- 追加された行はこの色です。
- 削除された行はこの色です。
* 予定のみ [#c00e6831]
-日時 :2006/6/26 (Mon)
-場所 :名大 理学部1号館(多元数理科学研究科) 307室
-時刻 :18:00〜19:30
-参加者:??名
#comment
*Chapter 6. Exercise続き [#t10e4420]
**Exercise 1 [#x1f19ca3]
**Exercise 6 (小笠原) [#x9271744]
**Exercise 7 [#vde22fdd]
**Exercise 8 [#k7994cd4]
**Exercise 8 (樋口)[#k7994cd4]
1,2,3,4からなる二分探索木を列挙し,それらを構成するためにaddに渡す要素の列を求めよ.~
まず,テキストより,2分木および,add, mem, preorderの定義があるとする.
type 'a tree = Lf | Br of 'a * 'a tree * 'a tree
let rec mem t x =
match t with
Lf -> false
|Br (y,left,right) -> if x = y then true else
if x < y then mem left x else mem right x
let rec add t x =
match t with
Lf -> Br (x,Lf,Lf)
|(Br (y,left,right) as whole) -> if x = y then whole else
if x < y then Br(y, add left x, right)
else Br(y, left, add right x)
let rec preorder t l =
match t with
Lf -> l
|Br(x, left, right) -> x :: (preorder left (preorder right l));;
基本方針は,[1;2;3;4]から順列を作り出し,
それらをaddして得られたtreeをpreorderでめぐり,
preorderがユニークなら新たな形の木として残し,
addへの入力した列を保存してゆく.
: makeUniqTreeInputs | excerciseの目標 [1;2;3;4] -> [(addへの入力のリスト)]
: makeUniqTrees | makeUniqTreeInputs と同様だが,返すものがaddへ入力後のtreeのリストを得る.
: makeSTree | 与えられたリストの要素を順にaddしtreeを得る.
: permutation | リストから順列を得る
: prefix | permutationで利用. ex. prefix [2;3] -> [[]; [2]; [2;3]]
: suffix | permutationで利用. ex. suffix [2;3] -> [[2;3]; [3]; []]
: interleave | permutationで利用. ex. interleave 1 [2;3] -> [[1;2;3]; [2;1;3]; [2;3;1]]
let rec prefix = function
[] -> [[]]
|x::xs -> [] :: (List.map (fun el -> x::el) (prefix xs))
let rec suffix = function
[] -> [[]]
|x::xs -> (x::xs)::(suffix xs)
let interleave el l =
let pl = (prefix l) and sl = (suffix l) in
(List.map2 (fun a b -> a @ el :: b) pl sl)
let rec permutation = function
[] -> [[]]
|x::xs -> List.flatten (List.map (fun e -> (interleave x e)) (permutation xs))
let rec makeSTree = function
[] -> Lf
|x::xs -> add (makeSTree xs) x
let makeUniqTreeInputs nodes =
let addTree input set =
let pord = (preorder (makeSTree input) []) in
if List.mem pord (fst set) then set else ((pord::(fst set)),input::(snd set))
in
let rec addTrees nodel set =
match nodel with
[] -> (set)
|x::xs -> addTree x (addTrees xs set)
in
snd (addTrees (permutation nodes) ([],[]))
let makeUniqTrees nodes =
let rec input2trees = function []->[]|x::xs -> (makeSTree x)::(input2trees xs) in
input2trees (makeUniqTreeInputs nodes);;
***実行結果 [#k19a8025]
# makeUniqTreeInputs [1;2;3;4];;
- : int list list =
[[1; 2; 3; 4]; [2; 1; 3; 4]; [2; 3; 1; 4]; [2; 3; 4; 1]; [3; 1; 2; 4];
[3; 2; 1; 4]; [3; 2; 4; 1]; [3; 4; 1; 2]; [3; 4; 2; 1]; [4; 1; 2; 3];
[4; 2; 1; 3]; [4; 2; 3; 1]; [4; 3; 1; 2]; [4; 3; 2; 1]]
# makeUniqTrees [1;2;3;4];;
- : int tree list =
[Br (4, Br (3, Br (2, Br (1, Lf, Lf), Lf), Lf), Lf);
Br (4, Br (3, Br (1, Lf, Br (2, Lf, Lf)), Lf), Lf);
Br (4, Br (1, Lf, Br (3, Br (2, Lf, Lf), Lf)), Lf);
Br (1, Lf, Br (4, Br (3, Br (2, Lf, Lf), Lf), Lf));
Br (4, Br (2, Br (1, Lf, Lf), Br (3, Lf, Lf)), Lf);
Br (4, Br (1, Lf, Br (2, Lf, Br (3, Lf, Lf))), Lf);
Br (1, Lf, Br (4, Br (2, Lf, Br (3, Lf, Lf)), Lf));
Br (2, Br (1, Lf, Lf), Br (4, Br (3, Lf, Lf), Lf));
Br (1, Lf, Br (2, Lf, Br (4, Br (3, Lf, Lf), Lf)));
Br (3, Br (2, Br (1, Lf, Lf), Lf), Br (4, Lf, Lf));
Br (3, Br (1, Lf, Br (2, Lf, Lf)), Br (4, Lf, Lf));
Br (1, Lf, Br (3, Br (2, Lf, Lf), Br (4, Lf, Lf)));
Br (2, Br (1, Lf, Lf), Br (3, Lf, Br (4, Lf, Lf)));
Br (1, Lf, Br (2, Lf, Br (3, Lf, Br (4, Lf, Lf))))]
***おまけ [#m4ad1608]
let trees2dot inputs =
let rec trees2edges inputs delta=
let rec tree2edges n = function
Lf -> "L" ^ (string_of_int n)
|Br(x,left,right) -> (string_of_int (x+n))^"[label="^(string_of_int x)^"];"^
match (left,right)with
(Lf,Lf) -> ""
|(Br(x1,_,_),Br(x2,_,_)) -> (string_of_int (x+n))^":sw ->"^(string_of_int (x1+n))^";"
^(string_of_int (x+n))^":se ->"^(string_of_int (x2+n))^";"
^(tree2edges n left) ^(tree2edges n right)
|(Br(x1,_,_),Lf) -> (string_of_int (x+n))^":sw ->"^(string_of_int (x1+n))^";"
^(tree2edges n left)
|(Lf,Br(x2,_,_)) -> (string_of_int (x+n))^":se ->"^(string_of_int (x2+n))^";"
^(tree2edges n right)
in match inputs with [] -> ""
|x::xs -> (tree2edges delta (makeSTree x) ) ^ " " ^ (trees2edges xs (delta+(List.length x)))
in "digraph forrest{ node [shape=box];"^(trees2edges inputs 0)^"}";;
trees2dot (makeUniqTreeInputs [1;2;3;4]);;
#ref(test.png)
**Exercise 9 (けいご) [#k464fb78]
*Chapter 7. Exercise [#dc67177c]
**Exercise 1 [#oe649801]
ref型を
type 'a ref = { mutable contents : 'a};;
こんな定義の更新可能レコードと見て,
関数ref, 前置演算子!, 中置演算子:= をレコード操作で書け.
-関数ref
# let ref x = { contents = x } ;;
val ref : 'a -> 'a ref = <fun>
-前置演算子!
# let ( ! ) x = x.contents ;;
val ( ! ) : 'a ref -> 'a = <fun>
-中置演算子:=
# let ( := ) x y = x.contents <- y;;
val ( := ) : 'a ref -> 'a -> unit = <fun>
**Exercise 2 [#u76c0a17]
整数の参照をインクリメントする関数incr
# let incr x = x := !x + 1;;
val incr : int ref -> unit = <fun>
**Exercise 3 [#fe92b783]
# let f = ref (fun y -> y+1)
let funny_fact x = if x = 1 then 1 else x * (!f(x-1));;
# f := funny_fact;;
# funny_fact 5;;
let fは単に関数の参照が用意したいだけで,funで定義された関数の中身に意味は無い.(例えば let f = ref (fun y -> 1) でもOK)~
f := funny_factによりfがfunny_factをさすようになる.~
その結果,funny_factの定義中の!fが自分自身(funny_fact)を呼ぶ事になり,~
階乗を素直に再帰的に定義した時と同じ形になっている.
**Exercise 4 [#v6d2c968]
# let fact_imp n =
let i = ref n and res = ref 1 in
while ( !i > 0 ) do
res := !res * !i;
i := !i - 1
done;
!res;;
val fact_imp : int -> int = <fun>
**Exercise 5 [#q5984fe9]
**Exercise 6 [#ke8a0983]
**Exercise 7 [#n2c3331e]
**Exercise 8 (末次) [#n6faefa9]
まず元の定義
let rec change = function
(_, 0) -> []
| ((c :: rest) as coins, total) ->
if c > total then change (rest, total)
else c :: change (coins, total - c) ;;
これだと
let us_coins = [25; 10; 5; 1]
and gb_coins = [50; 20; 10; 5; 2; 1];;
change (gb_coins, 43);;
change (us_coins, 43);;
は成功するが、
change ([5; 2], 16);;
Exception: Match_failure ("", 66, -211).
となって失敗する.これは大きい額から試していくので 5 で3回割ったあと、リストの最後 nil まで行って、 ([], 1) にマッチする規則が無いため.
そこで失敗したら戻って小さな額で割るようにすればよい.
let rec change = function
(_, 0) -> []
| ((c :: rest) as coins, total) ->
if c > total then change (rest, total)
else
(try
c :: change (coins, total - c)
with Failure "change" -> change (rest, total)) (* 失敗したら c で割るのを諦めて次に小さい数で続ける *)
| _ -> raise (Failure "change");; (* ([], 1以上) のときは例外を投げる *)
このようにすれば
change ([5; 2], 16);;
- : int list = [5; 5; 2; 2; 2]
と計算できる.