2-(v, 5; m) spontaneous emission error designs

Abstract

Quantum jump codes are quantum codes which correct errors caused by quantum jumps. A t-spontaneous emission error design (t-SEED) was introduced by Beth et al. in 2003 to construct quantum jump codes. The number of designs (dimension) in a t-SEED corresponds to the number of orthogonal basis states in a quantum jump code. Denote by $\overline{M}(t,k,v)$ the largest possible dimension m for which a nondegenerate t-(v, k; m) SEED exists. $\overline{M}(2,3,v)$ has been determined completely, which is based on a great deal of research on large sets of Steiner triple systems and large sets of pairwise disjoint compatible 2-(v, 3, 1) packings. For $k=4$, the upper bounds on dimensions of 2-(v, 4; m) SEEDs were also demonstrated and the corresponding leave graphs were investigated in Zhou and Chang (J Combin Des 24:439–460, 2016). In this paper we turn our attention to the case $t=2$ and $k=5$. We present general upper bounds on the dimensions of 2-(v, 5; m) SEEDs and describe the concrete leave graphs of the 2-SEEDs attaining the stated upper bounds.

Appendix: continued proof of Lemma 3.8

$\underline{(4) |E(G[Y])|=3.}$ Then G[Y] has degree sequence (2, 2, 1, 1) or (2, 2, 2, 0) by (i), corresponding to a path or a cycle of length 3. Let $E_Y^1=\{12,23,34\}$ and $E_Y^2=\{12,23,13\}$. Denote the degree sum $D_Z=\sum _{z\in Z}d_{G[Z]}(z)$. So $D_Z=12$ by (i). Since $|Z|=6$, G[Z] has all vertices of degree 2 or it contains at least one vertex of degree 1. We deal with these two possibilities.

Firstly let G[Z] be 2-regular. Note that this is impossible by (ii) if G[Y] has a vertex of degree 1. So we let $E(G[Y])=E_Y^2$. It is easy to know that G[Z] consists of two cycles of length 3 or a cycle of length 6. Let (a) $E(G[Z])=E_Z^1=\{ab,ac,bc,de,df,ef\}$, (b) $E(G[Z])=E_Z^2=\{ab,af,bc,cd,de,ef\}$. Note that $4z\in E$ for any $z\in Z$.

• (a) $E(G[Z])=E_Z^1$. We assume $N_{G}(a)\cap Y=\{1,2,4\}$. Then $Y_{a}=\{3,d,e,f\}$ and $de,df,ef\in E$. Moreover, the vertex 3 must be adjacent to at least one of d, e and f, that is, $G[Y_{a}]$ contains a vertex of degree 3, a contradiction to (i).

• (b) $E(G[Z])=E_Z^2$. For $i\in \{1,2,3\}$, let $Y_i=\{0,4,\alpha _i,\beta _i\}$ where $\alpha _i,\beta _i\in Z$. Note $0\alpha _i,0\beta _i,4\alpha _i,$$4\beta _i\in E$. We have $\alpha _i\beta _i\notin E$ by (i). Due to the symmetry of vertices in G[Z], we assume that (b1) $N_{G}(1)\cap Z=\{a,c,d,e\}$ or (b2) $N_{G}(1)\cap Z=\{a,c,d,f\}$. For (b1), $Y_{b}=\{1,d,e,f\}$ and $1e,de,ef\in E$, so $d_{G[Y_b]}(e)=3$, a contradiction to (i). Thus case (b2) holds. Then $2b,2e,3b,3e\in E$ by (vi). Now each of a, c, d, f has to be adjacent to exactly one of 2 and 3, two symmetric vertices. Three possibilities need to be considered: $\{2a,2c,3d,3f\}\subset E$, $\{2a,2d,3c,3f\}\subset E$, or $\{2a,2f,3c,3d\}\subset E$. By the very first reasoning of this case, $\{2a,2f,3c,3d\}\subset E$ is impossible by considering $Y_2,Y_3$ because $cd,af\in E$. Also note that if $\{2a,2c,3d,3f\}\subset E$ then $N_{G}(2)\cap Z=\{a,b,c,e\}$ is isomorphic to (b1) case. So we let $\{2a,2d,3c,3f\}\subset E$. Therefore, G[X] is isomorphic to $H_{17}$ and we check that it is a qualified subgraph.

Secondly we let G[Z] contain a vertex of degree 1, say f. Suppose that $d_{G[Z]}(e)=d_0$ and $ef\in E$. Based on the cases (1)–(3), we assume that the subgraph $G[Y_x]$ has degree sum 6 or 8 with degree sequence (2, 2, 1, 1), (2, 2, 2, 0), or (2, 2, 2, 2) for any vertex $x\in X$. Note that $G[Y_f]$ is the induced subgraph of G[Z] by $Z\setminus \{e,f\}$. We have the degree sum of $G[Y_f]$$D_{Y_f}=D_Z-d_0-1-(d_0-1)=12-2d_0$. Hence $D_{Y_f}=12-2d_0=6,8$, yielding $d_0=3,2$, respectively.

When $d_0=2$, $D_{Y_f}=8$, meaning that $G[Y_f]$ forms a 4-cycle by (i), say $(a\ b\ c\ d)$. Further let $ae\in E$. Then the edge set $E(G[Z])=\{ab,ad,ae,bc,cd,ef\}$. When G[Y] is a cycle of length 3, let $E(G[Y])=E_Y^2$. W.l.o.g. assume that $1a\in E$, then $Y_{a}=\{2,3,c,f\}$. Note that $23,2f,3f\in E$ and c must be adjacent to at least one of 2 and 3. So $d_{G[Y_a]}(2)= 3$ or $d_{G[Y_a]}(3)= 3$, contradicting to (i). When G[Y] is a path of length 3, let $E(G[Y])=E_Y^1$. In this case, $N_{G}(f)\cap Y=\{1,2,3,4\}$, $N_{G}(1)\cap Z=N_{G}(4)\cap Z=\{b,c,d,e,f\}$, and $2a,3a\in E$ by (ii) and (vi). Now each of vertices 2 and 3 has two neighbors undetermined and each of b, c, d, e has one. Let $Y_2=\{0,4,\alpha ,\beta \}$ where $\alpha ,\beta \in \{b,c,d,e\}$. Since $0\alpha ,0\beta ,4\alpha ,4\beta \in E$, we have $\alpha \beta \notin E$ by (i). Similarly, let $Y_3=\{0,4,\alpha ',\beta '\}$ where $\alpha ',\beta '\in \{b,c,d,e\}$. Then $\alpha '\beta '\notin E$. So we assume $2b,2d,3c,3e\in E$ and form a graph $H_{18}$, a qualified graph.

When $d_0=3$, $D_{Y_f}=6$, meaning that $G[Y_f]$ has degree sequence (2, 2, 2, 0) or (2, 2, 1, 1). So we let $E(G[Y_f])=E_f^1=\{ab,ac,bc\}$ or $E(G[Y_f])=E_f^2=\{ab,bc,cd\}$.

Firstly consider the case $E(G[Y_f])=E_f^1$. Then $de\in E$ by (i). Hence we assume $E(G[Z])=\{ab,ac,ae,bc,de,ef\}$. Then d and f are adjacent to every vertex of Y by (vi). If E(G[Y] is a cycle of length 3, then let $E(G[Y])=E_Y^2$. Obviously $Y_{a}=\{i,j,d,f\}$ where $\{i,j\}\subseteq \{1,2,3\}$ by (vi). Note that $id,jd,if,jf\in E$ and $ij\in E$, contradicting to (i). So we let G[Y] be a path of length 3 with edge set $E_Y^1$. The nonadjacent vertex of 1 and 4 in Z is a or e by (ii). We consider whether or not $N_{G}(1)\cap Z=N_{G}(4)\cap Z$. If $N_{G}(1)\cap Z=\{b,c,d,e,f\}$ and $N_{G}(4)\cap Z=\{a,b,c,d,f\}$. Then $2a,3e\in E$ by (ii) and hence $N_G(e)=\{0,1,3,a,d,f\}$ so that $Y_{e}=\{2,4,b,c\}$. Note that $4b,4c,bc\in E$ and 2 must be adjacent to one of b and c. This contradicts to (i). If $N_{G}(1)\cap Z=N_{G}(4)\cap Z=\{a,b,c,d,f\}$, then $Y_a=\{2,3,d,f\}$. But $23,2d,2f,3d,3f\in E$, contradicting to (i). Finally, if $N_{G}(1)\cap Z=N_{G}(4)\cap Z=\{b,c,d,e,f\}$, then $2a,3a\in E$. Now we form the edges 2b and 3c to form a 6-regular graph (noting the symmetry of b and c). However, considering the vertex 1 gives $E(G[Z_1])=\{2b,2d,2f,bc,de,ef\}$. Thus $d_{G[Z_1]}(c)=1$, $d_{G[Z_1]}(b)=2$, and $bc\in E$. An analogue of this has been considered in the previous subcase $d_0=2$.

Secondly we let $E(G[Y_f])=E_f^2$. We consider three possibilities of the two neighbors of e in Z other than f: (a) $ae,de\in E$, (b) $be,ce\in E$, (c) $ae,ce\in E$ or $ae,be\in E$.

• (a) $ae,de\in E$. If $E(G[Y])=E_Y^1$, by (ii), $1e,4e\notin E$ and $2e,3e\in E$. But then $Y_e=\{1,4,b,c\}$ and $1b,4b,bc\in E$, contradicting to (i). If $E(G[Y])=E_Y^2$, then $N_G(f)=Y\cup \{0,e\}$ and $N_G(4)=Z$ by (vi). W.l.o.g. suppose that $1b,2b\in E$, then $N_G(b)=\{0,1,2,4,a,c\}$ and $Y_{b}=\{3,d,e,f\}$. Note that $3f,de,ef\in E$ in $G[Y_{b}]$, so $3e\notin E$ by (i) and thus $N_G(3)\cap Z=\{a,c,d,f\}$. Now we only need to determine the last two neighbors in $\{a,c,d,e\}$ for each of 1 and 2. Note that for $i=1,2$ we have $Y_i=\{0,4,\alpha _i,\beta _i\}$ where $\alpha _i,\beta _i\in Z$ and $\{0\alpha _i,0\beta _i,4\alpha _i,4\beta _i\}\subset E$. Hence $\alpha _i\beta _i\notin E$ by (i). It follows that we may assume $Y_1=\{0,4,a,d\}$ and $Y_2=\{0,4,c,e\}$. Now we have that $Y_a=\{1,c,d,f\}$, $E(Y_a)=\{1c,1f,cd\}$, $Z_a=\{0,2,3,4,b,e\}$, and $G[Z_a]=\{0b,0e,23,2b,4b,4e\}$. Thus this is reduced to the parallel case $|E(Y)|=3$ and $d_0=2$, which has been considered previously.

• (b) $be,ce\in E$. In either case of $E(G[Y])=E_Y^1$ or $E(G[Y])=E_Y^2$, we may assume $Y_b=\{2,3,d,f\}$. This is obvious if $E(G[Y])=E_Y^2$. If $E(G[Y])=E_Y^1$, then $N_G(a)\cap Y=N_G(d)\cap Y=N_G(f)\cap Y=Y$ by (vi) and thus $|N_G(i)\cap \{b,c,e\}|=1$ for $i=2,3$. So at least one of $\{b,c,e\}$ is adjacent to neither 2 nor 3 and thence we may also assume $Y_b=\{2,3,d,f\}$. Clearly $23,2d,2f\in E$, contradicting to (i).

• (c) $ae,ce\in E$ or $ae,be\in E$. For the latter case, $Y_d=\{a,b,e,f\}$ and $d_{G[Y_d]}(e)=3$, contradicting to (i). Hence we let $ae,ce\in E$. When $E(G[Y])=E_Y^2$, we have $Y_{e}=\{i,j,b,d\}$ where $\{i,j\}\subseteq \{1,2,3\}$ and $id,jd,ij\in E$, but b must be adjacent to at least one of i and j, which contradicts to (i). When $E(G[Y])=E_Y^1$, the (only) nonadjacent vertex of 1 or 4 in G[Z] is c or e by (ii). Note that c and e are symmetric in G[Z]. If $N_{G}(1)\cap Z=N_{G}(4)\cap Z=\{a,b,c,d,f\}$, then $2e,3e\in E$ by (ii). Now $Y_c=\{2,3,a,f\}$. Note that $23,2f,3f\in E$ and a must be adjacent to one vertex of 2 and 3, which contradicts to (i). Thus we assume w.l.o.g. $N_{G}(1)\cap Z=\{a,b,d,e,f\}$ and $N_{G}(4)\cap Z=\{a,b,c,d,f\}$. Then $2c,3e\in E$, $N_G(d)\cap Y=N_G(f)\cap Y=Y$ by (ii) and (vi). Finally using (vi) leaves each vertex 2, 3, a, b only one neighbor to be determined. If $2b,3a\in E$, then $Y_2=\{0,4,a,e\}$ and we have a contradiction $d_{G[Y_2]}(a)=3$. So we must have $2a,3b\in E$. Therefore, G[X] is isomorphic to $H_{19}$, a qualified subgraph.

$\underline{(5) |E(G[Y])|=4.}$ Then G[Y] forms a 4-cycle, say $(1\ 2\ 3\ 4)$, by (i). Based on the cases (1)–(4), we assume that the subgraph $G[Y_x]$ has degree sequence (2, 2, 2, 2) for any vertex $x\in X$. Since $|Z|=6$, the minimum degree in G[Z] is 1 or 2. We deal with these two possibilities.

Firstly let G[Z] contain a vertex of degree 1, say f. Suppose that $d_{G[Z]}(e)=d_0$ and $ef\in E$. In particular we let $G[Y_f]$ has edge set $E'=\{ab,bc,cd,ad\}$. We have $8=\sum _{y\in Y_f}d_{G[Y_f]}(y)=\sum _{z\in Z}d_{G[Z]}(z)-d_0-1-(d_0-1)=14-2d_0$, yielding $d_0=3$. So we have two forms of G[Z] by adding two edges to $E'\cup \{ef\}$: (a) $ae,ce\in E$ or (b) $ce,de\in E$. For (a), it is easy to see that $Y_b=\{y,d,e,f\}$ for some $y\in \{1,2,3,4\}$ and $d_{G[Y_b]}(d)\le 1$, contradicting to our assumption. For (b), w.l.o.g. assume that $N_{G}(d)\cap Y=\{1,2\}$ or $N_{G}(d)\cap Y=\{1,3\}$. If $N_{G}(d)\cap Y=\{1,2\}$, then $Y_{d}=\{3,4,b,f\}$ and $34,3f,4f\in E$, so $G[Y_{d}]$ cannot be a 4-cycle, a contradiction. So we let $N_{G}(d)\cap Y=\{1,3\}$. Then $Y_{d}=\{2,4,b,f\}$. Note that $G[Y_{d}]$ is a 4-cycle and $2f,4f\in E$, then $2b,4b\in E$. Now b has to be adjacent to one of 1 and 3, two symmetric vertices, so w.l.o.g. assume $1b\in E$. Hence $Y_{b}=\{3,d,e,f\}$ and $G[Y_b]$ contains de, ef, 3d, 3f. Thus $3e\not \in E$ by (i) and the last two undetermined neighbors of 3 are a and c. According to $Y_{3}=\{0,1,b,e\}$ and $0b,0e,1b\in E$, we have $1e\in E$. Now $N_G(1)=\{2,4,b,d,e,f\}$. So $N_{G}(a)\cap Y=\{2,3,4\}$ as $1a\notin E$. Therefore, $Y_{a}=\{1,c,e,f\}$, yet $d_{G[Y_{a}]}(e)=3$, contradicting to (i).

Secondly we let the minimum degree of G[Z] be 2. It is easy to know that the degree sequence of G[Z] is (4, 2, 2, 2, 2, 2) or (3, 3, 2, 2, 2, 2).

If G[Z] has degree sequence (4, 2, 2, 2, 2, 2), we suppose $E(G[Z])=\{ab,ac,ad,ae,bc,df,ef\}$. Considering the symmetry of vertices in G[Y], w.l.o.g. suppose that $N_{G}(f)\cap Y=\{1,2,3\}$. Then $Y_{f}=\{4,a,b,c\}$ and $ab,ac,bc\in E$, so $G[Y_{f}]$ cannot be a 4-cycle, a contradiction.

If G[Z] has degree sequence (3, 3, 2, 2, 2, 2), then it is easy to show that there are four non-isomorphic G[Z]. The edge set of G[Z] is: (a) $\{ac,ad,af,bc,bd,be,ef\}$, (b) $\{ab,ac,ad,bc,be,df,ef\}$, (c) $\{ab,ac,ad,be,bf,cd,ef\}$, or (d) $\{ab,ac,ad,be,bf,cf,de\}$.

For (a), assume that $N_{G}(c)\cap Y=\{1,2,3\}$. Then $Y_{c}=\{4,d,e,f\}$. Note that $de,df\notin E$. Thus $d_{G[Y_c]}(d)\le 1$, a contradiction. For (b), assume $N_{G}(f)\cap Y=\{1,2,3\}$, then $Y_{f}=\{4,a,b,c\}$. Because $ab,ac,bc\in E$, $G[Y_{f}]$ cannot be a 4-cycle. Similarly to (b), we have that (c) is impossible. For (d), since $d_{G[Z]}(a)=3$, a has exactly two neighbors and two non-neighbors in Y. Then we may suppose (d1) $1a,2a\notin E$ or (d2) $1a,3a\notin E$. So $Y_1=\{0,3,a,z\}$ where $z\in Z$. Since $G[Y_1]$ is a 4-cycle and has edges 0a, 0z, we must have $3a,3z\in E$, meaning that (d2) does not hold. Hence we let $1a,2a\notin E$. But then $Y_a=\{1,2,e,f\}$ and $G[Y_a]$ cannot be a 4-cycle as $12\in E$ and $ef\notin E$, a contradiction.