4
$\begingroup$

Sidon sets are sets $A \subset \mathbb{N}$ such that for all $a_j,b_j \in A$ holds $$a_1+a_2=b_1+b_2 \iff \{a_1,a_2\}=\{b_1,b_2\}.$$ Thus if you know the sum of two elements, you know which elements were added in the first place.

The general case is known in literature as $B_k[g]$ sets. Every result of adding up at most $k$ elements will occur at most $g$ times in all possible combinations. Sidon sets are therefore called $B_2[1]$ sets.

An example of a $B_3[1]$ set is $\{1,7,11\}$ with $k = 3$, $n = \lvert B_k[1]\rvert=3$ elements, $m = \max B_k[1] = 11$ and ${n+k \choose k}={6 \choose 3}=20$ distinct sums:

0=0, 1=1, 1+1=2, 1+1+1=3, 7=7, 7+1=8, 7+1+1=9, 11=11, 11+1=12, 11+1+1=13, 7+7=14, 7+7+1=15, 11+7=18, 11+7+1=19, 7+7+7=21, 11+11=22, 11+11+1=23, 7+7+11=25, 11+11+7=29, 11+11+11=33.

I didn't find a $B_3[1]$ set with $n=3$ and a smaller $m$ than 11. However $\{1, 8, 11\}$ has similar properties.

My Question

Considering $B_k[1]$ sets (only one solution for any sum of at most $k$ elements). What is the smallest $m$ for given $k$ and $n$ and how to construct a set with this property?

If the general case for any given $k$ isn't known, we may set $k=3$.

Note: In a former question ("Set of small numbers with distinct $k$-sums") a construction by Javier Cilleruelo was given. However it doesn't manage to find sets with the smallest $m$ possible.

$\endgroup$
1
  • 1
    $\begingroup$ You might consider choosing a more informative title in order to make more people see this question. $\endgroup$
    – Stefan Kohl
    Oct 16, 2015 at 12:41

1 Answer 1

1
$\begingroup$

Since the question asks for exact smallest values, not bounds or asymptotics, it is likely to be a difficult problem.

First note that instead of taking sums of at most $k$ elements, we can take sums of exactly $k$ elements, if we include a zero element. So the three-element solution $\{1,7,11\}$ corresponds to a four-element solution $\{0,1,7,11\}$ with this interpretation. The result (minimum possible largest element) is not affected by this change, but the number of elements changes by 1.

For $k=3$ this is the OEIS sequence A227358, "Length of shortest Golomb-like (for sums of triples) ruler with n marks". Currently (last updated 2013) the values listed are:

$$ \begin{array}{lllllllll} n& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ \hline m& 0& 1& 4& 11& 23& 45& 82& 129& 208& 309 \end{array} $$

using the "exactly $k$ elements" definition (so the $n$ are off by one with respect to the current question). The OIES entry links to a C program that computes the solutions, answering the "how to construct" part.

For $k=3$ and $n=4=3+1$ the results of the OP are confirmed: There are exactly four minimal solutions with $m=11$, namely:

 0     1     7    11
 0     1     8    11
 0     3    10    11
 0     4    10    11

(The C program lists only the first two, because the other two are obtained by reversal symmetry, i.e. by the mapping $x \mapsto 11-x$.)

See also A227588 for a table where both $k$ and $n$ are varying.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.