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1 2013 年 5 月 31 日 Lisp プログラミング入門 西田豊明 Copyright 2013 Toyoaki Nishida All Rights Reserved.
2 今回あらすじ 1. Lisp の実践的な使い方を学習する. 2. Lisp インタープリタの動かし方, 電卓的使い方, 関数定義, 条件分岐,S 式の基本操作, プログラミング手法, プロトタイピング法などを中心に解説する. 3. チューリングマシンのシミュレータの構成方法を例にプログラミングの実際を学習する.
3 Lisp プログラミングの流れ 表現 人工知能プログラム 私の人工知能理論 解釈実行 表示 インタープリタ Gnu Common Lisp
4 情報の流れ 入力テープとヘッド位置の外部表現 チューリングマシンの状態遷移機械の外部表現 入力テープ左 ( 逆順 ) ヘッド位置のマス目の内容入力テープ右現在状態 シミュレータ チューリングマシンの状態表示
5 ((start q0) 初期状態がq0である (terminal q5) 停止状態はq5である. (states 以下では, 各状態における状態遷移規則を表す (q0 (0 (Right) q1)) 状態 q0で, 0を見たらヘッドを右に移動し, 状態 q1に遷移 (q1 (0 (Right) q1) 状態 q1で, 0を見たらヘッドを右に移動し, 状態 q1に遷移 (1 B q2)) 状態 q1で, 1を見たらBと書き換え, 状態 q2に遷移 (q2 (A (Left) q2) 状態 q2で, Aを見たらヘッドを左に移動し, 状態 q2に遷移 (B (Left) q2) 状態 q2で, Bを見たらBと書き換え, 状態 q2に遷移 (0 A q3)) 状態 q2で,0を見たらaと書き換え, 状態 q3に遷移 (q3 (1 B q2) 状態 q3で,1を見たらbと書き換え, 状態 q2に遷移 (A (Right) q3) 状態 q3で,aを見たらヘッドを右に移動し, 状態 q3に遷移 (B (Right) q3) 状態 q3で,bを見たらヘッドを右に移動し, 状態 q3に遷移 (_ (Left) q4)) (q4 (A (Left) q4) 状態 q4で,aを見たらヘッドを左に移動し, 状態 q4に遷移 (B (Left) q4) 状態 q4で,bを見たらヘッドを左に移動し, 状態 q4に遷移 ( q5)))) 状態 q4で, 空白 _ を見たら, 状態 q5に遷移 状態 q3 で, 空白 _ を見たらヘッドを左に移動し, 状態 q4 に遷移
6 変数 input 入力テープとヘッド位置の外部表現 tm チューリングマシンの状態遷移機械の外部表現 left-stack current-cell right-stack current-state 入力テープ左 ( 逆順 ) ヘッド位置のマス目の内容入力テープ右現在状態 test シミュレータ (print による副作用 ) チューリングマシンの状態表示
7 準備 1 input tm (let* ( [1] 入力テープとヘッド位置の外部表現 (current-state (second (assoc 'start tm))) ) ) チューリングマシンの状態遷移機械の外部表現 left-stack current-cell right-stack current-state 入力テープ左 ( 逆順 ) ヘッド位置のマス目の内容入力テープ右現在状態 tm=((start test q0) (terminal q5) ) [1] Current-state=q0 シミュレータ (print による副作用 ) チューリングマシンの状態表示
8 準備 2 input 入力テープとヘッド位置の外部表現 [1] (setq right-stack (cdr (member '* input))) [2] (cond (right-stack [2] (setq current-cell (car right-stack)) [2] (setq right-stack (cdr right-stack))) [2] (t (setq current-cell '_))) [3] (dolist tm (cell input) [3] (cond ((eq cell '*) (return))) [3] (push cell left-stack)) チューリングマシンの状態遷移機械の外部表現 left-stack current-cell right-stack current-state 入力テープ左 ( 逆順 ) ヘッド位置のマス目の内容入力テープ右現在状態 input=(1 test 2 3 * ); left-stack=(); current-cell=nil; right-stack= () [1] right-stack=( ) [2] シミュレータ current-cell=4; right-stack=(5 6 7) [3] left-stack=(3 2 1) (print による副作用 ) チューリングマシンの状態表示
9 中心部分 input 入力テープとヘッド位置の外部表現 tm チューリングマシンの状態遷移機械の外部表現 left-stack current-cell right-stack current-state 入力テープ左 ( 逆順 ) ヘッド位置のマス目の内容入力テープ右現在状態 test シミュレータ (print による副作用 ) チューリングマシンの状態表示
10 典型的な状況例 入力テープ チューリングマシンの場合 状態遷移機械 開始 1/B, L A/A, L B/B, L 0/A, R A/A, R B/B, R A/A, L B/B, L q 0 q 1 q 2 q 3 q 4 q 5 1/B, L _/_, L _/_, R
11 チューリングマシン 典型的な状況例 入力テープ B 1 1 状態遷移機械 開始 1/B, L A/A, L B/B, L 0/A, R A/A, R B/B, R A/A, L B/B, L q 0 q 1 q 2 q 3 q 4 q 5 1/B, L _/_, L _/_, R
12 中心部分 left-stack tm チューリングマシンの状態遷移機械の外部表現 current-cell right-stack current-state 入力テープ左 ( 逆順 ) ヘッド位置のマス目の内容入力テープ右現在状態 test シミュレータ tm=( (states (q1 (0 0 right q1) (1 B left q2)) ) ) left-stack=(0 0 0); current-cell=1; right-stack=(1 1) current-state=q1 [?] left-stack=(0 0); current-cell=0; right-stack=(b 1 1) current-state=q2
13 中心部分 tm=( (states (q1 (0 0 right q1) (1 B left q2)) ) ) left-stack=(0 0 0); current-cell=1; right-stack=(1 1) current-state=q1 [?] left-stack=(0 0); current-cell=0; right-stack=(b 1 1) current-state=q2 [1] (setq rules (cdr (assoc current-state states))) [2] (setq rule (assoc current-cell rules)) [3] (cond ((null rule) (return))) [4] (setq current-cell (second rule)) [5] (setq current-state (fourth rule)) [6] (case (third rule) [7] (right (cond [7] (right-stack [7] (setq left-stack `(,current-cell.,left-stack)) [7] (setq current-cell (car right-stack)) [7] (setq right-stack (cdr right-stack))) [7] (t (setq left-stack `(,current-cell.,left-stack)) [7] (setq current-cell '_)))) [7] (left (cond [7] (left-stack [7] (setq right-stack `(,current-cell.,right-stack)) [7] (setq current-cell (car left-stack)) [7] (setq left-stack (cdr left-stack))) [7] (t (setq right-stack `(,current-cell.,right-stack)) [7] (setq current-cell '_))))
14 プログラム test (defun test () (let* ((tm tm0) (input input-tm0) (left-stack nil) (right-stack) (current-state (second (assoc 'start tm))) (current-cell nil) (states (cdr (assoc 'states tm))) (terminal-states (cdr (assoc 'terminal tm))) (rules nil) (rule nil) (counter 1) (credits 1)) (setq right-stack (cdr (member '* input))) (cond (right-stack (setq current-cell (car right-stack)) (setq right-stack (cdr right-stack))) (t (setq current-cell '_))) (dolist (cell input) (cond ((eq cell '*) (return))) (push cell left-stack)) (loop (cond ((<= credits counter) (princ "How many more steps to go?") (terpri) (setq credits (read)) (cond ((<= credits 0) (return)) (t (setq counter 0)))) (t (setq counter (1+ counter)))) (princ "======================================") (terpri) (print-state left-stack current-cell right-stack current-state) (cond ((member current-state terminal-states) (princ "======================================") (terpri) (princ "*** Halt in a termnial state ***") (terpri) (princ "======================================") (terpri) (return))) ))) [1] (setq rules (cdr (assoc current-state states))) [2] (setq rule (assoc current-cell rules)) [3] (cond ((null rule) (return))) [4] (setq current-cell (second rule)) [5] (setq current-state (fourth rule)) [6] (case (third rule) [7] (right (cond [7] (right-stack [7] (setq left-stack `(,current-cell.,left-stack)) [7] (setq current-cell (car right-stack)) [7] (setq right-stack (cdr right-stack))) [7] (t (setq left-stack `(,current-cell.,left-stack)) [7] (setq current-cell '_)))) [7] (left (cond [7] (left-stack [7] (setq right-stack `(,current-cell.,right-stack)) [7] (setq current-cell (car left-stack)) [7] (setq left-stack (cdr left-stack))) [7] (t (setq right-stack `(,current-cell.,right-stack)) [7] (setq current-cell '_))))
15 ホームワーク 次のチューリングマシンの動作をシミュレートせよ 入力テープ 状態遷移機械 開始 0/X, R Y/Y, R 1/Y, L q 0 q 1 q 2 Y/Y, L 0/0, L X/X, R Y/Y, R _ / _, R q 3 q 4 Y/Y, R [Hopcroft 2001]
16 ホームワーク 次のチューリングマシンの動作を分析せよ 入力テープ _ 状態遷移機械 _ /_, R 1/1, R 開始 0/ _, R 1/1, R 0/1, L q 2 q 0 q 1 q 3 1/ _, R _ / _, L 0/0, L 1/1, L _ / _, R q 5 q 6 _ / 0, R q 4 0/ _, R 1/ _, R 0/ 0, L 1/ _, L [Hopcroft 2001]
17 ホームワーク 次のチューリングマシンの動作を分析せよ 入力テープ 状態遷移機械 開始 0/0, L _ /1, L q 0 q 1 _/ _, R q 2 _ / _, R q 11 1/1, R 0/0, L 1/1, L 0/ _, R 1/1, R q 3 q 4 0/X, R 1/1, R _ /0, L 0/0, L 1/1, L q 6 q 7 0/0, L q 5 1/1, L 1/1, R X/X, R 0/0, L 1/1, L q 8 q 9 q 10 X/0, L q 14 1/ _, R q 13 0/ _, R 1/ _, R q 12 _/ _, R [Hopcroft 2001]
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