Thursday, January 15, 2015

Demonstrating all the Headache that Can Happen When You Use the Wrong Method

Categorical Deduction for 64-Sq. Diagrams

Note: this paper shows the WRONG METHOD, although with a lot of promise. For the real method, see: https://www.academia.edu/10161352/Categorical_Deduction_for_64_Categories


[THIS METHOD REMAINS INCOMPLETE! (HERE BUT NOT THERE)]



Recall that the method for 16-SQUARE DIAGRAMS


Involved FOUR DEDUCTIONS


Those deductions were:

[A] ABFE-CDHG-KOPL-IMNJ
[B] ABFE-CGHD-KOPL-IJNM
[C] AEFB-CDHG-KLPO-IMNJ
[D] AEFB-CGHD-KLPO-IJNM


Each of the letters shown in the
previous diagram


DCBA
GHFE…

LKJI
PONM


Now refers to four squares in the new diagram.


The new diagram has:

 A first row:
8,7,6,5,4,3,2,1

A second row: 16,15,
14,13,12,11,10,9

A third row: 24,23,
22,21,20,19,18,17

A fourth row: 32, 31,
30,29,28,27,26,25

A fifth row: 40,39,
38,37,36,35,34,33

A sixth row: 48,47,
46,45,44,43,42,41

A seventh row: 56,55,
54,53,52,51,50,49

And,


An eighth row: 64, 63
62,61,60,59,58,57


Remember, there are other
methods

For showing the categories


The categories could be shown
cyclically


Within each quadrant for example.


I will keep the method I showed here


Because I feel it is the most objective.


Now remember, each of the
categories from 16 square


Corresponds to four of the categoiries
In 64 squares.


64 = 16 X 4


Otherwise the method would not
Reveal itself


So easily.


Now we know that:


‘A’ refers to
1.2,10,9


[Listed in cyclical order]

‘B’ refers to
3,4, 12,11

‘C’ refers to
5,6,14,13

‘D’ refers to
7,8, 16,15

‘E’ refers to
17,18,26,25

‘F’ refers to
19,20,28,27

‘G’ refers to
21,22,30,29

‘H’ refers to
23,24,32,31

‘I’ refers to
33,34,42,41

‘J’ refers to
35,36,44,43


‘K’ refers to
37,38,46,45

‘L’ refers to
39,40,48,47

‘M’ refers to
49,50,58,57

‘N’ refers to
51,52,60,59

‘O’ refers to
53,54,62,61

‘P’ refers to
55,56,64,63


Now, we know that opposite
numbers do not combine!


In terms of coherent quadra,


That means position A

Does not go with position C

Position B

Does not go with position D


That process was used once in
The 16-SQ diagram


Now we apply it again in the
64-SQ diagram


AGAIN,

The deductions for 16-SQ.
WERE:


[A] ABFE-CDHG-KOPL-IMNJ

[B] ABFE-CGHD-KOPL-IJNM

[C] AEFB-CDHG-KLPO-IMNJ

[D] AEFB-CGHD-KLPO-IJNM


So, the deductions for 64-SQ.


Merely involve:


Applying the cyclic order


For each quadrant


(within the numbers…)



SUCH THAT:


The two combinations for
Each cycle


Are maintained AT
EVERY SET LEVEL


Before we ascertain that,



We must find the quadra for every level.


The first two quadrant levels refer to the
16-SQ. diagram.


The third quadrant level refers to the
numbers.


We have already determined
the corresponding numbers


Quadrant A refers to:
ABFE

Quadrant B refers to:
CDHG

Quadrant C refers to:
KLPO

Quadrant D refers to:
IJNM


Now we substitute the
numbers:



QUADRANT A =

A: [1.2,10,9]
B: [3,4, 12,11]

F: [19,20,28,27]
E: [17,18,26,25]


QUADRANT B =



C: [5,6,14,13]
D: [7,8, 16,15]

H: [23,24,32,31]
G: [21,22,30,29]



QUADRANT C =

K: [37,38,46,45]
L: [39,40,48,47]

P: [55,56,64,63]
O: [53,54,62,61]


QUADRANT D =

I: [33,34,42,41]
J: [35,36,44,43]

N: [51,52,60,59]
M: [49,50,58,57]


NOW,


Part A of every level
Only relates with part B,D


Of every level


Part B of every level
Only relates with part C,A


Of every level




Part C of every level
Only relates with part D,A


Of every level


Part D of every level
Only relates with part A,C


Of every level


THEREFORE,


Nothing from quadrant A
Relates with quadrant C


Nothing from quadrant B
Relates with quadrant D


And vice versa



This includes the sections of
The numbers which



Correspond to those quadra



At the third set level.



Therefore, we take the 16-Sq.
Deductions:

[A] ABFE-CDHG-KOPL-IMNJ

[B] ABFE-CGHD-KOPL-IJNM

[C] AEFB-CDHG-KLPO-IMNJ

[D] AEFB-CGHD-KLPO-IJNM



The simplest answer is to  apply it


To each of the quadra.


This would leave us with eight
deductions, as predicted



Once the deduction is applied
to the overall quadra.


In QUADRANT A:


ABCDEFGH
IJKLMNOP



Refers to:

1,2,3,4
9,10,11,12

17,18,19,20
25,26,27,28




In QUADRANT B:

ABCDEFGH
IJKLMNOP

Refers to:
5,6,7,8
13,14,15,16

21,22,23,24
29,30,31,32



In QUADRANT C:

ABCDEFGH
IJKLMNOP

Refers to:

37,38,39,40
45,46,47,48

53,54,55,56
61,62,63,64



In QUADRANT D:

ABCDEFGH
IJKLMNOP

Refers to:

33,34,35,36
41,42,43,44

49,50,51,52
57,58,59,60



Now a 16-SQ deduction
for Quadrant A

[A] 1,2,10,9-3,4,12,11-
19,27,28,20-17,25,26,18

[B]1,2,10,9-3,11,12,4-
19,27,28,20-17,18,26,25

[C]1,9,10,2-3,4,12,11-
19,20,28,27-17,25,26,18

[D]1,9,10,2- 3,11,12,4-
19,20,28,27-17,18,26,25


Now a 16-SQ Deduction
for Quadrant B


[A] 5,6,14,13-7,8,16,15-
23,31,32,24-21,29,30,22

[B] 5,6,14,13-7,15,16,8-
23,31,32,24-21,22,30,29

[C] 5,13,14,6-7,8,16,15-
23,24,32,31-21,29,30,22

[D] 5,13,14,6-7,15,16,8-
23,24,32,31-21,22,30,29



Now a 16-SQ Deduction
for Quadrant C


[A] 37,38,46,45-39,40,48,47-
55,63,64,56-53,61,62,54

[B] 37,38,46,45-39,47,48,40-
-55,63,64,56-53,54,62,61

[C] 37,45,46,38-39,40,48,47-
-55,56,64,63-53,61,62,54

[D] 37,45,46,38-39,47,48,40-
55,56,64,63-53,54,62,61


Now a 16-SQ Deduction
for Quadrant D


[A] 33,34,42,41-35,36,44,43-
51,59,60,52-49,57,58,50

[B] 33,34,42,41-35,43,44,36-
-51,59,60,52-49,50,58,57

[C] 33,41,42,34-35,36,44,43-
51,52,60,59-49,57,58,50

[D] 33,41,42,34-35,43,44,36-
51,52,60,59-49,50,58,57




We are nearing our final solution!


Now we simply apply the
Formula:


ABCD and ADCB



On two levels!



In every quadrant
It also takes the order:


ADCB.


Thus, Quadrant A


Is not only:

[A] 1,2,10,9-3,4,12,11-
19,27,28,20-17,25,26,18

[B]1,2,10,9-3,11,12,4-
19,27,28,20-17,18,26,25

[C]1,9,10,2-3,4,12,11-
19,20,28,27-17,25,26,18

[D]1,9,10,2- 3,11,12,4-
19,20,28,27-17,18,26,25



But,

[A] 1,2,10,9-3,4,12,11-
19,27,28,20-17,25,26,18

[D]1,9,10,2- 3,11,12,4-
19,20,28,27-17,18,26,25

[C]1,9,10,2-3,4,12,11-
19,20,28,27-17,25,26,18

[B]1,2,10,9-3,11,12,4-
19,27,28,20-17,18,26,25


Quadrant B is not only:


[A] 5,6,14,13-7,8,16,15-
23,31,32,24-21,29,30,22

[B] 5,6,14,13-7,15,16,8-
23,31,32,24-21,22,30,29

[C] 5,13,14,6-7,8,16,15-
23,24,32,31-21,29,30,22

[D] 5,13,14,6-7,15,16,8-
23,24,32,31-21,22,30,29

But,

[A] 5,6,14,13-7,8,16,15-
23,31,32,24-21,29,30,22

[D] 5,13,14,6-7,15,16,8-
23,24,32,31-21,22,30,29

[C] 5,13,14,6-7,8,16,15-
23,24,32,31-21,29,30,22

[B] 5,6,14,13-7,15,16,8-
23,31,32,24-21,22,30,29


Quadrant C is not only:

[A] 37,38,46,45-39,40,48,47-
55,63,64,56-53,61,62,54

[B] 37,38,46,45-39,47,48,40-
-55,63,64,56-53,54,62,61

[C] 37,45,46,38-39,40,48,47-
-55,56,64,63-53,61,62,54

[D] 37,45,46,38-39,47,48,40-
55,56,64,63-53,54,62,61


But,

[A] 37,38,46,45-39,40,48,47-
55,63,64,56-53,61,62,54

[D] 37,45,46,38-39,47,48,40-
55,56,64,63-53,54,62,61

[C] 37,45,46,38-39,40,48,47-
-55,56,64,63-53,61,62,54

[B] 37,38,46,45-39,47,48,40-
-55,63,64,56-53,54,62,61



Quadrant D is not only:

[A] 33,34,42,41-35,36,44,43-
51,59,60,52-49,57,58,50

[B] 33,34,42,41-35,43,44,36-
-51,59,60,52-49,50,58,57

[C] 33,41,42,34-35,36,44,43-
51,52,60,59-49,57,58,50

[D] 33,41,42,34-35,43,44,36-
51,52,60,59-49,50,58,57


But,



[A] 33,34,42,41-35,36,44,43-
51,59,60,52-49,57,58,50

[D] 33,41,42,34-35,43,44,36-
51,52,60,59-49,50,58,57

[C] 33,41,42,34-35,36,44,43-
51,52,60,59-49,57,58,50

[B] 33,34,42,41-35,43,44,36-
-51,59,60,52-49,50,58,57


Thus, the overall set
takes the order


ABCD and ADCB


With alternation within
each category.


According to the above
dualities…


However, B must
remain opposite of D

And A must remain
Opposite of C…


Thus,

With A.A is
C.B…


With A.B is C.A…


With B.A is D.B…


With B.B is D.A…


With C.A is A.B…


With C.B is A.B…


With D.A is B.B…


With D.B is B.A…


3/4ths of these are
superfluous…


Thus, we have the
combinations:


A.A w/ C.B and
A.B w/ C.A and

B.A w/ D.B and
B.B w/ D.A.


The remaining half of
The categories


Are resolved by
the duality


In which A refers
to B or D…



So we have:

A.A (w/ C.B)
=

A.B (w/ C.A)
=


B.A (w/ D.B)
=


AND

B.B (w/ D.A)
=



Thus, the result is actually
equal to


2^2^2 as predicted…



However, notice, that in spite of the two levels of deductions, the 16-deduction level resulted in four SEPARATE DEDUCTIONS for each quadra, which is inadequate.


Note: this paper shows the WRONG METHOD, although with a lot of promise. For the real method, see: https://www.academia.edu/10161352/Categorical_Deduction_for_64_Categories


[THIS METHOD REMAINS INCOMPLETE (HERE, BUT NOT THERE)]

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