(*<*)
theory Trie imports Main begin
(*>*)
text\<open>
To minimize running time, each node of a trie should contain an array that maps
letters to subtries. We have chosen a
representation where the subtries are held in an association list, i.e.\ a
list of (letter,trie) pairs. Abstracting over the alphabet \<^typ>\<open>'a\<close> and the
values \<^typ>\<open>'v\<close> we define a trie as follows:
\<close>
datatype ('a,'v)trie = Trie "'v option" "('a * ('a,'v)trie)list"
text\<open>\noindent
\index{datatypes!and nested recursion}%
The first component is the optional value, the second component the
association list of subtries. This is an example of nested recursion involving products,
which is fine because products are datatypes as well.
We define two selector functions:
\<close>
primrec "value" :: "('a,'v)trie \<Rightarrow> 'v option" where
"value(Trie ov al) = ov"
primrec alist :: "('a,'v)trie \<Rightarrow> ('a * ('a,'v)trie)list" where
"alist(Trie ov al) = al"
text\<open>\noindent
Association lists come with a generic lookup function. Its result
involves type \<open>option\<close> because a lookup can fail:
\<close>
primrec assoc :: "('key * 'val)list \<Rightarrow> 'key \<Rightarrow> 'val option" where
"assoc [] x = None" |
"assoc (p#ps) x =
(let (a,b) = p in if a=x then Some b else assoc ps x)"
text\<open>
Now we can define the lookup function for tries. It descends into the trie
examining the letters of the search string one by one. As
recursion on lists is simpler than on tries, let us express this as primitive
recursion on the search string argument:
\<close>
primrec lookup :: "('a,'v)trie \<Rightarrow> 'a list \<Rightarrow> 'v option" where
"lookup t [] = value t" |
"lookup t (a#as) = (case assoc (alist t) a of
None \<Rightarrow> None
| Some at \<Rightarrow> lookup at as)"
text\<open>
As a first simple property we prove that looking up a string in the empty
trie \<^term>\<open>Trie None []\<close> always returns \<^const>\<open>None\<close>. The proof merely
distinguishes the two cases whether the search string is empty or not:
\<close>
lemma [simp]: "lookup (Trie None []) as = None"
apply(case_tac as, simp_all)
done
text\<open>
Things begin to get interesting with the definition of an update function
that adds a new (string, value) pair to a trie, overwriting the old value
associated with that string:
\<close>
primrec update:: "('a,'v)trie \<Rightarrow> 'a list \<Rightarrow> 'v \<Rightarrow> ('a,'v)trie" where
"update t [] v = Trie (Some v) (alist t)" |
"update t (a#as) v =
(let tt = (case assoc (alist t) a of
None \<Rightarrow> Trie None [] | Some at \<Rightarrow> at)
in Trie (value t) ((a,update tt as v) # alist t))"
text\<open>\noindent
The base case is obvious. In the recursive case the subtrie
\<^term>\<open>tt\<close> associated with the first letter \<^term>\<open>a\<close> is extracted,
recursively updated, and then placed in front of the association list.
The old subtrie associated with \<^term>\<open>a\<close> is still in the association list
but no longer accessible via \<^const>\<open>assoc\<close>. Clearly, there is room here for
optimizations!
Before we start on any proofs about \<^const>\<open>update\<close> we tell the simplifier to
expand all \<open>let\<close>s and to split all \<open>case\<close>-constructs over
options:
\<close>
declare Let_def[simp] option.split[split]
text\<open>\noindent
The reason becomes clear when looking (probably after a failed proof
attempt) at the body of \<^const>\<open>update\<close>: it contains both
\<open>let\<close> and a case distinction over type \<open>option\<close>.
Our main goal is to prove the correct interaction of \<^const>\<open>update\<close> and
\<^const>\<open>lookup\<close>:
\<close>
theorem "\<forall>t v bs. lookup (update t as v) bs =
(if as=bs then Some v else lookup t bs)"
txt\<open>\noindent
Our plan is to induct on \<^term>\<open>as\<close>; hence the remaining variables are
quantified. From the definitions it is clear that induction on either
\<^term>\<open>as\<close> or \<^term>\<open>bs\<close> is required. The choice of \<^term>\<open>as\<close> is
guided by the intuition that simplification of \<^const>\<open>lookup\<close> might be easier
if \<^const>\<open>update\<close> has already been simplified, which can only happen if
\<^term>\<open>as\<close> is instantiated.
The start of the proof is conventional:
\<close>
apply(induct_tac as, auto)
txt\<open>\noindent
Unfortunately, this time we are left with three intimidating looking subgoals:
\begin{isabelle}
~1.~\dots~{\isasymLongrightarrow}~lookup~\dots~bs~=~lookup~t~bs\isanewline
~2.~\dots~{\isasymLongrightarrow}~lookup~\dots~bs~=~lookup~t~bs\isanewline
~3.~\dots~{\isasymLongrightarrow}~lookup~\dots~bs~=~lookup~t~bs
\end{isabelle}
Clearly, if we want to make headway we have to instantiate \<^term>\<open>bs\<close> as
well now. It turns out that instead of induction, case distinction
suffices:
\<close>
apply(case_tac[!] bs, auto)
done
text\<open>\noindent
\index{subgoal numbering}%
All methods ending in \<open>tac\<close> take an optional first argument that
specifies the range of subgoals they are applied to, where \<open>[!]\<close> means
all subgoals, i.e.\ \<open>[1-3]\<close> in our case. Individual subgoal numbers,
e.g. \<open>[2]\<close> are also allowed.
This proof may look surprisingly straightforward. However, note that this
comes at a cost: the proof script is unreadable because the intermediate
proof states are invisible, and we rely on the (possibly brittle) magic of
\<open>auto\<close> (\<open>simp_all\<close> will not do --- try it) to split the subgoals
of the induction up in such a way that case distinction on \<^term>\<open>bs\<close> makes
sense and solves the proof.
\begin{exercise}
Modify \<^const>\<open>update\<close> (and its type) such that it allows both insertion and
deletion of entries with a single function. Prove the corresponding version
of the main theorem above.
Optimize your function such that it shrinks tries after
deletion if possible.
\end{exercise}
\begin{exercise}
Write an improved version of \<^const>\<open>update\<close> that does not suffer from the
space leak (pointed out above) caused by not deleting overwritten entries
from the association list. Prove the main theorem for your improved
\<^const>\<open>update\<close>.
\end{exercise}
\begin{exercise}
Conceptually, each node contains a mapping from letters to optional
subtries. Above we have implemented this by means of an association
list. Replay the development replacing \<^typ>\<open>('a * ('a,'v)trie)list\<close>
with \<^typ>\<open>'a \<Rightarrow> ('a,'v)trie option\<close>.
\end{exercise}
\<close>
(*<*)
(* Exercise 1. Solution by Getrud Bauer *)
primrec update1 :: "('a, 'v) trie \<Rightarrow> 'a list \<Rightarrow> 'v option \<Rightarrow> ('a, 'v) trie"
where
"update1 t [] vo = Trie vo (alist t)" |
"update1 t (a#as) vo =
(let tt = (case assoc (alist t) a of
None \<Rightarrow> Trie None []
| Some at \<Rightarrow> at)
in Trie (value t) ((a, update1 tt as vo) # alist t))"
theorem [simp]: "\<forall>t v bs. lookup (update1 t as v) bs =
(if as = bs then v else lookup t bs)"
apply (induct_tac as, auto)
apply (case_tac[!] bs, auto)
done
(* Exercise 2. Solution by Getrud Bauer *)
primrec overwrite :: "'a \<Rightarrow> 'b \<Rightarrow> ('a * 'b) list \<Rightarrow> ('a * 'b) list" where
"overwrite a v [] = [(a,v)]" |
"overwrite a v (p#ps) = (if a = fst p then (a,v)#ps else p # overwrite a v ps)"
lemma [simp]: "\<forall> a v b. assoc (overwrite a v ps) b = assoc ((a,v)#ps) b"
apply (induct_tac ps, auto)
apply (case_tac[!] a)
done
primrec update2 :: "('a, 'v) trie \<Rightarrow> 'a list \<Rightarrow> 'v option \<Rightarrow> ('a, 'v) trie"
where
"update2 t [] vo = Trie vo (alist t)" |
"update2 t (a#as) vo =
(let tt = (case assoc (alist t) a of
None \<Rightarrow> Trie None []
| Some at \<Rightarrow> at)
in Trie (value t) (overwrite a (update2 tt as vo) (alist t)))"
theorem "\<forall>t v bs. lookup (update2 t as vo) bs =
(if as = bs then vo else lookup t bs)"
apply (induct_tac as, auto)
apply (case_tac[!] bs, auto)
done
(* Exercise 3. Solution by Getrud Bauer *)
datatype ('a,dead 'v) triem = Triem "'v option" "'a \<Rightarrow> ('a,'v) triem option"
primrec valuem :: "('a, 'v) triem \<Rightarrow> 'v option" where
"valuem (Triem ov m) = ov"
primrec mapping :: "('a,'v) triem \<Rightarrow> 'a \<Rightarrow> ('a, 'v) triem option" where
"mapping (Triem ov m) = m"
primrec lookupm :: "('a,'v) triem \<Rightarrow> 'a list \<Rightarrow> 'v option" where
"lookupm t [] = valuem t" |
"lookupm t (a#as) = (case mapping t a of
None \<Rightarrow> None
| Some at \<Rightarrow> lookupm at as)"
lemma [simp]: "lookupm (Triem None (\<lambda>c. None)) as = None"
apply (case_tac as, simp_all)
done
primrec updatem :: "('a,'v)triem \<Rightarrow> 'a list \<Rightarrow> 'v \<Rightarrow> ('a,'v)triem" where
"updatem t [] v = Triem (Some v) (mapping t)" |
"updatem t (a#as) v =
(let tt = (case mapping t a of
None \<Rightarrow> Triem None (\<lambda>c. None)
| Some at \<Rightarrow> at)
in Triem (valuem t)
(\<lambda>c. if c = a then Some (updatem tt as v) else mapping t c))"
theorem "\<forall>t v bs. lookupm (updatem t as v) bs =
(if as = bs then Some v else lookupm t bs)"
apply (induct_tac as, auto)
apply (case_tac[!] bs, auto)
done
end
(*>*)