37 back to basics

[Drazen]

i disagree,,,i gave the paper what it was based off...i even gave you the tool how to read the streams tracker....no philiosophical crap....

You took this completely out of context. Please read it again. That part does not refer to anything you said or did. I appreciate everything you shared, for sure.

It's pure BS! But it doesn't surprise me coming from you...

Hmm... This is a very interesting part to refer to. I wish Mel could tell us what is behind this remark, but I sense that it will remain unexplained.

Cheers

[d80]

no it doesn,t...its only how to read it

 I appreciate what you have share to us.But please can you give more a tip of how read this tracker? its hard to us try apply it at a roulette.

[Herbyx]

Hi 6-th,
many thanks for your inputs here.

Its in the nature of the topic that questions arise will never stop.

My question: in your tracker, third columns have the headlines
streets: SOP
Lines: LOP

what do these accronyms mean ?  Street  ?  Position

third columns in the tracker: would you agree to say in short: derived of the derived ?

many thanks

[6th-sense]

yes herby...whats moved into those postitions..

[6th-sense]

I appreciate what you have share to us.But please can you give more a tip of how read this tracker? its hard to us try apply it at a roulette.

yes this part ...

The second story concerns my preparation of a lecture on the Pigeon-hole Principle: I wanted to show my students the most spectacular application of it I had ever encountered. I scratched my memory, and then I remembered: the lower bound for the length of the longest monotonic subsequence! I remembered my thrill, but had forgotten the argument, which I then set out to construct.

Let me state the problem first. We consider a sequence of numbers a.i with 0 ≤ i < N. We get a subsequence of length n by removing some N - n elements from the sequence and retaining the remaining ones in their original order. It is called "an upsequence" provided for any a.i and a.j in the subsequence we have

i < j ⇒ a.i ≤ a.j ;
in the case that we have
i < j ⇒ a.i ≥ a.j ;
it is called "a downsequence".
Let U be the length of a longest upsequence contained in the given sequence, and let D be the length of a longest downsequence contained in it. Prove that N ≤ U·D holds.

The argument I came up with went as follows. Let us construct U·D different labels; if we can now devise a regime that assigns to each element a distinct label, then N —being the number of labels used— is at most U·D —being the number of labels available.

How do we construct U·D different labels? The simplest way I can think of is all the integer pairs (u,d), with 1 ≤ u ≤ U and 1 ≤ d ≤ D. How do we devise a regime that assigns for i < j different labels to a.i and a.j? In order to relate the regime to up- and downsequences, we observe

i < j ⇒ a.i ≤ a.j ∨ a.i ≥ a.j
i.e. a.j can be used to extend either an upsequence or a downsequence ending at a.i. This observation reveals the regime: assign to a.i the pair (u,d) with u (and d) the maximum length of an upsequence (and a downsequence respectively) that ends at a.i. This guarantees
that distinct elements get distinct labels
that each label (u,d) needed satisfies 1 ≤ u ≤ U and 1 ≤ d ≤ D.
I was quite pleased with the reconstruction of this argument, until I realized that what used to be "a triumph of the Pigeon-hole Principle" no longer used the Pigeon-hole Principle at all!

I could reintroduce an appeal to the Pigeon-hole Principle, but only by a contorted rephrasing. [ Identify the elements with objects, the labels with compartments; assume N —the number of objects— to exceed U·D —the number of compartments—; then —PP— at least one compartment would contain more than one objects, which contradicts that distinct elements get distinct labels. Hence N does not exceed U·D. ]

Remark It is in this connection noteworthy that no one I asked formulated the Pigeon-hole Principle as follows: " With objects distributed into compartments such that each compartment contains at most one object, the number of objects is at most the number of compartments". It is logically equivalent to the original formulation, but looks quite different. And that, presumably, is precisely the trouble. (End of Remark.)

With its physical, object-oriented formulation, the classical Pigeon-hole Principle is very vivid, almost catchy. And there lie precisely its major shortcomings: the problem caused by such object-oriented formulation is that A ⇒ B and the logically equivalent "counter-positive" ¬B ⇒ ¬A invite completely different visualisations. The moral of the story seems to be that we should sharply distinguish between good mathematics and public relations.

[6th-sense]

this relates also to the partitions used....up sequence and down sequence...of partition lenght...in derived...

[Herbyx]

derived of the derived ?

Hi 6-th,
If I start the tracker, choose as first number: 3
It shows in the columns: 3 3 2

In the last position shouldn't there be a 3 ? 

Have a nice "Palmsonntag",  how would you name the sunday before easter sunday ?

P.S.: of course I study the Dijkstra text, I think it's not the end of the road

[ᶦ ᵃᵐ|Ä-łëx]

Hi 6-th,
If I start the tracker, choose as first number: 3
It shows in the columns: 3 3 2

In the last position shouldn't there be a 3 ? 

Have a nice "Palmsonntag",  how would you name the sunday before easter sunday ?

Herby that 2 is the next pos for the derived to repeat, if you input 2 again that a repeat of the derived. I think

[duchobor]

If I start the tracker, choose as first number: 3
It shows in the columns: 3 3 2

In the last position shouldn't there be a 3 ? 

2 is the number that took place (position 3) from the previous one - in that case, number 3.
1,2,3 ->
3,1,2

good luck

[Herbyx]

Hi duchobor
thanks and thanks to alex,

if you start number 3 comes from position 3 (=derived)
derived of derived should be 3 as well ?

I have to leave now, have all a nice day without war bastards.

[RayManZ]

The last column is the real number behind the position.

So yes. If your first number is 3. It position (derived) will be also 3.

But If you want the derived number to repeat. You have to bet 2. Because that is now in position 3.

[Blueprint]

Like so.

[Blueprint]

Or...

[nottophammer]

Person S
You don't need any trackers.
It's all in Dr Sir Anyone's repeat average

[Herbyx]

Thanks to RayManz and Blueprint