\) and \(\pi_a(G)\) for these graphs.
Definition 1.1.
The Cartesian product \(G \ \Box \ H\) of two graphs \(G\) and \(H\) is a graph with vertex set \(V(G)\times V(H)\) in which \((x_1,y_1)\) and \((x_2,y_2)\) are adjacent if one of the following condition holds:
Theorem 2.1 [2] For even integers \(m\) and \(n\) with \(4 \leq m \leq n\), the graph \(K_n-I\) decomposes cycles of length \(m\) if and only if the number of edges in \(K_n-I\) is a multiple of \(m\)
Lemma 2.2. [11] Let \(m \equiv 2(mod \ 4)\), \(n \equiv 1(mod \ 2)\) and \(6 \leq m \leq 2n\). Then \(C_m |K_{n,n}-I\) if and only if \(m|n(n-1)\).
Theorem 2.3. Given the graph \(K_n-I\), where \(n\) is even, the minimum path decomposition number for \(K_n-I\) is \(\frac{n-2}{2}\).
Proof. The graph \(K_n-I\) has \(n\) vertices and \(\frac{n(n-2)}{2}\) edges. The largest cycle which is a subgraph of \(K_n-I\) is a cycle of order \(n\). Now, by Theorem 2.1, \(C_n |K_n-I\). We only need to know the number of copies \(C_n\) that can be gotten from \(K_n-I\), which is \(\frac{n-2}{2}\). Thus, we have \(\frac{n-2}{2}\) copies of \(C_n\) in \(K_n-I\). Therefore, \(\pi (K_n-I)=\frac{n-2}{2}\).
Lemma 2.4. If \(n \geq 4\) and an even integer, then \(K_{n,n}-I\) is \(\Big ( \frac{n-2}{2} C_{2n},nP_2 \Big )\)-decomposable.
Proof.
Let \(X=\{1^1,2^1,3^1,…,n^1\}\) and \(Y=\{1^2,2^2,3^2,…,n^2\}\) form the column set of vertices in \(K_{n,n}-I\). Also, two vertices \(a^i\) and \(b^j\), has an edge in \(K_{n,n}-I\), if \(a \neq b\) and \(i \neq j\), \(i< j=2\). Since \(n\) is even, the degree of each vertex in \(K_{n,n}-I\) is odd.
Next, remove the edges $$
E(a^i, b^j)= \left\{\begin{array}{cc}
(a,n-a+1) ,& \; a=1,2 \\
(a,a-2), & \; a=3,4,5,…,n
\end{array}\right.
,a^i \in X, b^j \in Y
$$
which are exactly \(n\) number of \(P_2\)’s. By removal of these edges, each vertex in \(K_{n,n}-I\) would be of even degree. In total, we have \(n(n-2)\) edges. At this point, we need to show that the subgraph \((K_{n,n}-I) \setminus E(a^i,b^j)\) admits a \(C_{2n}\) decomposition.
Now, by \(C_{2n}^r\), \(r \leq 1\), we mean the \(r^{th}\) copy of \(C_{2n}\) in \((K_{n,n}-I) \setminus E(a^i,b^j)\). With exception of \(C_{2n}^1\), all other \(C_{2n}^r\), \(r>1\), follow a similar pattern. The construction of these cycles of order \(2n\) is given below.
\begin{align*}
C_{2n}^1 & =1^1,2^2,3^1,4^2,…,(n-1)^1,n^2,(n-2)^1,(n-3)^2,(n-4)^1,
(n-5)^2,\nonumber\\
& \ \ \ \ 2^1,1^2,n^1,(n-1)^2,1^1.
\end{align*}
For \(r=2,3,4,…,\frac{n-2}{2}\) we have that
\begin{align*}
C_{2n}^r & =1^1,(2r-1)^2,n^1,(2r-2)^2,(n-1)^1,
(2r-3)^2,(n-2)^1,…,1^2, \nonumber\\
& \ \ \ \ (n-2r+2)^1,(n-1)^2,(n-2r+1)^1,n^2,(n-2r)^1,(n-2)^2, \nonumber\\
& \ \ \ \ (n-2r-1)^1,(n-3)^2,(n-2r-2)^1,(n-4)^2,…,(2r)^2,1^1.
\end{align*}
From the above construction, we conclude that the graph \((K_{n,n}-I) \setminus E(a^i,b^j)\) admits a \(C_{2n}\) decomposition. Clearly, \(r=\frac{n-2}{2}\) and thus \(C_{2n}|\big\{K_{n,n} -I \setminus E(a^i,b^j)\big \}=(C_{2n} \oplus C_{2n} \oplus C_{2n} \oplus \cdots \oplus \frac{n-2}{2} C_{2n} )\). Finally, we have that \(K_{n,n}-I\) is \(\Big ( \frac{n-2}{2} C_{2n},nP_2 \Big )\)-decomposable. Hence the proof.
Theorem 2.5. For the complete bipartite graph \(K_{n,n}-I\), we have that $$ \pi (K_{n,n}-I)= \left\{\begin{array}{cc} \frac{n-1}{2} ,& \; if \ n \ is \ odd \\ \frac{3n-2}{2} ,& \; otherwise \end{array}\right. $$
Proof.
The graph \(K_{n,n}-I\) has \(2n\) vertices and \(n(n-1)\) edges. The largest cycle which is a subgraph of \(K_{n,n}-I\) is a cycle of order \(2n\). We now prove this theorem in two cases.
Case 1: when \(n\) is odd
By Lemma 2.2, \(C_{2n} |K_{n,n}-I\). We only need to know the number of copies of \(C_{2n}\) that can be obtained from \(K_{n,n}-I\), which is \(\frac{n-1}{2}\). Therefore, \(\pi (K_{n,n}-I)=\frac{n-1}{2}\).
Case 2: when \(n\) is even
By Lemma 2.4, the graph \(K_{n,n}-I\) can be decomposed into \(\frac{n-2}{2}\) copies of \(C_{2n}\) and \(n\) copies of \(P_2\). Since no vertex is repeated in these \(n\) copies of \(P_2\), we have that \(\pi (K_{n,n}-I)=\frac{3n-2}{2}\). The proof of this theorem is complete.
Remark 2.6. In [10], it was mentioned that every graph \(G\) which is Hamiltonian cycle decomposable attains the bound that \(\pi (G) \geq \lceil \frac{\Delta}{2} \rceil\). This is true as we see from Theorem 2.3 and Theorem 2.5 that the complete graph minus a one-factor and the complete bipartite graph \(K_{n,n}-I\), where \(n\) is odd, attains this bound. Now, when \(n\) is even in \(K_{n,n}-I\) we have \(\pi (K_{n,n}-I)=\frac{3\Delta+1}{2}\).
Theorem 3.1. Let \(m\) and \(n\) be positive integers then $$\pi(P_m \ \Box \ C_n )= \pi_a (P_m \ \Box \ C_n )=n$$.
Proof. First we give the construction of \(P_{mn}\) paths by constructing Hamilton paths of order \(n\) in each copy of \(C_n\) in \(P_m \ \Box \ C_n\). Let \(i\) be an odd number, in each copy of \(C_n^i\), join the end vertex of the Hamilton path in the \(i^{th}\) copy with the end vertex of the \(C_n^{i+1}\) copy of \(P_m \ \Box \ C_n\). Similarly, suppose \(i\) is even, in each copy of \(C_n^i\), join the first vertex of the Hamilton path in the \(i^{th}\) copy with the first vertex of the \(C_n^{i+1}\) copy of \(P_m \ \Box \ C_n\).
Next, for each internal vertex in the Hamilton path, join the vertices \(x_j^i\) and \(x_j^{i+1}\), \(1 \leq i \leq m\), \(i\) is calculated in modulo \(m\) and \(2 \leq j \leq n-1\). By this, we have \(n-2\) copies of \(P_m\) in \(P_m \ \Box \ C_n\).
Lastly, the left out edges which has not been covered by the path \(P_{mn}\) and the \(n-2\) copies of \(P_m\) form a path of order \(2m\). So we have that \(\pi(P_m \ \Box \ C_n)=\pi_a (P_m \ \Box \ C_n)=n\).
Remark 3.2. Since the Cartesian product of graph is commutative, the result in Theorem 3.1 holds for the graph \(C_m \ \Box \ P_n\) where \(m\) and \(n\) are positive integers.
Theorem 3.3. Let \(m\) and \(n\) be positive integers such that \(3 \leq n \leq m\), then \(\pi (C_m \ \Box \ C_n)=n\).
Proof. Since both \(m\) and \(n\) are positive integers, the proof of this theorem is split in two cases.
Case 1: when \(m\) is even and \(n \geq 3\).
First we give the construction of \(C_{mn}\) cycles by constructing Hamilton paths of order \(n\) in each copy of \(C_n\) in \(C_m \ \Box \ C_n\). Let \(i\) be an odd number, in each copy of \(C_n^i\), join the end vertex of the Hamilton path in the \(i^{th}\) copy with the end vertex of the \(C_n^{i+1}\) copy of \(C_m \ \Box \ C_n\). Similarly, suppose \(i\) is even, in each copy of \(C_n^i\), join the first vertex of the Hamilton path in the \(i^{th}\) copy with the first vertex of the \(C_n^{i+1}\) copy of \(C_m \ \Box \ C_n\).
Next, for each internal vertex in the Hamilton path, join the vertices \(x_j^i\) and \(x_j^{i+1}\), \(1 \leq i \leq m\), \(i\) is calculated in modulo \(m\) and \(2 \leq j \leq n-1\). By this, we have \(n-2\) copies of \(C_m \ \Box \ C_n\).
Now, notice that the left out edges which has not been covered by the cycle \(C_{mn}\) and the \(n-2\) copies of \(C_m\) form a cycle of order \(2m\). So we have that \(\pi (C_m \ \Box \ C_n)=n\).
Case 2: when \(m\) is odd and \(n \geq 3\).
Here, we first give the construction of \(C_{mn-1}\) cycles. For \(1 \leq i \leq m-2\), construct Hamilton paths of order \(n\) in each \(C_n^i\) copy in \(C_m \ \Box \ C_n\). Suppose \(i\) is odd, in each copy of \(C_n^i\) join the end vertex of the Hamilton path in the \(i^{th}\) copy with the end vertex of the \(C_n^{i+1}\) copy of \(C_m \ \Box \ C_n\). In the same way, if \(i\) is even, in each copy of \(C_n^i\) join the first vertex of the Hamilton path in the \(i^{th}\) copy of \(C_n^i\) with the first vertex of the \(C_n^{i+1}\) copy of \(C_m \ \Box \ C_n\). This gives a path of order \(n(m-2)\).
Next, let \(x\) be the first vertex in the \(C_n^{m-1}\) copy of \(C_m \ \Box \ C_n\). Now, construct a path \(P_{n-1}\) from \(C_n^{m-1} \setminus x\). Join the end vertex of \(C_n^{m-1}\) copy to the end vertex of \(C_n^{m-2}\) copy of \(C_m \ \Box \ C_n\). Since \(x\) is removed from \(C_n^{m-1}\), join the second vertex \(x_2^{m-1}\) of \(C_n^{m-1}\) to the second vertex \(x_2^m\) of \(C_n^m\) and then move in a clockwise direction to the first vertex in the \(m^{th}\) copy of \(C_m \ \Box \ C_n\). To get the desired \(C_{mn-1}\) cycle, join the first vertex \(x_1^m\) of \(C_n^m\) to \(x_1^1\) of \(C_n^1\) in the graph \(C_m \ \Box \ C_n\).
Furthermore, aside the second vertex, each internal vertex \(x_j^i\) and \(x_j^{i+1}\), \(1 \leq i \leq m\), \(i\) is calculated in modulo \(m\) and \(3 \leq j \leq n-1\) when joined in all other copies of \(C_n\) results to \(n-3\) copies of \(C_m\) in \(C_m \ \Box \ C_n\).
The left out edges which have not been covered by the cycle \(C_{mn-1}\) and the \(n-3\) copies of \(C_m\) form cycles \(C_{m+2}\) and \(C_{2m-1}\). We now give the construction of cycles \(C_{m+2}\) and \(C_{2m-1}\) as follows. By \(x_j^i\) we mean the \(j^{th}\) vertex of \(C_n\) in copy \(i\) of the graph \(C_m \ \Box \ C_n\).
\begin{align*}
C_{m+2}=x_2^1,x_2^2,x_2^3,…,x_2^{m-1},x_1^{m-1},x_1^m,x_2^m,x_2^1.\nonumber\\
C_{2m-1}=x_n^1,x_1^1,x_1^2,x_n^2,x_n^3,x_1^3,…,x_1^{m-1},x_n^{m-1},x_n^m,x_n^1.\nonumber
\end{align*}
Therefore we have that \(\pi (C_m \ \Box \ C_n)=n\). This completes the proof.
Remark 3.2. We note here in this section that although Arumugam et al. in [10] gave a relationship between the path decomposition number (or acyclic path decomposition number, as the case maybe) and the maximum degree \(\Delta\) of some graphs, we note that for the product \(G \ \Box \ H\) there is no such relationship since the parameters \(\pi (G \ \Box \ H)\) and \(\pi_a (G \ \Box \ H)\) do not depend on \(\Delta(G \ \Box \ H)\).