hurl3y4456
New member
- Joined
- Aug 31, 2018
- Messages
- 298
- MBTI Type
- SINE
I was wondering if anyone has the following pattern to figure out each function stack corresponding to all types. The following algorithm can be used:
Define each type according to indices x1,x2,....,x4.
Let (ISJ/ESP/INJ/ENP) correspond to case 1 & (INP/ENJ/ISP/ESJ) correspond to case 2
Now, form a square pattern by arranging inner two indices at bottom edge and corresponding opposites on top edge. If x1 =I, then apply I to x2 or x3. Otherwise, apply E to x2 or x3. For case 1, follow the sequence x2-->x3-->x3'-->x2' where " ' " designates the opposing perceiving or judging functions (i.e: S' = N and T' = F). Also, if the odd index is introverted, then let the even indices be extroverted and vice versa. Similarly, let case 3 follow the sequence: x3-->x2-->x2'-->x3'.
Arrange as follows:
Case 1 : Case 2
x2'↠x3' x2'→ x3'
↓ ↑ : ↑ ↓
x2 → x3 x2 ↠x3 S→ T
↑ ↓
Example (ENFJ = case 2) --> N ↠F Since x1 = E, sequence becomes Fe > NI > SE > TI
Define each type according to indices x1,x2,....,x4.
Let (ISJ/ESP/INJ/ENP) correspond to case 1 & (INP/ENJ/ISP/ESJ) correspond to case 2
Now, form a square pattern by arranging inner two indices at bottom edge and corresponding opposites on top edge. If x1 =I, then apply I to x2 or x3. Otherwise, apply E to x2 or x3. For case 1, follow the sequence x2-->x3-->x3'-->x2' where " ' " designates the opposing perceiving or judging functions (i.e: S' = N and T' = F). Also, if the odd index is introverted, then let the even indices be extroverted and vice versa. Similarly, let case 3 follow the sequence: x3-->x2-->x2'-->x3'.
Arrange as follows:
Case 1 : Case 2
x2'↠x3' x2'→ x3'
↓ ↑ : ↑ ↓
x2 → x3 x2 ↠x3 S→ T
↑ ↓
Example (ENFJ = case 2) --> N ↠F Since x1 = E, sequence becomes Fe > NI > SE > TI
