I am mainly posting because there’s been little activity and wondering what was going on with everyone. Is this still on?

I took out some of the motivation and put in too much detail.

Now I’m ready to think about the last 3 paragraphs of Tim’s comment on how to get from to . Unfortunately, I think I may need a little more direction. Any hints?

Define and rewrite this into a more balanced form

Let us, just to get some confidence in the result, take our record holder and and compute . has and thus the lhs is Therefore . Since and this yields (which is correct).

Assuming bounded discrepancy one would estimate the rhs. Control over the lhs is crucial to get a contradiction . The idea now is to observe the similarity of the lhs to and use/extend sieves to estimate it. I have no reason to be overly optimistic that this works and even results for discrepancy and special like (to establish f(2)=-1) or (to establish non-existence) would currently be a progress.

First, I dumped labelling the edges and just considered the edges to be from different sets. This allows me to clean up the terminology a little.

I added an acyclic condition because it is otherwise ambigious to find the discrepancy, and added the “paths only” condition because any trees can be simulated with multiple paths.

(General formulation)
Define a set of nodes with k sets of directed edges (for ) such that each set of edges (with corresponding nodes) forms an acyclic path.

Give each node a value of 1 or -1. Consider any traversal corresponding to a set that starts at the root node; define the absolute value of the sum of the values of the nodes visited to be . The largest value of for all possible values of j is the discrepancy.

Possible conditions given a set of nodes :

No subsets (NS): No set of edges is a subset of another set.

Omega set (OS): One labelled set of edges connects to for all i from 1 to n-1.

(Needs OS) Always increasing (AI): For any edge from to , i is less than j.

Isolated roots (IR): Each node can be the root node of only one labelled set of directed edges. [This condition is needed for the definition of completely multiplicative.]

It’d be best for conjectures to use a minimal number of conditions, but I don’t know which ones to choose yet.

Conjecture (I have a rough proof, but it needs checking):
Given only two sets of edges (and none of the extra conditions NS OS AI IR), with optimal placement of 1s and -1s the discrepancy can always be 1.

I picked a value for the length n which made sure that there are many different HAPs which include the value n, thereby restricting the value at n more.

Next I split the problem into subproblems by assigning valeus to the first few values. Each subproblem was then run for a fixed amount of time in an extended SAT-solver called Umsat, the source code for Umsat is available here http://abel.math.umu.se/~klasm/prog/Umsat/
The unsolved subcases were extended to include those forced variable assignment which the program had discovered during the run. Finally each remaining subcase was split into two subcases, by assigning the next unassigned variable.

Umsat, like most other modern SAT-solver uses non-chronological backtrack and clause learning. This means that it does not perform a simple depth first search, since the learning allows it to exclude large parts of the search space without explicitly searching through it. It also does not have a variable ordering which is constant during the search.

I should also mention that Daniel Andrén is working on a complete rewrite of Umsat which shoul make both faster and more flexible.

This computation required a lot more CPU time than I had expected. It has been running on several different machines, in the gaps between other things I am working on. A while back Tim wrote that this would be a “the first rigorous (and very computer-assisted) proof “. I don’t have an exact number right now but an estimate is that “very computer-assisted” translates into about 30 CPU-core years of computer time.

I’m not saying that I think that a homogeneous quasi-progression can be represented as a bounded combination of HAPs, but it might be worth understanding why it can’t. And even if we need too many HAPs to make a HQP, perhaps it can be done in such a way that the surplus HAPs cancel when you produce the representation.

Actually, while writing that I realize it’s got a serious problem to it, which is that presumably the HQP result produces a representation of the identity, and we know we can’t get that with HAPs.

The proof of this makes crucial use of Roth’s theorem on the discrepancy of APs.
The main observation is that any AP on can be realized as a quasi-progression. If the common difference of the AP is , the proof shows one can pick a quasi-progression of difference that coincides with the AP on the interval . By Roth’s theorem, the discrepancy of APs on an interval of size is . QED.

So the proof can be expressed in the language of diagonal representations by appealing to Tim’s diagonal representation version of Roth’s result, but I think Tim was hoping to extract something more interesting than this.

I’m vaguely wondering about a type-6 approach where one starts with a solution in some larger set system and then tweaks the set system until one reaches a system of HAPs.

Alantha Newman, Aleksander Nikolov and I have the following (not yet written up) result: Consider a system of sets on elements.
Then it is NP-hard to distinguish between set systems with discrepancy 0 and those with discrepancy (roughly speaking). Now given such a set system, we can always compute the best lower bound on discrepancy that can be proved by the representation of diagonals approach in time polynomial in (this involves solving an LP). This bound should not be able to distinguish between the discrepancy 0 and discrepancy case. Hence, for some set systems in the class of instances constructed by the NP-hardness proof, the representation of diagonals approach ought to give a bound of 0, while the actual discrepancy is . In fact, this must be true of any polynomial time computable bound (if ).