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**Naumberg****Guest**

*Let *

Hi forum!

Using the Erdos' lemma I can only deduce that

I would appreciate any further ideas!

Thanks for your help,

Michael

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi Naumberg;

Do you mean contiguous?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Naumberg****Guest**

Hi bobby,

what do you mean by "contiguous" here?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

You want a sequence that is next to each other or can we skip?

Example:

{x4,x5,x6,x7,x8} or {x2, x5, x11, x21, xn}

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Naumberg****Guest**

Ah! Yes, we skip!

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

As far as I know the Erdos bound is for integers that are permutations it does not apply to continuous data.

May I ask where you got both bounds you mention in post#1 from?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Naumberg****Guest**

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

*Let *

Now define another sequence

In our exercise we need to prove that

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi;

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:

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Naumberg****Guest**

Hi

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.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi;

Is the lecture online?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Naumberg****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi;

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Naumberg****Guest**

Mmh, I am really a newbie

1. Post a reply - ok

2. upload - how/where? I see no button and cannot use ""

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi;

Do you see this?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Naumberg****Member**- Registered: 2012-11-23
- Posts: 3

Logging in is sometimes helpful ...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi;

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Naumberg****Member**- Registered: 2012-11-23
- Posts: 3

Cool. Thanks in advance for your help!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

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.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Naumberg****Member**- Registered: 2012-11-23
- Posts: 3

Hi bobbym!

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

Cheers

Michael

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
}
}
}
}
return(Y)
```

Now let us estimate the expectation of $Y$,

\begin{alignat*}{1}

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{ " }\}}]

\end{alignat*}

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

\begin{alignat*}{1}

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}.

\end{alignat*}

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi;

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,532

Hi naumberg

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

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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