How to make the jumps between one level of complexity and another? We will use the concepts of 'folding' and 'unfolding'.

Complete unfolding - complete folding

Let's start with and example:

01 = 010 + 011 + 001 + 101 + (001 + 011)

Reading the above from the left part to the right part describes the results of completely unfolding the digram 01 into trigrams, whereas reading from right to left describes the complete folding which results in the digram 01.

How was this created?

To completely unfold 01, we add a line below it, which can be either be broken (0, yin) or full (1, yang) - this present the first pair of trigram. We also add a line above 01 (creating the second pair) and a line in the middle of 01 (creating the third pair).

Partial unfolding - partial folding

To use the same example again:

01 = 010 + 011

If we read from left to right, we have partially unfolded the digram 01 by adding a line below it, resulting in the trigrams 010 and 011.

By reading it from right to left, we can see the two trigrams matching in the first two places and different in the third place. If we fold the two trigrams (010 and 011), the third place balances out and thus we get the digram 01.

We can say that complete unfolding is done by doing all possible partial unfoldings.

When doing a partial unfolding from an n-gram to m-grams, if we wish, we can mark which partial unfolding we are using in an m-gram: the lines which are untouched are marked by 0, and the ones we unfold marked with 1.

For example, unfolding the digram 01 'from below' to trigrams is marked by the trigram 001 (that's where we have added a new line).

If we wish to unfold 01 'in the middle' to quadgrams, the quadgram marking the change would be 0110.

Let's take a look at the first example again:

01 = 010 + 011 + 001 + 101 + (001 + 011)

We can see that the unfolding resulted in all possible trigram combinations, and even resulted in one pair of trigram twice (001 and 011). We got an extra bit of 'residue'.

If we compare the first example and the second example:

01 = 010 + 011 + 001 + 101 + (001 + 011)

01 = 010 + 011

We can see that the second example is 'lacking'.

Full level unfolding - full level folding

An example; the full level unfolding of digrams to trigrams:

00 + 01 + 10 + 11 = 000 + 001 + 010 + 011 + 100 + 101 + 110 + 111

This example is the most accurate of them all.

The 'residue' we get when we completely unfold an n-gram, or the 'lack' we get when we partially unfold and n-gram, are both present because those equations are distortions of a full level unfolding.

For example, it is true that 010 and 011 complement each other in the third place, so for practical purposes we can partially fold them, but we shouldn't forget that the other trigrams exist as well and that these two trigrams acquire their meaning only by relation to those other trigrams.

(Two opposite trigrams complement each other, but all the trigrams together complement each other more fully.)

So, complete and partial unfoldings are distortions (anti-symmetry) compared to full level unfoldings.

This results in a tilted perception, an 'angle' of a system used - when compared to the full orthogonality of the binary system.