* 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 (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.