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1/3–2/3 conjecture

Unsolved problem on partial orders

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  • 6 chapters

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Example

The partial order formed by three elements a, b, and c with a single comparability relationship, a ≤ b, has three linear extensions, a ≤ b ≤ c, a ≤ c ≤ b, and c ≤ a ≤ b. In all three of these extensions, a is earlier than b. However, a is earlier than c in only two of them, and later than c in the third. Therefore, the pair of a and c have the desired property, showing that this partial order obeys the 1/3–2/3 conjecture.

This example shows that the constants 1/3 and 2/3 in the conjecture are tight; if q is any fraction strictly between 1/3 and 2/3, then there would not exist a pair x, y in which x is earlier than y in a number of partial orderings that is between q and 1 − q times the total number of partial orderings.

More generally, let P be any series composition of three-element partial orders and of one-element partial orders, such as the one in the figure. Then P forms an extreme case for the 1/3–2/3 conjecture in the sense that, for each pair x, y of elements, one of the two elements occurs earlier than the other in at most 1/3 of the linear extensions of P. Partial orders with this structure are necessarily series-parallel semiorders; they are the only known extreme cases for the conjecture and can be proven to be the only extreme cases with width two.

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