順序なしリストとしての集合
; 2.3.3 集合を表現する
; 順序なしリストとしての集合
; 集合にxが含まれているか？
(define (element-of-set? x set)
(cond
((null? set) #f)
((equal? x (car set)) #t)
(else (element-of-set? x (cdr set)))))
; 集合の積 intersection-set
(define (intersection-set set1 set2)
(cond
((or (null? set1) (null? set2)) '())
((element-of-set? (car set1) set2)
(cons (car set1) (intersection-set (cdr set1) set2)))
(else (intersection-set (cdr set1) set2))))
; Test
(define s1 (list 1 2 3))
(define s2 (list 1 3 5))
(print "s1:" s1)
(print "s2:" s2)
(print "要素が含まれているか")
(print "(element-of-set? 1 s1):" (element-of-set? 1 s1))
(print "(element-of-set? 10 s1):" (element-of-set? 10 s1))
s1:(1 2 3)
s2:(1 3 5)
要素が含まれているか
(element-of-set? 1 s1):#t
(element-of-set? 10 s1):#f
Ex 2.59
(print "===Ex 2.59===")
;練習問題 2.59: 順序なしリストとして表現した集合に対する
;union-set 演算を実装せよ。
(define (union-set base target)
;(print "base: " base "target: " target)
(cond
((null? target) base)
((not (element-of-set? (car target) base))
(union-set (cons (car target) base) (cdr target)))
(else (union-set base (cdr target))))
)
(print "(union-set s1 s2):" (union-set s1 s2))
base: (1 2 3)target: (1 3 5)
base: (1 2 3)target: (3 5)
base: (1 2 3)target: (5)
base: ((1 2 3) . 5)target: ()
(union-set s1 s2):(5 1 2 3)
Ex 2.60
従来の重複なしのリストで表現した場合
| 名前 | 意味 | 計算量 |
|---|---|---|
| element-of-set? | 含むか | O(n) |
| adjoin-set | 要素の追加 | O(n) |
| union-set | 要素の論理和 | O(n^2) |
| intersection-set | 要素の論理積 | O(n^2) |
重複ありのリストで表現した場合
| 名前 | 意味 | 計算量 |
|---|---|---|
| element-of-set? | 含むか | O(n) |
| adjoin-set | 要素の追加 | O(1) |
| union-set | 要素の論理和 | O(n) |
| intersection-set | 要素の論理積 | O(n^2) |
※ただしnがドンドンおっきくなってくよね・・・
(print "===Ex 2.60===")
;上の例では、集合は重複のないリストとして表現す
;るよう規定した。ここで、重複を許す場合について考えてみよう。そ
;の場合、例えば {1, 2, 3} という集合は (2 3 2 1 3 2 2) というリス
;トとして表現することもできる。この表現に対して演算を行う手続
;き element-of-set?, adjoin-set, union-set, intersection-set
;を設計せよ。それぞれの効率は、重複なし表現に対する手続きで
;それに対応するものと比べてどうだろうか。重複なしの表現より
;もこの表現のほうが向いているような応用はあるだろうか。
;adjoin-set: 要素の追加
;intersection: 積
;union-set: 和
(define (adjoin-set x set) (cons set x))
(define (union-set set1 set2)
(append set1 set2)
)
(print "(adjoin-set 5 s1): " (adjoin-set 5 s1))
(print "(adjoin-set s2 s1): " (adjoin-set s2 s1))
(adjoin-set 5 s1): ((1 2 3) . 5)
(adjoin-set s2 s1): ((1 2 3) 1 3 5)
雑談メモ
> 重複みないの、手抜きっぽいのに処理量へるのなんだかふしぎ〜〜
計算量は減りますけど
set(集合)の目的として、「〜は含まれているか」 の element-of-setが (edited)
メインで使われるものだと思うので (edited)
期待できる時間としては圧倒的にのびますよね。
結局union, intersectionしても、つかわなかったら意味ないですし・・・。
> なるほど。
結局、どの処理を１番メインに使うか。メインで使う関数の計算量を減らせる実装にするのが大事ですね〜＞＜
順序つきリストとしての集合
Ex 2.61
順序ありのリストで表現した場合
| 名前 | 意味 | 計算量 |
|---|---|---|
| element-of-set? | 含むか | O(n) |
| adjoin-set | 要素の追加 | O(n) |
| union-set | 要素の論理和 | O(n) |
| intersection-set | 要素の論理積 | O(n) |
element-of-set?
計算量としては変わらないけど
平均的には半分しか探索しないので、
実際の探索時間は半減が期待できる
論理和、論理積
順序付だと、element-of-setを使ってのチェックをしなくて済むので一気短くなる♪ O(n^2)-> O(n)
(print "===Ex2.61===")
;順序つき表現を使った adjoin-set を実装せよ。
;element-of-set? から類推して、順序つきであることの利点を生
;かして、順序なしの表現に比べて平均的に半分のステップを必要
;とする手続きを作るやり方を示せ。
; そもそもリストの真ん中に挿入はO(1)で出来る？
; 0~N までのリスト と N~Mまでのリストに分けられるっけ・・・
(define (adjoin-set x set)
(cond
((null? set) (list x))
((= x (car set)) set)
((< x (car set)) (cons x set))
(else
(cons (car set) (adjoin-set x (cdr set))))
)
)
(define s1 (list 1 2 3 4 5))
(define s2 (list 1 3 5 7 9))
(print "s1:" s1)
(print "s2:" s2)
(print "(adjoin-set 5 s1):" (adjoin-set 5 s1))
(print "(adjoin-set 4 s2):" (adjoin-set 4 s2))
s1:(1 2 3 4 5)
s2:(1 3 5 7 9)
(adjoin-set 5 s1):(1 2 3 4 5)
(adjoin-set 4 s2):(1 3 4 5 7 9)
メモ
Pairで連結されたやつの間に要素を追加する方法
(define s1 (list 1 2 3 4 5))
(print (cons (car s1) (cons 100 (cdr s1))))
; (1 100 2 3 4 5)
; これで１個目の要素のあとに新しい要素を追加できるっ
Ex 2.62
(print "===Ex2.62===")
; 順序付リストO(n)でunion-setを実装
(define (union-set base target)
(define (itr result l1 l2)
;(print result " " l1 (null? l1) " " l2 (null? l2))
(cond
((and (null? l1) (null? l2)) result)
((or
(null? l2)
(and (not (null? l1)) (<= (car l1) (car l2))))
(itr (cons (car l1) result) (cdr l1) l2))
;((or
; (null? l1)
; (and (not (null? l2)) (> (car l1) (car l2))))
(else
(itr (cons (car l2) result) l1 (cdr l2)))
)
)
(reverse (itr (list) base target)))
(define s1 (list 1 2 3 4 5))
(define s2 (list 1 3 5 7 9))
(print "s1:" s1)
(print "s2:" s2)
(print "(union-set s1 s2):" (union-set s1 s2))
===Ex2.62===
s1:(1 2 3 4 5)
s2:(1 3 5 7 9)
() (1 2 3 4 5)#f (1 3 5 7 9)#f
(1) (2 3 4 5)#f (1 3 5 7 9)#f
(1 1) (2 3 4 5)#f (3 5 7 9)#f
(2 1 1) (3 4 5)#f (3 5 7 9)#f
(3 2 1 1) (4 5)#f (3 5 7 9)#f
(3 3 2 1 1) (4 5)#f (5 7 9)#f
(4 3 3 2 1 1) (5)#f (5 7 9)#f
(5 4 3 3 2 1 1) ()#t (5 7 9)#f
(5 5 4 3 3 2 1 1) ()#t (7 9)#f
(7 5 5 4 3 3 2 1 1) ()#t (9)#f
(9 7 5 5 4 3 3 2 1 1) ()#t ()#t
(union-set s1 s2):(1 1 2 3 3 4 5 5 7 9)
hioさん別解
こっちのほうがしんぷるだ〜
(define (union-set set1 set2)
(cond
((null? set1) set2)
((null? set2) set1)
(else
(let ((x1 (car set1)) (x2 (car set2)))
(cond
((= x1 x2)
(cons x1 (union-set (cdr set1) (cdr set2))))
((< x1 x2)
(cons x1 (union-set (cdr set1) set2)))
((< x2 x1)
(cons x2 (union-set set1 (cdr set2)))))))))
二分木としての集合
Ex 2.63
(print "===Ex2.63===")
(define (tree->list-1 tree)
(if (null? tree)
'()
(append (tree->list-1 (left-branch tree))
(cons (entry tree)
(tree->list-1
(right-branch tree))))))
(define (tree->list-2 tree)
(define (copy-to-list tree result-list)
(if (null? tree)
result-list
(copy-to-list (left-branch tree)
(cons (entry tree)
(copy-to-list
(right-branch tree)
result-list )))))
(copy-to-list tree '()))
(define s (adjoin-set 5 '()))
(define s2 (adjoin-set 10 s))
(define s3 (adjoin-set 3 s2))
(define s4 (adjoin-set 4 s3))
(print s4)
;(5 (3 () (4 () ())) (10 () ()))
; 5
; /\
; 3 10
; \
; 4
(print (tree->list-1 s4))
(print (tree->list-2 s4))
;(3 4 5 10)
;(3 4 5 10)
(define s (adjoin-set 1 '()))
(define s (adjoin-set 2 s))
(define s (adjoin-set 3 s))
(define s (adjoin-set 4 s))
(define s (adjoin-set 5 s))
(define s (adjoin-set 6 s))
(define s (adjoin-set 7 s))
(print s)
(print "tree->list-1: " (tree->list-1 s))
(print "tree->list-2: " (tree->list-2 s))
;tree->list-1: (1 2 3 4 5 6 7)
;tree->list-2: (1 2 3 4 5 6 7)
a.
おんなじ
b.
1: O(nlogn) 毎回半分ずつappendするので
2: O(n)
サンプルの作り方
サンプルの作り方はこっちの方がいい!
(define s (fold adjoin-set '() (list 1 2 3 4 5 6 7)))
; (1 () (2 () (3 () (4 () (5 () (6 () (7 () ())))))))
memo
長さn のlistに別の長さnのlistをappendするのってOrder的にn
log n 回 tree->list-1が呼び出されて
その中のappendがnだから
1は O(n log n)
Ex 2.64
Quotient: 余り
(print "===Ex2.64===")
(define (list->tree elements)
(car (partial-tree elements (length elements))))
(define (partial-tree elts n)
;(print "partial-tree elts:" elts " n:" n)
(if (= n 0)
(cons '() elts)
(let
(
(left-size (quotient (- n 1) 2)); leftsize = n / 2
)
(let
(
(left-result (partial-tree elts left-size)) ;leftresult = elementのleft-size分
)
(let
(
(left-tree (car left-result)) ; left-tree = left-resultの1こめ
(non-left-elts (cdr left-result)) ; non-left-elts = leftresultののこり
(right-size (- n (+ left-size 1))) ; right-size = (n - leftsize) + 1
)
(let
(
(this-entry (car non-left-elts)) ; this entry = non-left-eltsの1こめ
(right-result ; right-result = left-resultの3こめ以降の木
(partial-tree
(cdr non-left-elts)
right-size))
)
(let
(
(right-tree (car right-result)) ; right-tree: right-resultの１番最初
(remaining-elts (cdr right-result)) ; のこり: right resultの2こ目以降
)
(cons
(make-tree
this-entry
left-tree
right-tree)
remaining-elts )
)))))))
(define l (list 1 2 3 4 5 6 7))
; 2
; /\
;1 3 (4 5 6 7)
; 2 5
; /\ /\
;1 3 と (4 ) と 6 7
;
; 4
; /\
; 2 5
; /\ /\
;1 3 6 7
挙動
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((1 () ()) 2 3 4 5 6 7)
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((3 () ()) 4 5 6 7)
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((2 (1 () ()) (3 () ())) 4 5 6 7)
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((5 () ()) 6 7)
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((7 () ()))
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((6 (5 () ()) (7 () ())))
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((4 (2 (1 () ()) (3 () ())) (6 (5 () ()) (7 () ()))))
(4 (2 (1 () ()) (3 () ())) (6 (5 () ()) (7 () ())))
;1-14
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((2 () ()) 3 4 5 6 7 8 9 10 11 12 13 14)
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((1 () (2 () ())) 3 4 5 6 7 8 9 10 11 12 13 14)
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((4 () ()) 5 6 7 8 9 10 11 12 13 14)
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((6 () ()) 7 8 9 10 11 12 13 14)
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((5 (4 () ()) (6 () ())) 7 8 9 10 11 12 13 14)
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((3 (1 () (2 () ())) (5 (4 () ()) (6 () ()))) 7 8 9 10 11 12 ...
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((8 () ()) 9 10 11 12 13 14)
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((10 () ()) 11 12 13 14)
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((9 (8 () ()) (10 () ())) 11 12 13 14)
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((12 () ()) 13 14)
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((14 () ()))
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((13 (12 () ()) (14 () ())))
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((11 (9 (8 () ()) (10 () ())) (13 (12 () ()) (14 () ()))))
#?="./2.3.3.scm":301:(cons (make-tree this-entry left-tree right-tree) remaining-elts)
#?- ((7 (3 (1 () (2 () ())) (5 (4 () ()) (6 () ()))) (11 (9 (8 () ...
(7 (3 (1 () (2 () ())) (5 (4 () ()) (6 () ()))) (11 (9 (8 () ()) (10 () ())) (13 (12 () ()) (14 () ()))))
- リストの真ん中をとってきて
- そこから真ん中以外の左と右にわける
- 1.繰り返し
みたいないめーじ
a.
(print (list->tree l))
; 4
; / \
; 2 6
; /\ /\
;1 3 5 7
b.
O (n)
Ex.2.65
(print "===Ex2.65===")
;(論理和)
(define (union-set set1 set2)
(define (union-set-i set l)
;(print set l)
(if
(null? l)
set
(let
(
(first (car l))
(lst (cdr l)))
(union-set-i (adjoin-set first set) lst)
)
)
)
(union-set-i set1 (tree->list-1 set2))
)
(define l1 (list 1 2 3 4 5 6 7))
(define l2 (list 1 3 5 7 9))
(print (union-set (list->tree l1) (list->tree l2)))
1個の要素追加にかかるコストがlog_n
なのでこれだと O(n log_n)
(print "===O(n)===")
;; ?
(define (union-set set1 set2)
(define (union-list base target)
; 最後の要素が同じでなければ追加
(define (add x l)
(cond
((null? l) (cons x l))
((= (car l) x) l)
(else (cons x l))
))
(define (itr result l1 l2)
(print result " " l1 (null? l1) " " l2 (null? l2))
(cond
((and (null? l1) (null? l2)) result)
;((null? l2) ())
((or
(null? l2)
(and (not (null? l1)) (<= (car l1) (car l2))))
(itr (add (car l1) result) (cdr l1) l2))
(else
(itr (add (car l2) result) l1 (cdr l2)))
)
)
(reverse (itr (list) base target)))
(list->tree (union-list (tree->list-1 set1) (tree->list-1 set2)))
)
(print (union-set (list->tree l1) (list->tree l2)))
;() (1 2 3 4 5 6 7)#f (1 3 5 7 9)#f
;(1) (2 3 4 5 6 7)#f (1 3 5 7 9)#f
;(1) (2 3 4 5 6 7)#f (3 5 7 9)#f
;(2 1) (3 4 5 6 7)#f (3 5 7 9)#f
;(3 2 1) (4 5 6 7)#f (3 5 7 9)#f
;(3 2 1) (4 5 6 7)#f (5 7 9)#f
;(4 3 2 1) (5 6 7)#f (5 7 9)#f
;(5 4 3 2 1) (6 7)#f (5 7 9)#f
;(5 4 3 2 1) (6 7)#f (7 9)#f
;(6 5 4 3 2 1) (7)#f (7 9)#f
;(7 6 5 4 3 2 1) ()#t (7 9)#f
;(7 6 5 4 3 2 1) ()#t (9)#f
;(9 7 6 5 4 3 2 1) ()#t ()#t
;(1 2 3 4 5 6 7 9)
;(4 (2 (1 () ()) (3 () ())) (6 (5 () ()) (7 () (9 () ()))))
(print "===intersection(積)===")
(define (intersection-set set1 set2)
(define (intersection-list base target)
(print base target)
(if (or (null? base) (null? target))
'()
(let ((x1 (car base)) (x2 (car target)))
(cond
((= x1 x2) (cons x1 (intersection-list (cdr base) (cdr target))))
((< x1 x2) (intersection-list (cdr base) target))
((< x2 x1) (intersection-list base (cdr target)))))
)
)
(list->tree (intersection-list (tree->list-1 set1) (tree->list-1 set2)))
)
(print (intersection-set (list->tree l1) (list->tree l2)))
;(1 2 3 4 5 6 7)(1 3 5 7 9)
;(2 3 4 5 6 7)(3 5 7 9)
;(3 4 5 6 7)(3 5 7 9)
;(4 5 6 7)(5 7 9)
;(5 6 7)(5 7 9)
;(6 7)(7 9)
;(7)(7 9)
;()(9)
;(3 (1 () ()) (5 () (7 () ())))
これだと2・nで実行できるので O(n) かな？
二分木で実装した場合の計算量
| 名前 | 意味 | 計算量(平均) | 計算量(最悪) |
|---|---|---|---|
| element-of-set? | 含むか | O(log n) | O(n) |
| adjoin-set | 要素の追加 | O(log n) | O(n) |
| union-set | 要素の論理和 | O(n) | O(n) |
| intersection-set | 要素の論理積 | O(n) | O(n) |
Ex.2.66
(print "===Ex2.66===")
(define (make-record k v) (list k v))
(define (key record) (car record))
(define (value record) (cadr record))
(define (lookup k records)
;(print records)
(if
(null? records)
#f
(let
((record (entry records)))
(cond
((= k (key record)) (value record))
((< k (key record)) (lookup k (left-branch records)))
((> k (key record)) (lookup k (right-branch records)))
))
)
)
(define _records
(list->tree
(list (make-record 1 "a") (make-record 2 "b") (make-record 3 "ab")))
)
(print (lookup 1 _records)) ;a
(print (lookup 3 _records)) ;ab