Testing triangular properties (1/2)
A few years ago, I worked together with Yves Verhoeven on the problem of testing triangular properties. We did some proofs, wrote a few pages but decided not to publish our work (since it is far from being optimal, and also far from being really interesting). However I think that this is worth a few lines in this blog. It is clearly more technical than the average post of my blog, so don’t force yourself into reading it.
Let be the complete graph on vertices and be a set. Let be a function, and let be a real number.
We are interested in triangular properties, that is properties of triangles (sets of 3 items). Our final goal is to decide if a set violates a triangular property.
definition
A triangular property is a function that is invariant under permutations. The function is said to have the property if for all triangles the value is equal to . The function is said to be at least -far from having property if for all function having the property the functions and differ on at least values.
Two triangles are said to be independent if their intersection is empty.
A triangle is said to be bad (with respect to ) if takes the value on it.
Let us now fix a triangular property .
Lemma 1
Suppose that is -far from having property . Let be a set of vertices so that induces a function that has property on the restriction of to . Then the cardinality of is greater than .
Proof
When one removes one element from , it removes edges from . When one removes a second element, it then removes an extra edges, and so on… So that when one removes elements, one removes edges form the graph, which is smaller than . As we want to remove at least edges, this leads to the following inequality
which implies .
Lemma 2
Suppose that there exists a like in lemma 1.
There exists a family of independent triangles in such that
- each triangle in is bad,
- .
Proof
Let us construct such a family . Suppose, without loss of generality, that is minimal (with respect to the cardinality).
In order to construct , we use the following greedy algorithm.
Input: a complete graph and a minimal set such that has property
Output: a family of independent bad triangles
- .
- .
- .
- WHILE there exists a bad triangle such that and
- pick such a triangle.
- restrict to .
- .
- add the triangle to .
- return .
Firstly, observe that by construction the family is made of independent triangles.
Now it remains to be proved that the second announced property holds, as we already know from the previous lemma that . Suppose this is not the case, that is . Using this we will construct a set verifying the same constraints as , and such that , thus contradicting the minimality of .
The fact that the algorithm stops comes from the fact that does not contain any bad triangle any more, so that satisfies the same conditions as . Let us examine the size of :
This concludes the proof.
We can now jump to the probabilistic part of our work. For a given family of triangles , we denote by the union of the triangles in .
Lemma 3
Let be a family of triangles as defined in lemma 2. If one randomly picks elements from then he gets at least one complete triangle in with a probability lower-bounded by a nonzero constant as goes to infinity.
I am pretty sure that you want the rest of this work as soon as possible (really, uh?), I will do as fast as I can to put the proof of lemma 3 and the remaining…