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Harborth's conjecture

Harborth's conjecture is a mathematics topic covered in the lgStudy science library. This page brings together a partial reference excerpt, illustrations, worked examples, real-world applications and a short study plan, so you can understand Harborth's conjecture rather than just read about it. In short: In mathematics, Harborth's conjecture states that every planar graph has a planar drawing in which every edge is a straight segment of integer length. This conjecture is named after Heiko Harborth, and (if true) would strengthen Fáry's theorem on the existence of straight-line drawings for every planar graph.

Harborth's conjecture — main illustration
Harborth's conjecture — illustration

Key takeaways

  • Harborth's conjecture belongs to mathematics; place it in that map before memorising details.
  • Learn the definition first, then one example that makes the definition concrete.
  • Connect Harborth's conjecture to a quantity you can measure, compute or draw — that is where exam questions come from.
  • Reproduce the core statement of Harborth's conjecture from memory before moving on to harder problems.

Reference excerpt

In mathematics, Harborth's conjecture states that every planar graph has a planar drawing in which every edge is a straight segment of integer length. This conjecture is named after Heiko Harborth, and (if true) would strengthen Fáry's theorem on the existence of straight-line drawings for every planar graph. For this reason, a drawing with integer edge lengths is also known as an integral Fáry embedding. Despite much subsequent research, Harborth's conjecture remains unsolved.

Special classes of graphs Although Harborth's conjecture is not known to be true for all planar graphs, it has been proven for several special kinds of planar graph. One class of graphs that have integral Fáry embeddings are the graphs that can be reduced to the empty graph by a sequence of operations of two types:

Removing a vertex of degree at most two. Replacing a vertex of degree three by an edge between two of its neighbors. (If such an edge already exists, the degree three vertex can be removed without adding another edge between its neighbors.) For such a graph, a rational Fáry embedding can be constructed incrementally by reversing this removal process, re-inserting the vertices that were removed. Re-inserting a degree-two vertex uses the fact that the set of points that are at a rational distance from two given points are dense in the plane. Re-inserting a degree-three vertex uses the fact that, if three points have rational distance between one pair and square-root-of-rational distance between the other two pairs, then the points at rational distances from all three are again dense in the plane. The distances in such an embedding can be made into integers by scaling the embedding by an appropriate factor. Based on this construction, the graphs known to have integral Fáry embeddings include the bipartite planar graphs, (2,1)-sparse planar graphs, planar graphs of treewidth at most 3, and graphs of degree at most four that either contain a diamond subgraph or are not 4-edge-connected. In particular, the graphs that can be reduced to the empty graph by the removal only of vertices of degree at most two (the 2-degenerate planar graphs) include both the outerplanar graphs and the series–parallel graphs. However, for the outerplanar graphs a more direct construction of integral Fáry embeddings is possible, based on the existence of infinite subsets of the unit circle in which all distances are rational. Additionally, integral Fáry embeddings are known for each of the five Platonic solids.

Related conjectures A stronger version of Harborth's conjecture, posed by Kleber (2008), asks whether every planar graph has a planar drawing in which the vertex coordinates as well as the edge lengths are all integers. It is known to be true for 3-regular graphs, for graphs that have maximum degree 4 but are not 4-regular, and for planar 3-trees. Another unsolved problem in geometry, the Erdős–Ulam problem, concerns the existence of dense subsets of the plane in which all distances are rational numbers. If such a subset existed, it would form a universal point set that could be used to draw all planar graphs with rational edge lengths (and therefore, after scaling them appropriately, with integer edge lengths). However, Ulam conjectured that dense rational-distance sets do not exist. According to the Erdős–Anning theorem, infinite non-collinear point sets with all distances being integers cannot exist. This does not rule out the existence of sets with all distances rational, but it does imply that in any such set the denominators of the rational distances must grow arbitrarily large.

See also Integer triangle, an integral Fáry embedding of the triangle graph Matchstick graph, a graph that can be drawn planarly with all edge lengths equal to 1 Erdős–Diophantine graph, a complete graph with integer distances that cannot be extended to a larger complete graph with the same property Euler brick, an integer-distance realization problem in three dimensions

References

Illustrations

Harborth's conjecture: Integral Fáry embedding of the octahedral graph K2,2,2
Integral Fáry embedding of the octahedral graph K2,2,2

Worked examples

Example 1 — a first encounter with Harborth's conjecture

Start with the simplest possible case. Write down what Harborth's conjecture claims or describes in one sentence, then invent the smallest concrete situation in which that sentence is true. In mathematics, the smallest case is usually a single object, a single equation or a single measurement. Check that every symbol or term in your sentence has a meaning in that case.

Example 2 — changing one variable

Take the situation from Example 1 and change exactly one quantity: double it, halve it, or set it to zero. Predict what should happen to Harborth's conjecture before you calculate. Comparing your prediction with the result is the fastest way to find out whether you understand the idea or only the words.

Example 3 — an exam-style question

Typical questions about Harborth's conjecture ask you to (a) state it precisely, (b) apply it to given data, and (c) explain a limitation. Practise writing all three answers in under five minutes; the third part is what separates a full-mark answer from an average one.

Applications of Harborth's conjecture

In research
Harborth's conjecture appears in mathematics research whenever the underlying quantities have to be modelled precisely. Papers usually cite it as a starting assumption and then explore where it breaks down.
In technology and industry
Engineering practice reuses Harborth's conjecture in design rules, simulations and safety margins. Knowing the idea lets you read a specification sheet and understand why the numbers look the way they do.
In the classroom
Harborth's conjecture is common in secondary-school and first-year university syllabi. It links to neighbouring topics Arithmetic problems of plane geometry, Conjectures, Statements about planar graphs, so understanding it makes those chapters shorter.
In everyday life
Look for Harborth's conjecture outside the textbook — in sport, cooking, traffic, electronics or the sky above you. An example you found yourself is remembered far longer than one you were given.
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How to study Harborth's conjecture in 20 minutes

  1. Read the reference excerpt below once, without taking notes.
  2. Close the page and write down what Harborth's conjecture means in your own words.
  3. Compare your version with the excerpt and mark what you missed.
  4. Work through the three examples above with pen and paper.
  5. Explain Harborth's conjecture out loud to somebody else — or to Teacher Smith in the lgStudy chat.

Frequently asked questions

What is Harborth's conjecture in simple terms?

In mathematics, Harborth's conjecture states that every planar graph has a planar drawing in which every edge is a straight segment of integer length. This conjecture is named after Heiko Harborth, and (if true) would strengthen Fáry's theorem on the existence of straight-line drawings for every pl…

Why does Harborth's conjecture matter?

Because it connects several mathematics ideas at once: it gives you a definition you can apply, a quantity you can calculate, and a way to check whether a result is plausible.

How should I study Harborth's conjecture?

Read the excerpt, restate it from memory, then work through the examples and applications listed on this page. The five-step study plan above takes about twenty minutes.

What does this page cover?

It gives you a compact reference excerpt plus original lgStudy explanations, examples, applications and study material on Harborth's conjecture.

Tags

  • Arithmetic problems of plane geometry
  • Conjectures
  • Statements about planar graphs
  • Unsolved problems in graph theory

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