Deeper than primes

[doronshadmi]

jsfisher, please look at this:

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?

 

[doronshadmi]

“(0,0) = ()”: “uncertainty” = 0, “redundancy”=0 ,

No, (0,0) = () = no measured uncertainty or redundancy.

Please look as 0x0 and 1,1 to get it.

Also let us examine this possibility:

Given your assertion that AB is a superposition of A and B it would not be the same "Identity" as A or B. Thus there would be no redundant "Identity" in (A,B,AB). Perhaps you are including the individual ’identities’ of the “Superposition” in your ‘redundant’ consideration. So (A,B,AB) would be one “Redundancy” and (A,B,AB) would be another. Making the total “Redundancy” of (A,B,AB) equal to 2. However that would also make (AA,BB) redundant by the same amount.

Since uncertainty is measured by a single branch then there is no “,” among the possible certain ids of this branch, which are in superposition.

Since redundancy is measured among several branches then there is “,” among the possible certain ids of these branches.

In that case (AB,AB) is 2-uncertainty x 2-redundancy tree, and we get this:

(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.

 

[The Man]

No, (0,0) = () = no measured uncertainty or redundancy.

Hence “uncertainty” = 0, “redundancy”=0

Please look as 0x0 and 1,1 to get it.

Please look at the lack of your ““uncertainty” and “redundancy” “to get it”.

Also let us examine this possibility:

A possibility you specifically renounced

Indeed there is no redundant id in DS (A,B,AB), but there is an uncertain id in DS (A,B,AB) because of AB superposition.

So are you reversing that stance now and AB is not a unique “id”?

Since uncertainty is measured by a single branch then there is no “,” among the possible certain ids of this branch, which are in superposition.

Again Doron you have specifically asserted that you “do not use” superposition in your “OM's superposition”.

Since redundancy is measured among several branches then there is “,” among the possible certain ids of these branches.

In that case (AB,AB) is 2-uncertainty x 2-redundancy tree, and we get this:

So now your are referring to (AB, AB) as one “redundancy” and (AB, AB) as another, but (AB, AB) is not one of your “redundancies”? Curious, as your “uncertainty” seems to be the only thing that is actually redundant, particularly considering that some degree of uncertainty is already what makes “x” a, well, variable).

(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.

Form now on why don’t you just try being specific and consistent about your “expression of k-uncertainty x k-redundancy” instead of just insisting and changing what you consider to be “the right expression of k-uncertainty x k-redundancy tree” because of your uncertainty of your own notions.

 

[jsfisher]

jsfisher, please look at this:
...
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?

Let's start by removing some of the agenda in what you have presented.

I'd start by ignoring your stars-dots-and-bars diagrams. They don't capture enough of the nuances of what you are presenting. Basically, in the 2X2 case, for instance, you are really dealing with objects that are nearly 2-tuples from a set of four symbols. I say "nearly 2-tuples" because you have abandoned the normal order requirement for tuples. The four symbols you used were A, B, AB, and blank. I'd use A, B, C, and D.

Your 2X2 case then becomes:

<A,A>
<A,B>  <B,B>
<A,C>  <B,C>  <C,C>
<A,D>  <B,D>  <C,D>  <D,D>

Curious shape, no? I wonder if it means anything?

In general, the kXk case deals with objects that are unordered k-tuples from a set of 2^k symbols.

The unordered stipulation makes it more complicated than it might have been, moving it from a simple (2^k)^k permutations of 2^k symbols selected k at a time with replacement to something uglier, but I'll let you contemplate some more on how it can be expressed. (Hint: "Combinations with replacement".)

 

[doronshadmi]

Hence “uncertainty” = 0, “redundancy”=0

No measurement is not the same as measurement 0.

So now your are referring to (AB, AB) as one “redundancy” and (AB, AB) as another, but (AB, AB) is not one of your “redundancies”?

By (2,2) = (AB,AB) we mean that 2 is the maxima of 2-Uncertanity or 2-Redundancy, and indeed (AB, AB) is under this maxima because it is only 1-redundancy since you compare between the uncertainty of the branches.

By using the notion of maxima, please explore 3x3 tree in http://www.internationalskeptics.com/forums/showpost.php?p=5942799&postcount=9866.

The four symbols you used were A, B, AB, and blank. I'd use A, B, C, and D.

DS (A,B,C,D) is a 0-Uncertanty x 0-Reduncany case under F (1,1,1,1) of 4-Uncertanty x 4-Reduncany tree.

You still try to reduce ON to 0-Uncertanty x 0-Reduncany case by doing this:

A = A
B = B
C = AB
D = blank

Your 2X2 case then becomes:

<A,A>
<A,B>  <B,B>
<A,C>  <B,C>  <C,C>
<A,D>  <B,D>  <C,D>  <D,D>

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.

 

[doronshadmi]

Here are the detailed examples of k=0 to 3:

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) = ()



3X3                                              
                                                 
A . . .                                          
  | | |                                          
B . . .                                          
  | | |                                          
C ._._.                                          
                                                 
(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) = ()

Maybe I have missed something in 3x3 , so 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 (where k is maxima, as shown in http://www.internationalskeptics.com/forums/showpost.php?p=5942653&postcount=9865).

 

[The Man]

By (2,2) = (AB,AB) we mean

Who is this "we"? Are you just redundant with respect to yourself?

that 2 is the maxima of 2-Uncertanity or 2-Redundancy, and indeed (AB, AB) is under this maxima because it is only 1-redundancy since you compare between the uncertainty of the branches.

Again (AB,AB) is one of your ‘redundancies’ (AB, AB) is another and now (AB, AB) is a third. All from your “uncertainty” only (AB,AB) configuration. So much for your maxima or that “compare between the uncertainty of the branches” excuse you just tried to posit. Doron you are simply making jsfisher's point, that you are so desperate to give any significance to your ‘connect the dot’ diagrams that you are just going to make up whatever excuse it takes (like your “Superposition” where you claim you “do not use” superposition) to make you think you can hold on to whatever imaginary significance you think they have.

 

[jsfisher]

The four symbols you used were A, B, AB, and blank. I'd use A, B, C, and D.

DS (A,B,C,D) is a 0-Uncertanty x 0-Reduncany case under F (1,1,1,1) of 4-Uncertanty x 4-Reduncany tree.

You completely missed it, didn't you? You are so tightly focused on the end point you absolutely refuse to see the path along the way.

I answered the question you posed. All this uncertainty/redundancy baggage is crap you have now added after the fact, after the question was answered.

I laid out a complete 2x2 pattern exactly equivalent to yours. You used the symbols {A, B, AB, blank} while I used {A, B, C, D}. Four symbols either way. Your AB symbol behaves exactly as does my C symbol, but you can't see that, can you?

 

[jsfisher]

Here are the detailed examples of k=0 to 3:
...

Consistency would have you use (3,1,1) instead of (1,1,3). The same is true for (2,1,1) vs. (1,1,2).

Then again, this is the same sloppiness you had the last time you presented this, isn't it? You also left out some valid combinations, just as you did last time.

 

[doronshadmi]

Again (AB,AB) is one of your ‘redundancies’ (AB, AB) is another and now (AB, AB) is a third.

In a k-Uncertainty x k-Redundancy tree you deal with the following questions?:

1) What is the maximum level of Uncertainty of a given branch under a given F (Frame)?

2) What is the maximum level of different redundancies under a given F (Frame)?

A given Redundancy is characterized by invariance under replacement and under F (2,2) there 2 different levels of invariance under replacement, which are (A,A) and (B,B).

(AB,AB) is taken as some case which is equivalent to (A,A) or (B,B) invariance under replacement because you take AB as a one id of invariance under replacement.

By doing that (AB,AB) is a one level Redundancy case (1-Redundancy) under F (2,2) , which is not added (there is no third level) to A and B redundancy levels under F (2,2).

 

[doronshadmi]

I laid out a complete 2x2 pattern exactly equivalent to yours. You used the symbols {A, B, AB, blank} while I used {A, B, C, D}. Four symbols either way. Your AB symbol behaves exactly as does my C symbol, but you can't see that, can you?

You laid out DS (A,B,C,D) that is a 0-Uncertanty x 0-Reduncany case under F (1,1,1,1) of 4-Uncertanty x 4-Reduncany tree.

 

[jsfisher]

You laid out DS (A,B,C,D) that is a 0-Uncertanty x 0-Reduncany case under F (1,1,1,1) of 4-Uncertanty x 4-Reduncany tree.

Repeating it doesn't make it true.

You little 2X2 collection, when stripped of all the nonsense you insist on adding, is nothing more than a set of unordered 2-tuples where each element of the pair is one of four symbols.

But you still can't see that, can you? It doesn't look like what you want it to look like, so you can't accept it. Too bad, that. You question has been answered; how to calculate the number of combinations has been exposed. And you don't get it.


By the way, how is it going correcting your 3X3 example?

 

[doronshadmi]

Repeating it doesn't make it true.

You little 2X2 collection, when stripped of all the nonsense you insist on adding, is nothing more than a set of unordered 2-tuples where each element of the pair is one of four symbols.

But you still can't see that, can you? It doesn't look like what you want it to look like, so you can't accept it. Too bad, that. You question has been answered; how to calculate the number of combinations has been exposed. And you don't get it.


By the way, how is it going correcting your 3X3 example?

Nothing is sripped off, you simply use DS (A,B,C,D) that is a 0-Uncertanty x 0-Reduncany case under F (1,1,1,1) of 4-Uncertanty x 4-Reduncany tree.

 

[The Man]

In a k-Uncertainty x k-Redundancy tree you deal with the following questions?:

No I don’t, as your “k-Uncertainty x k-Redundancy tree” is simply nonsense.

1) What is the maximum level of Uncertainty of a given branch under a given F (Frame)?

2) What is the maximum level of different redundancies under a given F (Frame)?

A given Redundancy is characterized by invariance under replacement and under F (2,2) there 2 different levels of invariance under replacement, which are (A,A) and (B,B).

(AB,AB) is taken as some case which is equivalent to (A,A) or (B,B) invariance under replacement because you take AB as a one id of invariance under replacement.

By doing that (AB,AB) is a one level Redundancy case under F (2,2) , which is not added (there is no third level) to A and B redundancy levels under F (2,2).

So as expected you are simply going to ignore that third “redundancy” to pretend your “maximum level of different redundancies under a given F (Frame)” for your (AB,AB) configuration is just 2. Again, I’m sure that surprises no one.

Firstly “(A,A) and (B,B)” are not your “F (2,2)” they are your ‘F (1,1)’ so your “(AB,AB) is taken as some case which is equivalent to (A,A) or (B,B)” cliam is demonstrably false as is your “invariance under replacement” nonsense (your “direct perception” has failed you again). By your own assertions “AB” is a “one id”, your “id” of “Uncertainty” in your “2-Uncertanty x 2-Reduncany tree”.


Again you're just making jsfisher’s point, stop wasting your time trying to get something, anything, to fit your predetermined 'conclusion' (which is just an extremely poor ‘connect the dots’ drawing) and try actually learning the principles involved. Then guess what, you won’t have to ask others to do your “(DS)”calculations for you or find that you have missed one of your own “Redundancies”.

 

[jsfisher]

Nothing is sripped off, you simply use DS (A,B,C,D) that is a 0-Uncertanty x 0-Reduncany case under F (1,1,1,1) of 4-Uncertanty x 4-Reduncany tree.

Oh, my apologies. It is well established you don't understand mappings, one-to-one correspondence, and isomorphic transformations. Of course you don't comprehend what I posted.

 

[doronshadmi]

Consistency would have you use (3,1,1) instead of (1,1,3). The same is true for (2,1,1) vs. (1,1,2)

Since order has no significance under F (or DS) then F (3,1,1) = F (1,3,1) = F (1,1,3) etc.

In order to get it please understand any given k-Uncertainty x k-Redundancy tree from a 3D view.

For example, let us take DS (A,B,C) case under F (1,1,1) of 3-Uncertainty x 3-Redundancy tree:

 

[doronshadmi]

Firstly “(A,A) and (B,B)” are not your “F (2,2)” they are your ‘F (1,1)

No, they are already used as different redundancy levels under F (2,2).

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.

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).

 

[doronshadmi]

one-to-one correspondence, and isomorphic transformations.

They are particular cases under ONs.

And in your particular case you simply use DS (A,B,C,D) that is a 0-Uncertanty x 0-Reduncany case under F (1,1,1,1) of 4-Uncertanty x 4-Reduncany tree.

 

[jsfisher]

Since order has no significance under F (or DS) then F (3,1,1) = F (1,3,1) = F (1,1,3) etc.

I never stated otherwise. I did, however, comment on your consistency of presentation. Well, inconsistency, actually.

How's it coming with fixing your errors in the presentation of the 3X3 case, by the way?

 

[jsfisher]

They are particular cases under ONs.

And in your particular case you simply use DS (A,B,C,D) that is a 0-Uncertanty x 0-Reduncany case under F (1,1,1,1) of 4-Uncertanty x 4-Reduncany tree.

Please point out one single difference between your 2X2 presentation and mine, other than the minor notational differences and choice of symbols.

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>​

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?