Erdős–Szekeres theorem

In mathematics, the Erdős–Szekeres theorem is a finitary result, which makes precise one of the corollaries of Ramsey's theorem. While Ramsey's theorem makes it easy to prove that any sequence of distinct real numbers contains "either" a monotonically increasing infinite subsequence, "or" a monotonically decreasing infinite subsequence, the result proved by Paul Erdős and George Szekeres goes further. For given "r", "s" they showed that any sequence of length at least

:"rs" − "r" − "s" + 2

contains "either" a monotonically increasing subsequence of length "r", "or" a monotonically decreasing subsequence of length "s". The proof appeared in the same 1935 paper that mentions the Happy Ending problem. Steele (1995) contains "six or more" proofs of the theorem.

Example

For "r" = 3 and "s" = 2, the formula tells us that any permutation of three numbers has an increasing subsequence of length three or a decreasing subsequence of length two. Among the six permutations of the numbers 1,2,3:
* 1,2,3 has an increasing subsequence consisting of all three numbers
* 1,3,2 has a decreasing subsequence 3,2
* 2,1,3 has a decreasing subsequence 2,1
* 2,3,1 has two decreasing subsequences, 2,1 and 3,1
* 3,1,2 has two decreasing subsequences, 3,1 and 3,2
* 3,2,1 has three decreasing length-2 subsequences, 3,2, 3,1, and 2,1.

Geometric interpretation

One can interpret the positions of the numbers in a sequence as "x"-coordinates of points in the Euclidean plane, and the numbers themselves as "y"-coordinates; conversely, for any point set in the plane, the "y"-coordinates of the points, ordered by their "x"-coordinates, forms a sequence of numbers. With this translation between sequences and point sets, the Erdős–Szekeres theorem can be interpreted as stating that in any set of at least "rs" − "r" − "s" + 2 points we can find a polygonal path of either "r" - 1 positive-slope edges or "s" - 1 negative-slope edges. For instance, taking "r" = "s" = 5, any set of at least 17 points has a four-edge path in which all slopes have the same sign.

An example of "rs" − "r" − "s" + 1 points without such a path, showing that this bound is tight, can be formed by applying a small rotation to an ("r" - 1) by ("s" - 1) grid.

Proof by the Pigeonhole principle

We prove the theorem by contradiction.We label each number, $$n_i, in the sequence with a pair $$(a_i,b_i), where $$a_i is the length of the longest monotonically increasing subsequence ending with $$n_i and $$b_i is the length of the longest monotonically decreasing subsequence ending with $$n_i. If we don't have a monotonically increasing or decreasing subsequence of the desired length, then each $$a_i is between 1 and $$r-1 and each $$b_i is between 1 and $$s-1. So, there are $$(r-1)(s-1)=rs-r-s+1 possible $$(a_i,b_i) pairs. Since there are $$rs-r-s+2 numbers in the sequence, the Pigeonhole principle guarantees that two numbers in the sequence, say $$n_j and $$n_k, have the same $$(a_i,b_i)pairs. However, this is impossible for the following reason. Assume $$j. If $$n_j leq n_k, then we can append $$n_k to the monotonically increasing subsequence ending at $$n_j to get a longer monotonic subsequence. If $$n_j geq n_k, then we can append $$n_k to the monotonically decreasing subsequence ending at $$n_j to get a longer monotonic subsequence. In each case, we reach a contradiction. So the assumption that we didn't have a monotonic subsequence of the desired length was false. This proves the theorem.

Proof via Dilworth's theorem

The Erdős–Szekeres theorem can be proved in several different ways; one of the proofs uses Dilworth's theorem on chain decompositions in partial orders, or its simpler dual.

To prove the theorem, define a partial ordering on the members of the sequence, in which "x" is less than or equal to "y" in the partial order if "x" ≤ "y" as numbers and "x" is not later than "y" in the sequence. A chain in this partial order is a monotonically increasing subsequence, and an antichain is a monotonically decreasing subsequence. By the dual of Dilworth's theorem, either there is a chain of length "r", or the sequence can be partitioned into at most "r" - 1 antichains; but in that case the largest of the antichains must form a decreasing subsequence with length at least:$$leftlceilfrac{rs-r-s+2}{r-1} ight ceil=s.

Alternatively, by Dilworth's theorem itself, either there is an antichain of length "s", or the sequence can be partitioned into at most "s" - 1 chains, the longest of which must have length at least "r".

See also

* Longest increasing subsequence problem

References

*cite journal
author = Erdős, Paul; Szekeres, George
title = A combinatorial problem in geometry
journal = Compositio Mathematica
volume = 2
pages = 463–470
year = 1935
url = http://www.numdam.org/item?id=CM_1935__2__463_0

*cite conference
author = Steele, J. Michael
title = Variations on the monotone subsequence problem of Erdős and Szekeres
booktitle = Discrete Probability and Algorithms
editor = Aldous, Diaconis, and Steele, eds.
pages = 111–132
publisher = Springer-Verlag
location = New York
year = 1995
url = http://www-stat.wharton.upenn.edu/~steele/Publications/PDF/VOTMSTOEAS.pdf

External links

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