Self-dual Labyrinths with even Numbers of Circuits

In labyrinths with even numbers of circuits, there are usually only two relatives, the base labyrinth and the dual (see: related posts, below). Here, I want to present another labyrinth with an even number of circuits and show, that first the calculation of the relatives also works in labyrinths with more axes than one and, second, show what results in a self-dual labyrinth with an even number of circuits. For this, I use a small labyrinth with two axes in order not to let the sequence of circuits grow too long.

Labyrinth with 2 Axes and 6 Circuits
Figure 1. Labyrinth with 2 Axes and 6 Circuits

In fig. 2, I perform the calculation following the usual method. The result is, that the transpose and the complementary sequences of circuits are the same. Both also result in the same figure, that is no labyrinth. In this figure, the pathway does not reach the center but ends in a dead-end.

If we write the complementary sequence of circuits in reverse order or complement the transpose sequence of circuits at each position to 7, we obtain the dual sequence of circuits. This is the same as the base sequence.

The Relatives of the Labyrinth with 2 Axes and 6 Circuits
Figure 2. The Relatives of the Labyrinth with 2 Axes and 6 Circuits

In labyrinths with even numbers of circuits, the group, thus, has only two members: the base labyrinth and the dual, or only one member in cases of self-dual labyrinths.

Related Posts:

Calculating the Relatives of the Classical Labyrinth Type

Now, I want to calculate the relatives of the basic labyrinth type / classical (Cretan) type. Those who are already familiar with the subject, will probably know the result. Nevertheless, I will conduct this calculation consequently once here. Figure 1 shows the step from the base labyrinth to the transpose. 

Figure 1.From the Base Labyrinth to the Transpose
Figure 1.From the Base Labyrinth to the Transpose

The calculation of the complement is illustrated in fig. 2. 

Figure 2. From the Base Labyrinth to the Complement
Figure 2. From the Base Labyrinth to the Complement

These directly calculated sequences of circuits of the transpose and the complement are the same. Now, let’s also derive the sequence of circuits of the dual in fig. 2 indirectly. As we know, this can be achieved by writing the sequence of circuits of the complement in reverse order. 

Figure 3. From the Complement to the Dual
Figure 3. From the Complement to the Dual

This results in the same sequence of circuits as for the base labyrinth. What means nothing else than that the labyrinth of the classical or base type is self-dual. In self-dual labyrinths also the two other relatives of them, the transpose and the complement coincide with each other because these are dual to each other and in this case they are also self-dual. 

Now (except for the “labyrinth” with one circuit) there exist no self-complementary labyrinths (see related posts below). Therefore, each group consists of either 2 or 4 different related labyrinths. This, however, applies only to labyrinths with an odd number of circuits, as will be shown in the next post. 

Related Post:

Calculating the Related Labyrinths

Based on the third arrangement of the labyrinths from the last post (see: related posts, below) the transpose and complementary labyrinths can be directly, and the dual labyrinth can be indirectly calculated quite simply. For this, the sequence of circuits of the base labyrinth is needed. 

Here, I want to exercise this with the example of the labyrinth depicted in fig. 1. 

Figure 1. Labyrinth from the 18. or 19. Century Carved on a Wooden Pillar of the Old Mosque at  Tal, in Northern Pakistan.
Figure 1. Labyrinth from the 18. or 19. Century Carved on a Wooden Pillar of the Old Mosque at Tal, in Northern Pakistan.
Source: Saward, p. 60°

The labyrinth is situated with the entrance on top and in anti-clockwise rotation. First, I redraw it such that the entrance lies at the bottom and the labyrinth rotates clockwise. By this, it presents itself in the form I always use for a comparison of labyrinths. This labyrinth with one axis and 9 circuits will serve as our base labyrinth. Its sequence of circuits is 5 4 3 2 1 6 9 8 7.

Figure 2. Labyrinth of Tal, Redrawn: Base Labyrinth
Figure 2. Labyrinth of Tal, Redrawn: Base Labyrinth

First, we write the sequence of circuits in reverse order

Base: 5 4 3 2 1 6 9 8 7 <—> 7 8 9 6 1 2 3 4 5: Transpose. 

This brings us to the transpose labyrinth (fig. 3).

Figure 3. The Transpose to the Labyrinth of Tal
Figure 3. The Transpose to the Labyrinth of Tal

Then, second, we complete the sequence of circuits of the base labyrinth to one greater than the number of circuits, i.e. to 10. 

Calculation

By this, we obtain the sequence of circuits 5 6 7 8 9 4 1 2 3 of the complementary labyrinth that is shown in fig. 4. 

Figure 4. The Complement to the Labyrinth of Tal
Figure 4. The Complement to the Labyrinth of Tal

And, if we then write the sequence of circuits of the complement in reverse order, we obtain indirectly the sequence of circuits of the dual labyrinth: 

Complement: 5 6 7 8 9 4 1 2 3 <—> 3 2 1 4 9 8 7 6 5: Dual 

The dual labyrinth is shown in fig. 5. 

Figure 5. The Dual to the Labyrinth of Tal
Figure 5. The Dual to the Labyrinth of Tal

Now we can even check this result by adding the sequences of circuits of the transpose and the dual labyrinth. These must add to 10 at each position, as the dual is the complementary of the transpose. 

Calculation

This examination confirms the result. The dual can also be calculated by completing the sequence of circuits of the transpose to 10. However, it is easier to just write the sequence of circuits of the complement in reverse order. 

Thus, we have to know the sequence of circuits of the base labyrinth. Then we write this in reverse order and obtain the transpose. We complete it to one greater than the number of circuits and obtain the complement. And finally we write the sequence of circuits of the complement in reverse order and obtain the dual. 

° Saward Jeff. Labyrinths & Mazes. The Definitive Guide to Ancient & Modern Traditions. Gaia Books: 2003.

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The Relatives of the Labyrinth Type Gossembrot 51 r

The labyrinth on folio 51 r is Gossembrot’s most important one. It is the earliest preserved example of a five-arm labyrinth at all. It’s course of the pathway using double barriers in all side-arms is unprecedented (see: related posts 1, below). However, it is not self-dual. Therefore, it can be expected that there exist three relatives of it (related posts 4).

I term as relatives of an (original) labyrinth the dual, complementary, and dual-complementary labyrinths of it (related posts 2 and 3). In fig. 1 the patterns of the Gossembrot 51 r-type labyrinth (a, original), the dual (b), the complementary (c), and the dual-complementary (d) of it are presented.

Figure 1. Patterns of the Relatives of Type Gossembrot 51 r

Figure 2 shows the labyrinths corresponding to the patterns in their basic form with the walls delimiting the pathway on concentric layout and in clockwise rotation.

Figure 2. The Relatives of Type Gossembrot 51 r in the Basic Form

These four related labyrinths all look quite similar. To me it seems, the dual (b) and the complementary (c) look somewhat less balanced than the original (a) and the dual-complementary (d). Presently, I am not aware of any existing examples of a relative to the Gossembrot 51 r-type labyrinth.

Related Posts:

  1. Sigmund Gossembrot / 2
  2. The Relatives of the Wayland’s House Type Labyrinth
  3. The Relatives of the Ravenna Type Labyrinth
  4. The Complementary versus the Dual Labyrinth