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$S_5$ is a set consisting of the following 5-length sequences $s$: (1) each digit of $s$ is $a$, $b$, or $c$; (2) $s$ has and only has one digit that is $c$.

$T_5$ is a set consisting of the following 5-length sequences $t$: (1) each digit of $t$ is $a$, $b$, or $c$; (2) $t$ has two digits that are $c$.

Is there a 3-partition $S_5= \bigcup_{j=0,1,2} A_j$, $A_j \cap A_i=\varnothing$ with the following property: for any $A_j$ and any $t$ $\in$ $T_5$, there is a $s$ $\in$ $A_j$ such that exsits a $n\in\{1,2,3,4,5\}$, $s_n \neq t_n$, $t_n=c$ and $s_m=t_m $ for any $m\neq n$, where $s_n$ (or $t_n$) is the n-th digit of $s$ (or $t$).

For example, $t=ccaab$ and $s=acaab$ where $n=1$.

And is there a 3-partition of $S_6$ with $T_6$ for 6-length sequences?

Thank you!

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Here is one such partition of $S_5$, obtained via integer linear programming: \begin{align} A_0 &= \{aaaac,aaacb,aabca,aacba,ababc,abbac,abcab,abcba,acaaa,acbbb,baabc, babac,bacaa,bacbb,bbaca,bbbbc,bbbcb,bbcaa,bcaab,bcbba,caaba,cabab,cabbb, cbaab,cbabb,cbbaa\} \\ A_1 &= \{aaaca,aabbc,aacab,abaac,abbca,abbcb,abcbb,acaba,acabb,acbaa,acbab, baaac,babca,babcb,bacba,bbabc,bbacb,bbbac,bbcab,bbcba,bcaaa,bcbbb,caaab, caabb,cabaa,cbaaa,cbbba\} \\ A_2 &= \{aaabc,aabac,aabcb,aacaa,aacbb,abaca,abacb,abbbc,abcaa,acaab,acbba, baaca,baacb,babbc,bacab,bbaac,bbbca,bbcbb,bcaba,bcabb,bcbaa,bcbab,caaaa, cabba,cbaba,cbbab,cbbbb\} \end{align}

And here is one for $S_6$: \begin{align} A_0 &= \{aaaaac,aaabca,aaacbb,aababc,aabacb,aabbac,aabcba,aacaaa,aacbab,aacbba, abaaac,abaacb,ababbc,abacaa,abbabc,abbbca,abbcab,abcaba,abcbaa,abcbbb,acaaba ,acaabb,acabab,acbaaa,acbaab,acbbbb,baaabc,baabcb,baacaa,babaac,babaca, babbbc,babcab,bacabb,bacbab,bacbba,bbaabc,bbaaca,bbabac,bbacbb,bbbaac,bbbbcb ,bbbcba,bbcaab,bbcbaa,bbcbbb,bcaaaa,bcaaab,bcabba,bcbaba,bcbabb,bcbbaa, caaaab,caaaba,caabaa,caabbb,cabbaa,cabbbb,cbabab,cbabba,cbbaaa,cbbabb,cbbbab ,cbbbba\} \\ A_1 &= \{aaaaca,aaabbc,aaacab,aabaac,aabbbc,aabbcb,aabcaa,aacaba,aacabb,aacbaa, abaabc,ababac,ababcb,abacba,abbaca,abbbac,abbcbb,abcaaa,abcaab,abcbba,acaaab ,acabaa,acabbb,acbabb,acbbab,acbbba,baaacb,baabac,baacba,bababc,babbac, babbca,babcbb,bacaaa,bacaab,bacbbb,bbaaac,bbabbc,bbabca,bbacab,bbbacb,bbbbbc ,bbbcaa,bbcaba,bbcabb,bbcbab,bcaaba,bcabaa,bcabbb,bcbaaa,bcbbab,bcbbba, caaaaa,caaabb,caabab,caabba,cabaab,cababa,cbaaaa,cbaabb,cbbaab,cbbaba,cbbbaa ,cbbbbb\} \\ A_2 &= \{aaaabc,aaaacb,aaabac,aaabcb,aaacaa,aaacba,aabaca,aabbca,aabcab,aabcbb, aacaab,aacbbb,abaaca,ababca,abacab,abacbb,abbaac,abbacb,abbbbc,abbbcb,abbcaa ,abbcba,abcabb,abcbab,acaaaa,acabba,acbaba,acbbaa,baaaac,baaaca,baabbc, baabca,baacab,baacbb,babacb,babbcb,babcaa,babcba,bacaba,bacbaa,bbaacb,bbabcb ,bbacaa,bbacba,bbbabc,bbbaca,bbbbac,bbbbca,bbbcab,bbbcbb,bbcaaa,bbcbba, bcaabb,bcabab,bcbaab,bcbbbb,cabaaa,cababb,cabbab,cabbba,cbaaab,cbaaba,cbabaa ,cbabbb\} \end{align}

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  • $\begingroup$ Thank you! I focused on the problem recently and used lingo to give an integer linear programming when I consider the 3-partition of 𝑆7 with 𝑇7 for 7-length sequences. However, the program ran for a long time without any results. Can you give the situation or suggestion at 7 so that I can try to give a general conclusion? $\endgroup$
    – 4869
    Commented Nov 24, 2022 at 15:11
  • $\begingroup$ Are you sure it is possible? One idea is to introduce a binary variable for each $(t,p)$ to indicate whether $t$ is uncovered in part $p$ and then minimize the sum of these binary variables. So far, the smallest sum I can find is $56$. $\endgroup$
    – RobPratt
    Commented Nov 24, 2022 at 15:39

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