OpenAI reports the Harary–Hill crossing-number formula

For every n≥3, the ordinary crossing number of Kₙ equals ¼⌊n/2⌋⌊(n−1)/2⌋⌊(n−2)/2⌋⌊(n−3)/2⌋.

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Understanding the problem

Read the background

For every n≥3, the ordinary crossing number of Kₙ equals ¼⌊n/2⌋⌊(n−1)/2⌋⌊(n−2)/2⌋⌊(n−3)/2⌋.

Why it matters

If correct, resolves the Harary–Hill conjecture for complete graphs of every order.

What counts as progress

Does every complete graph have the Harary–Hill crossing number?

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