1. ## Combining Reinin dichotomies

An INTj can be defined by using Jungian dichotomies:
INTj = introverted*intuitive*thinking*judging.

Using Reinin dichotomies there are several other possibilities of defining an INTj:
INTj = farsighted*obstinate*static*result
INTj = merry*judicious*result*strategic

I wonder how many possible definitions there are. To answer this question we must know which dichotmies are depending on each other. I found 13 dependencies:

farsighted * obstinate = democratic
farsighted * static = strategic
farsighted * merry = result
obstinate * static = emotivist
obstinate * judicious = result
static * negativist = result
democratic * strategic = emotivist
democratic * judicious = merry
strategic * negativist = merry
emotivist * negativist = judicious

Questions:
1.) Are combinations missing or wrong?
2.) What is the mathematical reason for the number 13?
3.) Does this have any use?

2. Originally Posted by JohnDo
An INTj can be defined by using Jungian dichotomies:
INTj = introverted*intuitive*thinking*judging.

Using Reinin dichotomies there are several other possibilities of defining an INTj:
INTj = farsighted*obstinate*static*result
INTj = merry*judicious*result*strategic

I wonder how many possible definitions there are. To answer this question we must know which dichotmies are depending on each other. I found 13 dependencies:

farsighted * obstinate = democratic
farsighted * static = strategic
farsighted * merry = result
obstinate * static = emotivist
obstinate * judicious = result
static * negativist = result
democratic * strategic = emotivist
democratic * judicious = merry
strategic * negativist = merry
emotivist * negativist = judicious

Questions:
1.) Are combinations missing or wrong?
2.) What is the mathematical reason for the number 13?
3.) Does this have any use?
You're missing a lot of it. The way that everything reinin work is logical equality and XOR logic.

Look at the wiki entry for the explaination.

You have 4 tiers of dichotomies.
1st, 2nd, 3rd, 4th.
0th Tier = Type
1. (null/nonnull)
1st Tier = Jungian Foundation
2. E (extroversion/introversion)
3. N (intuitive/sensing)
4. T (logical/ethical)
5. P (irrational/rational)
2nd Tier
6. EN (carefree/farsighted)
7. ET (obstinate/compliant)
8. EP (static/dynamic)
9. NT (aristocratic/democratic)
10. NP (tactical/strategic)
11. TP (constructivist/emotivist)
3rd Tier
12. ENT (positivist/negativist)
13. ENP (reasonable/resolute)
14. ETP (subjectivist/objectivist)
15. NTP (process/result)
4th Tier
16. ENTP (questioner/declarer)

Reinin dichotomies - Wikisocion

So take carefree and farsighted and the math would be.
EN or IS = Carefree
ES or IN = Farsighted

Mathmatically this would be
EN = 0 XOR 0 = 0
IS = 1 XOR 1 = 0

ES = 0 XOR 1 = 1
IN = 1 XOR 0 = 1

I hope this make things clearer for you.

Any 4 non-colliding reinin dichotomy can represent a type because something like.

ENT is simply EN * T.

So Carefree and Farsighted * Logical and Ethical = Positivist and Negativist.

So if someone is Positives or Negativist.
They are
Carefree + T = Positivist
Farsighted + F = Positivist

Carefree + F = Negativist
Farsighted + T = Negativist

Also the notation you see is 0 based so
ENTP = 0000
ISFJ = 1111

3. Originally Posted by hkkmr
You're missing a lot of it. The way that everything reinin work is logical equality and XOR logic.
I'm aware of that, of course.

When I write
farsighted*obstinate=democratic
I could also write
carefree/farsighted*yielding/obstinate=democratic/aristocratic
or to make it more obvious
carefree*yielding=democratic,
farsighted*obstinate=democratic,
carefree*obstinate=aristocratic,
farsighted*yielding=aristocratic
.

I thought that was obvious. So the question is not how the Reinin dichotomies work.

Nevertheless, your explanation is good so you are certainly able to think about the questions I ask in the OP...

4. Originally Posted by JohnDo
I'm aware of that, of course.

When I write
farsighted*obstinate=democratic
I could also write
carefree/farsighted*yielding/obstinate=democratic/aristocratic
or to make it more obvious
carefree*yielding=democratic,
farsighted*obstinate=democratic,
carefree*obstinate=aristocratic,
farsighted*yielding=aristocratic
.

I thought that was obvious. So the question is not how the Reinin dichotomies work.

Nevertheless, your explanation is good so you are certainly able to think about the questions I ask in the OP...
Sounds like a math problem... and it's pointless if you don't also include the 1st tier dichotomies.

It's some sort of factorial/combination problem likely. I'll do some research since I haven't done this sort of math for a bit.

It's probably something like this...

You can pick any of the 15 dichotomies initially, after that you can pick 14, at this point you can only pick from two more on the list because all the other one will be redundant. The last dichotomy will be only a single one.

So 15 * 14 * 2 * 1 = 420. I'm not sure I'm right, but it's something like this the solution, or it could be more complex.

5. Originally Posted by hkkmr
Sounds like a math problem... and it's pointless if you don't also include the 1st tier dichotomies.
Good point. Might be true...

farsighted * logical = negativist
farsighted * rational = judicious
farsighted * obstinate = democratic
farsighted * static = strategic
farsighted * merry = result
obstinate * intuitive = negativist
obstinate * rational = merry
obstinate * static = emotivist
obstinate * judicious = result
static * intuitive = judicious
static * logical = merry
static * negativist = result
democratic * introverted = negativist
democratic * rational = result
democratic * strategic = emotivist
democratic * judicious = merry
strategic * introverted = judicious
strategic * logical = result
strategic * negativist = merry
emotivist * introverted = merry
emotivist * intuitive = result
emotivist * negativist = judicious

So if I made no mistake there are 29 dependencies + the 11 original ones = 40.

Explanations...?

6. Originally Posted by hkkmr
You can pick any of the 15 dichotomies initially, after that you can pick 14, at this point you can only pick from two more on the list because all the other one will be redundant. The last dichotomy will be only a single one.

So 15 * 14 * 2 * 1 = 420. I'm not sure I'm right, but it's something like this the solution, or it could be more complex.
No, it is much more complicated. If you pick farsighted first you can't pick introverted or intuitive anymore: IN*I makes no sense...

The last dichotomy is not fix, either:
INTj = farsighted*obstinate*static*result
INTj = farsighted*obstinate*static*merry

7. Originally Posted by JohnDo
I wonder how many possible definitions there are.
product(16-2^x, for x=0 to 3)/4! = 15*14*12*8/24 = 840 possible definitions

That's including the first-tier dichotomies, though.

8. Originally Posted by JohnDo
An INTj can be defined by using Jungian dichotomies:
INTj = introverted*intuitive*thinking*judging.
I don't think this is completely accurate, but we've already had this discussion. Just for the sake of understanding where each other is coming from, I would say that type is better defined as in the leading function and in the creative, taking into account the rest of the IE placements that occur because of these first two's positions. This might include N and T but not necessarily I and J.

Originally Posted by JohnDo
3.) Does this have any use?
It's just more shorthand for figuring out or categorizing types. These are qualities that are taken out of context, and when assembled, do not imply its original context. So something like this might be useful to gain an idea of how the types are alike and different in certain aspects, but that's about it...

9. Perhaps with should replace "defined" with "identified." We are counting the ways to identify a type using different combinations of Reinin dichotomies.

Now for an explanation of my previous post:

If we include the null dichotomy, there are a total of 16 dichotomies, one of which (the null dichotomy) is automatically redundant - that is, with 0 base dichotomies, we can derive 1 Reinin dichotomy. With 1 base dichotomy, we can derive 2 Reinin dichotomies (null and that one dichotomy); with 2 base, we can derive 4. The dichotomies that are not yet redundant are all of the 16 that have not yet been derived. Thus, the number of non-redundant dichotomies is (16-2^(number of base dichotomies)). That's 8 dichotomies to choose from when choosing your 4th dichotomy.

Now, NIjT for INTj isn't really a different system, it's just rearranged. We need to account for this by dividing by the number of ways to rearrange 4 letters, which is 4 factorial, or 24.

10. Originally Posted by JohnDo
No, it is much more complicated. If you pick farsighted first you can't pick introverted or intuitive anymore: IN*I makes no sense...
The last dichotomy is not fix, either:
INTj = farsighted*obstinate*static*result
INTj = farsighted*obstinate*static*merry
You can, but you can't pick the both of them together.

You can pick Farsighted and Intuitive, but you can't pick farsighted intuitive and introverted together.

Because Farsighted and Intuitive means the same as Introverted Intuitive.

But you can pick both of those together, just the third and others are excluded at this point.

11. Originally Posted by Brilliand
product(16-2^x, for x=0 to 3)/4! = 15*14*12*8/24 = 840 possible definitions

That's including the first-tier dichotomies, though.

I think this is right.
Thanks for the math.

12. Originally Posted by hkkmr
I think picking the 2nd dichotomy, doesn't exclude all but to 2 of them but all but 4 of them.
If I pick E/I and N/S as my first two dichotomies, which 4 remain?

If E/I, N/S and T/F are my first three dichotomies, which 1 other one is the only one left? j/p? Static/Dynamic?

13. Originally Posted by Brilliand
If I pick E/I and N/S as my first two dichotomies, which 4 remain?

If E/I, N/S and T/F are my first three dichotomies, which 1 other one is the only one left? j/p? Static/Dynamic?
I think I was wrong there too, still trying to figure it out the other way. I was in the ballpark(combination/factorial), just didn't know the formula off the top of my head.

I have a problem in math where I usually have to derive the formula rather then memorize it.

14. Originally Posted by hkkmr
I think I was wrong there too, still trying to figure it out the other way. I was in the ballpark(combination/factorial), just didn't know the formula off the top of my head.

I have a problem in math where I usually have to derive the formula rather then memorize it.
I provided an explanation in the last post of the previous page, if that helps.

15. Originally Posted by Brilliand
I provided an explanation in the last post of the previous page, if that helps.
Got it.

16. Originally Posted by hkkmr
You can, but you can't pick the both of them together.

You can pick Farsighted and Intuitive, but you can't pick farsighted intuitive and introverted together.
You are right. farsighted*introverted=intuitive.

Originally Posted by Brilliand
product(16-2^x, for x=0 to 3)/4! = 15*14*12*8/24 = 840 possible definitions

That's including the first-tier dichotomies, though.
Yes, that's correct I think...

Originally Posted by Brilliand
Now for an explanation of my previous post:

If we include the null dichotomy, there are a total of 16 dichotomies, one of which (the null dichotomy) is automatically redundant - that is, with 0 base dichotomies, we can derive 1 Reinin dichotomy. With 1 base dichotomy, we can derive 2 Reinin dichotomies (null and that one dichotomy); with 2 base, we can derive 4. The dichotomies that are not yet redundant are all of the 16 that have not yet been derived. Thus, the number of non-redundant dichotomies is (16-2^(number of base dichotomies)). That's 8 dichotomies to choose from when choosing your 4th dichotomy.
Interesting. There is even a much easier way to explain it without null/nonnull:

First we pick one of the 15 dichotomies. Then we pick one of the remaining 14. One dichotomy can be derived from them so there are only 12 possibilities for the third dichotomy. With these 3 dichotomies we can derive 4. So 15-3-4=8 possibilities for the fourth dichotomy.

17. I thought it was more complicated. Nevertheless, let's see how to derive a dichotomy from two others...

Every dichotomy can be derived from 7 pairs of 2 dichotomies. 7*15 / 3 = 35

farsighted * introverted = intuitive
farsighted * logical = negativist
farsighted * rational = judicious
farsighted * obstinate = democratic
farsighted * static = strategic
farsighted * merry = result
obstinate * introverted = logical
obstinate * intuitive = negativist
obstinate * rational = merry
obstinate * static = emotivist
obstinate * judicious = result
static * introverted = rational
static * intuitive = judicious
static * logical = merry
static * negativist = result
democratic * intuitive = logical
democratic * introverted = negativist
democratic * rational = result
democratic * strategic = emotivist
democratic * judicious = merry
strategic * intuitive = rational
strategic * introverted = judicious
strategic * logical = result
strategic * negativist = merry
emotivist * logical = rational
emotivist * introverted = merry
emotivist * intuitive = result
emotivist * negativist = judicious