Let \(P_{k+1}\) denote a path of length \(k\), let \(S_{m}\) denote a star with \(m\) edges, and let \(K_{n}(\lambda)\) denote the complete multigraph on \(n\) vertices in which every edge is taken \(\lambda\) times. In this paper, we prove that the necessary conditions are also sufficient for a \(\{P_{4}, S_{4}\}\)-decomposition of \(K_{n}(\lambda)\).
All graphs considered here are finite and undirected, with no loops. For standard graph-theoretic terminology, the reader is referred to [1]. A simple graph in which each pair of distinct vertices is joined by an edge is called a complete graph. A complete graph on \(n\) vertices is denoted by \(K_{n}\). If more than one edge joining two vertices is allowed, the resulting object is called a multigraph. Let \(K_{n}(\lambda)\) denote the complete multigraph on \(n\) vertices in which every edge is taken \(\lambda\) times. A complete bipartite graph is a simple bipartite graph with bipartition \((X,Y)\) in which each vertex of \(X\) is joined to each vertex of \(Y\); if \(|X|=a\) and \(|Y|=b\), such a graph is denoted by \(K_{a,b}\). If \(a=b\), the complete bipartite graph is referred to as balanced. In \(K_{a,b}(\lambda)\), we label the vertices in the partite set \(X\) as \(1,2,\ldots,a\) and those in \(Y\) as \(a+1,a+2,\ldots,a+b\). A path is an open trail with no repeated vertex. A path with \(k\) edges is denoted by \(P_{k+1}\). The complete bipartite graph \(K_{1,m}\) is called a star and is denoted by \(S_{m}\). For \(m\geq3\), the vertex of degree \(m\) in \(S_{m}\) is called the center, and any vertex of degree \(1\) in \(S_{m}\) is called an end vertex.
Let \(G\) be a graph, and let \(G_{1}\) be a subgraph of \(G\). Then \(G\backslash G_{1}\) is obtained from \(G\) by deleting the edges of \(G_{1}\). Let \(G_{1}\) and \(G_{2}\) be subgraphs of \(G\). The union \(G_{1}\cup G_{2}\) of \(G_{1}\) and \(G_{2}\) is the graph with vertex set \(V(G_{1})\cup V(G_{2})\) and edge set \(E(G_{1})\cup E(G_{2})\). We say that \(G_{1}\) and \(G_{2}\) are edge-disjoint if they have no edge in common. If \(G_{1}\) and \(G_{2}\) are edge-disjoint, we denote their union by \(G_{1}+G_{2}\). A decomposition of a graph \(G\) is a collection of edge-disjoint subgraphs \(G_{1},G_{2},\ldots,G_{n}\) of \(G\) such that every edge of \(G\) is in exactly one \(G_{i}\). In this case, \(G\) is said to be decomposed or decomposable into \(G_{1},G_{2},\ldots,G_{n}\). If \(G\) has a decomposition into \(p_{1}\) copies of \(G_{1},\ldots,p_{n}\) copies of \(G_{n}\), then we say that \(G\) has a \(\{p_{1}G_{1},\ldots,p_{n}G_{n}\}\)-decomposition. If such a decomposition exists for all values of \(p_{1},\ldots,p_{n}\) satisfying the trivial necessary conditions, then we say that \(G\) has a \(\{G_{1},\ldots,G_{n}\}_{\{p_{1},\ldots,p_{n}\}}\)-decomposition or that \(G\) is fully \(\{G_{1},\ldots,G_{n}\}\)-decomposable. We say that \(G\) is decomposed into \(P_{4}\) and \(S_{4}\) if each \(G_{i}\simeq P_{4}\) or \(S_{4}\). We say that \(G\) is decomposed into \(H\), or that \(G\) has an \(H\)-decomposition, if each \(G_{i}\simeq H\).
In [10], Priyadharsini and Muthusamy gave necessary and sufficient conditions for the existence of a \(\{pG_{1},qG_{2}\}\)-decomposition of \(K_{n}(\lambda)\), when \[(G_{1},G_{2})\in\{(P_{n},S_{1,n-1}), (C_{n},S_{1,n-1}), (P_{n},C_{n})\}.\] In [9], Priyadharsini gave necessary and sufficient conditions for the existence of a \(\{pP_{n},\\ qS_{1,n-1}\}\)-decomposition of \(K_{n+1}(\lambda)\). In [11], Shyu gave the necessary conditions for a \(\{pP_{k+1},qS_{k}\}\)-decomposition of \(K_{n}\) and proved that \(K_{n}\) is fully \(\{P_{k+1},S_{k}\}\)-decomposable when \(n\geq4k\), such that either \(k\) is even and \(p\geq\frac{k}{2}\), or \(k\) is odd and \(p\geq k\); Shyu also settled the case \(k=3\) completely, i.e., for all \(n\geq6\). In [12], Shyu proved that \(K_{n}\) is fully \(\{P_{k+1},S_{k}\}\)-decomposable when \(n\geq4k\). In [7], Lee and Chen showed the existence of a \(\{pP_{k+1},qS_{k}\}\)-decomposition of \(K_{n}(\lambda)\) and \(K_{b,b}(\lambda)\). In [8], Lee and Chen gave the necessary conditions for a \(\{pF,qS_{3}\}\)-decomposition of \(K_{n}\) and proved that \(K_{n}\) is fully \(\{F,S_{3}\}\)-decomposable when \(F\in\{P_{n},C_{n}\}\). In [13], Shyu gave the necessary conditions for a \(\{pC_{k},qP_{k+1},rS_{k}\}\)-decomposition of \(K_{n}\) and proved that \(K_{n}\) is fully \(\{C_{4},P_{5},S_{4}\}\)-decomposable when \(n\) is odd. In [5], Ilayaraja and Muthusamy gave the necessary conditions for a \(\{pP_{k+1},qS_{m}\}\)-decomposition of \(K_{n}\) and proved that \(K_{n}\) is fully \(\{P_{4},S_{4}\}\)-decomposable. In [2], the authors gave the necessary conditions for a \(\{pP_{k+1},qS_{m}\}\)-decomposition of \(K_{n}(\lambda)\) and proved that \(K_{n}(\lambda)\) is fully \(\{P_{7},S_{4}\}\)-decomposable. In [3, 4], the authors proved that \(K_{n}(\lambda)\) is fully \(\{P_{7},S_{3}\}\)-decomposable and that \(K_{n}(\lambda)\) is fully \(\{P_{5},S_{3}\}\)-decomposable. In this paper, we prove that \(K_{n}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
For convenience, we denote \(V(K_{n}(\lambda))=\{x_{1},x_{2},\ldots,x_{n}\}\). The notation \(S(x_{1};x_{2},\ldots,x_{m})\) denotes an \(m\)-star with \(x_{1}\) as the center vertex and \(x_{2},\ldots,x_{m}\) as end vertices, and \((x_{1}x_{2}\ldots x_{k+1})\) denotes a path with vertices \(x_{1},x_{2},\ldots,x_{k+1}\) and edges \(x_{1}x_{2},x_{2}x_{3},\ldots,x_{k}x_{k+1}\).
We recall here some results on \(P_{k+1}\)– and \(S_{m}\)-decompositions that are useful for our proofs.
Theorem 2.1. [14] A necessary and sufficient condition for the existence of a \(P_{k+1}\)-decomposition of \(K_{n}(\lambda)\) is that \(\lambda{n\choose 2}\equiv 0\pmod{k}\) and \(n\geq k+1\).
Theorem 2.2. [15] A necessary and sufficient condition for the existence of an \(S_{m}\)-decomposition of \(K_{n}(\lambda)\) is that:
\(\lambda{n\choose 2}\equiv 0\pmod m\);
\(n\geq2m\) for \(\lambda=1\);
\(n\geq m+1\) for even \(\lambda\);
\(n\geq m+1+\frac{m}{\lambda}\) for odd \(\lambda\geq3\).
Theorem 2.3. [16] Let \(k\) be a positive integer, and let \(a\) and \(b\) be positive even integers such that \(a\geq b\). Then \(K_{a,b}(\lambda)\) has a \(P_{k+1}\)-decomposition if and only if \(a\geq\left\lceil\frac{k+1}{2}\right\rceil\), \(b\geq\left\lceil\frac{k}{2}\right\rceil\), and \(\lambda ab\equiv0 \pmod k\).
Theorem 2.4. [6] For positive integers \(a\) and \(b\) with \(a\geq b\), the complete bipartite multigraph \(K_{a,b}(\lambda)\) is \(S_{m}\)-decomposable if and only if \(a\geq m\) and \[\begin{cases} \lambda a\equiv 0\pmod m, & \text{if } b<m,\\ \lambda ab\equiv 0\pmod m, & \text{if } b\geq m. \end{cases}\]
In [2], the authors discussed the necessary conditions for a \(\{pP_{k+1},qS_{m}\}\)-decomposition of \(K_{n}(\lambda)\) when \(\lambda\geq1\), which are as follows.
Theorem 2.5. [2] Let \(\lambda\), \(n\), \(k\), and \(m\) be positive integers. Let \(p\) and \(q\) be non-negative integers. The necessary conditions for a \(\{pP_{k+1},qS_{m}\}\)-decomposition of \(K_{n}(\lambda)\) are \(pk+qm=\lambda{n\choose 2}\) and \(n\geq\max\{k+1,m+1\}\).
In this paper, we prove that the above necessary conditions are sufficient for a \(\{P_{4},S_{4}\}_{\{p,q\}}\)-decomposition of \(K_{n}(\lambda)\) in Theorem 3.4.
In this section, we discuss a \(\{P_{4},S_{4}\}_{\{p,q\}}\)-decomposition of \(K_{n}(\lambda)\) when \(\lambda\geq1\). Since \(K_{n}(\lambda)\) cannot be decomposed into \(P_{4}\) and \(S_{4}\) when \(n\leq 4\), we consider the decompositions for \(n\geq 5\).
In the following lemma, we discuss \(\{P_{4},S_{4}\}_{\{p,q\}}\)-decompositions of \(K_{4,6}\), which will be used later to decompose \(K_{n}(\lambda)\) into \(\{pP_{4},qS_{4}\}\).
Lemma 3.1. If \(p\) and \(q\) are non-negative integers such that \(3p+4q=24\), then \(K_{4,6}\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Proof. We have \((p,q)\in\{(8,0),(4,3),(0,6)\}\). By Theorem 2.3, \(K_{4,6}\) is \(\{8P_{4},0S_{4}\}\)-decomposable. The graph \(K_{4,6}\) can be decomposed into four copies of \(P_{4}\): \[(x_{8}x_{3}x_{5}x_{4}),\quad (x_{4}x_{6}x_{3}x_{10}),\quad (x_{1}x_{9}x_{2}x_{10}),\quad (x_{1}x_{7}x_{3}x_{9}),\] and three copies of \(S_{4}\): \[S(x_{1};x_{5},x_{6},x_{8},x_{10}),\quad S(x_{2};x_{5},x_{6},x_{7},x_{8}),\quad S(x_{4};x_{7},x_{8},x_{9},x_{10}).\] By Theorem 2.4, \(K_{4,6}\) is \(\{0P_{4},6S_{4}\}\)-decomposable. Therefore, \(K_{4,6}\) is fully \(\{P_{4},S_{4}\}\)-decomposable. ◻
Lemma 3.2. If \(p\) and \(q\) are non-negative integers such that \(3p+4q=24\), then \(K_{2,12}\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Proof. We have \((p,q)\in\{(8,0),(4,3),(0,6)\}\). By Theorem 2.3, \(K_{2,12}\) is \(\{8P_{4},0S_{4}\}\)-decomposable. The graph \(K_{2,12}\) can be decomposed into four copies of \(P_{4}\): \[(x_{3}x_{2}x_{7}x_{1}),\quad (x_{4}x_{2}x_{8}x_{1}),\quad (x_{5}x_{2}x_{9}x_{1}),\quad (x_{6}x_{2}x_{10}x_{1}),\] and three copies of \(S_{4}\): \[S(x_{1};x_{3},x_{4},x_{5},x_{6}),\quad S(x_{1};x_{11},x_{12},x_{13},x_{14}),\quad S(x_{2};x_{11},x_{12},x_{13},x_{14}).\] By Theorem 2.4, \(K_{2,12}\) is \(\{0P_{4},6S_{4}\}\)-decomposable. Therefore, \(K_{2,12}\) is fully \(\{P_{4},S_{4}\}\)-decomposable. ◻
Lemma 3.3. The graph \(K_{6}\) cannot be decomposed into one copy of \(P_{4}\) and three copies of \(S_{4}\).
Proof. Let \(V(K_{6})=\{x_{1},x_{2},\ldots,x_{6}\}\), and let \(P_{4}\) be the path \((x_{1}x_{2}x_{3}x_{4})\). Then \(x_{2}\) and \(x_{3}\) cannot be center vertices of any copy of \(S_{4}\). There are two cases.
Case 1. Suppose that we fix \(x_{1}\) as a center vertex. Then \(x_{4}\) cannot be a center vertex. Hence either \(x_{5}\) or \(x_{6}\) can be a center vertex of a copy of \(S_{4}\), but not both.
Case 2. Suppose that we fix \(x_{5}\) as a center vertex of a copy of \(S_{4}\). Then \(x_{6}\) is the only possible remaining center vertex of a copy of \(S_{4}\).
Hence, in both cases, we can have only two copies of \(S_{4}\). Therefore, \(K_{6}\) cannot be decomposed into one copy of \(P_{4}\) and three copies of \(S_{4}\). ◻
We now prove our main result.
Theorem 3.4. For any non-negative integers \(p\) and \(q\) and any integer \(n\geq5\), there exists a \(\{P_{4},S_{4}\}_{\{p,q\}}\)-decomposition of \(K_{n}(\lambda)\) if and only if \(3p+4q=\lambda{n\choose 2}\), except possibly when \((\lambda,n,p,q)=(1,6,1,3)\).
Proof. The necessary conditions are obvious. First, we prove the result for odd \(n\) with \(5\leq n\leq27\) and for even \(n\) with \(6\leq n\leq30\). Then we generalize it for any odd \(n>27\) and any even \(n>30\) by applying mathematical induction. Since we discuss the \(\{pP_{4},qS_{4}\}\)-decomposition of \(K_{n}(\lambda)\) for all possible choices of \(p\) and \(q\), we have the following cases.
Case 1. \(n=5\).
If \(\lambda=1\), then \((p,q)\in\{(2,1)\}\). The graph \(K_{5}\) can be decomposed into two copies of \(P_{4}\): \[(x_{1}x_{3}x_{4}x_{2}),\qquad (x_{3}x_{2}x_{1}x_{4}),\] and one copy of \(S_{4}\): \[S(x_{5};x_{1},x_{2},x_{3},x_{4}).\] Therefore, \(K_{5}\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
If \(\lambda=2\), then \((p,q)\in\{(0,5),(4,2)\}\). By Theorem 2.2, \(K_{5}(2)\) is \(\{0P_{4},5S_{4}\}\)-decomposable. We write \[K_{5}(2)=K_{5}+K_{5} =\{(2,1)\}+\{(2,1)\} =\{(4,2)\}.\] Therefore, \(K_{5}(2)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
If \(\lambda=3\), then \((p,q)\in\{(2,6),(6,3),(10,0)\}\). By Theorem 2.1, \(K_{5}(3)\) is \(\{10P_{4},0S_{4}\}\)-decomposable. We write \[K_{5}(3)=K_{5}(2)+K_{5} =\{(0,5),(4,2)\}+\{(2,1)\} =\{(2,6),(6,3)\}.\] Therefore, \(K_{5}(3)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
If \(\lambda=4\), then \((p,q)\in\{(0,10),(4,7),(8,4),(12,1)\}\). By Theorem 2.2, \(K_{5}(4)\) is \(\{0P_{4},10S_{4}\}\)-decomposable. We write \[K_{5}(4)=K_{5}(3)+K_{5} =\{(2,6),(6,3),(10,0)\}+\{(2,1)\} =\{(4,7),(8,4),(12,1)\}.\] Therefore, \(K_{5}(4)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
If \(\lambda=5\), then \((p,q)\in\{(2,11),(6,8),(10,5),(14,2)\}\). We write \[\begin{aligned} K_{5}(5)&=K_{5}(3)+K_{5}(2)\\ &=\{(2,6),(6,3),(10,0)\}+\{(0,5),(4,2)\}\\ &=\{(2,11),(6,8),(10,5),(14,2)\}. \end{aligned}\] Therefore, \(K_{5}(5)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
If \(\lambda=6\), then \[(p,q)\in\{(0,15),(4,12),(8,9),(12,6),(16,3),(20,0)\}.\] By Theorem 2.2, \(K_{5}(6)\) is \(\{0P_{4},15S_{4}\}\)-decomposable. We write \[\begin{aligned} K_{5}(6) &=K_{5}(3)+K_{5}(3)\\ &=\{(2,6),(6,3),(10,0)\}+\{(2,6),(6,3),(10,0)\}\\ &=\{(4,12),(8,9),(12,6),(16,3),(20,0)\}. \end{aligned}\] Therefore, \(K_{5}(6)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
If \(\lambda\geq7\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 6\), then \(K_{5}(\lambda)\) can be decomposed into \(\frac{\lambda}{6}\) copies of \(K_{5}(6)\). Therefore, \(K_{5}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
If \(\lambda\equiv1\pmod 6\), then we write \[K_{5}(\lambda)=K_{5}(\lambda-1)+K_{5}.\] The graph \(K_{5}(\lambda-1)\) can be decomposed into \(\frac{\lambda-1}{6}\) copies of \(K_{5}(6)\). Therefore, \(K_{5}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
If \(\lambda\equiv2\pmod 6\), then we write \[K_{5}(\lambda)=K_{5}(\lambda-2)+K_{5}(2).\] The graph \(K_{5}(\lambda-2)\) can be decomposed into \(\frac{\lambda-2}{6}\) copies of \(K_{5}(6)\). Therefore, \(K_{5}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
If \(\lambda\equiv3\pmod 6\), then we write \[K_{5}(\lambda)=K_{5}(\lambda-3)+K_{5}(3).\] The graph \(K_{5}(\lambda-3)\) can be decomposed into \(\frac{\lambda-3}{6}\) copies of \(K_{5}(6)\). Therefore, \(K_{5}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
If \(\lambda\equiv4\pmod 6\), then we write \[K_{5}(\lambda)=K_{5}(\lambda-4)+K_{5}(4).\] The graph \(K_{5}(\lambda-4)\) can be decomposed into \(\frac{\lambda-4}{6}\) copies of \(K_{5}(6)\). Therefore, \(K_{5}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
If \(\lambda\equiv5\pmod 6\), then we write \[K_{5}(\lambda)=K_{5}(\lambda-5)+K_{5}(5).\] The graph \(K_{5}(\lambda-5)\) can be decomposed into \(\frac{\lambda-5}{6}\) copies of \(K_{5}(6)\). Therefore, \(K_{5}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 2. \(n=6\).
If \(\lambda=1\), then \((p,q)\in\{(5,0)\}\), except when \((p,q)=(1,3)\). By Theorem 2.1, \(K_{6}\) is \(\{5P_{4},0S_{4}\}\)-decomposable.
If \(\lambda=2\), then \((p,q)\in\{(2,6),(6,3),(10,0)\}\). The graph \(K_{6}(2)\) can be decomposed into two copies of \(P_{4}\): \[(x_{4}x_{3}x_{5}x_{6}),\qquad (x_{1}x_{6}x_{3}x_{2}),\] and six copies of \(S_{4}\): \[\begin{aligned} &S(x_{1};x_{2},x_{3},x_{4},x_{5}),\quad S(x_{1};x_{2},x_{3},x_{4},x_{5}),\\ &S(x_{2};x_{3},x_{4},x_{5},x_{6}),\quad S(x_{4};x_{2},x_{3},x_{5},x_{6}),\\ &S(x_{5};x_{2},x_{3},x_{4},x_{6}),\quad S(x_{6};x_{1},x_{2},x_{3},x_{4}). \end{aligned}\] Also, \(K_{6}(2)\) can be decomposed into six copies of \(P_{4}\): \[\begin{aligned} &(x_{6}x_{5}x_{4}x_{3}),\quad (x_{6}x_{5}x_{4}x_{3}),\quad (x_{1}x_{5}x_{3}x_{6}),\\ &(x_{2}x_{1}x_{6}x_{4}),\quad (x_{1}x_{4}x_{2}x_{3}),\quad (x_{1}x_{3}x_{5}x_{2}), \end{aligned}\] and three copies of \(S_{4}\): \[S(x_{1};x_{2},x_{3},x_{4},x_{5}),\quad S(x_{2};x_{3},x_{4},x_{5},x_{6}),\quad S(x_{6};x_{1},x_{2},x_{3},x_{4}).\] By Theorem 2.1, \(K_{6}(2)\) is \(\{10P_{4},0S_{4}\}\)-decomposable.
If \(\lambda=3\), then \((p,q)\in\{(3,9),(7,6),(11,3),(15,0)\}\). The graph \(K_{6}(3)\) can be decomposed into three copies of \(P_{4}\): \[(x_{1}x_{6}x_{2}x_{5}),\qquad (x_{4}x_{5}x_{6}x_{3}),\qquad (x_{5}x_{6}x_{4}x_{2}),\] and nine copies of \(S_{4}\): \[\begin{aligned} &S(x_{1};x_{2},x_{3},x_{4},x_{5}),\quad S(x_{1};x_{2},x_{3},x_{4},x_{5}),\\ &S(x_{2};x_{1},x_{3},x_{4},x_{5}),\quad S(x_{3};x_{1},x_{2},x_{4},x_{5}),\\ &S(x_{3};x_{2},x_{4},x_{5},x_{6}),\quad S(x_{4};x_{1},x_{2},x_{3},x_{5}),\\ &S(x_{5};x_{1},x_{2},x_{3},x_{4}),\quad S(x_{6};x_{1},x_{2},x_{3},x_{4}),\\ &S(x_{6};x_{1},x_{2},x_{4},x_{5}). \end{aligned}\] By taking \(K_{6}(3)=K_{6}(2)+K_{6}\), we obtain all the other possible decompositions.
If \(\lambda=4\), then \[(p,q)\in\{(0,15),(4,12),(8,9),\ldots,(20,0)\},\] where the values of \(p\) increase by \(4\) and the values of \(q\) decrease by \(3\). By Theorem 2.2, \(K_{6}(4)\) is \(\{0P_{4},15S_{4}\}\)-decomposable. By taking \[K_{6}(4)=K_{6}(2)+K_{6}(2),\] we obtain all the above possible decompositions.
If \(\lambda=5\), then \[(p,q)\in\{(1,18),(5,15),(9,12),\ldots,(25,0)\}.\] The graph \(K_{6}(5)\) can be decomposed into one copy of \(P_{4}\): \[(x_{6}x_{5}x_{3}x_{4}),\] and eighteen copies of \(S_{4}\): \[\begin{aligned} &S(x_{1};x_{3},x_{4},x_{5},x_{6}),\quad S(x_{1};x_{2},x_{3},x_{4},x_{5}),\\ &S(x_{1};x_{2},x_{3},x_{4},x_{5}),\quad S(x_{2};x_{1},x_{3},x_{4},x_{5}),\\ &S(x_{2};x_{1},x_{3},x_{4},x_{6}),\quad S(x_{2};x_{1},x_{3},x_{4},x_{5}),\\ &S(x_{3};x_{1},x_{2},x_{4},x_{6}),\quad S(x_{3};x_{1},x_{2},x_{4},x_{6}),\\ &S(x_{4};x_{1},x_{2},x_{3},x_{6}),\quad S(x_{4};x_{1},x_{2},x_{3},x_{5}),\\ &S(x_{5};x_{1},x_{2},x_{3},x_{4}),\quad S(x_{5};x_{1},x_{3},x_{4},x_{6}),\\ &S(x_{5};x_{2},x_{3},x_{4},x_{6}),\quad S(x_{5};x_{2},x_{3},x_{4},x_{6}),\\ &S(x_{6};x_{1},x_{2},x_{3},x_{4}),\quad S(x_{6};x_{1},x_{2},x_{3},x_{4}),\\ &S(x_{6};x_{1},x_{2},x_{3},x_{4}),\quad S(x_{6};x_{1},x_{2},x_{4},x_{5}). \end{aligned}\] By taking \[K_{6}(5)=K_{6}(3)+K_{6}(2),\] we obtain all the other possible decompositions.
If \(\lambda\geq6\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 4\), then we write \[K_{6}(\lambda)=\frac{\lambda}{4}K_{6}(4).\]
If \(\lambda\equiv1\pmod 4\), then we write \[K_{6}(\lambda)=K_{6}(\lambda-5)+K_{6}(5) =\frac{\lambda-5}{4}K_{6}(4)+K_{6}(5).\]
If \(\lambda\equiv2\pmod 4\), then we write \[K_{6}(\lambda)=K_{6}(\lambda-2)+K_{6}(2) =\frac{\lambda-2}{4}K_{6}(4)+K_{6}(2).\]
If \(\lambda\equiv3\pmod 4\), then we write \[K_{6}(\lambda)=K_{6}(\lambda-3)+K_{6}(3) =\frac{\lambda-3}{4}K_{6}(4)+K_{6}(3).\] Therefore, \(K_{6}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 3. \(n=7\).
If \(\lambda=1\), then \((p,q)\in\{(3,3),(7,0)\}\). The graph \(K_{7}\) can be decomposed into three copies of \(P_{4}\): \[(x_{6}x_{5}x_{7}x_{4}),\qquad (x_{2}x_{5}x_{3}x_{4}),\qquad (x_{3}x_{2}x_{4}x_{5}),\] and three copies of \(S_{4}\): \[S(x_{1};x_{2},x_{3},x_{4},x_{5}),\quad S(x_{6};x_{1},x_{2},x_{3},x_{4}),\quad S(x_{7};x_{1},x_{2},x_{3},x_{6}).\] By Theorem 2.1, \(K_{7}\) is \(\{7P_{4},0S_{4}\}\)-decomposable.
If \(\lambda=2\), then \((p,q)\in\{(2,9),(6,6),(10,3),(14,0)\}\). The graph \(K_{7}(2)\) can be decomposed into two copies of \(P_{4}\): \[(x_{3}x_{2}x_{4}x_{7}),\qquad (x_{2}x_{3}x_{4}x_{6}),\] and nine copies of \(S_{4}\): \[\begin{aligned} &S(x_{1};x_{2},x_{3},x_{4},x_{5}),\quad S(x_{1};x_{2},x_{3},x_{4},x_{5}),\\ &S(x_{4};x_{2},x_{3},x_{6},x_{7}),\quad S(x_{5};x_{2},x_{3},x_{4},x_{6}),\\ &S(x_{5};x_{2},x_{3},x_{4},x_{6}),\quad S(x_{6};x_{1},x_{2},x_{3},x_{7}),\\ &S(x_{6};x_{1},x_{2},x_{3},x_{7}),\quad S(x_{7};x_{1},x_{2},x_{3},x_{5}),\\ &S(x_{7};x_{1},x_{2},x_{3},x_{5}). \end{aligned}\] By taking \[K_{7}(2)=K_{7}+K_{7},\] we obtain all the other possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(1,15),(5,12),(9,9),\ldots,(21,0)\}.\] The graph \(K_{7}(3)\) can be decomposed into one copy of \(P_{4}\): \[(x_{4}x_{7}x_{1}x_{6}),\] and fifteen copies of \(S_{4}\): \[\begin{aligned} &S(x_{1};x_{2},x_{3},x_{4},x_{5}),\quad S(x_{1};x_{2},x_{3},x_{4},x_{5}),\\ &S(x_{1};x_{2},x_{3},x_{4},x_{5}),\quad S(x_{2};x_{3},x_{4},x_{6},x_{7}),\\ &S(x_{2};x_{3},x_{4},x_{6},x_{7}),\quad S(x_{2};x_{3},x_{5},x_{6},x_{7}),\\ &S(x_{3};x_{4},x_{5},x_{6},x_{7}),\quad S(x_{3};x_{4},x_{5},x_{6},x_{7}),\\ &S(x_{4};x_{2},x_{3},x_{5},x_{6}),\quad S(x_{5};x_{2},x_{3},x_{4},x_{7}),\\ &S(x_{5};x_{2},x_{4},x_{6},x_{7}),\quad S(x_{6};x_{1},x_{4},x_{5},x_{7}),\\ &S(x_{6};x_{1},x_{3},x_{4},x_{5}),\quad S(x_{7};x_{1},x_{4},x_{5},x_{6}),\\ &S(x_{7};x_{1},x_{3},x_{4},x_{6}). \end{aligned}\] By taking \[K_{7}(3)=K_{7}(2)+K_{7},\] we obtain all the other possible decompositions.
If \(\lambda=4\), then \[(p,q)\in\{(0,21),(4,18),(8,15),\ldots,(28,0)\}.\] By Theorem 2.2, \(K_{7}(4)\) is \(\{0P_{4},21S_{4}\}\)-decomposable. By taking \[K_{7}(4)=K_{7}(2)+K_{7}(2),\] we obtain all the above possible decompositions.
If \(\lambda\geq5\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 4\), then we write \[K_{7}(\lambda)=\frac{\lambda}{4}K_{7}(4).\]
If \(\lambda\equiv1\pmod 4\), then we write \[K_{7}(\lambda)=K_{7}(\lambda-1)+K_{7} =\frac{\lambda-1}{4}K_{7}(4)+K_{7}.\]
If \(\lambda\equiv2\pmod 4\), then we write \[K_{7}(\lambda)=K_{7}(\lambda-2)+K_{7}(2) =\frac{\lambda-2}{4}K_{7}(4)+K_{7}(2).\]
If \(\lambda\equiv3\pmod 4\), then we write \[K_{7}(\lambda)=K_{7}(\lambda-3)+K_{7}(3) =\frac{\lambda-3}{4}K_{7}(4)+K_{7}(3).\] Therefore, \(K_{7}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 4. \(n=8\).
If \(\lambda=1\), then \((p,q)\in\{(0,7),(4,4),(8,1)\}\). By Theorem 2.2, \(K_{8}\) is \(\{0P_{4},7S_{4}\}\)-decomposable. By Theorem 2.1, \(K_{4}\) is \(\{2P_{4},0S_{4}\}\)-decomposable. By Theorem 2.4, \(K_{4,4}\) is \(\{0P_{4},4S_{4}\}\)-decomposable, and by Theorem 2.3, \(K_{4,3}\) is \(\{4P_{4},0S_{4}\}\)-decomposable. By taking \[K_{8}=K_{4}+K_{4}+K_{4,4},\] we obtain the decomposition when \((p,q)=(4,4)\), and by taking \[K_{8}=K_{5}+K_{4}+K_{4,3},\] we obtain the decomposition when \((p,q)=(8,1)\).
If \(\lambda=2\), then \[(p,q)\in\{(0,14),(4,11),(8,8),\ldots,(16,2)\}.\] By taking \[K_{8}(2)=K_{8}+K_{8},\] we obtain all the above possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(0,21),(4,18),(8,15),\ldots,(28,0)\}.\] By Theorem 2.1, \(K_{8}(3)\) is \(\{28P_{4},0S_{4}\}\)-decomposable. By taking \[K_{8}(3)=K_{8}(2)+K_{8},\] we obtain all the above possible decompositions.
If \(\lambda\geq4\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 3\), then we write \[K_{8}(\lambda)=\frac{\lambda}{3}K_{8}(3).\]
If \(\lambda\equiv1\pmod 3\), then we write \[K_{8}(\lambda)=K_{8}(\lambda-1)+K_{8} =\frac{\lambda-1}{3}K_{8}(3)+K_{8}.\]
If \(\lambda\equiv2\pmod 3\), then we write \[K_{8}(\lambda)=K_{8}(\lambda-2)+K_{8}(2) =\frac{\lambda-2}{3}K_{8}(3)+K_{8}(2).\] Therefore, \(K_{8}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 5. \(n=9\).
If \(\lambda=1\), then \((p,q)\in\{(0,9),(4,6),(8,3),(12,0)\}\). The graph \(K_{1,8}\) is \(\{0P_{4},2S_{4}\}\)-decomposable. By Theorem 2.1, \(K_{9}\) is \(\{12P_{4},0S_{4}\}\)-decomposable. By taking \[K_{9}=K_{8}+K_{1,8},\] we obtain all the other possible decompositions.
If \(\lambda\geq2\), then \(K_{9}(\lambda)\) can be decomposed into \(\lambda\) copies of \(K_{9}\). Therefore, \(K_{9}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 6. \(n=10\).
If \(\lambda=1\), then \((p,q)\in\{(3,9),(7,6),(11,3),(15,0)\}\). The graph \(K_{10}\) can be decomposed into three copies of \(P_{4}\): \[(x_{4}x_{2}x_{3}x_{6}),\qquad (x_{8}x_{6}x_{7}x_{5}),\qquad (x_{3}x_{4}x_{6}x_{2}),\] and nine copies of \(S_{4}\): \[\begin{aligned} &S(x_{1};x_{2},x_{3},x_{4},x_{5}),\quad S(x_{1};x_{6},x_{7},x_{8},x_{9}),\\ &S(x_{5};x_{2},x_{3},x_{4},x_{6}),\quad S(x_{7};x_{2},x_{3},x_{4},x_{8}),\\ &S(x_{8};x_{2},x_{3},x_{4},x_{5}),\quad S(x_{9};x_{2},x_{3},x_{4},x_{10}),\\ &S(x_{9};x_{5},x_{6},x_{7},x_{8}),\quad S(x_{10};x_{1},x_{2},x_{3},x_{4}),\\ &S(x_{10};x_{5},x_{6},x_{7},x_{8}). \end{aligned}\] The graph \(K_{4}\) is \(\{2P_{4},0S_{4}\}\)-decomposable. By taking \[K_{10}=K_{6}+K_{4}+K_{6,4},\] we obtain all the other possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(2,21),(6,18),(10,15),\ldots,(30,0)\}.\] By Theorem 2.4, \(K_{5,4}(2)\) is \(\{0P_{4},10S_{4}\}\)-decomposable. By taking \[K_{10}(2)=K_{6}(2)+K_{5}(2)+K_{5,4}(2),\] we obtain the decomposition when \((p,q)=(2,21)\). By taking \[K_{10}(2)=K_{10}+K_{10},\] we obtain all the other possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(1,33),(5,30),(9,27),\ldots,(45,0)\}.\] The graph \(K_{10}(3)\) can be decomposed into one copy of \(P_{4}\): \[(x_{1}x_{5}x_{4}x_{2}),\] and thirty-three copies of \(S_{4}\): \[\begin{aligned} &S(x_{1};x_{2},x_{3},x_{4},x_{5}),\quad S(x_{1};x_{2},x_{3},x_{4},x_{7}),\\ &S(x_{2};x_{1},x_{3},x_{4},x_{7}),\quad S(x_{3};x_{2},x_{4},x_{5},x_{6}),\\ &S(x_{3};x_{2},x_{4},x_{5},x_{6}),\quad S(x_{3};x_{1},x_{5},x_{6},x_{7}),\\ &S(x_{4};x_{1},x_{2},x_{3},x_{7}),\quad S(x_{5};x_{1},x_{2},x_{6},x_{7}),\\ &S(x_{5};x_{2},x_{4},x_{6},x_{7}),\quad S(x_{5};x_{2},x_{4},x_{6},x_{7}),\\ &S(x_{6};x_{1},x_{2},x_{4},x_{7}),\quad S(x_{6};x_{1},x_{2},x_{4},x_{7}),\\ &S(x_{6};x_{1},x_{2},x_{4},x_{7}),\quad S(x_{7};x_{1},x_{2},x_{3},x_{4}),\\ &S(x_{7};x_{1},x_{2},x_{3},x_{4}),\quad S(x_{8};x_{4},x_{5},x_{6},x_{7}),\\ &S(x_{8};x_{4},x_{5},x_{6},x_{7}),\quad S(x_{8};x_{4},x_{5},x_{6},x_{7}),\\ &S(x_{8};x_{1},x_{2},x_{3},x_{10}),\quad S(x_{8};x_{1},x_{2},x_{3},x_{10}),\\ &S(x_{8};x_{1},x_{2},x_{3},x_{10}),\quad S(x_{9};x_{1},x_{2},x_{3},x_{4}),\\ &S(x_{9};x_{1},x_{2},x_{3},x_{4}),\quad S(x_{9};x_{1},x_{2},x_{3},x_{4}),\\ &S(x_{9};x_{5},x_{6},x_{7},x_{8}),\quad S(x_{9};x_{5},x_{6},x_{7},x_{8}),\\ &S(x_{9};x_{5},x_{6},x_{7},x_{8}),\quad S(x_{10};x_{1},x_{2},x_{3},x_{4}),\\ &S(x_{10};x_{1},x_{2},x_{3},x_{4}),\quad S(x_{10};x_{1},x_{2},x_{3},x_{4}),\\ &S(x_{10};x_{5},x_{6},x_{7},x_{9}),\quad S(x_{10};x_{5},x_{6},x_{7},x_{9}),\\ &S(x_{10};x_{5},x_{6},x_{7},x_{9}). \end{aligned}\] By taking \[K_{10}(3)=K_{10}(2)+K_{10},\] we obtain all the other possible decompositions.
If \(\lambda=4\), then \[(p,q)\in\{(0,45),(4,42),(8,39),\ldots,(60,0)\}.\] By Theorem 2.2, \(K_{10}(4)\) is \(\{0P_{4},45S_{4}\}\)-decomposable. By taking \[K_{10}(4)=K_{10}(2)+K_{10}(2),\] we obtain all the above possible decompositions.
If \(\lambda\geq5\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 4\), then we write \[K_{10}(\lambda)=\frac{\lambda}{4}K_{10}(4).\]
If \(\lambda\equiv1\pmod 4\), then we write \[K_{10}(\lambda)=K_{10}(\lambda-1)+K_{10} =\frac{\lambda-1}{4}K_{10}(4)+K_{10}.\]
If \(\lambda\equiv2\pmod 4\), then we write \[K_{10}(\lambda)=K_{10}(\lambda-2)+K_{10}(2) =\frac{\lambda-2}{4}K_{10}(4)+K_{10}(2).\]
If \(\lambda\equiv3\pmod 4\), then we write \[K_{10}(\lambda)=K_{10}(\lambda-3)+K_{10}(3) =\frac{\lambda-3}{4}K_{10}(4)+K_{10}(3).\] Therefore, \(K_{10}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 7. \(n=11\).
If \(\lambda=1\), then \[(p,q)\in\{(1,13),(5,10),(9,7),\ldots,(17,1)\}.\] The graph \(K_{11}\) can be decomposed into one copy of \(P_{4}\): \[(x_{1}x_{11}x_{2}x_{10}),\] and thirteen copies of \(S_{4}\): \[\begin{aligned} &S(x_{1};x_{2},x_{3},x_{4},x_{5}),\quad S(x_{1};x_{6},x_{7},x_{8},x_{9}),\\ &S(x_{2};x_{3},x_{4},x_{5},x_{6}),\quad S(x_{4};x_{3},x_{5},x_{6},x_{7}),\\ &S(x_{5};x_{3},x_{6},x_{7},x_{8}),\quad S(x_{6};x_{3},x_{7},x_{8},x_{9}),\\ &S(x_{7};x_{2},x_{3},x_{8},x_{9}),\quad S(x_{8};x_{2},x_{3},x_{4},x_{9}),\\ &S(x_{9};x_{2},x_{3},x_{4},x_{5}),\quad S(x_{10};x_{1},x_{3},x_{4},x_{5}),\\ &S(x_{10};x_{6},x_{7},x_{8},x_{9}),\quad S(x_{11};x_{3},x_{4},x_{5},x_{6}),\\ &S(x_{11};x_{7},x_{8},x_{9},x_{10}). \end{aligned}\] By taking \[K_{11}=K_{7}+K_{5}+K_{6,4},\] we obtain all the other possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(2,26),(6,23),(10,20),\ldots,(34,2)\}.\] By taking \[K_{11}(2)=K_{11}+K_{11},\] we obtain all the above possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(3,39),(7,36),(11,33),\ldots,(55,0)\}.\] By Theorem 2.1, \(K_{11}(3)\) is \(\{55P_{4},0S_{4}\}\)-decomposable. By taking \[K_{11}(3)=K_{11}(2)+K_{11},\] we obtain all the above possible decompositions.
If \(\lambda=4\), then \[(p,q)\in\{(0,55),(4,52),(8,49),\ldots,(72,1)\}.\] By Theorem 2.2, \(K_{11}(4)\) is \(\{0P_{4},55S_{4}\}\)-decomposable. By taking \[K_{11}(4)=K_{11}(3)+K_{11},\] we obtain all the above possible decompositions.
If \(\lambda=5\), then \[(p,q)\in\{(1,68),(5,65),(9,62),\ldots,(89,2)\}.\] By taking \[K_{11}(5)=K_{11}(4)+K_{11},\] we obtain all the above possible decompositions.
If \(\lambda=6\), then \[(p,q)\in\{(2,81),(6,78),(10,75),\ldots,(110,0)\}.\] By Theorem 2.1, \(K_{11}(6)\) is \(\{110P_{4},0S_{4}\}\)-decomposable. By taking \[K_{11}(6)=K_{11}(4)+K_{11}(2),\] we obtain all the above possible decompositions.
If \(\lambda=7\), then \[(p,q)\in\{(3,94),(7,91),(11,88),\ldots,(127,1)\}.\] By taking \[K_{11}(7)=K_{11}(6)+K_{11},\] we obtain all the above possible decompositions.
If \(\lambda=8\), then \[(p,q)\in\{(0,110),(4,107),(8,104),\ldots,(144,2)\}.\] By taking \[K_{11}(8)=K_{11}(4)+K_{11}(4),\] we obtain all the above possible decompositions.
If \(\lambda=9\), then \[(p,q)\in\{(1,123),(5,120),(9,117),\ldots,(165,0)\}.\] By Theorem 2.1, \(K_{11}(9)\) is \(\{165P_{4},0S_{4}\}\)-decomposable. By taking \[K_{11}(9)=K_{11}(8)+K_{11},\] we obtain all the above possible decompositions.
If \(\lambda=10\), then \[(p,q)\in\{(2,136),(6,133),(10,130),\ldots,(182,1)\}.\] By taking \[K_{11}(10)=K_{11}(6)+K_{11}(4),\] we obtain all the above possible decompositions.
If \(\lambda=11\), then \[(p,q)\in\{(3,149),(7,146),(11,143),\ldots,(199,2)\}.\] By taking \[K_{11}(11)=K_{11}(10)+K_{11},\] we obtain all the above possible decompositions.
If \(\lambda=12\), then \[(p,q)\in\{(0,165),(4,162),(8,159),\ldots,(220,0)\}.\] By Theorem 2.1, \(K_{11}(12)\) is \(\{220P_{4},0S_{4}\}\)-decomposable. By taking \[K_{11}(12)=K_{11}(8)+K_{11}(4),\] we obtain all the above possible decompositions.
If \(\lambda\geq13\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod{12}\), then we write \[K_{11}(\lambda)=\frac{\lambda}{12}K_{11}(12).\]
If \(\lambda\equiv1\pmod{12}\), then we write \[K_{11}(\lambda)=K_{11}(\lambda-1)+K_{11} =\frac{\lambda-1}{12}K_{11}(12)+K_{11}.\]
If \(\lambda\equiv2\pmod{12}\), then we write \[K_{11}(\lambda)=K_{11}(\lambda-2)+K_{11}(2) =\frac{\lambda-2}{12}K_{11}(12)+K_{11}(2).\]
If \(\lambda\equiv3\pmod{12}\), then we write \[K_{11}(\lambda)=K_{11}(\lambda-3)+K_{11}(3) =\frac{\lambda-3}{12}K_{11}(12)+K_{11}(3).\]
If \(\lambda\equiv4\pmod{12}\), then we write \[K_{11}(\lambda)=K_{11}(\lambda-4)+K_{11}(4) =\frac{\lambda-4}{12}K_{11}(12)+K_{11}(4).\]
If \(\lambda\equiv5\pmod{12}\), then we write \[K_{11}(\lambda)=K_{11}(\lambda-5)+K_{11}(5) =\frac{\lambda-5}{12}K_{11}(12)+K_{11}(5).\]
If \(\lambda\equiv6\pmod{12}\), then we write \[K_{11}(\lambda)=K_{11}(\lambda-6)+K_{11}(6) =\frac{\lambda-6}{12}K_{11}(12)+K_{11}(6).\]
If \(\lambda\equiv7\pmod{12}\), then we write \[K_{11}(\lambda)=K_{11}(\lambda-7)+K_{11}(7) =\frac{\lambda-7}{12}K_{11}(12)+K_{11}(7).\]
If \(\lambda\equiv8\pmod{12}\), then we write \[K_{11}(\lambda)=K_{11}(\lambda-8)+K_{11}(8) =\frac{\lambda-8}{12}K_{11}(12)+K_{11}(8).\]
If \(\lambda\equiv9\pmod{12}\), then we write \[K_{11}(\lambda)=K_{11}(\lambda-9)+K_{11}(9) =\frac{\lambda-9}{12}K_{11}(12)+K_{11}(9).\]
If \(\lambda\equiv10\pmod{12}\), then we write \[K_{11}(\lambda)=K_{11}(\lambda-10)+K_{11}(10) =\frac{\lambda-10}{12}K_{11}(12)+K_{11}(10).\]
If \(\lambda\equiv11\pmod{12}\), then we write \[K_{11}(\lambda)=K_{11}(\lambda-11)+K_{11}(11) =\frac{\lambda-11}{12}K_{11}(12)+K_{11}(11).\] Therefore, \(K_{11}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 8. \(n=12\).
If \(\lambda=1\), then \[(p,q)\in\{(2,15),(6,12),(10,9),\ldots,(22,0)\}.\] By Theorem 2.1, \(K_{12}\) is \(\{22P_{4},0S_{4}\}\)-decomposable, and by Theorem 2.4, \(K_{7,4}\) is \(\{0P_{4},7S_{4}\}\)-decomposable. The graph \(K_{1,4}\) is \(\{0P_{4},1S_{4}\}\)-decomposable. By taking \[K_{12}=K_{8}+K_{5}+K_{6,4}+K_{1,4},\] we obtain the decompositions when \((p,q)\in\{(14,6),(18,3)\}\). By taking \[K_{12}=K_{8}+K_{5}+K_{7,4},\] we obtain all the other possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(0,33),(4,30),(8,27),\ldots,(44,0)\}.\] By Theorem 2.2, \(K_{12}(2)\) is \(\{0P_{4},33S_{4}\}\)-decomposable. By taking \[K_{12}(2)=K_{12}+K_{12},\] we obtain all the above possible decompositions.
If \(\lambda\geq3\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 2\), then we write \[K_{12}(\lambda)=\frac{\lambda}{2}K_{12}(2).\]
If \(\lambda\equiv1\pmod 2\), then we write \[K_{12}(\lambda)=K_{12}(\lambda-1)+K_{12} =\frac{\lambda-1}{2}K_{12}(2)+K_{12}.\] Therefore, \(K_{12}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 9. \(n=13\).
If \(\lambda=1\), then \[(p,q)\in\{(2,18),(6,15),(10,12),\ldots,(26,0)\}.\] By Theorem 2.1, \(K_{13}\) is \(\{26P_{4},0S_{4}\}\)-decomposable. The graph \(K_{1,12}\) is \(\{0P_{4},3S_{4}\}\) – decomposable. By taking \[K_{13}=K_{12}+K_{1,12},\] we obtain all the other possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(0,39),(4,36),(8,33),\ldots,(52,0)\}.\] By Theorem 2.2, \(K_{13}(2)\) is \(\{0P_{4},39S_{4}\}\)-decomposable. By taking \[K_{13}(2)=K_{13}+K_{13},\] we obtain all the above possible decompositions.
If \(\lambda\geq3\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 2\), then we write \[K_{13}(\lambda)=\frac{\lambda}{2}K_{13}(2).\]
If \(\lambda\equiv1\pmod 2\), then we write \[K_{13}(\lambda)=K_{13}(\lambda-1)+K_{13} =\frac{\lambda-1}{2}K_{13}(2)+K_{13}.\] Therefore, \(K_{13}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 10. \(n=14\).
If \(\lambda=1\), then \[(p,q)\in\{(1,22),(5,19),(9,16),\ldots,(29,1)\}.\] The graph \(K_{14}\) can be decomposed into one copy of \(P_{4}\): \[(x_{5}x_{3}x_{2}x_{13}),\] and twenty-two copies of \(S_{4}\): \[\begin{aligned} &S(x_{1};x_{2},x_{3},x_{7},x_{13}),\quad S(x_{4};x_{1},x_{2},x_{3},x_{5}),\\ &S(x_{5};x_{1},x_{2},x_{6},x_{7}),\quad S(x_{6};x_{1},x_{2},x_{3},x_{4}),\\ &S(x_{7};x_{2},x_{3},x_{4},x_{6}),\quad S(x_{8};x_{1},x_{2},x_{3},x_{9}),\\ &S(x_{8};x_{4},x_{5},x_{6},x_{7}),\quad S(x_{9};x_{1},x_{2},x_{3},x_{11}),\\ &S(x_{9};x_{4},x_{5},x_{6},x_{7}),\quad S(x_{10};x_{4},x_{5},x_{6},x_{7}),\\ &S(x_{10};x_{1},x_{2},x_{3},x_{11}),\quad S(x_{10};x_{8},x_{9},x_{13},x_{14}),\\ &S(x_{11};x_{5},x_{6},x_{7},x_{8}),\quad S(x_{11};x_{1},x_{2},x_{3},x_{4}),\\ &S(x_{12};x_{1},x_{2},x_{3},x_{13}),\quad S(x_{12};x_{4},x_{5},x_{6},x_{7}),\\ &S(x_{12};x_{8},x_{9},x_{10},x_{11}),\quad S(x_{13};x_{3},x_{5},x_{6},x_{7}),\\ &S(x_{13};x_{4},x_{8},x_{9},x_{11}),\quad S(x_{14};x_{2},x_{3},x_{4},x_{13}),\\ &S(x_{14};x_{5},x_{6},x_{7},x_{8}),\quad S(x_{14};x_{1},x_{9},x_{11},x_{12}). \end{aligned}\] By taking \[K_{14}=K_{8}+K_{6}+K_{4,6}+K_{4,6},\] we obtain all the above possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(2,44),(6,41),(10,38),\ldots,(58,2)\}.\] By taking \[K_{14}(2)=K_{14}+K_{14},\] we obtain all the above possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(3,66),(7,63),(11,60),\ldots,(91,0)\}.\] By Theorem 2.1, \(K_{14}(3)\) is \(\{91P_{4},0S_{4}\}\)-decomposable. By taking \[K_{14}(3)=K_{14}(2)+K_{14},\] we obtain all the above possible decompositions.
If \(\lambda=4\), then \[(p,q)\in\{(0,91),(4,88),(8,85),\ldots,(120,1)\}.\] By Theorem 2.2, \(K_{14}(4)\) is \(\{0P_{4},91S_{4}\}\)-decomposable. By taking \[K_{14}(4)=K_{14}(3)+K_{14},\] we obtain all the above possible decompositions.
If \(\lambda=5\), then \[(p,q)\in\{(1,113),(5,110),(9,107),\ldots,(149,2)\}.\] By taking \[K_{14}(5)=K_{14}(4)+K_{14},\] we obtain all the above possible decompositions.
If \(\lambda=6\), then \[(p,q)\in\{(2,135),(6,132),(10,129),\ldots,(182,0)\}.\] By Theorem 2.1, \(K_{14}(6)\) is \(\{182P_{4},0S_{4}\}\)-decomposable. By taking \[K_{14}(6)=K_{14}(4)+K_{14}(2),\] we obtain all the above possible decompositions.
If \(\lambda=7\), then \[(p,q)\in\{(3,157),(7,154),(11,151),\ldots,(211,1)\}.\] By taking \[K_{14}(7)=K_{14}(4)+K_{14}(3),\] we obtain all the above possible decompositions.
If \(\lambda=8\), then \[(p,q)\in\{(0,182),(4,179),(8,176),\ldots,(240,2)\}.\] By taking \[K_{14}(8)=K_{14}(4)+K_{14}(4),\] we obtain all the above possible decompositions.
If \(\lambda=9\), then \[(p,q)\in\{(1,204),(5,201),(9,198),\ldots,(273,0)\}.\] By Theorem 2.1, \(K_{14}(9)\) is \(\{273P_{4},0S_{4}\}\)-decomposable. By taking \[K_{14}(9)=K_{14}(8)+K_{14},\] we obtain all the above possible decompositions.
If \(\lambda=10\), then \[(p,q)\in\{(2,226),(6,223),(10,220),\ldots,(302,1)\}.\] By taking \[K_{14}(10)=K_{14}(6)+K_{14}(4),\] we obtain all the above possible decompositions.
If \(\lambda=11\), then \[(p,q)\in\{(3,248),(7,245),(11,242),\ldots,(331,2)\}.\] By taking \[K_{14}(11)=K_{14}(8)+K_{14}(3),\] we obtain all the above possible decompositions.
If \(\lambda=12\), then \[(p,q)\in\{(0,273),(4,270),(8,267),\ldots,(364,0)\}.\] By Theorem 2.1, \(K_{14}(12)\) is \(\{364P_{4},0S_{4}\}\)-decomposable. By taking \[K_{14}(12)=K_{14}(8)+K_{14}(4),\] we obtain all the above possible decompositions.
If \(\lambda\geq13\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod{12}\), then we write \[K_{14}(\lambda)=\frac{\lambda}{12}K_{14}(12).\]
If \(\lambda\equiv1\pmod{12}\), then we write \[K_{14}(\lambda)=K_{14}(\lambda-1)+K_{14} =\frac{\lambda-1}{12}K_{14}(12)+K_{14}.\]
If \(\lambda\equiv2\pmod{12}\), then we write \[K_{14}(\lambda)=K_{14}(\lambda-2)+K_{14}(2) =\frac{\lambda-2}{12}K_{14}(12)+K_{14}(2).\]
If \(\lambda\equiv3\pmod{12}\), then we write \[K_{14}(\lambda)=K_{14}(\lambda-3)+K_{14}(3) =\frac{\lambda-3}{12}K_{14}(12)+K_{14}(3).\]
If \(\lambda\equiv4\pmod{12}\), then we write \[K_{14}(\lambda)=K_{14}(\lambda-4)+K_{14}(4) =\frac{\lambda-4}{12}K_{14}(12)+K_{14}(4).\]
If \(\lambda\equiv5\pmod{12}\), then we write \[K_{14}(\lambda)=K_{14}(\lambda-5)+K_{14}(5) =\frac{\lambda-5}{12}K_{14}(12)+K_{14}(5).\]
If \(\lambda\equiv6\pmod{12}\), then we write \[K_{14}(\lambda)=K_{14}(\lambda-6)+K_{14}(6) =\frac{\lambda-6}{12}K_{14}(12)+K_{14}(6).\]
If \(\lambda\equiv7\pmod{12}\), then we write \[K_{14}(\lambda)=K_{14}(\lambda-7)+K_{14}(7) =\frac{\lambda-7}{12}K_{14}(12)+K_{14}(7).\]
If \(\lambda\equiv8\pmod{12}\), then we write \[K_{14}(\lambda)=K_{14}(\lambda-8)+K_{14}(8) =\frac{\lambda-8}{12}K_{14}(12)+K_{14}(8).\]
If \(\lambda\equiv9\pmod{12}\), then we write \[K_{14}(\lambda)=K_{14}(\lambda-9)+K_{14}(9) =\frac{\lambda-9}{12}K_{14}(12)+K_{14}(9).\]
If \(\lambda\equiv10\pmod{12}\), then we write \[K_{14}(\lambda)=K_{14}(\lambda-10)+K_{14}(10) =\frac{\lambda-10}{12}K_{14}(12)+K_{14}(10).\]
If \(\lambda\equiv11\pmod{12}\), then we write \[K_{14}(\lambda)=K_{14}(\lambda-11)+K_{14}(11) =\frac{\lambda-11}{12}K_{14}(12)+K_{14}(11).\] Therefore, \(K_{14}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 11. \(n=15\).
If \(\lambda=1\), then \[(p,q)\in\{(3,24),(7,21),(11,18),\ldots,(35,0)\}.\] By taking \[K_{15}=K_{9}+K_{7}+K_{4,6}+K_{4,6},\] we obtain all the above possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(2,51),(6,48),(10,45),\ldots,(70,0)\}.\] By taking \[K_{15}(2)=K_{9}(2)+K_{7}(2)+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6},\] we obtain all the above possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(1,78),(5,75),(9,72),\ldots,(105,0)\}.\] By taking \[\begin{aligned} K_{15}(3) &=K_{9}(3)+K_{7}(3)+K_{4,6}+K_{4,6}+K_{4,6}\\ &\quad +K_{4,6}+K_{4,6}+K_{4,6}, \end{aligned}\] we obtain all the above possible decompositions.
If \(\lambda=4\), then \[(p,q)\in\{(0,105),(4,102),(8,99),\ldots,(140,0)\}.\] By Theorem 2.2, \(K_{15}(4)\) is \(\{0P_{4},105S_{4}\}\)-decomposable. By taking \[K_{15}(4)=K_{15}(2)+K_{15}(2),\] we obtain all the above possible decompositions.
If \(\lambda\geq5\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 4\), then we write \[K_{15}(\lambda)=\frac{\lambda}{4}K_{15}(4).\]
If \(\lambda\equiv1\pmod 4\), then we write \[K_{15}(\lambda)=K_{15}(\lambda-1)+K_{15} =\frac{\lambda-1}{4}K_{15}(4)+K_{15}.\]
If \(\lambda\equiv2\pmod 4\), then we write \[K_{15}(\lambda)=K_{15}(\lambda-2)+K_{15}(2) =\frac{\lambda-2}{4}K_{15}(4)+K_{15}(2).\]
If \(\lambda\equiv3\pmod 4\), then we write \[K_{15}(\lambda)=K_{15}(\lambda-3)+K_{15}(3) =\frac{\lambda-3}{4}K_{15}(4)+K_{15}(3).\] Therefore, \(K_{15}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 12. \(n=16\).
If \(\lambda=1\), then \[(p,q)\in\{(0,30),(4,27),(8,24),\ldots,(40,0)\}.\] By Theorem 2.2, \(K_{16}\) is \(\{0P_{4},30S_{4}\}\)-decomposable, and by Theorem 2.4, \(K_{2,8}\) is \(\{0P_{4},4S_{4}\}\)-decomposable. By taking \[K_{16}=K_{8}+K_{8}+K_{6,4}+K_{6,4}+K_{2,8},\] we obtain the decomposition when \((p,q)=(4,27)\). By Theorems 2.3 and 2.4, \(K_{6,6}\) is both \(\{12P_{4},0S_{4}\}\)-decomposable and \(\{0P_{4},9S_{4}\}\)-decomposable. By taking \[K_{16}=K_{10}+K_{6}+K_{4,6}+K_{6,6},\] we obtain all the above possible decompositions.
If \(\lambda\geq2\), then \(K_{16}(\lambda)\) can be decomposed into \(\lambda\) copies of \(K_{16}\). Therefore, \(K_{16}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 13. \(n=17\).
If \(\lambda=1\), then \[(p,q)\in\{(0,34),(4,31),(8,28),\ldots,(44,1)\}.\] By Theorem 2.4, \(K_{2,8}\) is \(\{0P_{4},4S_{4}\}\)-decomposable, and by Theorems 2.3 and 2.4, \(K_{6,6}\) is both \(\{12P_{4},0S_{4}\}\)-decomposable and \(\{0P_{4},9S_{4}\}\)-decomposable. By taking \[K_{17}=K_{11}+K_{7}+K_{4,6}+K_{6,6},\] we obtain the decomposition when \((p,q)=(44,1)\), and by taking \[K_{17}=K_{9}+K_{9}+K_{6,4}+K_{6,4}+K_{2,8},\] we obtain all the other possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(0,68),(4,65),(8,62),\ldots,(88,2)\}.\] By taking \[K_{17}(2)=K_{17}+K_{17},\] we obtain all the above possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(0,102),(4,99),(8,96),\ldots,(136,0)\}.\] By Theorem 2.1, \(K_{17}(3)\) is \(\{136P_{4},0S_{4}\}\)-decomposable. By taking \[K_{17}(3)=K_{17}(2)+K_{17},\] we obtain all the above possible decompositions.
If \(\lambda\geq4\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 3\), then we write \[K_{17}(\lambda)=\frac{\lambda}{3}K_{17}(3).\]
If \(\lambda\equiv1\pmod 3\), then we write \[K_{17}(\lambda)=K_{17}(\lambda-1)+K_{17} =\frac{\lambda-1}{3}K_{17}(3)+K_{17}.\]
If \(\lambda\equiv2\pmod 3\), then we write \[K_{17}(\lambda)=K_{17}(\lambda-2)+K_{17}(2) =\frac{\lambda-2}{3}K_{17}(3)+K_{17}(2).\] Therefore, \(K_{17}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 14. \(n=18\).
If \(\lambda=1\), then \[(p,q)\in\{(3,36),(7,33),(11,30),\ldots,(51,0)\}.\] By Theorem 2.4, \(K_{10,8}\) is \(\{0P_{4},20S_{4}\}\)-decomposable. By taking \[K_{18}=K_{10}+K_{8}+K_{10,8},\] we obtain the decomposition when \((p,q)=(3,36)\), and by taking \[K_{18}=K_{12}+K_{6}+K_{4,6}+K_{4,6}+K_{4,6},\] we obtain all the other possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(2,75),(6,72),(10,69),\ldots,(102,0)\}.\] By taking \[\begin{aligned} K_{18}(2) &=K_{12}(2)+K_{6}(2)+K_{4,6}+K_{4,6}+K_{4,6}\\ &\quad +K_{4,6}+K_{4,6}+K_{4,6}, \end{aligned}\] we obtain all the above possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(1,114),(5,111),(9,108),\ldots,(153,0)\}.\] By Theorem 2.4, \(K_{4,8}\) is \(\{0P_{4},8S_{4}\}\)-decomposable. By taking \[\begin{aligned} K_{18}(3) &=K_{10}(3)+K_{8}(3)+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}\\ &\quad +K_{6,4}+K_{6,4}+K_{4,8}+K_{4,8}+K_{4,8}, \end{aligned}\] we obtain the decomposition when \((p,q)=(1,114)\), and by taking \[\begin{aligned} K_{18}(3) &=K_{12}(3)+K_{6}(3)+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}\\ &\quad +K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}, \end{aligned}\] we obtain all the other possible decompositions.
If \(\lambda=4\), then \[(p,q)\in\{(0,153),(4,150),(8,147),\ldots,(204,0)\}.\] By Theorem 2.2, \(K_{18}(4)\) is \(\{0P_{4},153S_{4}\}\)-decomposable. By taking \[K_{18}(4)=K_{18}(2)+K_{18}(2),\] we obtain all the above possible decompositions.
If \(\lambda\geq5\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 4\), then we write \[K_{18}(\lambda)=\frac{\lambda}{4}K_{18}(4).\]
If \(\lambda\equiv1\pmod 4\), then we write \[K_{18}(\lambda)=K_{18}(\lambda-1)+K_{18} =\frac{\lambda-1}{4}K_{18}(4)+K_{18}.\]
If \(\lambda\equiv2\pmod 4\), then we write \[K_{18}(\lambda)=K_{18}(\lambda-2)+K_{18}(2) =\frac{\lambda-2}{4}K_{18}(4)+K_{18}(2).\]
If \(\lambda\equiv3\pmod 4\), then we write \[K_{18}(\lambda)=K_{18}(\lambda-3)+K_{18}(3) =\frac{\lambda-3}{4}K_{18}(4)+K_{18}(3).\] Therefore, \(K_{18}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 15. \(n=19\).
If \(\lambda=1\), then \[(p,q)\in\{(1,42),(5,39),(9,36),\ldots,(57,0)\}.\] By Theorem 2.4, \(K_{4,8}\) is \(\{0P_{4},8S_{4}\}\)-decomposable. By taking \[K_{19}=K_{11}+K_{9}+K_{6,4}+K_{6,4}+K_{4,8},\] we obtain the decomposition when \((p,q)=(1,42)\), and by taking \[K_{19}=K_{13}+K_{7}+K_{4,6}+K_{4,6}+K_{4,6},\] we obtain all the other possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(2,84),(6,81),(10,78),\ldots,(114,0)\}.\] By taking \[K_{19}(2)=K_{19}+K_{19},\] we obtain all the above possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(3,126),(7,123),(11,120),\ldots,(171,0)\}.\] By taking \[K_{19}(3)=K_{19}(2)+K_{19},\] we obtain all the above possible decompositions.
If \(\lambda=4\), then \[(p,q)\in\{(0,171),(4,168),(8,165),\ldots,(228,0)\}.\] By Theorem 2.2, \(K_{19}(4)\) is \(\{0P_{4},171S_{4}\}\)-decomposable. By taking \[K_{19}(4)=K_{19}(2)+K_{19}(2),\] we obtain all the above possible decompositions.
If \(\lambda\geq5\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 4\), then we write \[K_{19}(\lambda)=\frac{\lambda}{4}K_{19}(4).\]
If \(\lambda\equiv1\pmod 4\), then we write \[K_{19}(\lambda)=K_{19}(\lambda-1)+K_{19} =\frac{\lambda-1}{4}K_{19}(4)+K_{19}.\]
If \(\lambda\equiv2\pmod 4\), then we write \[K_{19}(\lambda)=K_{19}(\lambda-2)+K_{19}(2) =\frac{\lambda-2}{4}K_{19}(4)+K_{19}(2).\]
If \(\lambda\equiv3\pmod 4\), then we write \[K_{19}(\lambda)=K_{19}(\lambda-3)+K_{19}(3) =\frac{\lambda-3}{4}K_{19}(4)+K_{19}(3).\] Therefore, \(K_{19}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 16. \(n=20\).
If \(\lambda=1\), then \[(p,q)\in\{(2,46),(6,43),(10,40),\ldots,(62,1)\}.\] By taking \[K_{20}=K_{12}+K_{8}+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4},\] we obtain all the above possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(0,95),(4,92),(8,89),\ldots,(124,2)\}.\] By Theorem 2.2, \(K_{20}(2)\) is \(\{0P_{4},95S_{4}\}\)-decomposable. By taking \[K_{20}(2)=K_{20}+K_{20},\] we obtain all the above possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(2,141),(6,138),(10,135),\ldots,(190,0)\}.\] By Theorem 2.1, \(K_{20}(3)\) is \(\{190P_{4},0S_{4}\}\)-decomposable. By taking \[K_{20}(3)=K_{20}(2)+K_{20},\] we obtain all the above possible decompositions.
If \(\lambda=4\), then \[(p,q)\in\{(0,190),(4,187),(8,184),\ldots,(252,1)\}.\] By Theorem 2.2, \(K_{20}(4)\) is \(\{0P_{4},190S_{4}\}\)-decomposable. By taking \[K_{20}(4)=K_{20}(3)+K_{20},\] we obtain all the above possible decompositions.
If \(\lambda=5\), then \[(p,q)\in\{(2,236),(6,233),(10,230),\ldots,(314,2)\}.\] By taking \[K_{20}(5)=K_{20}(3)+K_{20}(2),\] we obtain all the above possible decompositions.
If \(\lambda=6\), then \[(p,q)\in\{(0,285),(4,282),(8,279),\ldots,(380,0)\}.\] By Theorem 2.2, \(K_{20}(6)\) is \(\{0P_{4},285S_{4}\}\)-decomposable. By taking \[K_{20}(6)=K_{20}(3)+K_{20}(3),\] we obtain all the above possible decompositions.
If \(\lambda\geq7\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 6\), then we write \[K_{20}(\lambda)=\frac{\lambda}{6}K_{20}(6).\]
If \(\lambda\equiv1\pmod 6\), then we write \[K_{20}(\lambda)=K_{20}(\lambda-1)+K_{20} =\frac{\lambda-1}{6}K_{20}(6)+K_{20}.\]
If \(\lambda\equiv2\pmod 6\), then we write \[K_{20}(\lambda)=K_{20}(\lambda-2)+K_{20}(2) =\frac{\lambda-2}{6}K_{20}(6)+K_{20}(2).\]
If \(\lambda\equiv3\pmod 6\), then we write \[K_{20}(\lambda)=K_{20}(\lambda-3)+K_{20}(3) =\frac{\lambda-3}{6}K_{20}(6)+K_{20}(3).\]
If \(\lambda\equiv4\pmod 6\), then we write \[K_{20}(\lambda)=K_{20}(\lambda-4)+K_{20}(4) =\frac{\lambda-4}{6}K_{20}(6)+K_{20}(4).\]
If \(\lambda\equiv5\pmod 6\), then we write \[K_{20}(\lambda)=K_{20}(\lambda-5)+K_{20}(5) =\frac{\lambda-5}{6}K_{20}(6)+K_{20}(5).\] Therefore, \(K_{20}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 17. \(n=21\).
If \(\lambda=1\), then \[(p,q)\in\{(2,51),(6,48),(10,45),\ldots,(70,0)\}.\] By taking \[K_{21}=K_{13}+K_{9}+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4},\] we obtain all the above possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(0,105),(4,102),(8,99),\ldots,(140,0)\}.\] By Theorem 2.2, \(K_{21}(2)\) is \(\{0P_{4},105S_{4}\}\)-decomposable. By taking \[K_{21}(2)=K_{21}+K_{21},\] we obtain all the above possible decompositions.
If \(\lambda\geq3\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 2\), then we write \[K_{21}(\lambda)=\frac{\lambda}{2}K_{21}(2).\]
If \(\lambda\equiv1\pmod 2\), then we write \[K_{21}(\lambda)=K_{21}(\lambda-1)+K_{21} =\frac{\lambda-1}{2}K_{21}(2)+K_{21}.\] Therefore, \(K_{21}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 18. \(n=22\).
If \(\lambda=1\), then \[(p,q)\in\{(1,57),(5,54),(9,51),\ldots,(77,0)\}.\] By Theorem 2.4, \(K_{14,8}\) is \(\{0P_{4},28S_{4}\}\)-decomposable. By taking \[K_{22}=K_{14}+K_{8}+K_{14,8},\] we obtain the decomposition when \((p,q)=(1,57)\). By taking \[K_{22}=K_{16}+K_{6}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6},\] we obtain all the above possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(2,114),(6,111),(10,108),\ldots,(154,0)\}.\] By taking \[K_{22}(2)=K_{22}+K_{22},\] we obtain all the above possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(3,171),(7,168),(11,165),\ldots,(231,0)\}.\] By taking \[K_{22}(3)=K_{22}(2)+K_{22},\] we obtain all the above possible decompositions.
If \(\lambda=4\), then \[(p,q)\in\{(0,231),(4,228),(8,225),\ldots,(308,0)\}.\] By Theorem 2.2, \(K_{22}(4)\) is \(\{0P_{4},231S_{4}\}\)-decomposable. By taking \[K_{22}(4)=K_{22}(2)+K_{22}(2),\] we obtain all the above possible decompositions.
If \(\lambda\geq5\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 4\), then we write \[K_{22}(\lambda)=\frac{\lambda}{4}K_{22}(4).\]
If \(\lambda\equiv1\pmod 4\), then we write \[K_{22}(\lambda)=K_{22}(\lambda-1)+K_{22} =\frac{\lambda-1}{4}K_{22}(4)+K_{22}.\]
If \(\lambda\equiv2\pmod 4\), then we write \[K_{22}(\lambda)=K_{22}(\lambda-2)+K_{22}(2) =\frac{\lambda-2}{4}K_{22}(4)+K_{22}(2).\]
If \(\lambda\equiv3\pmod 4\), then we write \[K_{22}(\lambda)=K_{22}(\lambda-3)+K_{22}(3) =\frac{\lambda-3}{4}K_{22}(4)+K_{22}(3).\] Therefore, \(K_{22}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 19. \(n=23\).
If \(\lambda=1\), then \[(p,q)\in\{(3,61),(7,58),(11,55),\ldots,(83,1)\}.\] By taking \[K_{23}=K_{17}+K_{7}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6},\] we obtain all the above possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(2,125),(6,122),(10,119),\ldots,(166,2)\}.\] By taking \[\begin{aligned} K_{23}(2) &=K_{17}(2)+K_{7}(2)+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}\\ &\quad +K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}, \end{aligned}\] we obtain all the above possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(1,189),(5,186),(9,183),\ldots,(253,0)\}.\] By taking \[\begin{aligned} K_{23}(3) &=K_{17}(3)+K_{7}(3)+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}\\ &\quad +K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}\\ &\quad +K_{4,6}+K_{4,6}, \end{aligned}\] we obtain all the above possible decompositions.
If \(\lambda=4\), then \[(p,q)\in\{(0,253),(4,250),(8,247),\ldots,(336,1)\}.\] By Theorem 2.2, \(K_{23}(4)\) is \(\{0P_{4},253S_{4}\}\)-decomposable. By taking \[K_{23}(4)=K_{23}(3)+K_{23},\] we obtain all the above possible decompositions.
If \(\lambda=5\), then \[(p,q)\in\{(3,314),(7,311),(11,308),\ldots,(419,2)\}.\] By taking \[K_{23}(5)=K_{23}(3)+K_{23}(2),\] we obtain all the above possible decompositions.
If \(\lambda=6\), then \[(p,q)\in\{(2,378),(6,375),(10,372),\ldots,(506,0)\}.\] By taking \[K_{23}(6)=K_{23}(3)+K_{23}(3),\] we obtain all the above possible decompositions.
If \(\lambda=7\), then \[(p,q)\in\{(1,442),(5,439),(9,436),\ldots,(589,1)\}.\] By taking \[K_{23}(7)=K_{23}(4)+K_{23}(3),\] we obtain all the above possible decompositions.
If \(\lambda=8\), then \[(p,q)\in\{(0,506),(4,503),(8,500),\ldots,(672,2)\}.\] By taking \[K_{23}(8)=K_{23}(4)+K_{23}(4),\] we obtain all the above possible decompositions.
If \(\lambda=9\), then \[(p,q)\in\{(3,567),(7,564),(11,561),\ldots,(759,0)\}.\] By taking \[K_{23}(9)=K_{23}(6)+K_{23}(3),\] we obtain all the above possible decompositions.
If \(\lambda=10\), then \[(p,q)\in\{(2,631),(6,628),(10,625),\ldots,(842,1)\}.\] By taking \[K_{23}(10)=K_{23}(6)+K_{23}(4),\] we obtain all the above possible decompositions.
If \(\lambda=11\), then \[(p,q)\in\{(1,695),(5,692),(9,689),\ldots,(925,2)\}.\] By taking \[K_{23}(11)=K_{23}(8)+K_{23}(3),\] we obtain all the above possible decompositions.
If \(\lambda=12\), then \[(p,q)\in\{(0,759),(4,756),(8,753),\ldots,(1012,0)\}.\] By Theorem 2.2, \(K_{23}(12)\) is \(\{0P_{4},759S_{4}\}\)-decomposable. By taking \[K_{23}(12)=K_{23}(6)+K_{23}(6),\] we obtain all the above possible decompositions.
If \(\lambda\geq13\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod{12}\), then we write \[K_{23}(\lambda)=\frac{\lambda}{12}K_{23}(12).\]
If \(\lambda\equiv1\pmod{12}\), then we write \[K_{23}(\lambda)=K_{23}(\lambda-1)+K_{23} =\frac{\lambda-1}{12}K_{23}(12)+K_{23}.\]
If \(\lambda\equiv2\pmod{12}\), then we write \[K_{23}(\lambda)=K_{23}(\lambda-2)+K_{23}(2) =\frac{\lambda-2}{12}K_{23}(12)+K_{23}(2).\]
If \(\lambda\equiv3\pmod{12}\), then we write \[K_{23}(\lambda)=K_{23}(\lambda-3)+K_{23}(3) =\frac{\lambda-3}{12}K_{23}(12)+K_{23}(3).\]
If \(\lambda\equiv4\pmod{12}\), then we write \[K_{23}(\lambda)=K_{23}(\lambda-4)+K_{23}(4) =\frac{\lambda-4}{12}K_{23}(12)+K_{23}(4).\]
If \(\lambda\equiv5\pmod{12}\), then we write \[K_{23}(\lambda)=K_{23}(\lambda-5)+K_{23}(5) =\frac{\lambda-5}{12}K_{23}(12)+K_{23}(5).\]
If \(\lambda\equiv6\pmod{12}\), then we write \[K_{23}(\lambda)=K_{23}(\lambda-6)+K_{23}(6) =\frac{\lambda-6}{12}K_{23}(12)+K_{23}(6).\]
If \(\lambda\equiv7\pmod{12}\), then we write \[K_{23}(\lambda)=K_{23}(\lambda-7)+K_{23}(7) =\frac{\lambda-7}{12}K_{23}(12)+K_{23}(7).\]
If \(\lambda\equiv8\pmod{12}\), then we write \[K_{23}(\lambda)=K_{23}(\lambda-8)+K_{23}(8) =\frac{\lambda-8}{12}K_{23}(12)+K_{23}(8).\]
If \(\lambda\equiv9\pmod{12}\), then we write \[K_{23}(\lambda)=K_{23}(\lambda-9)+K_{23}(9) =\frac{\lambda-9}{12}K_{23}(12)+K_{23}(9).\]
If \(\lambda\equiv10\pmod{12}\), then we write \[K_{23}(\lambda)=K_{23}(\lambda-10)+K_{23}(10) =\frac{\lambda-10}{12}K_{23}(12)+K_{23}(10).\]
If \(\lambda\equiv11\pmod{12}\), then we write \[K_{23}(\lambda)=K_{23}(\lambda-11)+K_{23}(11) =\frac{\lambda-11}{12}K_{23}(12)+K_{23}(11).\] Therefore, \(K_{23}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 20. \(n=24\).
If \(\lambda=1\), then \[(p,q)\in\{(0,69),(4,66),(8,63),\ldots,(92,0)\}.\] By Theorem 2.2, \(K_{24}\) is \(\{0P_{4},69S_{4}\}\)-decomposable. By taking \[K_{24}=K_{12}+K_{12}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6},\] we obtain all the above possible decompositions.
If \(\lambda\geq2\), then \(K_{24}(\lambda)\) can be decomposed into \(\lambda\) copies of \(K_{24}\). Therefore, \(K_{24}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 21. \(n=25\).
If \(\lambda=1\), then \[(p,q)\in\{(0,75),(4,72),(8,69),\ldots,(100,0)\}.\] By Theorem 2.2, \(K_{25}\) is \(\{0P_{4},75S_{4}\}\)-decomposable. By taking \[K_{25}=K_{13}+K_{13}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6},\] we obtain all the above possible decompositions.
If \(\lambda\geq2\), then \(K_{25}(\lambda)\) can be decomposed into \(\lambda\) copies of \(K_{25}\). Therefore, \(K_{25}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 22. \(n=26\).
If \(\lambda=1\), then \[(p,q)\in\{(3,79),(7,76),(11,73),\ldots,(107,1)\}.\] By taking \[K_{26}=K_{18}+K_{8}+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4},\] we obtain all the above possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(2,161),(6,158),(10,155),\ldots,(214,2)\}.\] By taking \[\begin{aligned} K_{26}(2) &=K_{18}(2)+K_{8}(2)+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}\\ &\quad +K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}\\ &\quad +K_{6,4}+K_{6,4}, \end{aligned}\] we obtain all the above possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(1,243),(5,240),(9,237),\ldots,(325,0)\}.\] By taking \[\begin{aligned} K_{26}(3) &=K_{18}(3)+K_{8}(3)+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}\\ &\quad +K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}\\ &\quad +K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}\\ &\quad +K_{6,4}+K_{6,4}, \end{aligned}\] we obtain all the above possible decompositions.
If \(\lambda=4\), then \[(p,q)\in\{(0,325),(4,322),(8,319),\ldots,(432,1)\}.\] By Theorem 2.2, \(K_{26}(4)\) is \(\{0P_{4},325S_{4}\}\)-decomposable. By taking \[K_{26}(4)=K_{26}(3)+K_{26},\] we obtain all the above possible decompositions.
If \(\lambda=5\), then \[(p,q)\in\{(3,404),(7,401),(11,398),\ldots,(539,2)\}.\] By taking \[K_{26}(5)=K_{26}(4)+K_{26},\] we obtain all the above possible decompositions.
If \(\lambda=6\), then \[(p,q)\in\{(2,486),(6,483),(10,480),\ldots,(650,0)\}.\] By taking \[K_{26}(6)=K_{26}(3)+K_{26}(3),\] we obtain all the above possible decompositions.
If \(\lambda=7\), then \[(p,q)\in\{(1,568),(5,565),(9,562),\ldots,(757,1)\}.\] By taking \[K_{26}(7)=K_{26}(4)+K_{26}(3),\] we obtain all the above possible decompositions.
If \(\lambda=8\), then \[(p,q)\in\{(0,650),(4,647),(8,644),\ldots,(864,2)\}.\] By taking \[K_{26}(8)=K_{26}(4)+K_{26}(4),\] we obtain all the above possible decompositions.
If \(\lambda=9\), then \[(p,q)\in\{(3,729),(7,726),(11,723),\ldots,(975,0)\}.\] By taking \[K_{26}(9)=K_{26}(6)+K_{26}(3),\] we obtain all the above possible decompositions.
If \(\lambda=10\), then \[(p,q)\in\{(2,811),(6,808),(10,805),\ldots,(1082,1)\}.\] By taking \[K_{26}(10)=K_{26}(6)+K_{26}(4),\] we obtain all the above possible decompositions.
If \(\lambda=11\), then \[(p,q)\in\{(1,893),(5,890),(9,887),\ldots,(1189,2)\}.\] By taking \[K_{26}(11)=K_{26}(8)+K_{26}(3),\] we obtain all the above possible decompositions.
If \(\lambda=12\), then \[(p,q)\in\{(0,975),(4,972),(8,969),\ldots,(1300,0)\}.\] By Theorem 2.1, \(K_{26}(12)\) is \(\{1300P_{4},0S_{4}\}\)-decomposable. By taking \[K_{26}(12)=K_{26}(8)+K_{26}(4),\] we obtain all the above possible decompositions.
If \(\lambda\geq13\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod{12}\), then we write \[K_{26}(\lambda)=\frac{\lambda}{12}K_{26}(12).\]
If \(\lambda\equiv1\pmod{12}\), then we write \[K_{26}(\lambda)=K_{26}(\lambda-1)+K_{26} =\frac{\lambda-1}{12}K_{26}(12)+K_{26}.\]
If \(\lambda\equiv2\pmod{12}\), then we write \[K_{26}(\lambda)=K_{26}(\lambda-2)+K_{26}(2) =\frac{\lambda-2}{12}K_{26}(12)+K_{26}(2).\]
If \(\lambda\equiv3\pmod{12}\), then we write \[K_{26}(\lambda)=K_{26}(\lambda-3)+K_{26}(3) =\frac{\lambda-3}{12}K_{26}(12)+K_{26}(3).\]
If \(\lambda\equiv4\pmod{12}\), then we write \[K_{26}(\lambda)=K_{26}(\lambda-4)+K_{26}(4) =\frac{\lambda-4}{12}K_{26}(12)+K_{26}(4).\]
If \(\lambda\equiv5\pmod{12}\), then we write \[K_{26}(\lambda)=K_{26}(\lambda-5)+K_{26}(5) =\frac{\lambda-5}{12}K_{26}(12)+K_{26}(5).\]
If \(\lambda\equiv6\pmod{12}\), then we write \[K_{26}(\lambda)=K_{26}(\lambda-6)+K_{26}(6) =\frac{\lambda-6}{12}K_{26}(12)+K_{26}(6).\]
If \(\lambda\equiv7\pmod{12}\), then we write \[K_{26}(\lambda)=K_{26}(\lambda-7)+K_{26}(7) =\frac{\lambda-7}{12}K_{26}(12)+K_{26}(7).\]
If \(\lambda\equiv8\pmod{12}\), then we write \[K_{26}(\lambda)=K_{26}(\lambda-8)+K_{26}(8) =\frac{\lambda-8}{12}K_{26}(12)+K_{26}(8).\]
If \(\lambda\equiv9\pmod{12}\), then we write \[K_{26}(\lambda)=K_{26}(\lambda-9)+K_{26}(9) =\frac{\lambda-9}{12}K_{26}(12)+K_{26}(9).\]
If \(\lambda\equiv10\pmod{12}\), then we write \[K_{26}(\lambda)=K_{26}(\lambda-10)+K_{26}(10) =\frac{\lambda-10}{12}K_{26}(12)+K_{26}(10).\]
If \(\lambda\equiv11\pmod{12}\), then we write \[K_{26}(\lambda)=K_{26}(\lambda-11)+K_{26}(11) =\frac{\lambda-11}{12}K_{26}(12)+K_{26}(11).\] Therefore, \(K_{26}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 23. \(n=27\).
If \(\lambda=1\), then \[(p,q)\in\{(1,87),(5,84),(9,81),\ldots,(117,0)\}.\] By taking \[K_{27}=K_{19}+K_{9}+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4}+K_{6,4},\] we obtain all the above possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(2,174),(6,171),(10,168),\ldots,(234,0)\}.\] By taking \[K_{27}(2)=K_{27}+K_{27},\] we obtain all the above possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(3,261),(7,258),(11,255),\ldots,(351,0)\}.\] By taking \[K_{27}(3)=K_{27}(2)+K_{27},\] we obtain all the above possible decompositions.
If \(\lambda=4\), then \[(p,q)\in\{(0,351),(4,348),(8,345),\ldots,(468,0)\}.\] By Theorem 2.2, \(K_{27}(4)\) is \(\{0P_{4},351S_{4}\}\)-decomposable. By taking \[K_{27}(4)=K_{27}(2)+K_{27}(2),\] we obtain all the above possible decompositions.
If \(\lambda\geq5\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 4\), then we write \[K_{27}(\lambda)=\frac{\lambda}{4}K_{27}(4).\]
If \(\lambda\equiv1\pmod 4\), then we write \[K_{27}(\lambda)=K_{27}(\lambda-1)+K_{27} =\frac{\lambda-1}{4}K_{27}(4)+K_{27}.\]
If \(\lambda\equiv2\pmod 4\), then we write \[K_{27}(\lambda)=K_{27}(\lambda-2)+K_{27}(2) =\frac{\lambda-2}{4}K_{27}(4)+K_{27}(2).\]
If \(\lambda\equiv3\pmod 4\), then we write \[K_{27}(\lambda)=K_{27}(\lambda-3)+K_{27}(3) =\frac{\lambda-3}{4}K_{27}(4)+K_{27}(3).\] Therefore, \(K_{27}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 24. \(n=28\).
If \(\lambda=1\), then \[(p,q)\in\{(2,93),(6,90),(10,87),\ldots,(126,0)\}.\] By taking \[K_{28}=K_{16}+K_{12}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6} +K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6},\] we obtain all the above possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(0,189),(4,186),(8,183),\ldots,(252,0)\}.\] By Theorem 2.2, \(K_{28}(2)\) is \(\{0P_{4},189S_{4}\}\)-decomposable. By taking \[K_{28}(2)=K_{28}+K_{28},\] we obtain all the above possible decompositions.
If \(\lambda\geq3\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 2\), then we write \[K_{28}(\lambda)=\frac{\lambda}{2}K_{28}(2).\]
If \(\lambda\equiv1\pmod 2\), then we write \[K_{28}(\lambda)=K_{28}(\lambda-1)+K_{28} =\frac{\lambda-1}{2}K_{28}(2)+K_{28}.\] Therefore, \(K_{28}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Case 25. \(n=30\).
If \(\lambda=1\), then \[(p,q)\in\{(145,0),(141,3),(137,6),\ldots,(1,108)\}.\] By Theorem 2.4, \(K_{22,8}\) is \(\{0P_{4},44S_{4}\}\)-decomposable. By taking \[K_{30}=K_{22}+K_{8}+K_{22,8},\] we obtain the decomposition when \((p,q)=(1,108)\), and by taking \[K_{30}=K_{24}+K_{6}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6}+K_{4,6},\] we obtain all the above possible decompositions.
If \(\lambda=2\), then \[(p,q)\in\{(290,0),(286,3),(282,6),\ldots,(2,216)\}.\] By taking \[K_{30}(2)=K_{30}+K_{30},\] we obtain all the above possible decompositions.
If \(\lambda=3\), then \[(p,q)\in\{(435,0),(431,3),(427,6),\ldots,(3,324)\}.\] By taking \[K_{30}(3)=K_{30}(2)+K_{30},\] we obtain all the above possible decompositions.
If \(\lambda=4\), then \[(p,q)\in\{(580,0),(576,3),(572,6),\ldots,(0,435)\}.\] By Theorem 2.2, \(K_{30}(4)\) is \(\{0P_{4},435S_{4}\}\)-decomposable. By taking \[K_{30}(4)=K_{30}(2)+K_{30}(2),\] we obtain all the above possible decompositions.
If \(\lambda\geq5\), then the proof is divided into the following cases.
If \(\lambda\equiv0\pmod 4\), then we write \[K_{30}(\lambda)=\frac{\lambda}{4}K_{30}(4).\]
If \(\lambda\equiv1\pmod 4\), then we write \[K_{30}(\lambda)=K_{30}(\lambda-1)+K_{30} =\frac{\lambda-1}{4}K_{30}(4)+K_{30}.\]
If \(\lambda\equiv2\pmod 4\), then we write \[K_{30}(\lambda)=K_{30}(\lambda-2)+K_{30}(2) =\frac{\lambda-2}{4}K_{30}(4)+K_{30}(2).\]
If \(\lambda\equiv3\pmod 4\), then we write \[K_{30}(\lambda)=K_{30}(\lambda-3)+K_{30}(3) =\frac{\lambda-3}{4}K_{30}(4)+K_{30}(3).\] Therefore, \(K_{30}(\lambda)\) is fully \(\{P_{4},S_{4}\}\)-decomposable.
Now we prove the result for odd \(n>27\) and even \(n>30\). Let \(n=2r\) or \(n=2r+1\), where \(r\geq2\). We prove the result by mathematical induction on \(n\), splitting the proof into two cases as follows.
Case A. \(n\equiv0\pmod 2\).
Let \(n=2r\), where \(r\geq3\). If \(3\leq r\leq15\), then the result follows from Cases 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, and 25. Now, for some \(t>15\), assume that there exists a \(\{P_{4},S_{4}\}_{\{p,q\}}\)-decomposition of \(K_{2r}(\lambda)\) for all \(r\) with \(3\leq r<t\). We write \[\begin{aligned} K_{2t}(\lambda) &=K_{2(t-12)}(\lambda)+K_{24}(\lambda)+K_{2(t-12),24}(\lambda)\\ &=K_{2(t-12)}(\lambda)+K_{24}(\lambda)+(t-12)K_{2,24}(\lambda)\\ &=K_{2(t-12)}(\lambda)+K_{24}(\lambda)+2(t-12)K_{2,12}(\lambda). \end{aligned}\] By the induction hypothesis, there exists a \(\{P_{4},S_{4}\}_{\{p,q\}}\)-decomposition of \(K_{2(t-12)}(\lambda)\). By Case 20, there exists a \(\{P_{4},S_{4}\}_{\{p,q\}}\)-decomposition of \(K_{24}(\lambda)\). By Lemma 3.2, \(K_{2,12}\) is fully \(\{P_{4},S_{4}\}\)-decomposable, and hence \(K_{2,12}(\lambda)\) is decomposable by taking \(\lambda\) copies. Therefore, a \(\{P_{4},S_{4}\}_{\{p,q\}}\)-decomposition of \(K_{2t}(\lambda)\) exists.
Case B. \(n\equiv1\pmod 2\).
Let \(n=2r+1\), where \(r\geq2\). If \(2\leq r\leq13\), then the result follows from Cases 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, and 23. Now, for some \(t>13\), assume that there exists a \(\{P_{4},S_{4}\}_{\{p,q\}}\)-decomposition of \(K_{2r+1}(\lambda)\) for all \(r\) with \(2\leq r<t\). We write \[\begin{aligned} K_{2t+1}(\lambda) &=K_{2(t-12)+1}(\lambda)+K_{25}(\lambda)+K_{2(t-12),24}(\lambda)\\ &=K_{2(t-12)+1}(\lambda)+K_{25}(\lambda)+(t-12)K_{2,24}(\lambda)\\ &=K_{2(t-12)+1}(\lambda)+K_{25}(\lambda)+2(t-12)K_{2,12}(\lambda). \end{aligned}\] By the induction hypothesis, there exists a \(\{P_{4},S_{4}\}_{\{p,q\}}\)-decomposition of \(K_{2(t-12)+1}(\lambda)\). By Case 21, there exists a \(\{P_{4},S_{4}\}_{\{p,q\}}\)-decomposition of \(K_{25}(\lambda)\). By Lemma 3.2, \(K_{2,12}\) is fully \(\{P_{4},S_{4}\}\)-decomposable, and hence \(K_{2,12}(\lambda)\) is decomposable by taking \(\lambda\) copies. Therefore, a \(\{P_{4},S_{4}\}_{\{p,q\}}\)-decomposition of \(K_{2t+1}(\lambda)\) exists, and the result follows by mathematical induction. ◻