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--- lfa:dfa [2018/07/23 12:37]
pdmatei
+++ lfa:dfa [2020/10/19 15:36] (current)
pdmatei
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 ====== Deterministic automata ======
-===== Motivation =====
-In the last lecture, we have shown that, for **each** regular expression $math[e], we can construct an NFA $math[M], such that $math[L(e) = L(M)].
-While NFAs are a computational instrument for establishing acceptance, they **are not an ideal one**. Nondeterminism is difficult to translate per. se. in a programming language.
-Suppose we enforce the following restriction on the **transition relation** $math[\Delta]:
- * $math[\Delta : K \times \Sigma \rightarrow K],
- * the function $math[\Delta] is **total** (i.e. it is defined for all possible values in its input).
-Thus:
-  * each transition must occur on a **symbol**,
-  * exactly **one** transition is possible for each state-symbol combination
 $def[DFA]
-A **Deterministic Finite Automata** is an NFA, where $math[\Delta : K \times \Sigma \rightarrow K] is a **total** function. In what follows, we write $math[\delta] instead of $math[\Delta] to refer to the transition function of a DFA.
+A **Deterministic Finite Automata** is a tuple $math[(K,\Sigma,\delta,q_0,F)], where $math[K] is a set of states, $math[\Sigma] is the alphabet, $math[\Delta : K \times \Sigma \rightarrow K] is a **total** transition function.
 $end
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 We say a word $math[w] is **accepted** by a DFA $math[M] iff $math[(q_0,w)\vdash_M^*(q,\epsilon)] and $math[q\in F] ($math[q] is a final state).
 $end
-Example(s)
 ===== NFA to DFA transformation =====
-Let $math[M=(K,\Sig
+Let $math[M=(K,\Sigma, \Delta, q_0, F)] be an NFA. We assume $math[M] **does not contain transitions on words of length larger than 1**. If $math[(q,w,q')\in\Delta] for some $math[w=c_1\ldots c_n] of size 2 or more, we construct intermediary states $math[q^1, \ldots, q^{n+1}] as well as transitions $math[(q^1,c_1,q^2), \ldots (q^n,c_n,q^{n+1})], where $math[q=q^1] and $math[q'=q^n]
-ma, \Delta, q_0, F)] be an NFA. We assume $math[M] **does not contain transitions on words of length larger than 1**. If $math[(q,w,q')\in\Delta] for some $math[w=c_1\ldots c_n] of size 2 or more, we construct intermediary states $math[q^1, \ldots, q^{n+1}] as well as transitions $math[(q^1,c_1,q^2), \ldots (q^n,c_n,q^{n+1})], where $math[q=q^1] and $math[q'=q^n]
 We denote by $math[E_M(q) = \{p\in K\mid (q,\epsilon)\vdash^*_M (p,\epsilon)\}], the **$math[\epsilon]-closure** of state $math[q]. In effect $math[E_M(q)] contains **all states reachable from $math[q] via $math[\epsilon]-transitions**. When the automaton $math[M] is understood from the context, we omit the subscript and simply write $math[E(q)].
@@ Line 55: / Line 38: @@
 $prop[1]
-For all $math[q,p], $math[(q,w)\vdash^*_M (p,\epsilon)] iff $math[(E(q),w)\vdash^*_{M'}(P,\epsilon)] with $math[p\in P].
+For all $math[q,p\in K], $math[(q,w)\vdash^*_M (p,\epsilon)] iff $math[(E(q),w)\vdash^*_{M'}(P,\epsilon)], for some $math[P] such that $math[p\in P].
 $end
@@ Line 81: / Line 64: @@
 The proof is by **induction** over the length of the word $math[w].
-** Basis case: $math[w=\epsilon]**
+** Basis step**: $math[\mid w\mid=0] that is $math[w=\epsilon]
   * //direction $math[\implies]://
     - Suppose $math[(q,\epsilon) \vdash^*_M(p,\epsilon)].
     - From the definition of $math[E] and 1., we have that $math[p\in E(q)].
-    - Since $math[\vdash^*_{M'}] is **reflexive**, we have $math[(E(q)=P]
+    - Since $math[\vdash^*_{M'}] is **reflexive**, we have $math[(E(q),\epsilon)\vdash^*_{M'}(E(q),\epsilon)].
-    - From 2. and 3. we have $math[E(q),\epsilon)\vdash^*_{M'}(P,\epsilon)] with $math[p\in P].
+    - Therefore, we have $math[E(q),\epsilon)\vdash^*_{M'}(P,\epsilon)] with $math[p\in P]: $math[P] is actually $math[E(q)].
   * //direction $math[\impliedby]://
@@ Line 93: / Line 76: @@
     - By the definition of $math[E], we have that $math[(q,\epsilon)\vdash^*_{M}(p,\epsilon)] for any $math[p\in E(q)].
-** Induction step: $math[w'=wa]** (hence $math[a] is the last symbol of $math[w']).
+** Induction hypothesis**: suppose that the claim is true for all strings w such that $math[\mid w\mid\leq k] for $math[k\geq0]
+** Induction step**: we prove for any string $math[w] of length $math[k+1]; let $math[w'=wa] (hence $math[a] is the last symbol of $math[w']).
   * //direction $math[\implies]://
     - Suppose $math[(q,wa)\vdash^*_{M} (p,\epsilon)].
-    - By the definition of $math[\vdash^*], we have: $math[(q,wa)\vdash^*_{M} (r_1,a) \vdash_{M} (r_2,\epsilon)\vdash^*_{M} (p,epsilon)]. In words, there is a path from $math[q] which takes us to $math[r_1] by consuming $math[w], then to $math[r_2] via a **one-step** transition, then to $math[p] in zero or more $math[\epsilon]-transitions. Notice that $math[p] may be equal to $math[r_2], which is taken into account since $math[\vdash^*_{M}] is reflexive.
+    - By the definition of $math[\vdash^*], we have: $math[(q,w)\vdash^*_{M} (r_1,a) \vdash_{M} (r_2,\epsilon)\vdash^*_{M} (p,\epsilon)]. In other words, there is a //path// from $math[q] which takes us to $math[r_1] by consuming $math[w], then to $math[r_2] via a **one-step** transition, then to $math[p] in zero or more $math[\epsilon]-transitions. Notice that $math[p] may be equal to $math[r_2], which is taken into account since $math[\vdash^*_{M}] is reflexive.
     - By the construction of $math[\vdash^*_{M}], we also have $math[(q,w)\vdash^*_{M}(r_1,\epsilon)]
     - From 3. by **induction hypothesis**, we have $math[(E(q),w)\vdash^*_{M'}(R_1,\epsilon)] with $math[r_1 \in R_1]
     - By construction of $math[\vdash^*_{M'}], we have $math[(E(q),wa)\vdash^*_{M'}(R_1,a)]
-    - Since $math[(r_1,a) \vdash_{M} (r_2,\epsilon)], by the definition of $math[\delta], we have $math[(R_1,a)\vdash_{M'}(R_2,\epsilon)] and $math[E(r_2)\subseteq R_2]. Also, $math[r_2\in E(r_2)].
+    - Since $math[(r_1,a) \vdash_{M} (r_2,\epsilon)], by the definition of $math[\delta], we have $math[E(R_2) \subseteq \delta(R_1,a)].
-    - By the definition of $math[E] (and $math[(r_2,\epsilon)\vdash^*_{M} (p,epsilon)]), also $math[p \in E(r_2)].
+    - Since $math[(r_2,\epsilon) \vdash_{M} (p,\epsilon)] it follows that $math[p \in E(r_2)], and therefore $math[p \in \delta(R_1,a)].
     - In effect, from 5. and 7. we have shown that $math[(E(q),wa)\vdash^*_{M'}(R_1,a)\vdash_{M'}(R_2,\epsilon)] with $math[p\in R_2], which concludes our proof.