[英]How to store the value of a recursive function in a list in prolog?
我的任務是:
給定函數f的以下定義,創建一個Prolog程序來計算所有0 <i <32的f(i)。
- f(0)= 0
- f(1)= 1
- 對於n> 1,f(n)= f(n-2)+ 2 * f(n-1)
到目前為止我的代碼是:
problemThree(0, 0).
problemThree(1, 1).
problemThree(N, NF) :-
N > 1,
A is N - 2,
B is N - 1,
problemThree(A, AF),
problemThree(B, BF),
NF is AF + 2*BF.
它正在發揮作用,但它將永遠顯示N> 20的值。
請告訴我如何將值存儲在列表中以使程序更快。
這是一個DCG方法,它將序列生成為列表:
prob3(1, F0, F1) --> [F0, F1].
prob3(N, F0, F1) --> {N > 1, F2 is 2*F1 + F0, N1 is N-1}, [F0], prob3(N1, F1, F2).
prob3(0, [0]).
prob3(N, FS) :-
phrase(prob3(N, 0, 1), FS).
?- prob3(10, L).
L = [0, 1, 2, 5, 12, 29, 70, 169, 408, 985] ;
false.
?- prob3(169, L).
L = [1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025,
..., 17280083176824678419775054525017769508908307965108250063833395641] ;
false
?- time((prob3(1000, L),false)).
% 3,011 inferences, 0.005 CPU in 0.005 seconds (100% CPU, 628956 Lips)
false.
?- prob3(20, L).
L = [0, 1, 2, 5, 12, 29, 70, 169, 408|...] [write] % PRESSED 'w' here
L = [0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025, 470832, 1136689, 2744210, 6625109, 15994428] ;
false
?-
這只是SWI Prolog通過大量輸出不會弄亂屏幕的方式。 在這里,你可以用w
回答,它將給出整個結果:
?- prob3(20, L). L = [0, 1, 2, 5, 12, 29, 70, 169, 408|...] [write] % PRESSED 'w' here L = [0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025, 470832, 1136689, 2744210, 6625109, 15994428] ; false ?-
請參閱, 幫助:我想要整個答案 。
無需存儲超過前兩個數字!
這是我從臀部快速而骯臟的鏡頭:
p3(N,F) :- ( N =:= 0 -> F = 0 ; N =:= 1 -> F = 1 ; N > 1 -> N0 is N-2, p3_(N0,0,1,F) ). p3_(N,F0,F1,F) :- F2 is F0 + 2*F1, ( N =:= 0 -> F2 = F ; N0 is N-1, p3_(N0,F1,F2,F) ).
示例查詢:
?- between(25,35,N), p3(N,F). N = 25, F = 1311738121 ; N = 26, F = 3166815962 ; N = 27, F = 7645370045 ; N = 28, F = 18457556052 ; N = 29, F = 44560482149 ; N = 30, F = 107578520350 ; N = 31, F = 259717522849 ; N = 32, F = 627013566048 ; N = 33, F = 1513744654945 ; N = 34, F = 3654502875938 ; N = 35, F = 8822750406821.
更大的東西一點點 :
?- p3(111,F).
F = 1087817594842494380941469835430214208491185.
?- p3(123,F).
F = 42644625325266431622582204734101084193553730205.
?- p3(169,F).
F = 17280083176824678419775054525017769508908307965108250063833395641.
夠快嗎?
?- time((between(0,1000,N), p3(N,_), false)).
% 2,006,005 inferences, 0.265 CPU in 0.265 seconds (100% CPU, 7570157 Lips)
false.
雖然它比其他答案慢得多,但我喜歡這個具有功能精神的人:
:- use_module(library(lambda)).
f(N, FN) :-
cont_f(N, _, FN, \_^Y^_^U^(U = Y)).
cont_f(N, FN1, FN, Pred) :-
( N < 2 ->
call(Pred, 0, 1, FN1, FN)
;
N1 is N - 1,
P = \X^Y^Y^U^(U is X + 2*Y),
cont_f(N1, FNA, FNB, P),
call(Pred, FNA, FNB, FN1, FN)
).
記憶是有用的,我把它用於使用Erathostenes篩的加速計算
?- time((between(0,1000,N), prob3(N,_), false)).
% 10,939 inferences, 0.011 CPU in 0.012 seconds (99% CPU, 951780 Lips)
:- dynamic memo/2.
prob3(0, 0).
prob3(1, 1).
prob3(N, R) :- memo(N, R), !.
prob3(N, R) :-
N > 1, N2 is N-2, N1 is N-1, prob3(N2,R2), prob3(N1,R1), R is R2+2*R1, assertz(memo(N, R)).
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.