DeepAI

# Improved bounds on the size of the smallest representation of relation algebra 32_65 with the aid of a SAT solver

In this paper, we shed new light on the spectrum of relation algebra 32_65. We show that 1024 is in the spectrum, and no number smaller than 20 is in the spectrum. In addition, we derive upper and lower bounds on the smallest member of the spectra of an infinite class of algebras derived from 32_65 via splitting.

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08/03/2021

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## 1 Introduction

Relation algebra has atoms , , , and , all symmetric, with all diversity cycles not involving forbidden. The atom is flexible, and has the mandatory cycles required to make flexible and no others.

Relation algebra was shown in [1] to be representable over a finite set, namely a set of 416,714,805,914 points. This was reduced in [4] to 63,432,274,896 points, which was later reduced to 8192 by the first and fifth authors (unpublished), and finally to 3432 in [2]. Here, we give the smallest known representation, over 1024 points.

There are few published lower bounds in the literature. Most can be found in [3], where the spectrum of every relation algebra with three or fewer atoms. Going up to four atoms increases the difficulty considerably. To the best of our knowledge, the only finitely representable symmetric integral RA on four atoms for which the spectrum is known is , and .

No lower bound on the size of representations of has been published. We give a non-trivial such bound for an infinite class of algebras in Section 3.

###### Definition 1.

Let be a finite relation algebra. Then

 Spec(A)={α≤ω:A has a representation over a % set of cardinality α}.

Let denote the integral symmetric relation algebra with atoms , , , …, , where a diversity cycle is mandatory if and only if it involves the atom . (So is .) Let

 f(n)=min(Spec(An)).

It was shown in [1] that is finite for all .

Because representing finite integral relation algebras amounts to edge-coloring complete graphs with the diversity atoms, we will use the language of graph theory. So that we can use colors to make pretty pictures, we will refer to as red, to as light blue, and to as dark blue.

## 2 An upper bound on f(n)

In this section, we give a representation of over 1024 points, and then generalize to give representations of all the s.

Consider , and consider the elements as bitstrings. Define

 R={x∈G:x has between one and six 1s}
 B={x∈G:x has at least seven 1s}

This defines a group representation of , which is a subalgebra of . There exists a way of splitting into and so that:

• , ;

• , ;

• .

This yields a group representation of over points, improving the previous smallest-known representation over points [2]. We note that while the representation given here is smaller, the representation over points in [2] has a nice, compact description.

The split was found in the following way. The first author checked several million random splits. None of them worked, but some got “close”. He took one of the close ones and tinkered with it for about three hours until it worked. The curious can view the process in the Jupyter notebook 32_65 splitting.ipynb at https://github.com/algorithmachine/RA-32-of-65. Also there can be found the happily-named 32_65-VICTORY.txt, which contains the members of encoded as integers between 1 and 1023.

We generalize this argument as follows:

###### Theorem 2.

For all , is representable over for sufficiently large . In particular, for , it suffices to take .

Theorem 2 tells us that is at most exponential in . In contrast, the most you could say from [1] was that was bounded above by (roughly) . See Figure 2.

###### Proof.

We will proceed via a probabilistic argument. Consider , and for , let denote the coordinate of . Let denote . The key idea is the following partition of into two sets and .

Let and

Then , , and . As we will see below, is a sum-free set with high additive energy.

We actually prove a stronger result, namely that we can split both the “red” and “blue” atoms of into atoms and find a representation over a finite set. So we will now split and into parts and uniformly at random. We need to count the “witnesses” to the “needs” of each element. We will show that each need is witnessed at least times.

First we count witnesses for . Let , For , we show that there are at least witnesses to . Let be the support of . We construct randomly so that . For each , flip a coin to choose between

 x(i)=1 and y(i)=0 OR x(i)=0 and y(i)=1

(This is flips.)

For the left-most indices , flip a coin to choose between

 x(j) =1=y(j) OR x(j) =0=y(j)

(This is flips.)

For all remaining indices, let . Thus , and since there were a total of flips, there are at least ordered pairs such that .

Now let , let , and let be the support of . For the least indices , flip a coin to choose between

 x(i)=1 and y(i)=0 OR x(i)=0 and y(i)=1

For the remaining indices , let and , or and in such a way that ensures that both and have at least 1’s in coordinates in .

Then for all indices , let . There were flips, so there are at least witnesses.

Now let us consider witnesses to . Fix with , and let . We randomly construct , so that .

Let be the support of . For the indices of least index, set and . For the remaining indices , flip a coin to choose between

 x(i)=1 and y(i)=0 OR x(i)=0 and y(i)=1.

For each index , flip a coin to choose between

 x(j) =1=y(j) OR x(j) =0=y(j).

Again, there were flips, so we have at least witnesses.

Next, let us consider witnesses to . Let , and let be the support of . We construct so that . For every , set . For the snallest indices , flip a coin to choose between

 x(i)=1 and y(i)=0 OR x(i)=0 and y(i)=1.

For the remaining indices , choose either

 x(i)=1 and y(i)=0 OR x(i)=0 and y(i)=1.

in such a way that ensures that neither nor receives more than 1’s. Clearly, there are at least witnesses.

Now we consider witnesses for .

Let , and let be the support of . Suppose . We build , so that . First, consider the case where . For , set and . For , choose of the indices, and set . [Set all others to .] Since , we have at least witnesses. Now consider the case where . For the smallest indices , set and . For the remaining indices , flip a coin to choose between

 x(i)=1 and y(j)=0 OR x(i)=0 and y(j)=1.

Now we choose of the remaining indices . There are choices, which ranges between and . Therefore there are at least witnesses. It is not hard to check that for , , and therefore .

Finally, we consider witnesses for . Let with support with . We construct so that . First, consider the case where . For each , set and . Then for the smallest indices outside of , choose of then. For each such selected , set , and otherwise. This gives at least witnesses.

For the case where , for each , flip a coin to choose between

 x(i)=1 and y(i)=0 OR x(i)=0 and y(i)=1.

This gives witnesses.

Now we are ready to compute the probability that our random partition

and fails to be a representation. Let . If , then has “needs”:

• .

If , then has “needs”:

• .

So is a bound on the number of “needs”. Given fixed , the probability that some need is not satisfied is

 Pr[z has an unsatisfied need]≤3n2(1−1n2)2k,

and

 Pr[∃z with an unsatisfied need] ≤∑z3n2(1−1n2)2k (1) =23k+1⋅3n2(1−1n2)2k (2) ≤23(k+1)⋅n2(1−1n2)2k. (3)

We want (3) to be less than 1, which is equivalent to its logarithm being less than zero:

 ⟺3(k+1)log2+2logn+2klog(n2−1n2)<0 ⟺3(k+1)log2+2logn<2klog(n2n2−1)

Now

 3(k+1)log2+2logn <3(k+1)+k ≤5k

so long as , and

 2k⋅log(n2n2−1) =2k[log(n2)−log(n2−1)] >2k⋅1n2

So we need . Setting , we have , which holds for all . Hence taking gives a non-zero probability that a random partition yields a representation.

## 3 A lower bound

In this section, we consider representations of as edge-colorings of with all mandatory triangles present and no all-blue triangles. To be more precise, let , where , be a representation. Then label the verticies of with , and let the color of edge be the atom such that .

.

###### Proof.

There must be some red edge . Any red edge has nine “needs”. There must be nine points that witness these needs, which together with and make a total of 11 points. ∎

We can easily obtain a slight improvement using the classical Ramsey number .

.

###### Proof.

We know that at least 11 points are required. Since , and there are no all-blue triangles, there must be a red . Let be an edge in this red . Then must have its red-red need met twice, hence there must be ten points besides and . ∎

###### Lemma 5.

In any representation of , for every red edge there is a red that is vertex-disjoint from it. In particular, off of every red edge one can find the configuration depicted in Figure 4.

###### Proof.

Let be red, with witnesses to all needs as in Figure 3. Then induce a red , since any blue edge among them would create an all-blue triangle with (and also with ). Furthermore, any edge running from any of to any of must be red, since any such blue edge would create an all-blue triangle with either (for and ) or (for and ). Thus we have the configuration depicted in Figure 4. ∎

.

###### Proof.

Consider the configuration depicted in Figure 4. The edge is red. Then and both witness the light-blue-dark-blue need, while and all witness the red-red need. There are seven needs yet unsatisfied. The remaining vertex could witness some need, but vertices through will have to be added. Thus there are at least 17 points. See Figure 5. ∎

Lemma 6 generalizes nicely as follows.

###### Theorem 7.

For all , .

Note that the trivial bound is , roughly half the bound in Theorem 7.

###### Proof.

Call the shades of blue through . Fix a red edge . Let denote the set of vertices that witness a blue-blue need for , and let denote the set of vertices that witness either a red-blue need or a blue-red need for . induces a red clique, and all edges from to are red. Note that and . This gives the trivial lower bound of .

Let witness - for and let witness - for . The edge is red, hence has needs. Both and witness the same - need, and all points in and (besides and ) witness the red-red need. Hence there must be at least points outside of . Hence there are at least points. ∎

###### Corollary 8.

In any representation of , the clique number of the red subgraph of the underlying graph of the representation is at least .

## 4 SAT solver results

In this section, we improve the lower bound on using a SAT solver.

.

###### Proof.

We build an unsatisfiable boolean formula whose satisfiability is a necessary condition for to be representable over 17 points. For all and , define a boolean . We interpret being TRUE to mean that is red, being TRUE to mean that is light blue, and being TRUE to mean that is dark blue. Then define

 Φ0=⋀i

Then asserts that for each , exactly one of , , and is TRUE.

Consider the subgraph depicted in Figure 4. Let

Define

 Φ1=⎛⎝⋀(i,j)∈Rϕi,j,0⎞⎠∧⎛⎝⋀(i,j)∈Blϕi,j,1⎞⎠∧⎛⎝⋀(i,j)∈Bdϕi,j,2⎞⎠

Then asserts that any edges colored in Figure 4 are colored correctly.

Finally, define

 Φ2=⋀(i,j)∈R⎡⎣⋀c1,c2∈{0,1,2}⎛⎝⋁i≠k≠jϕi,k,c1∧ϕk,j,c2⎞⎠⎤⎦

Then asserts that every red edge in Figure 4 has its needs satisfied.

Let .

has been verified by SAT solver to be unsatisfiable when there are 17 points.

###### Corollary 10.

In any representation of , the clique number of the red subgraph of the underlying graph of the representation is at least six.

###### Proof.

From the previous lemma, we see that at least 18 points are required to represent . Since , there must be a red . ∎

Thus we can always find the subgraph depicted in Figure 6.

Unfortunately, is satisfiable on 18 or more points. We must add more clauses to make unsatisfiable. First, expand to include the red as in Figure 6. Second, we add clauses to forbid all-blue triangles:

 Φ3=⋀i
###### Lemma 11.

Let . Then on 18 and on 19 points, is unsatisfiable. Hence .

###### Proof.

We have verified the unsatisfiability of via SAT solver. ∎

## 5 Summary and open problems

We summarize our work as follows.

###### Theorem 12.

We have , and

 2n2+4n+1≤f(n)≤23n+1

for all .

Is ?

###### Problem 2.

Is for some , or are exponentially many points required?

###### Problem 3.

Can some modification of the technique used in [2] give a smaller representation of ? The most obvious thing to try – replacing by – doesn’t work.

###### Problem 4.

Which has the smaller minimal representation, or ? While has atoms , all symmetric, with all-blue triangles forbidden, has atoms , with all-blue triangles forbidden. The atom is flexible in both cases. The lower bound proven in Theorem 7 applies to representations of as well.

## References

• [1] J. Alm, R. Maddux, and J. Manske, Chromatic graphs, Ramsey numbers and the flexible atom conjecture, Electron. J. Combin. 15 (2008), no. 1, Research paper 49, 8. MR 2398841 (2009a:05202)
• [2] Jeremy F. Alm and David A. Andrews, A reduced upper bound for an edge-coloring problem from relation algebra, Algebra Universalis 80 (2019), no. 2, Art. 19, 11. MR 3951643
• [3] Hajnal Andréka and Roger D. Maddux, Representations for small relation algebras, Notre Dame J. Formal Logic 35 (1994), no. 4, 550–562. MR 1334290
• [4] L. Dodd and R. Hirsch, Improved lower bounds on the size of the smallest solution to a graph colouring problem, with an application to relation algebra, Journal on Relational Methods in Computer Science 2 (2013), 18–26.