Supervisory Control - The Unknown Component Problem: Theory and Applications - page 238

Notice that the if part of the previous statement holds because the smallest solution
exists, and of course it is Pref . Instead, in general the smallest solution does not

exist, and it is not true that every language contained in the largest solution of a
given equation is itself a solution of the equation.
The largest controller solution can be seen as the collection of all particular
solutions, out of which the “best” one may be extracted according to a given
criterion.

.S /

P D S Df ; b.a C b/ ? g , i.e., all words

Example 15.8. Consider ˙ Df a; b g ,

starting by b . The equation P \ X D S is satisfied by X D S ( P \ S D S ); S is

the s mallest controller solution, because it is pre fix -closed .
P

Pref is the set of all
words. So the largest solution is the set of all words over the alphabet

is the set of all words that do not start by b , P [ S D .P [ S/

˙ Df a; b g ,

which may yield a simpler implementation.

The followin g e xample shows an equation with no solution; as expected, both
Pref

Pref

.S / and .P [ S/ fail the solvability test.

Example 15.9. Consider ˙ Df a; b g , P Df a; ab g , S Df ab g . The equation P \

X D S ). However

S Df ab g is not a controller solution, because it is not prefix-closed. The smallest
prefix-closed language that contains

is satisfied by X D S ( P \ S Df a; ab g\f ab gDf ab gD S

S is Pref .S / Df ; a; ab g .However Pref .S /

is no t a controller solution, because P \ Pref .S / Df a; ab g¤ S Df ab g .

P is the set of all words except a and ab , P [ S is the set of all words except a ,

Pref

.P [ S/ is the set of all words that do not start by a , i.e., those that start by b .

Pref

So .P [ S/ Df ; b; bb ; ba ;::: g .

Pref Pref

However .P [ S/ is not a controller solution, because P \ .P [ S/ D

f a; ab g\f ; b; bb ; ba ;::: gD;¤ S Df ab g .

Example 15.10. Consider ˙ Df a; b; c g , P Df a; abc g , S Df a g . The smallest

controller solution is Pref .S / Df ; a g , indeed P \ Pref .S / Df a; abc g\f ; a gD

f a gD S .

P is the set of all words except a and abc , P [ S is the set of all words except abc ,

Pref

.P [ S/ is the set of all words that do not start by abc . So the largest controller

Pref

Df .a C b/ C .a C c/ C .b C c/ solution is .P [ S/ C a.a C c/.a C

b C c/ C ab.a C b/.a C b C c/ C b.a C b C c/ C c.a C b C c/ g , indeed

Pref

P \ .P [ S/ Df a; abc g\f words that do not start by abc gDf a gD S .

Example 15.11. Consider a plant P and a specification S shown respectively in

Fig. 15.2 a,b. The smallest so lu tion C m D Pref .S / is shown in Fig. 15.2 c, whereas

Pref obtained by BALM is shown in Fig. 15.3 .
Finally, Fig. 15.2 d shows an intermediate solution

the largest solution C M D .P [ S/

C m C i C M ;
notice that aaa 2 C i , aaa 62 C m ,andthat aaba 2 C M , aaba 62 C i .
If

C i such that

Pref Pref

P \ .P [ S/ D P \ Pref .S / S ,then P \ .P [ S/ S is the supremal

Pref

controllable sublanguage of S , meaning that the language P \ .P [ S/ is the

largest language that can be controlled.