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Relative t-designs in Johnson association schemes for P-polynomial structure


Relative t-designs are defined in both P- and Q-polynomial association schemes. In this paper, we investigate relative t-designs in Johnson association schemes J(vk) for P-polynomial structure. It is known that each nontrivial shell of J(vk) is identified with the product of two smaller Johnson association schemes. We prove that relative t-designs in J(vk) supported by one shell are equivalent to weighted \(\mathcal T\)-designs in the shell (as product of association schemes) for \(\mathcal T=\{(t_1,t_2) \mid 0\le t_1,t_2\le t\}\). We study the existence problem of tight relative t-designs on one shell of J(vk) for \(t=2,3\). We propose an algorithm to explicitly construct a family of non-trivial tight relative 2-designs. In addition, we obtain tight relative 3-designs for some special parameters.

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The authors sincerely thank Eiichi Bannai, Etsuko Bannai, Jack H. Koolen, William J. Martin and Hajime Tanaka for many valuable discussions and helpful comments. The first author is supported by NSFC Grant No. 11801353 and China Postdoctoral Science Foundation No. 2018M632078.

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Correspondence to Naoki Watamura.

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Communicated by J. H. Koolen.

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Define the point set \(V_1=\{1,2,3,4,5,6,7\}\) and the blocks \(B_\ell ,\ \ell \in \{1, 2, 3, \ldots , 15\}\) as follows.

$$\begin{aligned}&B_1=\{1, 2, 3\}, \ B_2=\{1, 2, 7\},\ B_3=\{1, 3, 6\},\ B_4=\{1, 4, 5\},\ B_5=\{1, 6, 7\}, \\&B_6=\{2, 3, 5\},\ B_7=\{2, 4, 6\},\ B_8=\{2, 4, 7\},\ B_9=\{2, 5, 6\},\ B_{10}=\{2, 5, 7\},\\&B_{11}=\{3, 4, 6\},\ B_{12}=\{3, 4, 7\},\ B_{13}=\{3, 5, 6\},\ B_{14}=\{3, 5, 7\},\ B_{15}=\{5, 6, 7\}. \end{aligned}$$

In Step 1 of the algorithm, we fix one 2-(7, 3, 7) design as the left projection \(Y^{(L)}\). In this example, we take the second 2-(7, 3, 7) design on the homepage of Spence [13] as the left projection \(Y^{(L)}\):

$$\begin{aligned} Y^{(L)}=\{y_1^{(L)},y_2^{(L)},\ldots , y_{49}^{(L)}\}, \end{aligned}$$


$$\begin{aligned}&y_i^{(L)}=B_1, \ 1\le i\le 7; \qquad y_i^{(L)}=B_4, \ 8\le i\le 14; \qquad y_i^{(L)}=B_5, \ 15\le i\le 21; \nonumber \\&y_i^{(L)}=B_7, \ 22\le i\le 27; \quad y_i^{(L)}=B_{10}, \ 30\le i\le 35; \quad y_i^{(L)}=B_{12}, \ 37\le i\le 42; \qquad \nonumber \\&y_i^{(L)}=B_{13}, \ 43\le i\le 48; \quad y_{28}^{(L)}=B_8, \quad y_{29}^{(L)}=B_9, \quad y_{36}^{(L)}=B_{11}, \quad y_{49}^{(L)}=B_{14}. \end{aligned}$$

From here, we follow each step of the algorithm by constructing an example of tight (2, 2)-(7, 3, 7, 3, 1) design. We represent each step as Step m or Step m -n . The number m represents the step given in the algorithm. In the algorithm, we go back to previous steps and repeat them, which we refer as loop. The number n represents the loop count, which shows the number of repetition.

  1. Step 1

    We choose one 2-(7, 3, 7) design from Database 1 as the left projection. In this example, we use the one specific 2-(7, 3, 7) design \(\tilde{Y}^{(L)}=({y_1^{(L)},y_2^{(L)},\ldots , y_{49}^{(L)}})\) which is given in Eq. (6.1) as (ordered) left projection.

  2. Step 2

    Set \(\mathcal {S}_1=\{c=(\emptyset , \cdots , \emptyset )\}\) where c is a sequence of length 49 and \(\mathcal {S}_2=\emptyset \).

  3. Step 3

    We choose two points from \(V_1\). In this example, we choose here first \(\{1,2\}\subset V_1\). We obtain \(\mathrm {Ind}(1,2)=\{1,2,3,4,5,6,7\}\).

  4. Step 4

    We pick one element from \(\mathcal {S}_1\) and obtain the partial sequence of it. When we do this step for the first time, we have only one element \(c \in \mathcal {S}_1\) to choose, since \(\mathcal {S}_1\) is a single set. Therefore \(\rho (1,2;c)=(\emptyset , \emptyset , \emptyset , \emptyset , \emptyset , \emptyset , \emptyset )\) in this example.

  5. Step 5

    We want to find a 2-(7, 3, 1) design that matches with the partial sequence \(\rho (1,2;c)\) where it is not empty set \(\emptyset \). However, in the first loop, all the entries of the partial sequence are empty sets. Thus we can take any 2-(7, 3, 1) design. In this example, we take \(\mathcal D^{(1)}=(B_2,B_3,B_4,B_6,B_7,B_{12},B_{15})\). We define a new sequence \(c^{(1)}=(c_1^{(1)},c_2^{(1)},\ldots ,c_{49}^{(1)})\) as

    $$\begin{aligned} c^{(1)}_{i_j}=\left\{ \begin{array}{ll} \left( \mathcal D^{(1)} \right) _j &{}\text {if}~ i_j \in \mathrm {Ind}(1,2),\\ c_{i_j} ( = \emptyset ) &{}\text {otherwise.} \end{array}\right. \end{aligned}$$

    Equivalently, \(c^{(1)}=(B_2,B_3,B_4,B_6,B_7,B_{12},B_{15},\emptyset ,\cdots ,\emptyset )\). We add this element to \(\mathcal {S}_2\). Thus, here we set \(\mathcal S_2=\{c^{(1)}\}\).

  6. Step 6

    We do the previous step for all 2-(7, 3, 1) designs in Database 2. Remark: Since here we just give an example instead of exhausting all designs in Database 2, we stop this step with \(c^{(1)}\).

  7. Step 7

    Since we have only one element in \({\mathcal S}_1\), we have nothing to do and we go to next step.

  8. Step 8

    Replace all the elements of \(\mathcal S_1\) by that of \({\mathcal S}_2\) and redefine \(\mathcal S_2\) as empty set. In this example, we set \(\mathcal S_1=\{c^{(1)}\}\) and \(\mathcal S_2=\emptyset \) in this step.

  9. Step 9

    We repeat the algorithm from Step 3 for all choices of two points from set \(V_1\).

We start with the second loop:

  1. Step 3-2

    Choose other two pairs \(\{1,3\}, \{2,3\}\subset V_1\). We can check that \(\mathrm {Ind}(1,3)=\mathrm {Ind}(2,3)=\{1,2,3,4,5,6,7\}\).

Remark: Since \(\mathrm {Ind}(1,2)=\mathrm {Ind}(1,3)=\mathrm {Ind}(2,3)\), there are no changes in \(\mathcal S_1\) and \(\mathcal S_2\).

We continue with the third loop:

  1. Step 3-3

    Choose two points \(\{1,4\}\subset V_1\) and obtain \(\mathrm {Ind}(1,4)=\{8,9,10,11,12,13,14\}\).

  2. Step 4-3

    In this step, we have again only one element \(c^{(1)} \in \mathcal S_1\) to consider.

  3. Step 5-3

    Since we obtain again \(\rho (1,4;c^{(1)})=(\emptyset , \emptyset , \emptyset , \emptyset , \emptyset , \emptyset , \emptyset )\), we can choose any 2-(7, 3, 1) design from Database 2. In this example, we choose a 2-(7, 3, 1) design \(\mathcal D^{(2)}=(B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13})\). Since all the components of \(\rho (1,4;c^{(1)})\) are empty sets, we define \(c^{(2)}\) as follows

    $$\begin{aligned} c^{(2)}=(B_2,B_3,B_4,B_6,B_7,B_{12},B_{15},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},\emptyset ,\cdots ,\emptyset ). \end{aligned}$$
  4. Step 6-3

    Since we are showing here an example, we do not exhaust the Database 2 and do not repeat Step 5.

  5. Step 7-3

    We have only one element in \(\mathcal {S}_1\), so we go to the next step.

  6. Step 8-3

    Redefine \(\mathcal S_1, \mathcal S_2\) in the same way, i.e., \(\mathcal S_1=\{c^{(2)}\}\) and \(\mathcal S_2=\emptyset \).

We repeat from Step 3 to Step 8 for all choices of two points from \(V_1\). In the following, we list the pairs we choose, the designs from Database 2 we use, the resulting \(c^{(\ell )}\), \({\mathcal S}_1\) and \({\mathcal S}_2\), where \(\ell \) corresponds to the number of loop.

Remark: If we choose pairs \(\{1,5\}, \{4,5\}\subset V_1\), then we can check that \(\mathrm {Ind}(1,4)=\mathrm {Ind}(1,5)=\mathrm {Ind}(4,5)\). Thus \(\mathcal S_1\) and \(\mathcal S_2\) will not be changed.

  1. Step 3-4

    We choose \(\{1,6\}\subset V_1\). Then we get

    • \(\mathrm {Ind}(1,6)=\{15,16,17,18,19,20,21\}\).

    • \(\rho (1,6;c^{(2)})=(\emptyset , \emptyset , \emptyset , \emptyset , \emptyset , \emptyset , \emptyset )\).

    • \(\mathcal D^{(2)}=(B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13})\).

    • \(c^{(3)}=(B_2,B_3,B_4,B_6,B_7,B_{12},B_{15},B_1,B_4,B_5,B_7,B_{10},\)\(B_{12},B_{13},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},\emptyset ,\cdots ,\emptyset )\).

    • \(\mathcal S_1=\{c^{(3)}\}\) and \(\mathcal S_2=\emptyset \).

Remark: For pairs \(\{1,7\}, \{6,7\}\subset V_1\), we can check that \(\mathrm {Ind}(1,6)=\mathrm {Ind}(1,7)=\mathrm {Ind}(6,7)\). Thus these two paris will not change \(\mathcal S_1\) and \(\mathcal S_2\).

  1. Step 3-5

    We choose \(\{2,4\}\subset V_1\).

    • \(\mathrm {Ind}(2,4)=\{22,23,24,25,26,27,28\}\).

    • \(\rho (2,4;c^{(3)})=(\emptyset , \emptyset , \emptyset , \emptyset , \emptyset , \emptyset , \emptyset )\).

    • \(\mathcal D^{(2)}=(B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13})\).

    • \(c^{(4)}= (B_2,B_3,B_4,B_6,B_7,B_{12},B_{15},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},B_1,B_4,B_5, B_7,\)\(B_{10},B_{12},B_{13},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},\emptyset ,\cdots ,\emptyset ). \)

    • \(\mathcal S_1=\{c^{(4)}\}\) and \(\mathcal S_2=\emptyset \).

  2. Step 3-6

    We choose \(\{2,7\}\subset V_1\). Then we get

    • \(\mathrm {Ind}(2,7)=\{28,30,31,32,33,34,35\}\).

    • \(\rho (2,7;c^{(4)})=(B_{13},\emptyset , \emptyset , \emptyset , \emptyset , \emptyset , \emptyset )\).

    • Since the first component of \(\rho (2,7;c^{(4)})\) is \(B_{13}\) which is determined in the previous step, we have to choose a design whose first block is \(B_{13}\). Thus we take \(\mathcal D^{(3)}=(B_{13},B_1,B_4,B_5,B_7,B_{10},B_{12})\).

    • \(c^{(5)}= (B_2,B_3,B_4,B_6,B_7,B_{12},B_{15},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},B_1,B_4,B_5,B_7,\)\(B_{10},B_{12},B_{13}, B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},\emptyset ,B_1,B_4,B_5,B_7,B_{10},B_{12},\emptyset ,\cdots ,\)\(\emptyset )\),

    • \(\mathcal S_1=\{c^{(5)}\}\) and \(\mathcal S_2=\emptyset \).

  3. Step 3-7

    We choose \(\{2,5\}\subset V_1\). Then we get

    • \(\mathrm {Ind}(2,5)=\{29,30,31,32,33,34,35\}\).

    • \(\rho (2,5;c^{(5)})=(\emptyset ,B_1,B_4,B_5,B_7,B_{10},B_{12})\).

    • \(\mathcal D^{(3)}=(B_{13},B_1,B_4,B_5,B_7,B_{10},B_{12})\).

    • \(c^{(6)}= (B_2,B_3,B_4,B_6,B_7,B_{12},B_{15},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},B_1,B_4,B_5,B_7,\)\(B_{10},B_{12},B_{13}, B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},B_{13},B_1,B_4,B_5,B_7,B_{10},B_{12},\emptyset ,\cdots ,\)\(\emptyset )\).

    • \(\mathcal S_1=\{c^{(6)}\}\) and \(\mathcal S_2=\emptyset \).

  4. Step 3-8

    We choose \(\{3, 4\}\subset V_1\). Then we get

    • \(\mathrm {Ind}(3,4)=\{36,37,38,39,40,41,42\}\).

    • \(\rho (3,4;c^{(6)})=(\emptyset ,\emptyset , \emptyset , \emptyset , \emptyset , \emptyset , \emptyset )\).

    • \(\mathcal D^{(3)}=(B_{13},B_1,B_4,B_5,B_7,B_{10},B_{12})\).

    • \(c^{(7)}= (B_2,B_3,B_4,B_6,B_7,B_{12},B_{15},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},B_1,B_4,B_5,B_7,\)\(B_{10},B_{12},B_{13},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},B_{13},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},B_1,\)\(B_4,B_5,B_7,B_{10},B_{12},\emptyset ,\emptyset ,\emptyset ,\emptyset ,\emptyset ,\emptyset ,\emptyset )\).

    • \(\mathcal S_1=\{c^{(7)}\}\) and \(\mathcal S_2=\emptyset \).

  5. Step 3-9

    We choose \(\{3, 7\}\subset V_1\).

    • \(\mathrm {Ind}(3,7)=\{37,38,39,40,41,42,49\}\).

    • \(\rho (3,7;c^{(7)})=(B_1,B_4,B_5,B_7,B_{10},B_{12},\emptyset )\).

    • \(\mathcal D^{(2)}=(B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13})\).

    • \(c^{(8)}= (B_2,B_3,B_4,B_6,B_7,B_{12},B_{15},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},B_1,B_4,B_5,B_7,\)\(B_{10},B_{12},B_{13},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},B_{13},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},B_1,\)\(B_4,B_5,B_7,B_{10},B_{12},\emptyset ,\emptyset ,\emptyset ,\emptyset ,\emptyset ,\emptyset ,B_{13})\),

    • \(\mathcal S_1=\{c^{(8)}\}\) and \(\mathcal S_2=\emptyset \).

  6. Step 3-10

    We choose \(\{3, 5\}\subset V_1\).

    • \(\mathrm {Ind}(3,5)=\{43,44,45,46,47,48,49\}\).

    • \(\rho (3,5;c^{(8)})=(\emptyset ,\emptyset ,\emptyset ,\emptyset ,\emptyset ,\emptyset ,B_{13})\).

    • \(\mathcal D^{(2)}=(B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13})\).

    • \(c^{(9)}= (B_2,B_3,B_4,B_6,B_7,B_{12},B_{15},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},B_1,B_4,B_5,B_7,\)\(B_{10},B_{12},B_{13},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},B_{13},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13},B_1,\)\(B_4,B_5,B_7,B_{10},B_{12},B_1,B_4,B_5,B_7,B_{10},B_{12},B_{13})\),

    • \(\mathcal S_1=\{c^{(9)}\}\) and \(\mathcal S_2=\emptyset \).

Therefore, the right projection is \(\tilde{Y}^{(R)}=c^{(9)}\). The tight 2-(7, 3, 7, 3, 1) design Y is obtained by

$$\begin{aligned} Y&= \left\{ \left( y^{(L)}_i, y^{(R)}_i \right) | 1 \le i \le 49 \right\} \end{aligned}$$

where, \(y^{(L)}_i\) and \(y^{(R)}_i\) are the i-th element of \(\tilde{Y}^{(L)}\) and \(\tilde{Y}^{(R)}\), respectively. And we can check that the distance set is

$$\begin{aligned} A(Y)=\{(3,2),(2,3),(2,2),(2,1),(1,2),(2,0),(0,2)\} \end{aligned}$$

in this case.

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Zhu, Y., Watamura, N. Relative t-designs in Johnson association schemes for P-polynomial structure. Des. Codes Cryptogr. 88, 2101–2118 (2020).

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  • Relative t-design
  • Johnson association scheme
  • P-polynomial association scheme
  • Tight design
  • Product association scheme

Mathematics Subject Classification

  • 05B05
  • 05B30
  • 05E30