#@#@##!!!!!!!!!

[Jules]

Grrr yeah have exams today and monday so couldn't go Missed it last yr due to uni exams too!

 

[Skeet]

Originally posted by Skyhawk
err sarcasium???

Is that Latin?

 

[Pinchaaaay!]

Sam.......i'm with ya all the way bro'

I pulled my ligament in my knee (in spectacular style of course ) at the p.a league and it has stopped me from going to the dam!

Anyone need help revising?



"Pinchaaay!"

 

[Bunker Imp]

Originally posted by rev. pinchmeister
Sam.......i'm with ya all the way bro'

I pulled my ligament in my knee (in spectacular style of course ) at the p.a league and it has stopped me from going to the dam!

Anyone need help revising?



"Pinchaaay!"

ill take u up on that mr pinchaaay!!

know anything about supply and demand

or any good cheating methods

 

[John C]

Originally posted by rev. pinchmeister

Anyone need help revising?



"Pinchaaay!"

Pinchy you are a hero.

Consider a digital search tree (DST) built over $m$
independently generated binary strings with ''$0$'' occurring with`
probability $p$ and ''$1$'' with probability
$q=1-p$.

Limiting Distribution of the Profile.

Let us fixed $k$ (that may depend on $m$), and define
$B_m^k$ as the number of nodes in the DST at level $k$.
This is called the profile of DST.
We are interested in obtaining the limiting distribution of
$B_m^k$ for any $k$.
Aldous and Shields [AS] established it for the symmetric case.
In the asymmetric case the limiting distribution is unknown.
It is easy to derive a recurrence for $B_m^k$.
Define $B_m^k(u)=Eu^{B_m^k}$. Then
$$
B_{m+1}^k(u)=\sum_{l=0}^m {m \choose l} p^lq^{m-l} B_l^{k-1}(u)
B_{m-l}^{k-1}(u)~,
$$
with $B_0^0(u)=1$. Let now $B^k(z,u)=\sum_{m=0}^\infty B_m^k(u)
{z^m \over m!}$. Then, the above becomes
$$
{\partial B^k(z,u) \over \partial z} =
B^{k-1}(pz,u) B^{k-1}(qz,u)
$$
with $B^0(z,u)=u(e^z-1)+1$. We conjecture that for
$k=O(\log m)$ the limiting distribution of $B_m^k$ is normal.
The average profile was established in [LS]

Limiting Distribution of the Height.

Let $H_n^k$ denote the probability that the {\it height} in a DST
with $m$ nodes is smaller equal to $k$. Define
$H^k(z)=\sum_{m\geq 0} H_m^k {z^m \over m!}$. Then, one
easily see that for $k\geq 1$
$$
{d H^k(z) \over d z} = H^{k-1}(zp) H^{k-1}(zq)
$$
with $H^0(z)=1+z$. What is the limiting distribution of
the height? From [Pittel] we only know
that $H_m /\log m \to 1/\log (\min\{1/p, 1/q\})$
almost surely.




Ill give you half an hour...

 

[Orange Peel]

Ha ha ha!! Exams! Jeez, I remember being a student. It never stopped me from playing 'ball.... well, it shows really

Most of my guys are in the 'dam right now. Not too sure about Stevie P though, rumor has it he fled the airport and got a coach down to the Eurostar without telling anyone??

No 'dam for me.... I just quit smoking (cigarettes) so it'd be too much of a temptation....

 

[Bunker Imp]

Originally posted by John C
Pinchy you are a hero.

Consider a digital search tree (DST) built over $m$
independently generated binary strings with ''$0$'' occurring with`
probability $p$ and ''$1$'' with probability
$q=1-p$.

Limiting Distribution of the Profile.

Let us fixed $k$ (that may depend on $m$), and define
$B_m^k$ as the number of nodes in the DST at level $k$.
This is called the profile of DST.
We are interested in obtaining the limiting distribution of
$B_m^k$ for any $k$.
Aldous and Shields [AS] established it for the symmetric case.
In the asymmetric case the limiting distribution is unknown.
It is easy to derive a recurrence for $B_m^k$.
Define $B_m^k(u)=Eu^{B_m^k}$. Then
$$
B_{m+1}^k(u)=\sum_{l=0}^m {m \choose l} p^lq^{m-l} B_l^{k-1}(u)
B_{m-l}^{k-1}(u)~,
$$
with $B_0^0(u)=1$. Let now $B^k(z,u)=\sum_{m=0}^\infty B_m^k(u)
{z^m \over m!}$. Then, the above becomes
$$
{\partial B^k(z,u) \over \partial z} =
B^{k-1}(pz,u) B^{k-1}(qz,u)
$$
with $B^0(z,u)=u(e^z-1)+1$. We conjecture that for
$k=O(\log m)$ the limiting distribution of $B_m^k$ is normal.
The average profile was established in [LS]

Limiting Distribution of the Height.

Let $H_n^k$ denote the probability that the {\it height} in a DST
with $m$ nodes is smaller equal to $k$. Define
$H^k(z)=\sum_{m\geq 0} H_m^k {z^m \over m!}$. Then, one
easily see that for $k\geq 1$
$$
{d H^k(z) \over d z} = H^{k-1}(zp) H^{k-1}(zq)
$$
with $H^0(z)=1+z$. What is the limiting distribution of
the height? From [Pittel] we only know
that $H_m /\log m \to 1/\log (\min\{1/p, 1/q\})$
almost surely.




Ill give you half an hour...

aww m8 thats easy. the answer is.... actually ill let u do it for urself otherwise u will not learn

 

[John C]

Originally posted by Bunker Imp
aww m8 thats easy. the answer is.... actually ill let u do it for urself otherwise u will not learn

Gee thanks Bunker Imp, you really have my education at heart

 

[sjt19]

Just finished one exam, and have one more to do tomorrow morning, then i fly to the dam tomorrow afternoon.

Dam saturday exams!

 

[Baca Loco]

Originally posted by sjt19
then i fly to the dam tomorrow afternoon.

Why?