Deeper than primes

Status
Not open for further replies.
jsfisher said:
I am using w through z for my symbols since my previous choice may have confused you. The correspondence of symbols between yours and mine is:

blank <=> w, A <=> x, B <=> y, AB <=> z​


So, naysayer, where are the differences?
Click to expand...

You have no uncertainty in your system.

Also you do not understand that 2X2 is a partial case under 3x3 (DS under F (1,1,1)) , 4x4 (DS under F (1,1,1,1)) , ... etc, where your w,x,y,z case is a partial case under 4x4 (DS under F (1,1,1,1)).

In general, you transtale ON to a collection of distinct elements, where

(2,2) = ( (AB,AB) , (AB,A) , (AB,B) , (AB) , (A,A) , (B,B) , (A,B) , (A) , (B) , () ) can be taken both at-once (in parallel), by step-by-step (is serial, as you do) and by any state between parallel and serial, where each DS is both local and global state of the entire k x k tree.
 
Last edited:
doronshadmi said:
You have no uncertainty in your system.
Click to expand...


Go back and re-read the question you posed, the question that I answered. "Uncertainty" -- which you still have been unable to define -- played no part in the question. There was no implicit meaning in your collection of things, just As and Bs and ABs and blanks.

Besides, if z and AB are equivalent under the mapping I stipulated, than any "uncertainty" you attribute to AB applies to z as well. They are just different symbols for the same thing.

You wanted to know how to calculate the number of objects in the general kXk case. I answered that question.
 
jsfisher said:
"Uncertainty" -- which you still have been unable to define -- played no part in the question
Click to expand...
Wrong, Uncertainty is a superposition between more than one certain possibilities, as long as it is not collapsed into one of the certain possibilities.

jsfisher said:
than any "uncertainty" you attribute to AB applies to z as well.
Click to expand...

Not at all.

z is a certain value which is used to notate AB, where AB is a notation for uncertain value. So uncertain AB value does apply to z certain value, and vice versa.

For example: z is a quantum element (this is a certain knowledge), but its certain place or its certain momentum are not known as long as we do not use the appropriate measurement method in order to measure its certain place or its certain momentum.

Another example: z is certain person in a cross AB roads but the knowledge about certain person z taking A read or B road is uncertain, as long as z does not take one of the roads. So the person is certain but his possible choices are AB uncertain.

More example: z that is a certain notation that is used to notate AB uncertain, where AB uncertain is a superposition of A certain notation and B certain notation. So uncertain AB notation does apply to certain z notation, and vice versa.

-----------------------------

Again, your (w,x,y,z) is a partial case under 4x4 (DS under F (1,1,1,1) as follows ( just a partial case is given ( also there is no uncertainty under F (1,1,1,1) ) ):

(1,1,1,1) =

(w,w,w,w) , (x,x,x,x) , (y,y,y,y) , (z,z,z,z)
(w,w,w,x) , (x,x,x,w) , (y,y,y,z) , (z,z,z,y)
(w,w,w,y) , (x,x,x,z) , (y,y,y,x) , (z,z,z,w)
(w,w,x,x) , (y,y,z,z) , (z,z,w,w) , (w,w,y,y)

…

Your (w,x,y,z)

…
 
Last edited:
doronshadmi said:
z is a certain value which is used to notate AB, where AB is a notation for uncertain value. So uncertain AB value does apply to z certain value, and vice versa.
Click to expand...


Excellent! Now that's the twisted sort of gibberish that keeps them coming back for more. Well done, doron. Although, for someone that belittles notation as meaningless tricks, you certainly put a lot of stock into your own.


At any rate, your question was answered: You have all you need, and far more than you deserve, to calculate the number of objects in any of your kXk collections. Have fun with that.
 
...and lest we forget, how come you haven't fixed your 3X3 presentation, yet? So many errors in such a simple example. We are all embarrassed for you.
 
doronshadmi said:
No, they are already used as different redundancy levels under F (2,2).
Click to expand...

Well if they were (2,2) they wouldn’t be “different redundancy levels under F (2,2)”

doronshadmi said:
Actually F (2,1) , (2,0) , (1,1) , (1,0) and (0,0) are forms under F (2,2) that are used as tools to capture better the different DS under (2,2), so actually 2-Uncertainty x 2-Redundancy tree can be written as:
Click to expand...

Again, hence they are not your “form” “F (2,2)”.

doronshadmi said:
(2,2) = ( (AB,AB) , (AB,A) , (AB,B) , (AB) , (A,A) , (B,B) , (A,B) , (A) , (B) , () ) where their order has no significance.
Click to expand...

The order isn’t the only thing that “has no significance”

doronshadmi said:
Frames are nothing but tools that may be used to define the general formula that returns the number of DS of a given k-Uncertainty x k-Redundancy tree (where k=0 to n, and n is any natural number).
Click to expand...

Unlikely, as this is all just some whimsical flight of fancy on your part. Before it was…

doronshadmi said:
(2,2) = (AB,AB) has 2-uncertainty x 2-redundancy

(2,1) = (AB,A),(AB,B) have 1-uncertainty x 1-redundancy

(2,0) = (AB) has 1-uncertainty x 0-redundancy

(1,1) = (A,A),(B,B) have 0-uncertainty x 1-redundancy ,(A,B) has 0-uncertainty x 0-redundancy

(1,0) = (A),(B) have 0-uncertainty x 0-redundancy

(0,0) = () has no-uncertainty x no-redundancy

Form now on let us take this as the right expression of k-uncertainty x k-redundancy tree.
Click to expand...

Now it is just…

doronshadmi said:
(2,2) = ( (AB,AB) , (AB,A) , (AB,B) , (AB) , (A,A) , (B,B) , (A,B) , (A) , (B) , () ) where their order has no significance.
Click to expand...

The fact of the matter is that you just don’t know what you want to represent with your “Frames".
 
jsfisher said:
Excellent! Now that's the twisted sort of gibberish that keeps them coming back for more. Well done, doron. Although, for someone that belittles notation as meaningless tricks, you certainly put a lot of stock into your own.


At any rate, your question was answered: You have all you need, and far more than you deserve, to calculate the number of objects in any of your kXk collections. Have fun with that.
Click to expand...

As we see 2x2 tree has 10 DS:
doronshadmi said:
Actually F (2,1) , (2,0) , (1,1) , (1,0) and (0,0) are forms under F (2,2) that are used as tools to capture better the different DS under (2,2), so actually 2-Uncertainty x 2-Redundancy tree can be written as:

(2,2) = ( (AB,AB) , (AB,A) , (AB,B) , (AB) , (A,A) , (B,B) , (A,B) , (A) , (B) , () ) where their order has no significance.
Click to expand...


Also we see that there are also 10 DS under F (1,1,0,0) of 4x4 tree:
doronshadmi said:
jsfisher said:
Your 2X2 case then becomes:
Code: 
<A,A>
<A,B>  <B,B>
<A,C>  <B,C>  <C,C>
<A,D>  <B,D>  <C,D>  <D,D>

...

For your reference, here are your 10 objects:

() (A) (B) (AB) (A,A) (A,B) (B,B) (AB,A) (AB,B) (AB,AB)

...and here are mine:

<w,w> <x,w> <y,w> <z,w> <x,x> <x,y> <y,y> <z,x> <z,y> <z,z>
Click to expand...
All you did is to use the case of F (1,1,0,0) under 4-Uncertainy x 4-Rrundancy tree, which has 10 DS.
Click to expand...

In other words jsfisher, two times you used F (1,1,0,0) under 4-Uncertainy x 4-Rrundancy tree (you just changed the notations for certainty), so you did not answer my question, which is:
Code: 
0x0

(0)=()



1x1                                        
                                           
A * .                                        
                                           
(1) = (A)
(0) = ()

                                          
                                 
2X2

(AB,AB) (AB,A)  (AB,B)  (AB)    (A,A)   (B,B)   (A,B)   (A)     (B)     ()

A * *   A * *   A * .   A * .   A * *   A . .   A * .   A * .   A . .   A . .
  | |     | |     | |     | |     | |     | |     | |     | |     | |     | |
B *_*   B *_.   B *_*   B *_.   B ._.   B *_*   B ._*   B ._.   B *_.   B ._.

(2,2) = (AB,AB)
(2,1) = (AB,A),(AB,B)
(2,0) = (AB)
(1,1) = (A,A),(B,B),(A,B)
(1,0) = (A),(B)
(0,0) = ()

By following 0x0,1x1 and 2x2 detailed examples, can you define 3x3 and 4x4 and then the general formula that returns the number of any KxK where K=0 to n, where n is a natural number?

A general formula of k=0 to n (where n is some natural number) actually points out that more DS or F must be defined under a given k-Uncertainty x k-Redundancy tree.
 
Last edited:
The Man said:
The fact of the matter is that you just don’t know what you want to represent with your “Frames".
Click to expand...

The fact is that F (2,1) , (2,0) , (1,1) , (1,0) and (0,0) are frames under F (2,2) such that F (2,2) is the maxima which has the "room" for all the DS of the rest frames.

Let us look at 2x2 tree from its minimal frame to its maximal frame:

F (0,0) = () has only no-uncertainty x no-redundancy.

F (1,0) = (A),(B) have 0-uncertainty x 0-redundancy, and also a "room" for the DS of F (0,0).

F (1,1) = (A,A),(B,B) have 0-uncertainty x 1-redundancy, (A,B) has 0-uncertainty x 0-redundancy, and also a "room" for the DS of F (0,0) and F (1,0).

F (2,0) = (AB) has 1-uncertainty x 0-redundancy, and also a "room" for the DS of F (0,0) and F (1,0).

F (2,1) = (AB,A),(AB,B) have 1-uncertainty x 1-redundancy, and also a "room" for the DS of F (0,0) , F (1,0) and F (1,1).

F (2,2) = (AB,AB) has 2-uncertainty x 2-redundancy, and also a "room" for the DS of F (0,0) , F (1,0) , F (1,1) and F (2,1).

In this representation
F (2,2) = (AB,AB) has 2-uncertainty x 2-redundancy

F (2,1) = (AB,A),(AB,B) have 1-uncertainty x 1-redundancy

F (2,0) = (AB) has 1-uncertainty x 0-redundancy

F (1,1) = (A,A),(B,B) have 0-uncertainty x 1-redundancy ,(A,B) has 0-uncertainty x 0-redundancy

F (1,0) = (A),(B) have 0-uncertainty x 0-redundancy

F (0,0) = () has no-uncertainty x no-redundancy
Click to expand...
I simply wrote only the DS that have the maximum Uncertainty/Redundancy under a given frame of 2x2 tree.

The example of 2x2 tree can be expanded to any k x k tree.
 
Last edited:
doronshadmi said:
so you did not answer my question.
Click to expand...


I did answer your questions. It does not surprise me, though, that you missed it because you are too busy misconstruing everything around the answers.

How are those corrections coming for your 3X3 presentation?
 
jsfisher said:
I did answer your questions.
Click to expand...

I asked for a general formula, instead you have found that a partial case of some k x k tree has the same DS amount of some previous k x k tree, as follows:

Your method works like this:

0x0 tree

F (0) = ()



1X1 tree

F (1) = (A)
F (0) = ()

Now your method is used like this:

You take some DS of 1x1, called (A), that is under F (1), and use it in order to define the amount of DS of 0x0 tree:

A < = = > ()


2X2

(2,2) = (AB,AB)
(2,1) = (AB,A),(AB,B)
(2,0) = (AB)
(1,1) = (A,A),(B,B),(A,B)
(1,0) = (A),(B)
(0,0) = ()

Now you are taking some DS of 4x4 tree called (A,B,C,D), that is under F (1,1,1,1), and use it in order to define the amount of DS of 2x2 tree:

A < = = > (AB)
B < = = > (A)
C < = = > (B)
D < = = > ()

Then you are using 2 places to define the combinations with repetition of 4 symbols, and by doing that you are using F (1,1,0,0) (which is a partial case under 4x4 tree) and show that this partial case has the same amount of 2x2 tree, as follows:

<A,A> < = = > (AB,AB)
<A,B> < = = > (AB,A)
<B,B> < = = > (AB,B)
<A,C> < = = > (AB)
<B,C> < = = > (A,A)
<C,C> < = = > (B,B)
<A,D> < = = > (A,B)
<B,D> < = = > (A)
<C,D> < = = > (B)
<D,D> < = = > ()

Now let us continue to 3x3 tree (I am not sure that I wrote all its DS):

3X3

(3,3,3) = (ABC,ABC,ABC)
(3,3,2) = (ABC,ABC,AB),(ABC,ABC,AC),(ABC,ABC,BC)
(3,3,1) = (ABC,ABC,A),(ABC,ABC,B),(ABC,ABC,C)
(3,3,0) = (ABC,ABC)
(3,2,2) =
(ABC,AB,AB),(ABC,AB,AC),(ABC,AB,BC)
(ABC,AC,AC),(ABC,BC,BC)
(3,2,1) =
(ABC,AB,A),(ABC,AB,B),(ABC,AB,C)
(ABC,AC,A),(ABC,AC,B),(ABC,AC,C)
(ABC,BC,A),(ABC,BC,B),(ABC,BC,C)
(3,2,0) = (ABC,AB),(ABC,AC),(ABC,BC)
(2,2,2) =
(AB,AB,AB),(AB,AC,AB),(AB,BC,AB)
(AC,AC,AC),(AC,AB,AC),(AC,BC,AC)
(BC,BC,BC),(BC,AB,BC),(BC,AC,BC)
(2,2,1) =
(AB,AB,A),(AB,AB,B),(AB,AB,C)
(AB,AC,A),(AB,AC,B),(AB,AC,C)
(AB,BC,A),(AB,BC,B),(AB,BC,C)
(2,2,0) = (AB,AB),(AB,AC),(BC,BC)
(1,1,3) =
(A,A,ABC),(B,B,ABC),(A,B,ABC)
(A,C,ABC),(B,C,ABC)
(3,1,0) = (ABC,A),(ABC,B),(ABC,C)
(3,0,0) = (ABC)
(1,1,2) =
(A,A,AB),(A,A,AC),(A,A,BC)
(B,B,AB),(B,B,AC),(B,B,BC)
(C,C,AB),(C,C,AC),(C,C,BC)
(A,B,AB),(A,B,AC),(A,B,BC)
(A,C,AB),(A,C,AC),(A,C,BC)
(B,C,AB),(B,C,AC),(B,C,BC)
(2,1,0) = (AB,A),(AB,B),(AB,C),(AC,A),(AC,B),(AC,C),(BC,A),(BC,B),(BC,C)
(2,0,0) = (AB),(AC),(BC)
(1,1,1) =
(A,A,A),(B,B,B),(C,C,C)
(A,A,B),(A,A,C),(B,B,A)
(B,B,C),(C,C,A),(C,C,B),(A,B,C)
(1,1,0) = (A,A),(B,B),(C,C),(A,B),(A,C),(C,B)
(1,0,0) = (A),(B),(C)
(0,0,0) = ()

Now let us define what symbols you are using by your method, which are is a 1-1 correspondence with (as you wrongly get it) the distinct values of 3x3 tree:

A < = = > (ABC)
B < = = > (AB)
C < = = > (AC)
D < = = > (BC)
E < = = > (A)
F < = = > (B)
G < = = > (C)
H < = = > ()

So you are using (A,B,C,D,E,F,G,H) that is a particular DS of F (1,1,1,1,1,1,1,1) under 8x8 tree.

Then you are using 3 places to define the combinations with repetition of 8 symbols, and by doing that you are using F (1,1,1,0,0,0,0,0) that is a partial case under 8x8 tree and show that this partial case has the same amount of 2x2 tree, as follows:

Code: 
<A,A,A> < = = > (ABC,ABC,ABC)
<A,A,B> < = = > (ABC,ABC,AB)
<A,A,C> < = = > (ABC,ABC,AC) 
<A,A,D> < = = > (ABC,ABC,BC)
<A,A,E> < = = > (ABC,ABC,A)
<A,A,F> < = = > (ABC,ABC,B)
<A,A,G> < = = > (ABC,ABC,C)
<A,A,H> < = = > (ABC,ABC)
<B,B,A> < = = > (ABC,AB,AB)
<B,B,B> < = = > (ABC,AB,AC)
<B,B,C> < = = > (ABC,AB,BC)                 
<B,B,D> < = = > (ABC,AC,AC)
<B,B,E> < = = > (ABC,BC,BC)                 
<B,B,F> < = = > (ABC,AB,A)
<B,B,G> < = = > (ABC,AB,B)
<B,B,H> < = = > (ABC,AB,C)
<C,C,A> < = = > (ABC,AC,A) 
<C,C,B> < = = > (ABC,AC,B)
<C,C,C> < = = > (ABC,AC,C)
<C,C,D> < = = > (ABC,BC,A) 
<C,C,E> < = = > (ABC,BC,B)
<C,C,F> < = = > (ABC,BC,C)
<C,C,G> < = = > (ABC,AB)
<C,C,H> < = = > (ABC,AC)
<D,D,A> < = = > (ABC,BC)
<D,D,B> < = = > (AB,AB,AB)
<D,D,C> < = = > (AB,AC,AB)
<D,D,D> < = = > (AB,BC,AB)                 
<D,D,E> < = = > (AC,AC,AC)
<D,D,F> < = = > (AC,AB,AC)
<D,D,G> < = = > (AC,BC,AC)                 
<D,D,H> < = = > (BC,BC,BC)
<E,E,A> < = = > (BC,AB,BC)
<E,E,B> < = = > (BC,AC,BC)                 
<E,E,C> < = = > (AB,AB,A)
<E,E,D> < = = > (AB,AB,B)
<E,E,E> < = = > (AB,AB,C)                    
<E,E,F> < = = > (AB,AC,A)
<E,E,G> < = = > (AB,AC,B)
<E,E,H> < = = > (AB,AC,C)                    
<F,F,A> < = = > (AB,BC,A)
<F,F,B> < = = > (AB,BC,B)
<F,F,C> < = = > (AB,BC,C)
<F,F,D> < = = > (AB,AB)
<F,F,E> < = = > (AB,AC) 
<F,F,F> < = = > (BC,BC)
<F,F,G> < = = > (A,A,ABC)
<F,F,H> < = = > (B,B,ABC)
<G,G,A> < = = > (A,B,ABC)                    
<G,G,B> < = = > (A,C,ABC)
<G,G,C> < = = > (B,C,ABC)
<G,G,D> < = = > (ABC,A)
<G,G,E> < = = > (ABC,B)
<G,G,F> < = = > (ABC,C)
<G,G,G> < = = > (ABC)
<G,G,H> < = = > (A,A,AB)
<H,H,A> < = = > (A,A,AC)
<H,H,B> < = = > (A,A,BC)                       
<H,H,C> < = = > (B,B,AB)
<H,H,D> < = = > (B,B,AC)
<H,H,E> < = = > (B,B,BC) 
<H,H,F> < = = > (C,C,AB)
<H,H,G> < = = > (C,C,AC)
<H,H,H> < = = > (C,C,BC)                        
<A,B,C> < = = > (A,B,AB)
<A,B,D> < = = > (A,B,AC)
<A,B,E> < = = > (A,B,BC)                       
<A,B,F> < = = > (A,C,AB)
<A,B,G> < = = > (A,C,AC)
<A,B,H> < = = > (A,C,BC)                       
<A,C,D> < = = > (B,C,AB)
<A,C,E> < = = > (B,C,AC)
<A,C,F> < = = > (B,C,BC)                       
<A,C,G> < = = > (AB,A)
<A,C,H> < = = > (AB,B)
<A,D,E> < = = > (AB,C)
<A,D,F> < = = > (AC,A)
<A,D,G> < = = > (AC,B)
<A,D,H> < = = > (AC,C)
<A,E,F> < = = > (BC,A)
<A,E,G> < = = > (BC,B)
<A,E,H> < = = > (BC,C)
<A,F,G> < = = > (AB)
<A,F,H> < = = > (AC)
<B,C,D> < = = > (BC)                  
<B,C,E> < = = > (A,A,A)
<B,C,F> < = = > (B,B,B)
<B,C,G> < = = > (C,C,C)
<B,C,H> < = = > (A,A,B)
<B,D,E> < = = > (A,A,C)
<A,D,F> < = = > (B,B,A)
<B,D,G> < = = > (B,B,C)
<B,D,H> < = = > (C,C,A)
<B,E,F> < = = > (C,C,B)
<B,E,G> < = = > (A,B,C)
<B,E,H> < = = > (A,A)
<B,F,G> < = = > (B,B)
<B,F,H> < = = > (C,C)
<C,D,E> < = = > (A,B)
<C,D,F> < = = > (A,C)
<C,D,G> < = = > (C,B)
<C,D,H> < = = > (A)
<C,E,F> < = = > (B)
<C,E,G> < = = > (C)
<C,E,H> < = = > ()
<C,F,G> < = = >
<C,F,H> < = = >
<D,F,H> < = = >

After 2x2 it is very hard to find DS by hand, whether they are under F (3,3) of 3x3 tree (where only 3 levels of Uncertainty/Redundancy are allowed) or they are under F (1,1,1,0,0,0,0,0) of 8x8 tree (where only Redundancy is allowed under combinations with repetition of 3 places for 8 distinct symbols).

So jsfisher:

1) You are using a partial case of DS of greater k x k trees in order to define the amount of DS of smaller k x k trees, by using only combinations with repetition, where uncertainty is ignored.

2) By doing that you actually reinforce OMs claim that the amount of X is not sufficient enough to get X complexity as both Uncertainty/Redundancy phenomenon.

3) By your symbolic method, there is a 1-to-1 correspondence between distinct z and the entire universe, for example:

z < = = > The entire universe

which does not provide enough knowledge in order to get the entire universe's Complexity.

This is of course an extreme example, but it starts form this simple example:

z < = = > (AB)

which does not provide enough knowledge in order to get (AB) Complexity (and in this case (AB) uncertainty).
 
Last edited:
doronshadmi said:
I asked for a general formula
Click to expand...

...and I told you rather clearly how to calculate the size of each of your tree sets. You, however, are so focused on notation and some imagined higher meaning you absolutely could not accept an alternate to your own.

For the purposes of counting the number of objects generated in one of your kXk thingies, it doesn't matter one iota what the objects represent in your barren imagination. If I choose to use the symbol X to mean 3Y^72, then that's what X will mean. If I choose the symbol z to mean what you interpret AB to mean, well, that's ok, too. But the point is the choice of symbols doesn't change the final count.

...
After 2x2 it is very hard to find DS by hand
Click to expand...

Nonsense. It is trivial my way. Your way, a little more clumsy because of the unnecessary notational baggage, but still trivial. It does, however, get very laborious beginning at 4X4.



Any guesses about the significance of the number, 131,254,487,936?
 
...and are you ever going to correct your 3X3 case?

How can anyone take you seriously if everything you do is so riddled with such simple errors? Moreover how important can these uncertain/redundant thingies be when parts can go missing and not be noticed?

Your blunders tell us clearly that you just don't get this OM stuff.
 
doronshadmi said:
Which is not a formula.
Click to expand...

You can't make the leap between the two? How sad for you.


You can't because z is certain and AB is uncertain.
Click to expand...

Did you not understand this part: "For the purposes of counting the number of objects generated in one of your kXk thingies, it doesn't matter one iota what the objects represent"?

Remember the question you posed, the question that I have answered? It was all about "how many".


Now, go and fix your 3X3 presentation. Your blunders continue to embarrass you.
 
jsfisher said:
Moreover how important can these uncertain/redundant thingies be when parts can go missing and not be noticed?
Click to expand...

The important thing here is to understand Uncertainty/Redundancy Complexity.

ONs are too complex to be drawn by hand.
 
jsfisher said:
You can't make the leap between the two? How sad for you.




Did you not understand this part: "For the purposes of counting the number of objects generated in one of your kXk thingies, it doesn't matter one iota what the objects represent"?

Remember the question you posed, the question that I have answered? It was all about "how many".


Now, go and fix your 3X3 presentation. Your blunders continue to embarrass you.
Click to expand...
jsfisher you are a mathematician, so please finish the work and define the general formula "For the purposes of counting " by your combinations with repetition method.
 
Last edited:
doronshadmi said:
The important thing here is to understand Uncertainty/Redundancy Complexity.
Click to expand...

Complexity isn't that hard to understand. At least that's been my experience in my few discussions with him.

ONs are too complex to be drawn by hand.
Click to expand...

No, actually, your so-called organic numbers have defied definition because of all the contradiction you've introduced. You've confused contraction with complexity. Not a common error, but understandable, I guess.
 
Last edited:
doronshadmi said:
jsfisher you are a mathematician, so please finish the work and define the general formula for your combinations with repetition method.
Click to expand...


It's all finished, but you've told me my work was all wrong. You don't want something that's all wrong, do you? That would just increase the wrongness you've already introduced on your own.

Besides, you can do your own homework.
 
jsfisher said:
It's all finished, but you've told me my work was all wrong.
Click to expand...
It is wrong because it uses F and DS of greater k x k tree in order to define the amount of DS of smaller k x k tree by ignoring uncertainty, which appears also in the greater k x k tree and so on ... ad infinitum.

So your view of this subject is limited.

Any way you did not define the general formula "For the purposes of counting" by your combinations with repetition method.
 
jsfisher said:
Are you suggesting Complexity is in the Q category of GLBTQ? I don't think so. He has been clear on his preferences.

If you are uncertain, perhaps you should seek his council.
Click to expand...

jsfisher said:
Your AB symbol behaves exactly as does my C symbol, but you can't see that, can you?
Click to expand...
Your C symbol does not behave exactly as does my AB symbol, because AB is a symbol for Uncertainty and C is a symbol fo Certainty.

Certainty and Uncertainty are two different qualities that are not the same exactly as a point and a line are two qualities that are not the same, for example:

Uncertainty ≠ Certainty

Line ≠ Point

What you are doing is this:

Uncertainty < = = > Certainty

Line < = = > Point

But:

(Uncertainty < = = > Certainty) ≠ (Uncertainty ≠ Certainty)

(Line < = = > Point) ≠ (Line ≠ Point)

and you still do not get it.
 
Last edited:
doronshadmi said:
Your C symbol does not behave exactly as does my AB symbol, because AB is a symbol for Uncertainty and C is a symbol fo Certainty.

Certainty and Uncertainty are two different qualities that are not the same exactly as a point and a line are two qualities that are not the same, for example:

Uncertainty ≠ Certainty

Line ≠ Point

What you are doing is this:

Uncertainty < = = > Certainty

Line < = = > Point

But:

(Uncertainty < = = > Certainty) ≠ (Uncertainty ≠ Certainty)

(Line < = = > Point) ≠ (Line ≠ Point)

and you still do not get it.
Click to expand...

He is so focused on his crap. Can't even understand a joke or maybe it is English he does not understand. Very sad for him anyway.
 
sympathic said:
He is so focused on his crap. Can't even understand a joke or maybe it is English he does not understand. Very sad for him anyway.
Click to expand...


It is a combination of sharp focus and an overwhelming desire to contradict. His stuff must stand as correct, and everything else must be dismissed as wrong...at all cost.

With "it's wrong" or "you're wrong" as a primary goal, there is little use wasting time with reading comprehension. He already knows the conclusion he wants.
 
jsfisher said:
...and are you ever going to correct your 3X3 case?
Click to expand...

Already corrected (by adding (C,C,AB), (C,C,AC) and (C,C,BC)) both in
http://www.internationalskeptics.com/forums/showpost.php?p=5934321&postcount=9831 and http://www.internationalskeptics.com/forums/showpost.php?p=5953576&postcount=9891 .

Yet there are more DS under F (1,1,1,0,0,0,0,0) of 8x8 tree than all the DS under 3x3 tree, as can be seen in http://www.internationalskeptics.com/forums/showpost.php?p=5953576&postcount=9891 .

You are invited to provide the missing DS of 3x3 tree in order to show the 1-to-1 correspondence between all the DS under F (1,1,1,0,0,0,0,0) of 8x8 tree and all the DS under 3x3 tree.

Another option is that I have added too many (or even missed some of the) DS of F (1,1,1,0,0,0,0,0) of 8x8 tree.

If nothing is wrong in the mapping in http://www.internationalskeptics.com/forums/showpost.php?p=5953576&postcount=9891 , then your method fails at 3x3 case.

So please try to show the correct detailed 1-to-1 correspondence between all the DS under F (1,1,1,0,0,0,0,0) of 8x8 tree and all the DS under 3x3 tree, by using http://www.internationalskeptics.com/forums/showpost.php?p=5953576&postcount=9891.
 
Last edited:
doronshadmi said:
Already corrected (by adding (C,C,AB), (C,C,AC) and (C,C,BC)) both in
http://www.internationalskeptics.com/forums/showpost.php?p=5934321&postcount=9831 and http://www.internationalskeptics.com/forums/showpost.php?p=5953576&postcount=9891 .
Click to expand...


Is that all you are going to fix?

One would have to conclude that being right or wrong doesn't matter...and that would be evidence this whole kXk blunder of yours is without significance.

Will you be reaching this conclusion anytime soon? The rest of us are waiting for you to catch up.
 
jsfisher said:
Is that all you are going to fix?

One would have to conclude that being right or wrong doesn't matter...and that would be evidence this whole kXk blunder of yours is without significance.

Will you be reaching this conclusion anytime soon? The rest of us are waiting for you to catch up.
Click to expand...

Here is the last fixed version of 3x3 tree ( I missed DS under F (2,2,2) , F (2,2,1) and F (2,2,0) ). But now there are less elements according to your method. Can you find them?:

Code: 
<A,A,A> < = = > (ABC,ABC,ABC)
<A,A,B> < = = > (ABC,ABC,AB)
<A,A,C> < = = > (ABC,ABC,AC)
<A,A,D> < = = > (ABC,ABC,BC)
<A,A,E> < = = > (ABC,ABC,A)
<A,A,F> < = = > (ABC,ABC,B)
<A,A,G> < = = > (ABC,ABC,C)
<A,A,H> < = = > (ABC,ABC)
<B,B,A> < = = > (ABC,AB,AB)
<B,B,B> < = = > (ABC,AB,AC)
<B,B,C> < = = > (ABC,AB,BC)
<B,B,D> < = = > (ABC,AC,AC)
<B,B,E> < = = > (ABC,BC,BC)
<B,B,F> < = = > (ABC,AB,A)
<B,B,G> < = = > (ABC,AB,B)
<B,B,H> < = = > (ABC,AB,C)
<C,C,A> < = = > (ABC,AC,A)
<C,C,B> < = = > (ABC,AC,B)
<C,C,C> < = = > (ABC,AC,C)
<C,C,D> < = = > (ABC,BC,A)
<C,C,E> < = = > (ABC,BC,B)
<C,C,F> < = = > (ABC,BC,C)
<C,C,G> < = = > (ABC,AB)
<C,C,H> < = = > (ABC,AC)
<D,D,A> < = = > (ABC,BC)
<D,D,B> < = = > (AB,AB,AB)
<D,D,C> < = = > (AB,AB,AC)
<D,D,D> < = = > (AB,AC,AC)
<D,D,E> < = = > (AC,AC,AC)
<D,D,F> < = = > (AB,AB,BC)
<D,D,G> < = = > (AB,AC,BC)
<D,D,H> < = = > (AB,BC,BC)
<E,E,A> < = = > (AC,AC,BC)
<E,E,B> < = = > (AC,BC,BC)
<E,E,C> < = = > (BC,BC,BC)
<E,E,D> < = = > (AB,AB,A)
<E,E,E> < = = > (AB,AC,A)
<E,E,F> < = = > (AB,BC,A)
<E,E,G> < = = > (AC,AC,A)
<E,E,H> < = = > (AC,BC,A)
<F,F,A> < = = > (BC,BC,A)
<F,F,B> < = = > (AB,AB,B)
<F,F,C> < = = > (AB,AC,B)
<F,F,D> < = = > (AB,BC,B)
<F,F,E> < = = > (AC,AC,B)
<F,F,F> < = = > (AC,BC,B)
<F,F,G> < = = > (BC,BC,B)
<F,F,H> < = = > (AB,AB,C)
<G,G,A> < = = > (AB,AC,C)
<G,G,B> < = = > (AB,BC,C)
<G,G,C> < = = > (AC,AC,C)
<G,G,D> < = = > (AC,BC,C)
<G,G,E> < = = > (BC,BC,C)
<G,G,F> < = = > (AB,AB)
<G,G,G> < = = > (AB,AC)
<G,G,H> < = = > (AB,BC)
<H,H,A> < = = > (AC,AC)
<H,H,B> < = = > (AC,BC)
<H,H,C> < = = > (BC,BC)
<H,H,D> < = = > (A,A,ABC)
<H,H,E> < = = > (B,B,ABC)
<H,H,F> < = = > (A,B,ABC)
<H,H,G> < = = > (A,C,ABC)
<H,H,H> < = = > (B,C,ABC)
<A,B,C> < = = > (ABC,A)
<A,B,D> < = = > (ABC,B)
<A,B,E> < = = > (ABC,C)
<A,B,F> < = = > (ABC)
<A,B,G> < = = > (A,A,AB)
<A,B,H> < = = > (A,A,AC)
<A,C,D> < = = > (A,A,BC)
<A,C,E> < = = > (B,B,AB)
<A,C,F> < = = > (B,B,AC)
<A,C,G> < = = > (B,B,BC)
<A,C,H> < = = > (C,C,AB)
<A,D,E> < = = > (C,C,AC)
<A,D,F> < = = > (C,C,BC)
<A,D,G> < = = > (A,B,AB)
<A,D,H> < = = > (A,B,AC)
<A,E,F> < = = > (A,B,BC)
<A,E,G> < = = > (A,C,AB)
<A,E,H> < = = > (A,C,AC)
<A,F,G> < = = > (A,C,BC)
<A,F,H> < = = > (B,C,AB)
<B,C,D> < = = > (B,C,AC)
<B,C,E> < = = > (B,C,BC)
<B,C,F> < = = > (AB,A)
<B,C,G> < = = > (AB,B)
<B,C,H> < = = > (AB,C)
<B,D,E> < = = > (AC,A)
<A,D,F> < = = > (AC,B)
<B,D,G> < = = > (AC,C)
<B,D,H> < = = > (BC,A)
<B,E,F> < = = > (BC,B)
<B,E,G> < = = > (BC,C)
<B,E,H> < = = > (AB)
<B,F,G> < = = > (AC)
<B,F,H> < = = > (BC)
<C,D,E> < = = > (A,A,A)
<C,D,F> < = = > (B,B,B)
<C,D,G> < = = > (C,C,C)
<C,D,H> < = = > (A,A,B)
<C,E,F> < = = > (A,A,C)
<C,E,G> < = = > (B,B,A)
<C,E,H> < = = > (B,B,C)
<C,F,G> < = = > (C,C,A)
<C,F,H> < = = > (C,C,B)
<D,F,H> < = = > (A,B,C)
                (A,A)
                (B,B)
                (C,C)
                (A,B)
                (A,C)
                (C,B)
                (A)
                (B)
                (C)
                ()

3X3 tree:

(3,3,3) = (ABC,ABC,ABC)
(3,3,2) = (ABC,ABC,AB),(ABC,ABC,AC),(ABC,ABC,BC)
(3,3,1) = (ABC,ABC,A),(ABC,ABC,B),(ABC,ABC,C)
(3,3,0) = (ABC,ABC)
(3,2,2) =
(ABC,AB,AB),(ABC,AB,AC),(ABC,AB,BC)
(ABC,AC,AC),(ABC,BC,BC)
(3,2,1) =
(ABC,AB,A),(ABC,AB,B),(ABC,AB,C)
(ABC,AC,A),(ABC,AC,B),(ABC,AC,C)
(ABC,BC,A),(ABC,BC,B),(ABC,BC,C)
(3,2,0) = (ABC,AB),(ABC,AC),(ABC,BC)
(2,2,2) =
(AB,AB,AB)
(AB,AB,AC)
(AB,AC,AC)
(AC,AC,AC)
(AB,AB,BC)
(AB,AC,BC)
(AB,BC,BC)
(AC,AC,BC)
(AC,BC,BC)
(BC,BC,BC)
(2,2,1) =
(AB,AB,A)
(AB,AC,A)
(AB,BC,A)
(AC,AC,A)
(AC,BC,A)
(BC,BC,A)
(AB,AB,B)
(AB,AC,B)
(AB,BC,B)
(AC,AC,B)
(AC,BC,B)
(BC,BC,B)
(AB,AB,C)
(AB,AC,C)
(AB,BC,C)
(AC,AC,C)
(AC,BC,C)
(BC,BC,C)
(2,2,0) =
(AB,AB)
(AB,AC)
(AB,BC)
(AC,AC)
(AC,BC)
(BC,BC)
(1,1,3) =
(A,A,ABC),(B,B,ABC),(A,B,ABC)
(A,C,ABC),(B,C,ABC)
(3,1,0) = (ABC,A),(ABC,B),(ABC,C)
(3,0,0) = (ABC)
(1,1,2) =
(A,A,AB),(A,A,AC),(A,A,BC)
(B,B,AB),(B,B,AC),(B,B,BC)
(C,C,AB),(C,C,AC),(C,C,BC)
(A,B,AB),(A,B,AC),(A,B,BC)
(A,C,AB),(A,C,AC),(A,C,BC)
(B,C,AB),(B,C,AC),(B,C,BC)
(2,1,0) = (AB,A),(AB,B),(AB,C),(AC,A),(AC,B),(AC,C),(BC,A),( BC,B),(BC,C)
(2,0,0) = (AB),(AC),(BC)
(1,1,1) =
(A,A,A),(B,B,B),(C,C,C)
(A,A,B),(A,A,C),(B,B,A)
(B,B,C),(C,C,A),(C,C,B),(A,B,C)
(1,1,0) = (A,A),(B,B),(C,C),(A,B),(A,C),(C,B)
(1,0,0) = (A),(B),(C)
(0,0,0) = ()
 
Last edited:
sympathic said:
that's putting it mildldy. He is obsessed. That's why we will never be able to explain how wrong he is.
Click to expand...

sympathic you are obsessed to reduce everything into sharp forms.

As a result all you get is sums.

Since you are an expert of sharp forms, maybe you can halp jsfisher to find the missing forms in my latest correction (now there are 11 DS of 3x3 tree that are not covered by the DS of F (1,1,1,0,0,0,0,0) under 8x8 tree):

Here is the last fixed version of 3x3 tree ( I missed DS under F (3,2,2) , F (2,2,2) , F (2,2,1) and F (2,2,0) ). But now there are less elements according to jsfisher's method. Can you find them?:

Code: 
<A,A,A> < = = > (ABC,ABC,ABC)
<A,A,B> < = = > (ABC,ABC,AB)
<A,A,C> < = = > (ABC,ABC,AC)
<A,A,D> < = = > (ABC,ABC,BC)
<A,A,E> < = = > (ABC,ABC,A)
<A,A,F> < = = > (ABC,ABC,B)
<A,A,G> < = = > (ABC,ABC,C)
<A,A,H> < = = > (ABC,ABC)  
<B,B,A> < = = > (ABC,AB,AB)
<B,B,B> < = = > (ABC,AB,AC)
<B,B,C> < = = > (ABC,AB,BC)
<B,B,D> < = = > (ABC,AC,AC)
<B,B,E> < = = > (ABC,AC,BC)
<B,B,F> < = = > (ABC,BC,BC)
<B,B,G> < = = > (ABC,AB,A) 
<B,B,H> < = = > (ABC,AB,B) 
<C,C,A> < = = > (ABC,AB,C) 
<C,C,B> < = = > (ABC,AC,A) 
<C,C,C> < = = > (ABC,AC,B) 
<C,C,D> < = = > (ABC,AC,C) 
<C,C,E> < = = > (ABC,BC,A) 
<C,C,F> < = = > (ABC,BC,B) 
<C,C,G> < = = > (ABC,BC,C) 
<C,C,H> < = = > (ABC,AB)   
<D,D,A> < = = > (ABC,AC)   
<D,D,B> < = = > (ABC,BC)   
<D,D,C> < = = > (AB,AB,AB) 
<D,D,D> < = = > (AB,AB,AC) 
<D,D,E> < = = > (AB,AC,AC) 
<D,D,F> < = = > (AC,AC,AC) 
<D,D,G> < = = > (AB,AB,BC) 
<D,D,H> < = = > (AB,AC,BC) 
<E,E,A> < = = > (AB,BC,BC) 
<E,E,B> < = = > (AC,AC,BC) 
<E,E,C> < = = > (AC,BC,BC) 
<E,E,D> < = = > (BC,BC,BC) 
<E,E,E> < = = > (AB,AB,A)  
<E,E,F> < = = > (AB,AC,A)  
<E,E,G> < = = > (AB,BC,A)  
<E,E,H> < = = > (AC,AC,A)  
<F,F,A> < = = > (AC,BC,A)  
<F,F,B> < = = > (BC,BC,A)  
<F,F,C> < = = > (AB,AB,B)  
<F,F,D> < = = > (AB,AC,B)  
<F,F,E> < = = > (AB,BC,B)  
<F,F,F> < = = > (AC,AC,B)  
<F,F,G> < = = > (AC,BC,B)  
<F,F,H> < = = > (BC,BC,B)  
<G,G,A> < = = > (AB,AB,C)  
<G,G,B> < = = > (AB,AC,C)  
<G,G,C> < = = > (AB,BC,C)  
<G,G,D> < = = > (AC,AC,C)  
<G,G,E> < = = > (AC,BC,C)  
<G,G,F> < = = > (BC,BC,C)  
<G,G,G> < = = > (AB,AB)    
<G,G,H> < = = > (AB,AC)    
<H,H,A> < = = > (AB,BC)    
<H,H,B> < = = > (AC,AC)    
<H,H,C> < = = > (AC,BC)    
<H,H,D> < = = > (BC,BC)    
<H,H,E> < = = > (A,A,ABC)  
<H,H,F> < = = > (B,B,ABC)  
<H,H,G> < = = > (A,B,ABC)  
<H,H,H> < = = > (A,C,ABC)  
<A,B,C> < = = > (B,C,ABC)  
<A,B,D> < = = > (ABC,A)    
<A,B,E> < = = > (ABC,B)    
<A,B,F> < = = > (ABC,C)    
<A,B,G> < = = > (ABC)      
<A,B,H> < = = > (A,A,AB)   
<A,C,D> < = = > (A,A,AC)   
<A,C,E> < = = > (A,A,BC)   
<A,C,F> < = = > (B,B,AB)   
<A,C,G> < = = > (B,B,AC)   
<A,C,H> < = = > (B,B,BC)   
<A,D,E> < = = > (C,C,AB)   
<A,D,F> < = = > (C,C,AC)   
<A,D,G> < = = > (C,C,BC)   
<A,D,H> < = = > (A,B,AB)   
<A,E,F> < = = > (A,B,AC)   
<A,E,G> < = = > (A,B,BC)   
<A,E,H> < = = > (A,C,AB)   
<A,F,G> < = = > (A,C,AC)   
<A,F,H> < = = > (A,C,BC)   
<B,C,D> < = = > (B,C,AB)   
<B,C,E> < = = > (B,C,AC)   
<B,C,F> < = = > (B,C,BC)   
<B,C,G> < = = > (AB,A)     
<B,C,H> < = = > (AB,B)     
<B,D,E> < = = > (AB,C)     
<A,D,F> < = = > (AC,A)     
<B,D,G> < = = > (AC,B)     
<B,D,H> < = = > (AC,C)     
<B,E,F> < = = > (BC,A)     
<B,E,G> < = = > (BC,B)     
<B,E,H> < = = > (BC,C)     
<B,F,G> < = = > (AB)       
<B,F,H> < = = > (AC)       
<C,D,E> < = = > (BC)       
<C,D,F> < = = > (A,A,A)    
<C,D,G> < = = > (B,B,B)    
<C,D,H> < = = > (C,C,C)    
<C,E,F> < = = > (A,A,B)    
<C,E,G> < = = > (A,A,C)    
<C,E,H> < = = > (B,B,A)    
<C,F,G> < = = > (B,B,C)    
<C,F,H> < = = > (C,C,A)    
<D,F,H> < = = > (C,C,B)    
                (A,B,C)    
                (A,A)      
                (B,B)      
                (C,C)      
                (A,B)      
                (A,C)      
                (C,B)      
                (A)        
                (B)        
                (C)        
                ()


3X3 tree:

(3,3,3) = (ABC,ABC,ABC)
(3,3,2) = (ABC,ABC,AB),(ABC,ABC,AC),(ABC,ABC,BC)
(3,3,1) = (ABC,ABC,A),(ABC,ABC,B),(ABC,ABC,C)
(3,3,0) = (ABC,ABC)
(3,2,2) =
(ABC,AB,AB)
(ABC,AB,AC)
(ABC,AB,BC)
(ABC,AC,AC)
(ABC,AC,BC)
(ABC,BC,BC)
(3,2,1) =
(ABC,AB,A),(ABC,AB,B),(ABC,AB,C)
(ABC,AC,A),(ABC,AC,B),(ABC,AC,C)
(ABC,BC,A),(ABC,BC,B),(ABC,BC,C)
(3,2,0) = (ABC,AB),(ABC,AC),(ABC,BC)
(2,2,2) =
(AB,AB,AB)
(AB,AB,AC)
(AB,AC,AC)
(AC,AC,AC)
(AB,AB,BC)
(AB,AC,BC)
(AB,BC,BC)
(AC,AC,BC)
(AC,BC,BC)
(BC,BC,BC)
(2,2,1) =
(AB,AB,A)
(AB,AC,A)
(AB,BC,A)
(AC,AC,A)
(AC,BC,A)
(BC,BC,A)
(AB,AB,B)
(AB,AC,B)
(AB,BC,B)
(AC,AC,B)
(AC,BC,B)
(BC,BC,B)
(AB,AB,C)
(AB,AC,C)
(AB,BC,C)
(AC,AC,C)
(AC,BC,C)
(BC,BC,C)
(2,2,0) =
(AB,AB)
(AB,AC)
(AB,BC)
(AC,AC)
(AC,BC)
(BC,BC)
(1,1,3) =
(A,A,ABC),(B,B,ABC),(A,B,ABC)
(A,C,ABC),(B,C,ABC)
(3,1,0) = (ABC,A),(ABC,B),(ABC,C)
(3,0,0) = (ABC)
(1,1,2) =
(A,A,AB),(A,A,AC),(A,A,BC)
(B,B,AB),(B,B,AC),(B,B,BC)
(C,C,AB),(C,C,AC),(C,C,BC)
(A,B,AB),(A,B,AC),(A,B,BC)
(A,C,AB),(A,C,AC),(A,C,BC)
(B,C,AB),(B,C,AC),(B,C,BC)
(2,1,0) = (AB,A),(AB,B),(AB,C),(AC,A),(AC,B),(AC,C),(BC,A),( BC,B),(BC,C)
(2,0,0) = (AB),(AC),(BC)
(1,1,1) =
(A,A,A),(B,B,B),(C,C,C)
(A,A,B),(A,A,C),(B,B,A)
(B,B,C),(C,C,A),(C,C,B),(A,B,C)
(1,1,0) = (A,A),(B,B),(C,C),(A,B),(A,C),(C,B)
(1,0,0) = (A),(B),(C)
(0,0,0) = ()
 
Last edited:
Status
Not open for further replies.

Back
Top Bottom