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Topic review (newest first)

2012-11-28 01:27:21

Hi naumberg

You can use the math tag {math}{/math} except that the brackets are square and not curly...

2012-11-27 18:17:56


I am sorry but I can not read that very well.

2012-11-27 11:16:29

Hi bobbym!

I solved the problem. Please find my tex code below (compile yourself for better reading, sry!)


We want to determine a strategy to select an increasing subsequence of $x_1,...,x_n$. Let $Y$ be the length of our increasing subsequence. As $X$ is defined as the longest increasing subsequence we surely have $E[X] \ge E[Y]$. Depending on how good our strategy is we hope to get $E[Y] \ge (1-o(1))(1-\frac{1}{e})\sqrt{n}$ which would complete the proof.

Let us assume that $m:= \sqrt{n}$ is an integer and partition our random sequence in blocks $L_j=(x_{(j-1)m+1},...,x_{jm} )$ for $j=1,...,m$ (we can assume this as we look at asymptotics in $n$ in the end).

The strategy picks the first number $y_1$ out of $x_1,...,x_n$ that is $\le \frac{1}{m}$ and skips to the next block. It then continuous to pick a number $y_i$ in each of the remaining blocks if $y_{i-1} \le y_i \le y_{i-1} + \frac{1}{m}$ and skips this block otherwise. At the end we receive an increasing subsequence $y_1,...,y_Y$ of length $Y$.


input: sequence x(1),...x(n)
output: length Y of an increasing subsequence y(1)<=...<=y(Y)

Y = 0         \\ counting the length of the subsequence
s = zero array     \\ storing the subsequence here

\\ go through intervals elements of L_j
for j = 1 to m   
   \\ boolean helper to implement stopping time, i.e. breaking condition for the loop
   success == 0
   while (success == 0) do
      \\ go through elements of L_j
      for k = (j-1)*m+1 to j*m 
         \\ find a larger element which is still small enough
         if (s(Y) <= x_k <= s(Y)+1/m)     
            Y = Y+1       \\ length of subsequence ++
            s(Y) = x_k      \\ store element
            success == 1   \\ stop searching in L_j 
                     \\ and go to next interval


Now let us estimate the expectation of $Y$,

E[Y] &= E[\sum_{j=1}^{m} \mathds{1}_{\{ \text{ "found suitable number in }L_j\text{ " }\}}]\\
&= \sum_{j=1}^{m}E[ \mathds{1}_{\{ \text{ "found suitable number in }L_j\text{ " }\}}]

As our "tolerance of increase" $\frac{1}{m}$ stays the same for all numbers we search for, all $x_i$ are independently and uniformly distributed and all parts of the sequence $L_j$ contain the same amount of numbers $m$ we get that

E[Y] &= m P[ \text{ "found suitable number in }L_1 \text{ " }]\\
&= m (1- P[\forall i=1,...,m : x_i > \frac{1}{m}]) \\
&= m (1- (P[x_1 > \frac{1}{m}])^m) \\
&= m (1- (1- \frac{1}{m})^m) \\
&= \sqrt{n} (1- (1- \frac{1}{\sqrt{n}})^{\sqrt{n}}) \\
&= (1-o(1))(1-\frac{1}{e})\sqrt{n}.

So our strategy gives the lower bound $E[X] \ge E[Y]$.

2012-11-24 16:06:09

Hi Naumberg;

I have looked through 3 papers on this exact subject and the original paper by Erdos and Szekeres. In all of them it seems that the bound

was a major achievement. I could not find anyway to tie in what they were doing with the sharper bound you are interested in. I would very much like to see your answer when you find it.

2012-11-24 10:37:20

Cool. Thanks in advance for your help!

2012-11-24 10:30:38


I will look at it and post if I solve it. Thanks for the images.

2012-11-24 10:12:20

Logging in is sometimes helpful ...

2012-11-24 09:38:05


Do you see this?

2012-11-24 09:30:17

Mmh, I am really a newbie

1. Post a reply - ok
2. upload - how/where? I see no button and cannot use "[img][/img]"


2012-11-24 09:21:05


Yes, just use image upload in Post reply. Do not use quick post.

2012-11-24 09:15:17

Unfortunately not. But I could scan you the respective page from the notes. Is there a possibility to upload it here?

2012-11-24 08:42:42


Is the lecture online?

2012-11-24 08:39:06


I cross-checked the paper and this is clearly way to complicated as we usually solve those exercises with one page maximum. One can attend this lecture from the third year or higher.

2012-11-24 08:28:01


I am looking at a paper that discusses this problem but I am not understanding where he is going. Maybe you can follow it better: … p;type=pdf

2012-11-24 08:15:49

Sure. In the lecture we had the following version of Erdos/Szerkeres which applies to continuous data as well:


be a seuqence of n distinct numbers. Then this sequence contains an increasing subsequence or a decreasing subsequence of length

Now define another sequence
. If
is the length of the longest increasing subsequence in
, then
are identically distributed and
leads to
. So this lower bound is proven. In the lecture we even showed that
and that

In our exercise we need to prove that
. Does this help?

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