The Inclusive OR (IOR) and summing hexagrams

[lightofdarkness]

Just to cover the 'bit' representation rather than the 'wave' representation of lines (the latter covered in http://members.iimetro.com.au/~lofting/myweb/WaveStruct.html) and how we can 'add' hexagrams through their lines, we need to focus on the inclusive OR operator of logic. What this does is show the result of 'summing' two or more hexagrams by analogy with some other hexagram.

Using the focus on hexagrams 15 and 16, the 'sum' is 62. This is NOT an "AND" operation since AND-ing 15 and 16 gives:

001000
000100
------ AND
000000

The focus of AND is on sameness in BOTH.

We could use XOR (one or the other but not both) to give:

001000
000100
------ XOR
001100 - that works

We can use IOR (one or the other or both):

001000
000100
------
001100 - and so does that

We not there is no apparent difference here between XOR and IOR. However we know that for some relationship to be consistent so the relationship of two WHOLES must also be manifest in the relationship of their PARTS. In previous work I have identified getting the spectrum of a hexagram using the XOR operation. Lets use 27-ness as an example. The 27-ness of 15 is 101001. The 27-ness of 16 is 100101. The 27-ness of 62 is 101101. So we need to derive the 27-ness of 62 from the sum of the 27-ness of 15 and 16, so lets apply the operators:

101001 (27-ness of 15)
100101 (27-ness of 16)
------ (XOR)
001100 (wrong, we want the 27-ness of this)

101001
100101
------ (AND)
100001 (wrong)

101001
100101
------ (IOR)
101101 (RIGHT - this is the 27-ness of 62))


SO ... the correct methodology in 'summing' hexagrams is the use of the IOR operator where we sum all of the yang lines across the hexagrams ( and so the notion of one or the other or both)

The overall idea is to express in I Ching terms the summing of hexagrams in that the sum is in the form of another hexagram serving as a source of analogy (e.g. if we sum all of the hexagrams the result will be 01 - reflecting the 'total actualisation' of all hexagrams as the characteristics of 01.)

If we focus on the personality properties of hexagrams so there is scope for generic relational analysis (limited to a language of 64 categories - and so a possible need to move to dodecagram levels to elicit hexagrams with moving lines)

This - XOR methods give us a the spectrum, a parts list of a hexagram, whereas IOR allows us to sum hexagrams and derive, by analogy to some other hexagram, a description of that 'mix'.

Chris

[martin]

Not sure what you intend here.
(27 xor 15) ior (27 xor 16) = 27 xor (15 ior 16)?
(27 xor 15) ior (27 xor 16) = 27 xor (15 xor 16)?

In this case it doesn't matter because 15 ior 16 = 15 xor 16. And both equations are true.
But in general both are false.

(a xor b) ior (a xor c) <> a xor (b ior c)

Take a=1, b=1, c=0;
(1 xor 1) ior (1 xor 0) = 0 ior 1 = 1
but 1 xor (1 ior 0) = 1 xor 1 = 0

(a xor b) ior (a xor c) <> a xor (b xor c)

Take a=0, b=1, c=1;
(0 xor 1) ior (0 xor 1) = 1 ior 1 =1
but 0 xor (1 xor 1) = 0 xor 0 = 0

http://www.onlineclarity.co.uk/I_Ching_community/clipart/happy.gif

[lightofdarkness]

I get your point .... the focus is on SINGLE line additions as covered in the wave material. All we have to work with are 24, 7, 15, 16, 8, and 23 and so assertions such as:

15+16 = 62
07+08 = 29
23+24 = 27
07+24 = 19
08+23 = 20
....
....
24+07+15+16+08+23 = 01

QUALITATIVELY this appears to work out (e.g. the containment/control of 29 derived from the 'king' position of 5 with the 'supervisor' position of 2 etc; or the passive attracting of 08 with a particular spiritual aspect of 23 gives us the nature of 20 with admiration of a particular, someone 'being' an example (the negative being someone being made an example of...) so we need to uncover the quantitative aspect in the form of logic operators.

Each line represents a frequency of 2^n cps(where n = 1 to 6) working bottom-up.

now, if we assert that 08 + 23 = 20 then it must be the case that 27-of-08 + 27-of-23 = 27-of-20 but which operator allows for this?

We know that XOR is focused on parts extraction so it is not XOR.

IOR test:

08 = 000010; 27-ness = 100011
23 = 000001; 27-ness = 100000
20 = 000011; 27-ness = 100010

100011 (27 of 08)
100000 (27 of 23)
------ IOR
100011 (!!wrong - we need 100010)

ok... i was 'hasty' so lets use the other operators:

AND

this wont work as its focus is on sameness and so only gives us shared yang lines.

COIN (EQV)

This is 'double' AND in that 11 = 1 as does 00 = 1 but it does not give us what we want since:

100011 (27 of 08)
100000 (27 of 23)
------ COIN (EQV)
111100 (Wrong)


IMP
This one is sneaky in that it is asymmetric and works as X <= Y.

so..

27-of-08 - 100011
27-of-23 - 100000
-----------
27 of 20 - 111100 (34) wrong

whereas

27-of-23 - 100000
27-of-08 - 100011
-----------
27-of-20 - 111111 (01) wrong

Then come NAND and NOR.


Implicit here is a composite logic operator - will think about it (as you can too! ;-))

Chris.

[martin]

Okay, suppose we have a line/bit operator op so that for all a,b and c:

(a xor b) op (a xor c) = a xor (b op c) (*)

Substitute a=1 in (*) and you get:
(not b) op (not c) = not (b op c) [1]

Substitute b=0, c=1 in [1] and the result is:
1 op 0 = not (0 op 1) [2]
This implies that op is not symmetric (not commutative)

Substitute b=1, c=1 in [1] and you get:
0 op 0 = not (1 op 1) [3]


There are 16 possible operators op (16 truth tables) but [2] and [3] reduce the possibilities to 4, because if you know the first 2 values (for x y = 0 0 and x y = 1 0) you also know the others.
The possible op's p,q,r and s are:

x y p q r s
0 0 1 1 0 0
1 0 1 0 1 0
0 1 0 1 0 1
1 1 0 0 1 1

p = not y, q = not x, r = x, s = y
All these op's leave either x or y out and are therefore useless.

So the conclusion is that there is no "interesting " line/bit operator that fulfills the condition (*)

We will have to "mix" lines somehow ..


Btw, a note on ior:
(a xor (b ior c)) imp ((a xor b) ior (a xor c)) for all a, b and c.
But is that useful?

[martin]

An equality that might be useful:
a and (b xor c) = (a and b) xor (a and c)

?