Lecture Notes in Computer Science 3472

578 Therese Berg and Harald Raffelt
Assuming that U is a language accepted by a DFA, the suffixes in sets Es must
include evidence (see the second item on the list of conditions on an observation
pack) that the access strings belong each to a different equivalence class in the
right congruence associated with any regular set U agreeing with the pack. These
classes correspond to states of the minimal deterministic finite automaton (DFA)
for U: access strings are used to reach the states, hence the name. There can not
be any more access strings than states in such an automaton.
Lemma 19.11. Let O be a pack, S its set of access strings, U a regular language
which agrees with O, andM the minimal DFA that recognizes U. Then |S| ≤
|M|.
Proof. Let δ U and q U 0 be the transition function and the initial state of M,
respectively. Let the mapping f map S into the states of M in the natural way:
from s to δ U (q U 0 , s). We prove that this mapping is injective.
Let s and s ′ be two access strings in the pack and s �= s ′ . Assume that they
both are mapped by f to the same state via the the transition function δ U , i.e.
f is not injective. So δ U (q U 0 , s) =qi and δ U (q U 0 , s′ )=qi. LetF U be the set of
accepting states for M.
According to the properties of the observation pack, there exists a word
w ∈ Es ∩ Es ′ such that sw ∈ U ⇐⇒ s′ w �∈ U. Either,δ U (qi , w) ∈ F U or
δ U (qi, w) �∈ F U . This means
(1) δ U (qi, w) ∈ F U =⇒ sw ∈ U and s ′ w ∈ U, or
(2) δ U (qi, w) �∈ F U =⇒ sw �∈ U and s ′ w �∈ U
But this is a contradiction to sw ∈ U ⇐⇒ s ′ w �∈ U, sof is injective. ⊓⊔
Definition 19.12. Let O be a pack, with access strings S, andU a set agreeing
with O. We say that a word z is like s ∈ S for U if and only if ∀ w ∈ Es
sw ∈ U ⇐⇒ zw ∈ U.
To find out whether z is like s we first use the information that sw is labeled
+ in the pack if sw ∈ U, otherwise−. Secondly, for zw we can conduct a
membership query and see if zw ∈ U or not.
Lemma 19.13. For every word z there is at most one word s ∈ S such that z
is like s for U.
Proof. Let s �= s ′ ,bothinS, and assume z is like s, i.e., ∀ w ∈ Es sw ∈ U ⇐⇒
zw ∈ U. The pack provides a word w ∈ Es ∩ Es ′ such that sw ∈ U ⇐⇒ s′ w �∈ U.
Thus, zw ∈ U ⇐⇒ s ′ w �∈ U, sothatz is not like s ′ for U. ⊓⊔
Let γ O,U : Σ ∗ → S be the partial function that maps each z to the single
access string it is like for U, if there is one; it remains undefined if z is not like
any access string for U. From now on, we will use this function in a context in
which both O and U are fixed, so we omit the superscripts and use only γ.