Spectrum of Sizes for Perfect Deletion-Correcting Codes

One peculiarity with deletion-correcting codes is that perfect $t$-deletion-correcting codes of the same length over the same alphabet can have different numbers of codewords, because the balls of radius $t$ with respect to the Levenshte\u{\i}n distance may be of different sizes. There is interest, therefore, in determining all possible sizes of a perfect $t$-deletion-correcting code, given the length $n$ and the alphabet size~$q$. In this paper, we determine completely the spectrum of possible sizes for perfect $q$-ary 1-deletion-correcting codes of length three for all $q$, and perfect $q$-ary 2-deletion-correcting codes of length four for almost all $q$, leaving only a small finite number of cases in doubt.Comment: 23 page

Designs for graphs with six vertices and nine edges

The design spectrum has been determined for eleven of the 21 graphs with six vertices and nine edges. In this paper we completely solve the design spectrum problem for the remaining ten graphs

On the minisymposium problem

The generalized Oberwolfach problem asks for a factorization of the complete graph $K_v$ into prescribed $2$-factors and at most a $1$-factor. When all $2$-factors are pairwise isomorphic and $v$ is odd, we have the classic Oberwolfach problem, which was originally stated as a seating problem: given $v$ attendees at a conference with $t$ circular tables such that the $i$th table seats $a_i$ people and ${\sum_{i=1}^t a_i = v}$, find a seating arrangement over the $\frac{v-1}{2}$ days of the conference, so that every person sits next to each other person exactly once. In this paper we introduce the related {\em minisymposium problem}, which requires a solution to the generalized Oberwolfach problem on $v$ vertices that contains a subsystem on $m$ vertices. That is, the decomposition restricted to the required $m$ vertices is a solution to the generalized Oberwolfach problem on $m$ vertices. In the seating context above, the larger conference contains a minisymposium of $m$ participants, and we also require that pairs of these $m$ participants be seated next to each other for $\left\lfloor\frac{m-1}{2}\right\rfloor$ of the days. When the cycles are as long as possible, i.e.\ $v$, $m$ and $v-m$, a flexible method of Hilton and Johnson provides a solution. We use this result to provide further solutions when $v \equiv m \equiv 2 \pmod 4$ and all cycle lengths are even. In addition, we provide extensive results in the case where all cycle lengths are equal to $k$, solving all cases when $m\mid v$, except possibly when $k$ is odd and $v$ is even.Comment: 25 page