Functional Programming
09 Examples
Tower of Hanoi
A
B
C
Tower of Hanoi
Tower of Hanoi
 希望執行結果
 > (tower-of-hanoi ‘(3 2 1) ‘a ‘b ‘c) ;三個盤子 柱子a b c
7
 第一步 寫外殼
 (defun tower-of-hanoi (disks from to spare)
)
 第二步 決定遞迴
 何時停止？ 盤子搬完
 還有盤子就搬並計算次數 （繼續遞迴）
 (if (endp disks)
xxx ;沒有盤子就return 0 (需要零次)
ooo ;有盤子就繼續搬並計算次數
)
Tower of Hanoi
 第三步 決定次數計算方式
 xxx?
 0 (沒盤子就不必搬了,所以直接return 0)
 ooo?
 先把上面n-1個搬到暫存區spare
要幾次?
(tower-of-hanoi (rest disks ) from spare to)
 再把最底下那個搬到目的地to
要幾次?
1
 再把暫存區的n-1個搬到目的地to
要幾次?
(tower-of-hanoi (rest disks ) spare to from)
Tower of Hanoi
(defun tower-of-hanoi (disks from to spare)
(if (endp disks)
0
(+
(tower-of-hanoi (rest disks) from spare to)
1
(tower-of-hanoi (rest disks) spare to from)
)
)
)
Tower of Hanoi
 希望執行結果 要包含搬法
 > (tower-of-hanoi ‘(3 2 1) ‘a ‘b ‘c) ;三個環 柱子a b c
Move 1 from A to B.
Move 2 from A to C.
Move 1 from B to C.
Move 3 from A to B.
Move 1 from C to A.
Move 2 from C to B.
Move 1 from A to B.
NIL
Tower of Hanoi
 僅須修改剛剛的ooo
 先把上面n-1個搬到暫存區spare
要幾次?
(tower-of-hanoi (rest disks ) from spare to)
 再把最底下那個搬到目的地to
要幾次?
1
 再把暫存區的n-1個搬到目的地to
要幾次?
(tower-of-hanoi (rest disks ) spare to from)
->把1 改成
(format t “Move ~A from ~A to ~A.~%” (car disks) from to)
把+去掉改成
(progn
)
但是如果盤子空了要作什麼？
沒要作什麼..所以把if 改成unless..那麼prog可以去掉
Tower of Hanoi
(defun tower-of-hanoi (disks from to spare)
(unless (endp disks)
(tower-of-hanoi (rest disks) from spare to)
(format t "Move ~A from ~A to ~A.~%" (car disks) from to)
(tower-of-hanoi (rest disks) spare to from)
)
)
Inference - Matching
 > (match ‘(p a b c a) ‘(p ?x ?y c ?x))





((?Y . B) (?X . A))
T
> (match ‘(p ?x b ?y a) ‘(p ?y b c a))
((?Y . C) (?X . ?Y))
T
> (match ‘(a b c) ‘(a a a))
NIL
> (match ‘(p ?x) ‘(p ?x))
NIL
T
> (match ‘(p ?v b ?x d (?z ?z))
‘(p a ?w c ?y (e e))
‘((?v . a) (?w . b)))
((?Z . E) (?Y . D) (?X . C) (?V . A) (?W . B))
T
> (match ‘(?x a) ‘(?y ?y))
((?Y . A) (?X . ?Y))
T
Inference - Matching
 第一步 想match的步驟
 If x and y are eql, then match; otherwise,
 If x is a variable that has a binding, they match if it
matches y; otherwise,
 If y is a variable that has a binding, they match if it
matches x; otherwise,
 If x is a variable(without a binding), they match and
thereby establish a binding for it; otherwise,
 If y is a variable(without a binding), they match and
thereby establish a binding for it; otherwise,
 They match if they are both conses, and the cars match,
and the cdrs match with the bindings generated thereby.
Inference - Matching
 第二步 寫一個可判斷是否為變數的函數
 是否為symbol
(symbolp x)
 Symbol的字串中第一個字元是否為’?’
(eql (char (symbol-name x) 0) #\?)
 (defun var? (x)
(and (symbolp x)
(eql (char (symbol-name x) 0) #\?)))
Inference - Matching
 回顧 assoc
 > (setq x (cons ‘a ‘b))
(A . B)
 > (setq y (cons ‘c ‘d))
(C . D)
 > (setq q (list x y))
((A . B) (C . D))
 > (assoc ‘a q)
(A . B)
Inference - Matching
 If x and y are eql, then match;
((eql x y) (values binds t))
 If x is a variable that has a binding, they match if it matches y;
((assoc x binds) (match (binding x binds) y binds))
 If y is a variable that has a binding, they match if it matches x;
((assoc y binds) (match x (binding y binds) binds))
 If x is a variable (without a binding), they match and thereby establish a
binding for it;
((var? x) (values (cons (cons x y) binds) t)) ;把新的bind(關於x)加入binds
 If y is a variable (without a binding), they match and thereby establish a
binding for it;
((var? y) (values (cons (cons y x) binds) t)) ;把新的bind (關於y)加入binds
Inference - Matching
 They match if they are both conses, and the cars match, and the
cdrs match with the bindings generated thereby.
(when (and (consp x) (consp y)) ;x與y都是cons
(multiple-value-bind (b2 yes) ;把b2與yes bind 到
(match (car x) (car y) binds) ;(car x)與(car y)的match
(and yes (match (cdr x) (cdr y) b2))))
;若(car x) 與(car y)已經match,也就是yes為T,則再接
;著match (cdr x) (cdr y),且已經將binding更新為b2
Inference - Matching
(defun match (x y &optional binds)
(cond
((eql x y) (values binds t))
((assoc x binds) (match (binding x binds) y binds))
((assoc y binds) (match x (binding y binds) binds))
((var? x) (values (cons (cons x y) binds) t))
((var? y) (values (cons (cons y x) binds) t))
(t
(when (and (consp x) (consp y))
(multiple-value-bind (b2 yes)
(match (car x) (car y) binds)
(and yes (match (cdr x) (cdr y) b2)))))))
Inference - Matching
 第三步 找某個變數的binding方式
(defun binding (x binds)
(let ((b (assoc x binds))) ;找binds裡頭有x的,例如 (x, □)
(if b
;如果有找到,知道被bind到(cdr b)
(or (binding (cdr b) binds) ;繼續找看看(cdr b) bind 到誰,例如(□, ◎)
(cdr b))))) ;往下沒有可繼續bind的話,則 就是(cdr b),例如 □
Inference - Answering Queries
 第四步 回答誰誰誰符合某種關係
 > (parent ?x ?y)
(((?x . donald) (?y . nancy))
 呼叫<-以加入rule或fact並且return目前關於某個predicate
的rule有幾個
 > (<- (parent donald nancy))
1
 > (<- (child ?x ?y) (parent ?y ?x))
1
Inference - Answering Queries
 (defvar *rules* (make-hash-table)) ;定義一個全域變數為hash table
 (defmacro <- (con &optional ant) ;定義一個巨集<-
`(length (push (cons (cdr ',con) ',ant)
(gethash (car ',con) *rules*))))
 If you prefix a comma to something within a ` (backquoated)
expression, it will be evaluated
 > `(a b c)
(A B C)
 > (setf a 1 b 2)
2
 > `(a is ,a and b is ,b)
(A IS 1 AND B IS 2)
Inference - Answering Queries
(defun prove (expr &optional binds) ;expr 可能是(parent ?x ?y)或 (and ( ) ( ) ( ))
(case (car expr) ;看是否為rule(有and or not)
(and (prove-and (cdr expr) binds))
(or (prove-or (cdr expr) binds))
(not (prove-not (cadr expr) binds))
(t (prove-simple (car expr) (cdr expr) binds)))) ;不是rule,例如(parent ?x ?y)
(defun prove-simple (pred args binds) ; parent (?x ?y) binds
(mapcan #'(lambda (r)
(multiple-value-bind (b2 yes) ;分別把match的binding結果以及
(match args (car r) ; 成功與否給b2與yes
binds)
(when yes ;如果有成功binding
(if (cdr r) ;如果是rule,即 r是像 (and (male ?x) ( ) )
(prove (cdr r) b2) ;要以b2為新的binds,繼續prove
(list b2))))) ;return (b2)
(mapcar #‘change-vars
;把以pred為key的那些rule找出來,
(gethash pred *rules*)))) ;並且把其中的變數換成新產生的
Inference - Answering Queries
 > (prove-simple 'parent ' (donald nancy) nil)
(NIL) ;沒有binding可以prove所問的
 > (prove-simple 'child '(?x ?y) nil)
(((#:?6 . NANCY) (#:?5 . DONALD) (?Y . #:?5) (?X . #:?6)))
;有binding可以促成所詢問的關係
Inference - Answering Queries
(defun change-vars (r)
(sublis (mapcar #'(lambda (v) (cons v (gensym "?")))
(vars-in r))
r)) ;把r中的變數都換成新產生的?xxxx
(defun vars-in (expr) ;找expr中所有的變數
(if (atom expr)
(if (var? expr) (list expr))
(union (vars-in (car expr))
(vars-in (cdr expr)))))
Inference - Answering Queries
(defun prove-and (clauses binds)
(if (null clauses) ;若遞迴到clauses為null
(list binds)
;直接return目前的binds
(mapcan #‘ (lambda (b)
(prove (car clauses) b)) ;第一個clause作prove
(prove-and (cdr clauses) binds)))) ;後面剩下的clause作prove-and
(defun prove-or (clauses binds)
(mapcan #‘(lambda (c) (prove c binds)) ;對每一個clause都
clauses))
;繼續做prove
(defun prove-not (clause binds) ;clause只有一句
(unless (prove clause binds) ;繼續prove,找新的binds
(list binds)))
;prove不出來就用舊的binds
Inference - Answering Queries
(defmacro with-answer (query &body body)
(let ((binds (gensym))) ;產生一個新變數,假設是g1
‘(dolist (,binds (prove ’,query)) ;將prove完的binding結果
(let ,(mapcar #’(lambda (v) ;依序給g1
‘(,v (binding ',v ,binds)))
(vars-in query))
,@body))))
;query: 想問的問題,如 (parent ?x ?y)
;body : 找出binding後想把它怎麼處理
;,@會把body evaluate出來並且把所有elements拆解出來
 > (prove '(parent ?x ?y))
(((?Y . DEBBIE) (?X . DONALD)) ((?Y . NANCY) (?X .
DONALD)))
Inference - Answering Queries
(with-answer (p ?x ?y)
(f ?x ?y))
is macroexpanded into:
(dolist (#:g1 (prove ‘(p ?x ?y)))
(let ((?x (binding ‘?x #:g1))
(?y (binding ‘?y #:g1)))
(f ?x ?y)))
 > (with-answer (parent ?x ?y)
(format t “~A is the parent of ~A. ~%” ?x ?y))
DONALD is the parent of NANCY.
NIL
Inference - Analysis
 If we do a (clrhash *rules*) and then define the following
rules and facts,
定義一堆事實與規則到hash table *rule*中
(<- (parent donald nancy))
(<- (parent donald debbie))
(<- (male donald))
(<- (father ?x ?y) (and (parent ?x ?y) (male ?x)))
(<- (= ?x ?x))
(<- (sibling ?x ?y) (and (parent ?z ?x)
(parent ?z ?y)
(not (= ?x ?y))))
Inference - Analysis
我們最後想做到:
 we will be able to make inferences like the following:
 > (with-answer (father ?x ?y)
(format t "~A is the father of ~A. ~%” ?x ?y))
DONALD is the father of DEBBIE.
DONALD is the father of NANCY.
NIL
 > (with-answer (sibling ?x ?y)
(format t "~A is the sibling of ~A. ~%” ?x ?y))
DEBBIE is the sibling of NANCY.
NANCY is the sibling of DEBBIE.
NIL