The first research level math problem I seriously worked on (without much success) was the sum-free subset problem. A sum-free set (in an additive group) is a set that contains no solution to the equation . The sum-free subset problem concerns the following theorem of Erdős, which is one of the early applications of the probabilistic method.
Theorem: (Erdős, 1965) Let A be a finite set of natural numbers. There exists a sum-free subset such that .
The natural problem here is to determine how much one can improve on this result. Given the simplicity of the proof, it seems that one should be able to better. However, the best result to date is (for ) due to Bourgain (1995) using Fourier analysis. It seems likely that there is a function such that one may take in the above theorem, however proving this seems to be quite challenging. On the other hand proving a good upper bound also was a longstanding open problem, namely deciding if the constant could be replaced by a larger constant. This problem, in fact, has received a fair amount of attention over the years. In his 1965 paper Erdos states that Hinton proved that the constant is at most and that Klarner improved this to . In 1990 Alon and Kleitman improved this further to , and Malouf in her thesis (as well as Furedi) improved this further to . A couple years ago I improved this to (I recently learned that Erdős claimed the same bound without a proof in a handwritten letter from 1992). More recently, Alon improved this to (according to Eberhard, Green and Manners, Alon’s paper doesn’t appear to be electronically available. UPDATE: Alon’s paper can be found here). All of these results (with the exception of Alon’s) proceed simply be constructing a set and proving its largest sum-free subset is at most a certain size (then an elementary argument can be used to construct arbitrarily large sets with the same constant: see this blog post of Eberhard). On the other hand, it is obvious from the statement of the theorem above that one can’t have a set whose largest sum-free subset is exactly , thus a simple proof by example is hopeless. The futile industry of constructing examples aside, I spent a lot of effort thinking about this problem and never got very far.
About a month ago Sean Eberhard, Ben Green and Freddie Manners solved this problem, showing that is indeed the optimal constant! I don’t yet fully understand their proof, and will refer the reader to their introduction for a summary of their ideas. However, having learned the problem’s difficulty firsthand, I am very pleased to see it solved.
In the wake of this breakthrough on the upper bound problem, I thought I’d take this opportunity to highlight a toy case of the lower bound problem.
Definition: For a natural number define the sum-free subset number Sum-free to be the largest number such that every set of natural numbers is guaranteed to contain a sum-free subset of size at least Sum-free.
Thus by Bourgain’s theorem we have that Sum-free (and the Eberhard-Green-Manners theorem states Sum-free). It turns out that Bourgain’s result is sharp for sets of size and smaller. However, Bourgain’s theorem only tells us that a set of size must contain a sum-free subset of size while it seems likely (from computer calculations) that every set of size must in fact contain a sum-free subset of size . Indeed, I’ll even
While the asymptotic lower bound problem (showing that there ) seems very hard and is connected with subtle questions in harmonic analysis (such as the now solved Littewood conjecture on exponential sums), this problem is probably amenable to elementary combinatorics (although, it certainly does not seem easy!). Below is a chart showing what I know about the first sum-free subset numbers.
|Bourgain’s lower bound
|Best known upper
bound on Sum-free
|Upper bound set example|
I recently arXiv’ed a short note titled An Improved Upper Bound for the Sum-free Subset Constant. In this post I will briefly describe the result.
We say a set of natural numbers is sum-free if there is no solution to the equation with . The following is a well-known theorem of Erdős.
Theorem Let be a finite set of natural numbers. There exists a sum-free subset such that .
The proof of this theorem is a common example of the probabilistic method and appears in many textbooks. Alon and Kleitman have observed that Erdős’ argument essentially gives the theorem with the slightly stronger conclusion . Bourgain has improved this further, showing that the conclusion can be strengthened to . Bourgain’s estimate is sharp for small sets, and improving it for larger sets seems to be a difficult problem. There has also been interest in establishing upper bounds for the problem. It seems likely that the constant cannot be replaced by a larger constant, however this is an open problem. In Erdős’ 1965 paper, he showed that the constant could not be replaced by a number greater than by considering the set . In 1990, Alon and Kleitman improved this to . In a recent survey of open problems in combinatorics, it is reported that Malouf has shown the constant cannot be greater than . While we have not seen Malouf’s proof, we note that this can be established by considering the set . In this note we further improve on these results by showing that the optimal constant cannot be greater than .
Update (May 2, 2010): After posting this preprint, Stefan Neuwirth informed us that Rudin’s question had been previously answered by Y. Meyers in 1968. It appears that Meyers’ construction doesn’t, however, say anything about the anti-Freiman problem. Indeed Meyers’ set (and all of its subsets) contains a set of density . Hence, the construction of a set that doesn’t contain a large set still appears to be new. A revised version of the paper has been posted reflecting this information. Most notably, we have changed the title to “On the Structure of Sets of Large Doubling”.
Allison Lewko and I recently arXiv’ed our paper “Sets of Large Doubling and a Question of Rudin“. The paper (1) answers a question of Rudin regarding the structure of sets (2) negatively answers a question of O’Bryant about the existence of a certain “anti-Freiman” theorem (3) establishes a variant of the (solved) Erdös-Newman conjecture. I’ll briefly describe each of these results below.
— Structure of sets —
Before describing the problem we will need some notation. Let and define to be the number of unordered solutions to the equation with . We say that is a set if for all . There is a similar concept with sums replaced by differences. Since this concept is harder to describe we will only introduce it in the case . For we define to be the number of solutions to the equation with . If for all nonzero we say that is a set.
Let be a subset of the integers , and call an -polynomial if it is a trigonometric polynomial whose Fourier coefficients are supported on (i.e. if ). We say that is a set (for ) if
holds for all -polynomials where the constant only depends on and . If is an even integer, we can expand out the norm in 1. This quickly leads to the following observation: If is a set then is also an set (, ). One can also easily show using the triangle inequality that the union of two sets is also a set. It follows that the finite union of sets is a set. In 1960 Rudin asked the following natural question: Is every set is a finite union of sets?
In this paper we show that the answer is no in the case of sets. In fact, we show a bit more than this. One can easily show that a set is also a set. Our first counterexample to Rudin’s question proceeded (essentially) by constructing a set which wasn’t the finite union of sets. This however raised the following variant of Rudin’s question: Is every set the mixed finite union of and sets? We show that the answer to this question is no as well. To do this we construct a set, A, which isn’t a finite union of sets, and a set, , which isn’t the finite union of sets. We then consider the product set which one can prove is a subset of . It isn’t hard to deduce from this that is a subset of that isn’t a mixed finite union of and sets. Moreover, one can (essentially) map this example back to while preserving all of the properties stated above. Generalizing this further, we show that there exists a set that doesn’t contain (in a sense that can be made precise) a large or . This should be compared with a related theorem of Pisier which states that every Sidon set contains a large independent set (it is conjectured that a Sidon set is a finite union of independent sets, however this is open).
We have been unable to extend these results to sets for . Very generally, part of the issue arises from the fact that the current constructions hinges on the existence of arbitrary large binary codes which can correct strictly more than a fraction of errors. To modify this construction (at least in a direct manner) to address the problem for, say, sets it appears one would need arbitrary large binary codes that can correct strictly more than a fraction of errors. However, one can show that such objects do not exist.
— Is there an anti-Freiman theorem? —
Let be a finite set of integers and denote the sumset of as . A trivial inequality is the following
In fact, it isn’t hard to show that equality only occurs on the left if is an arithmetic progression and only occurs on the right if is a set. A celebrated theorem of Freiman states that if then is approximately an arithmetic progression. More precisely, if is a finite set satisfying for some constant , then is contained in a generalized arithmetic progression of dimension and size where and depend only on and not on .
It is natural to ask about the opposite extreme: if , what can one say about the structure of as a function only of ? A first attempt might be to guess that if for some positive constant , then can be decomposed into a union of sets where and depend only on . This is easily shown to be false. For example, one can start with a of elements contained in the interval and take its union with the arithmetic progression . It is easy to see that regardless of . However, the interval cannot be decomposed as the union of sets with and independent of .
There are two ways one might try to fix this problem: first, we might ask only that contains a set of size , where and depend only on . (This formulation was posed as an open problem by O’Bryant here). Second, we might ask that hold for all subsets for the same value of . Either of these changes would rule out the trivial counterexample given above. In this paper we show that even applying both of these modifications simultaneously is not enough to make the statement true. We provide a sequence of sets where holds for all of their subsets for the same value of , but if we try to locate a set, , of density in then must tend to infinity with the size of . As above, our initial construction of such a sequence of ‘s turned out to be sets. This leads us to the even weaker anti-Freiman conjecture:
(Weak Anti-Freiman) Suppose that satisfies and for all subsets . Then contains either a set or a set of size , where and depend only on .
We conclude by showing that even this weaker conjecture fails. The constructions are the same as those used in the results above. The two problems are connected by the elementary observation that if is a subset of a set then holds where only depends on the constant of the set .
— A variant of the Erdös-Newman conjecture —
In the early 1980’s Erdös and Newman independently made the following conjecture: For every there exists a that isn’t a finite union of sets for any . This conjecture was later confirmed by Erdös for certain values of using Ramsey theory, and finally resolved completely by Nešetřil and Rödl using Ramsey graphs. One further application of our technique is the following theorem which can be viewed as an analog of the Erdös-Newman problem with the roles of the union size and reversed.
Theorem 1 For every there exists a union of sets that isn’t a finite union of sets for any .
In this post I will present one of my favorite logic puzzles. As with many great logic puzzles (see the blue-eyed islanders puzzle, for example), the mathematics is embedded in a gruesome storyline. The puzzle can be stated as follows:
There are 100 prisoners. The warden places 100 wooden boxes in a row in an empty room. He writes the name of each prisoner on 100 separate slips of paper and places one of these slips into each of the 100 boxes, in a manner of his choosing.
Each prisoner is lead into the room, one at a time. The prisoner is allowed to open up to 50 boxes. If the prisoner does not find his own name, all prisoners are executed. If a prisoner finds his own name, the prisoner must reset the boxes exactly as he found them. The process then continues onto the next prisoner. If all 100 prisoners find their own name, all of the prisoners will be released the next morning.
The prisoners are allowed to agree on a strategy before this process starts, however they may not communicate in any way after the process has started. What strategy should the prisoners use to minimize the chance of execution?
Notice that if each of the prisoners open 50 randomly selected boxes, then each prisoner will find his own name with probability . Thus the prisoners will escape execution with probability
This is roughly 1 divided by the number of grams of matter composing the Earth. Below the fold we present a strategy that succeeds with probability about .307 (highlight to see probability). (more…)