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

[fordanaldino]

to get hhhhhhhhhiiiiiiiiiiiiiiiiiiiigggggggggggghhhhhhhhh

 

[sjt19]

Originally posted by Baca Loco
Why?

Why not? Flight costed me £30. I want to be there to support my team, and just go to the event, last years event was awesome. Its only £30 so its not gonna break the bank. That ok Paul?

And Fordham....thats not the reason...and you know it.

 

[fordanaldino]

Originally posted by sjt19
And Fordham....thats not the reason...and you know it.

yeah i know. i just didnt want to announce to the world that you were gettin a prostitute..............

 

[craigpil]

lol i am guess that there was a maths question somewhere in that load of garbage John

and am also guessing that is degree level or even higher though

 

[DarWood]

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...

Dude the word 'Pittel' doesnt have no meaning, and doesnt exist according to the internet!
The only thing found on Pittel is 'Harvey Pittel', a world-renowned saxophonist.
Other than that that whole context is just aload of variable's, multiplied, or divided by each other. The $ signs are there to confuse people.

Damn i wish i could watch the x-ball games

Peace
Darren

 

[ANDY-501]

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...

Is the answer six ?
Is the limiting of the profile refering to a paintball equasion on snapshooting ?
Did you get this from the russion legion top secret training website ?

 

[Baca Loco]

Originally posted by sjt19
Its only £30 so its not gonna break the bank. That ok Paul?

Sure. Not a problem. You go on and "support the team"-- nudge, nudge, say no more--all ya like.

 

[Skeet]

Originally posted by fordanaldino
yeah i know. i just didnt want to announce to the world that you were gettin a prostitute..............

Only one...christ, if your gonna do it, might as well make it worth while and kill two birds with one stone...so to speak...(did any of you hear that joke?!)

 

[Pinchaaaay!]

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...

Damn it, i just tried to cheat and put all that into google.......and my p.c just threw up

What sort of freaky **** is that??

Tell me what answer you got, and i'll tell u if it's right

"Pinchaaay!"

 

[John C]

Its analysis of computer algorithms. Finding upper and lower limits and average cases.

Darwood: Your almost nearly a bit close . Pittel is a book. If you read the top bit its a digital search tree not a maths problem. Go type "Binary Tree" into google once you understand that type "Digital Search Tree".

Pinchy: I dont have the answer, its too hard