Drawing planar graphs with integer edge lengths - MathOverflow

Drawing planar graphs with integer edge lengths

2011-02-02T15:57:29Z

It is well known that every planar graph has an embedding such that every edge is drawn as a straight line segment (Fáry's Theorem). Kemnitz and Harborth made the following stronger conjecture

Conjecture 1. Every planar graph has a straight line embedding with integer edge lengths.

I was wondering if it is possible to attack this problem with the following approach.

Conjecture 2. Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear. For any $\epsilon >0$, there exists $X':=\{x_1', \dots, x_n'\}$ such that for all $i, j \in [n]$

$d(x_i, x_i') < \epsilon$,
$d(x_i', x_j') \in \mathbb{Q}$, and
no three points of $X'$ are colinear.

To prove Conjecture 2, it suffices to prove the following conjecture.

Conjecture 3. Let $X:=\{ x_1, \dots, x_n \}$ be a finite set of points in the plane such that no three points of $X$ are colinear and all pairwise distances are rational. Then the set of points which are at rational distance from all points in $X$ is a dense subset of the plane.

Conjecture 3 is trivial for $n=1$ and easy for $n=2$. Almering proved it for $n=3$, and I think it is open for $n>3$. Note that Conjecture 3 is (essentially) a weakening of:

Conjecture 4. There exists a dense subset of the plane with all pairwise distances rational.

This question was posed by Ulam in 1945 (see this mathoverflow question for more background). So, the reason I like Conjecture 3 is that it is still strong enough to prove Conjecture 1, but appears much weaker than Conjecture 4. Unfortunately, Conjecture 3 is beyond my limited area of expertise. Hence:

Question. What are the prospects for proving Conjecture 3? A proof or disproof would be fantastic. However, even arguments suggesting that it is true/false but say beyond current technology would be most welcome.

Answer by Igor Rivin for Drawing planar graphs with integer edge lengths

2011-02-02T16:28:20Z

Not quite an answer, but:

The Kemnitz/Harborth conjecture was proved for cubic planar graphs in:

Straight line embeddings of cubic planar graphs with integer edge lengths Jim Geelen1, Anjie Guo2,†, David McKinnon3 (Journal of Graph Theory, 2008)

They state a condition which would imply Kemnitz/Harborth (property 3.1 in their paper).

They cite the following theorem, which is related to, but not the same as, what you conjecture:

Theorem 2.1 (Berry 1992, Acta Arith). If $A, B, C \in \mathbb{R}^2$ are non-collinear points such that $dist(A, B), dist(A, C)^2,$ and $dist(B, C)^2$ are rational, then the set of points that are a rational distance from each of $A, B, C$ forms a dense subset of $\mathbb{R}^2.$

Answer by Gerry Myerson for Drawing planar graphs with integer edge lengths

2011-02-02T22:14:45Z

Problem D19 on pages 283-287 of Guy, Unsolved Problems In Number Theory, asks, "Is there a point all of whose distances from the corners of the unit square are rational?" This suggests that even a very weak form of Conjecture 3 is wide open (or was, as of 2004).

Answer by Holowitz for Drawing planar graphs with integer edge lengths

2012-03-07T17:36:55Z

I think conjecture 3 is actually stronger then conjecture 4.

I prove $C_3\implies C_4$:

Pick any sequence of integers $a_n$, which contains all integers infinite times.

Pick any enumeration of all squares $s_n$ in the plane with corners at rational coordinates.

Then assuming conjecture 3, at step $n$ we can find a point with rational distances in $s_{a_n}$ which is not collinear to any previously used rationals, since there are only finitely many straight lines in our set so far.

At step $\omega$, we have $\omega$ many rationals in every square, so a dense set of all rational distances.