(*<*) theory Trie imports Main begin (*>*) text‹ 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 🍋‹'a› a
values 🍋‹'v› we define a trie as follows: ›
text‹\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: ›
text‹\noindent Association lists come with a generic lookup function. Its result involves type ‹option› b ›
primrec assoc :: "('key * 'val)list ==> 'key ==> '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‹ 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: ›
primrec lookup :: "('a,'v)trie ==> 'a list ==> 'v option"where "lookup t [] = value t" | "lookup t (a#as) = (case assoc (alist t) a of None ==> None | Some at ==> lookup at as)"
text‹ As a first simple property we prove that looking up a string in the empty trie 🍋‹Trie None []› a
distinguishes the two cases whether the search string is empty or not: ›
text‹ 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: ›
primrec update:: "('a,'v)trie ==> 'a list ==> 'v ==> ('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 ==> Trie None [] | Some at ==> at) in Trie (value t) ((a,update tt as v) # alist t))"
text‹\noindent The base case is obvious. In the recursive case the subtrie 🍋‹tt› a
recursively updated, andthen placed in front of the association list.
The old subtrie associated with🍋‹a›is still in the association list
but no longer accessible via 🍋‹assoc›. Clearly, there is room here for
optimizations!
Before we start on any proofs about 🍋‹update› we tell the simplifier to
expand all ‹let›s andto split all ‹case›-constructs over
options: ›
declare Let_def[simp] option.split[split]
text‹\noindent The reason becomes clear when looking (probably after a failed proof attempt) at the body of 🍋‹update›: ‹let›and a case distinction over type ‹option›.
Our main goal isto prove the correct interaction of 🍋‹update›and 🍋‹lookup›: ›
theorem"∀t v bs. lookup (update t as v) bs = (if as=bs then Some v else lookup t bs)"
txt‹\noindent Our plan is to induct on 🍋‹as›;
quantified. From the definitions it is clear that induction on either 🍋‹as› or 🍋‹bs›is required. The choice of 🍋‹as›is
guided by the intuition that simplification of 🍋‹lookup› might be easier if🍋‹update› has already been simplified, which can only happen if 🍋‹as›is instantiated.
The start of the proofis conventional: › apply(induct_tac as, auto)
txt‹\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 🍋‹bs› a
well now. It turns out that instead of induction, case distinction
suffices: › apply(case_tac[!] bs, auto) done
text‹\noindent \index{subgoal numbering}% All methods ending in ‹tac› t
specifies the range of subgoals they are applied to, where‹[!]› means
all subgoals, i.e.\ ‹[1-3]›in our case. Individual subgoal numbers,
e.g. ‹[2]› 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 proofstates are invisible, and we rely on the (possibly brittle) magic of ‹auto› (‹simp_all› will not do --- try it) to split the subgoals
of the induction up in such a way that case distinction on 🍋‹bs› makes
sense and solves the proof.
\begin{exercise}
Modify 🍋‹update› (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 🍋‹update› that does not suffer from the
space leak (pointed out above) caused by not deleting overwritten entries from the association list. Prove the main theoremfor your improved 🍋‹update›. \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 🍋‹('a * ('a,'v)trie)list› with🍋‹'a ==> ('a,'v)trie option›. \end{exercise}
›
(*<*)
(* Exercise 1. Solution by Getrud Bauer *)
primrec update1 :: "('a, 'v) trie ==> 'a list ==> 'v option ==> ('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 ==> Trie None [] | Some at ==> at) in Trie (value t) ((a, update1 tt as vo) # alist t))"
theorem [simp]: "∀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 ==> 'b ==> ('a * 'b) list ==> ('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]: "∀ 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 ==> 'a list ==> 'v option ==> ('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 ==> Trie None [] | Some at ==> at) in Trie (value t) (overwrite a (update2 tt as vo) (alist t)))"
theorem"∀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
primrec valuem :: "('a, 'v) triem ==> 'v option"where "valuem (Triem ov m) = ov"
primrec mapping :: "('a,'v) triem ==> 'a ==> ('a, 'v) triem option"where "mapping (Triem ov m) = m"
primrec lookupm :: "('a,'v) triem ==> 'a list ==> 'v option"where "lookupm t [] = valuem t" | "lookupm t (a#as) = (case mapping t a of None ==> None | Some at ==> lookupm at as)"
primrec updatem :: "('a,'v)triem ==> 'a list ==> 'v ==> ('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 ==> Triem None (λc. None) | Some at ==> at) in Triem (valuem t) (λc. if c = a then Some (updatem tt as v) else mapping t c))"
theorem"∀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 (*>*)
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