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A set of integers ${\displaystyle A=\{a_{1},a_{2},...,a_{m}\}}$ where ${\displaystyle a_{1}<a_{2}<...<a_{m}}$ is a Golomb ruler if and only if $$ \text{for all } i,j,k,l \in \{1,2,...,m\} {\text{ such that }} i\neq j {\text{ and }} k\neq l, a_{i}-a_{j}=a_{k}-a_{l}\iff i=k{\text{ and }}j=l. $$

A modular Goloumb ruler works the same, not on a line but on a circle of length $q$. I wonder how these rulers are connected to sets $B=\{b_1,b_2,...,b_m\}$, where $b_1<b_2<...<b_m$ and the sum $b^\star=\sum_{k=1}^m b_k \bmod q$ is unique, in the sense, that every other sum of $m$ elements of $B$ (e.g. $b^{\text{other}_1}=2b_1+b_3+...b_m \bmod q$), has a value other than $b^\star$. So I don't bother if $b^{\text{other}_k}=b^{\text{other}_l}$, only $b^\star$ shall be unique.

Can Goloumb Ruler help to generate the sets $B$ as well?

They were for example given in this answer, to a question that looks similar to my question...

draks ...
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  • The sets from the linked answer are related to Golomb rulers, but they differ from sets with unique sums. The sequences to sum consist of $m$ elements for the latter, but of consecutive elements for the former. – Alex Ravsky Jul 06 '20 at 03:09

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Each set $B$ with unique sums $\mod q$ is a Golomb ruler $\mod q$. Indeed, if there exists $i,j,k,l \in \{1,2,...,m\}$ such that $i\neq j$ and $k\neq l$ and ($i\ne k$) or $j\ne l$) but $b_{i}-b_{j}=b_{k}-b_{l} \pmod q $ then $b^*=b^*+b_i+b_l-b_j-b_k \pmod q$, a contradiction.

The condition for $B$ to be a set with unique sums ($\mod q$) is much stronger than to be a (modular) Golomb ruler and the former sets are much sparse than the latter.

Indeed, if $B=\{b_1,\dots, b_m\}$ is a set with unique sums $\mod q$ and $m’=\lfloor m/2\rfloor$ then all ${m\choose m’}$ sums of subsets of $B$ of size $m’$ are distinct, which follows ${m\choose m’}\le q$. By Robbins’ bounds for each $m\ge 2$ we have $$\frac {2^{m-1/2}}{\sqrt{\pi (m+\delta)}}\exp\left(-\frac {1}{4(m+\delta)-1}\right)<{m\choose m’}<\frac {2^{m-1/2}}{\sqrt{\pi (m+\delta)}}\exp\left(-\frac {1}{4(m+\delta)+1}\right),$$ where $\delta$ equals $0$, if $m$ is even and $1$, is $m$ is odd.

On the other hand, the constructions at the page you linked suggest that at least for some $q$ there exist modular Golomb rulers of size close to $\sqrt{q}$.

Alex Ravsky
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    +1, thanks, are you saying that, when I ask, for $\b^\star$ being unique, among all other sums with the same number of terms (dupes allowed), I get a exponentially large set of unique sums of $B$ for free? Concerning the other hand: Would you suggest to just scan Singer/Bose/Ruzka constructions for my additional condition? – draks ... Jul 06 '20 at 06:49
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    @draks... Yes for the first question. For the other hand, I don’t see how modular Golomb rules can help to construct a set with unique sums $\mod q$. Or what additional condition do you mean? PS. The upper bound above for a size of a set with unique sums $\mod q$ is rather tight, because if $m\ge 2$ and $(m-2)2^{m-1}+1<q$ then a set ${2^0,2^1,\dots, 2^{m-1}}$ has unique sums $\mod q$. – Alex Ravsky Jul 06 '20 at 08:01
  • Ok, but what then do mean by "the page you linked suggests that at least for some $q$ there exist modular Golomb rulers of size close to $\sqrt{q}$."? – draks ... Jul 06 '20 at 08:13
  • @draks... I guess the lenght of the modular Golomb rulers for the constructions mentioned there is about $q$. I tried to check this looking at “James B. Shearer's Golomb Ruler page with a piece on modular Golomb Rulers” linked there, but the page doesn’t exist. – Alex Ravsky Jul 06 '20 at 09:14
  • Hi, I found a cached version of the linked page at https://web.archive.org/web/20171225101048/http://www.research.ibm.com/people/s/shearer/grule.html but it is not very helpful in my opinion... – draks ... Jul 08 '20 at 07:28
  • @draks... OK. But each Golomb ruler of length $L$ is also a Golomb ruler $\mod q$ for each $q>2L$. Therefore for each natural $n$ and each $q>2G(n)$, where $G(n)$ is a shortest lenght of a Golomb ruler with $m$ marks. In particular, Erdős–Turán construction shows that $G(p)\le 2p^2-2p+1$ for each odd prime $p$. – Alex Ravsky Jul 08 '20 at 17:38
  • The respective bounds can be adapted for non-prime $n$ using the approximations from this my question. – Alex Ravsky Jul 08 '20 at 17:38
  • @draks... I fixed a constant in the approximation of ${m\choose m’}$. Sorry for the error. – Alex Ravsky Dec 02 '20 at 13:01