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:

3 thoughts on “Calculating the Relatives of the Classical Labyrinth Type

  1. Pingback: How to sort a Labyrinth Group | blogmymaze

  2. Pingback: How to repair the Mistakes in Historical Scandinavian Labyrinths, Part 5 | blogmymaze

  3. Pingback: The Relatives of Labyrinths With an Even Number of Circuits | blogmymaze

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