Browse Groups

• ## [Computational Complexity] Monotone Sequence Theorem: literature thing

(1)
• NextPrevious
• Sometimes the literature gives a proof, and then a reference, but the reference is not really to that proof. Here is an example. The theorem in question is the
Message 1 of 1 , Nov 14, 2008
View Source
Sometimes the literature gives a proof, and then a reference, but the reference is not really to that proof. Here is an example. The theorem in question is the following:
For any seq of n2+1 distinct reals there is either a dec subseq of length n or an inc subseq of length n+1.
In Proofs from the book the authors attribute the following argument to Erdos-Szekeres (paraphrased):
Let the seq be a(1),...,a(n2+1). Associate to each a(i) the number t(i) which is the length of a longest inc subseq starting at a(i). If there is an i such that t(i) &ge n+1 then we are done. If not then the function mapping a(i) to t(i) maps a set of size n2+1 to a set of size n. Hence there is some s such that n+1 elements of the seq map to s. These n+1 elements form a dec seq.
The actual proof in the Erdos-Szekeres paper is this (paraphased):
Let f(n) be the minimum number of points out of which we can select n inc or dec subseq. We show f(n+1) = f(n)+2n-1. Let a(1),...,a(f(n)+2n-1) be a sequence of distinct reals. Out of the first f(n) of them there is an inc or dec subseq of length n. Call the last elements of that subseq b(1). Remove it. (There is now a seq of length f(n)+2n-2.) Repeat the process to obtain b(1), b(2), ..., b(2n). Let A be the b(i)'s that are from inc subseq. Let B be the b(i)'s that are from dec subseq. It is easy to show that A is itself a dec subseq and that B is itself an inc subseq. If A or B has n+1 elements then we are done. The only case left is that A and B each have n elements. Let a be the last element in A and b be the last element in b. It is easy to show that a=b. But one of them was removed before the other, so this is impossible.
Thoughts:
1. I suspect that in talks Erdos gave he did the proof now atributed to Erdos-Szekeres and hence people naturally assumed that this is the paper it was in.
2. I am surprised that Proofs from the book gave this proof. There is, IMHO, a better proof. I do not know whose it is.
Let the seq be a(1),...,a(n2+1). Map every 1 \le i \le n2+1 to (x,y) where x (y) is the length of the longest inc (dec) subseq than ends with a(i). It is easy to show that this map is 1-1. If all of the x,y that are mapped to are in [n]x[n] then the domain is bigger than the range, contradiction.
3. Erdos-Szekeres first proved this theorem in 1935. Martin and Joseph Kruskal proved it 15 years later without knowing that Erdos-Szekeres had proven it; though by the time Joesph Kruskal published it he knew. I have gathered up 5 proofs of the theorem that I know here.
4. In those days it was harder to find out if someone else had done what you had done since they didn't have google, but it may have (in some cases) been easier since there were so many fewer researchers- you could just call them. OH- but long distance was expensive back then.

--
Posted By GASARCH to Computational Complexity at 11/14/2008 01:51:00 PM
Your message has been successfully submitted and would be delivered to recipients shortly.
• Changes have not been saved
Press OK to abandon changes or Cancel to continue editing
• Your browser is not supported
Kindly note that Groups does not support 7.0 or earlier versions of Internet Explorer. We recommend upgrading to the latest Internet Explorer, Google Chrome, or Firefox. If you are using IE 9 or later, make sure you turn off Compatibility View.