51 lines
1.8 KiB
OCaml
51 lines
1.8 KiB
OCaml
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open Lambda
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(* Input: a list of (key, action) pairs, where keys are integers. *)
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(* Output: a table of (low, high, offset) triples for Ktranslate
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an array of actions for Kswitch *)
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let make_decision_tree casei =
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(* Sort the cases by increasing keys *)
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let cases =
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Sort.list (fun (key1,act1) (key2,act2) -> key1 <= key2) casei in
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(* Extract the keys and the actions *)
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let keyv = Array.of_list (List.map fst cases)
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and actv = Array.of_list (List.map snd cases) in
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let n = Array.length keyv in
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(* Partition the set of keys keyv into maximal dense enough segments.
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A segment is dense enough if its span (max point - min point) is
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less than four times its size (number of points). *)
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let rec partition start =
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if start >= n then [] else
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let stop = ref (n-1) in
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while keyv.(!stop) - keyv.(start) > 4 * (!stop - start) do
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decr stop
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done;
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(* We've found a dense enough segment.
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In the worst case, !stop = start and the segment is a single point *)
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(* Record the segment and continue *)
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(start, !stop) :: partition (!stop + 1) in
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let part = partition 0 in
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(* Compute the length of the switch table.
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Slot 0 is reserved and always contains Lstaticfail. *)
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let switchl = ref 1 in
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List.iter
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(fun (start, stop) -> switchl := !switchl + keyv.(stop) - keyv.(start) + 1)
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part;
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(* Build the two tables *)
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let transl = Array.new (List.length part) (0, 0, 0)
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and switch = Array.new !switchl Lstaticfail in
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let tr_pos = ref 0
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and sw_ind = ref 1 in
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List.iter
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(fun (start, stop) ->
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transl.(!tr_pos) <- (keyv.(start), keyv.(stop), !sw_ind);
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for i = start to stop do
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switch.(!sw_ind + keyv.(i) - keyv.(start)) <- actv.(i)
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done;
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incr tr_pos;
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sw_ind := !sw_ind + keyv.(stop) - keyv.(start) + 1)
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part;
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(transl, switch)
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