By A Kostovskii

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**Extra resources for Geometrical Constructions Using Compasses Only (Popular Lectures in Mathematics)**

**Example text**

Suppose there exists a vertex v in D but not in D'. Then there exist a shortest path from v to a vertex q\ in D' and a shortest path from a vertex q2 in D' to v. If there exists a vertex u in V but not in D' which occurs in both of these paths, then let u be the closest to D' having this property. If there does not exist any vertex u having the above property, then we identify u with the vertex v. If q1 = q2, then two paths u q1 and q2 u form a cycle C = (C, EC) where distinctness of the vertices follows from minimality.

By (V o , w) E, the transformation e : V V with is a D-compatible collapsing that acts as the identity on V \ V. Then P" = e is also an element of S(D) such that In addition, if q : V'\ {w} V' \ {w} is another arbitrary bijection, we have corresponding q' : V and in S(D (l) ) constructed as for p. Thus (u) = q(u) for M V \ {w}, (v0) = (w}, and acts as the identity on V \ V'. Then we have for p" = ep and q" = eq 40 Chapter 2. Directed Graphs, Automata, and Automata Networks that Let P be the set of all functions p" : V V with such that p : V \ {w} V \ {w;}is a bijection.

In this case, we also say / is allowed. , f(n)), if f(i) = f(j), i, j { 1 , . . , n — 1. Of course, the identity F ( l , . . 10 However, we have the next fact. 5. Let F2 and F1 be configuration transformations generated from compatible ones. I f F 2 is not allowed, then F 2 F 1 also is not. Proof. , fi(n)) for i = 1, 2. 7), we may work with f1 and f2 but must consider products in the reverse order. Thus, we will show that f1 f2 is not allowed whenever f2 is not allowed. , n}\ n - 1, i = 1, 2.