37 back to basics

[6th-sense]

And you also haven't explained the different php the professor was speaking about.  So we can't even say yea the tracker shows this because we don't know what we looking for

and this my friend is how members feel about what you are saying....

the difference here is that YES i will explain it openely....i can do this without showing how to bet...and can do this for the concrete evidence that herby and person s want...that at least a portion of the fantasy is viable...or seen

[MoneyT101]

and this my friend is how members feel about what you are saying....

the difference here is that YES i will explain it openely....i can do this without showing how to bet...and can do this for the concrete evidence that herby and person s want...that at least a portion of the fantasy is viable...or seen

Ive explained openly and I wish I can delete it, honestly!

For example I explained the whole php 5 hits in 13 spins and you told me it's not php lol

So if I explain something and you don't understand it.  How is that my fault?

You have yet to explain anything so it's not the same.  I explained many things.  Not my fault ppl can't comprehend. I did my part in explaining.  The only other thing is for me to sit down and tell you what to do in a step by step, right down to the last detail.

[MoneyT101]

I have it all wrong and you win.

I'm going back to being a student and learn! 


Everyone ignore everything I said! I don't know anything and I am ready to learn

[6th-sense]

For example I explained the whole php 5 hits in 13 spins and you told me it's not php lol

i only said this is not the dyslexic php based concept....its just a rehash of normal php which can,t be used in this game...

[6th-sense]

The only other thing is for me to sit down and tell you what to do in a step by step, right down to the last detail.

i think i have to...though its not rocket science

[6th-sense]

I just want you to keep in mind this is nothing new!

Dykselic said... "Roulette has NOTHING to do with numbers if you replaced the numbers with pictures of bunny rabbits, this mathematical principle would STILL hold true"

Dykselic said... "RNG Roulette as we all KNOW consists of 37 SEPARATE numbers. However, as I previously explained, the numbers are really just 'PLACE' holders,
 and could easily be replaced with ANY other 37 'PLACE' holders (e.g 37 colours, 37 animals, 37 pictures of 'Forum Haters' etc."

you should not be posting this as though you know...

[6th-sense]

now its time to school you...why this above has been said

[6th-sense]

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.

Austin, 19 September 1986

prof. dr. Edsger W. Dijkstra
Department of Computer Sciences
The University of Texas at Austin
Austin, TX 78750–1188
United States of America

transcribed by Martijn van der Veen
revised Thu, 27 May 2010

[6th-sense]

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.

[6th-sense]

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

your derived

[6th-sense]

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!

[6th-sense]

right lets just go through what is being said..the purified pigeonhole concept...

YOU cannot use the normal php ...how can you

[6th-sense]

Of course I know the Pigeon-hole Principle, but I never.....", and then wandering thoughts prevented him from completing the sentence. Was he still looking for the objects and the compartments? The way he had been introduced to the principle and all the imaginations that went with it was obviously a barrier to its straight-forward application. I felt that I had had a glimpse of something frightening.

[6th-sense]

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

[6th-sense]

how can you assign a php principle to objects already in there pigeonholes...or make compartments from what you have?

for us its easy....i can explain this to a 5 year old....the compartments are ALREADY preset in roulette...they are called partitions