Page 29 - 《软件学报》2020年第12期

赵世忠 等:循环迭代程序的一种可信计算算法 3695

否则,若式(33)不成立,则由假设知:总可以获得一个新的计算值 Y ,使得由其得到的 z 满足式(33).这样,新获
z
得的 z 的函数值 e ε/2 能够满足式(32).证明完毕. □
算法 9. AccurateEnough_Exp( {( , ) |1y ε i ≤≤ n }, ( , ,...,f x x 2 x n ),ε ).

i
i
1
Require: {( , ) | yε i i = y ,1≤≤ } n ,f(x 1 ,x 2 ,…,x n ),ε>0;
y
i
i
i
i ε
Ensure:返回一个 4 元组 D = (, , ,y k ε 0 enough ) .若 enough=true,则 | ye− ( fy 1 , y 2 ,..., n y ) | ε< ;否则,当 enough=false 时,
y 的精度ε k 不能满足计算的要求,其精度需要提高至ε 0 .
k
1: ε′⇐0.1;
2: Calculate and Judge ({( , ) |1y ε ≤≤ n }, , )f ε′ ;

i
_
_
i i
3: ε″⇐0.2;

+
4: y ⇐ e f ε ′ ε ′′ ;
5: ε″′⇐min{0.1,δ(ε/(2(y+ε″)))};

)
i
_
_
6: Calculate and Judge ({( , ) |1y ε i ≤≤ n }, ,f ε′′′ ;
i
7: y ⇐ e f ε /2 ;
8: return (y,0,0,true);
2.1.3 递归计算子算法
首先,对于下面式(34)中的函数,我们有式(35)中对应的等式:
cos( ),tan( ),cot( ),sec( ),csc( ),arcsin( ),arccos( ),arccot( ),logf 1 f 1 f 1 f 1 f 1 f 1 f 1 f 1 1 f f 2 ,sinh( ),cosh( )f 1 f 1 (34)
⎧ sin( / 2π − f ),sin( )/cos( ),cos( )/sin( ),1/cos( ),1/sin( ),2arctan( /(1f f f f f f f + 1− f 2 ))
⎪
⎨ 1 1 1 1 1 1 1 1 1 (35)
⎪ ⎩ π / 2 arcsin( ), / 2 arctan( ),ln( )/ln( ),(f π − 1 − f 1 f 1 f 2 e − 1 f e − 1 f )/(2 e + , 1 f e − 1 f ) 2/
然后,根据前面的两个子算法,我们可以得出递归计算的子算法.其中,f 为−f 1 , ∑ 1 n , f f 2 f 或式(34)中之一.
i= 1 i 1

算法 10. Sub 3_ AccurateEnough ({( , ) |1y ε i ≤≤ n }, ( , ,...,f x x 2 x n ), ).ε
i
i
1
1: if f=−f 1 then

2: D ⇐ AccurateEnough ({( , ) |1y ε i ≤≤ n }, , )f ε ;

i
1
1
i

3: D ⇐− (D 1 ),Snd (D 1 ),Thd (D 1 ),Fth (D 1 )) ;
( Fst
4: else if f = ∑ 1 n f then
i= 1 i
5: for i=1 to n 1 do
6: Calculate and Judge ({( , ) |1y ε i i i n }, , ( / ))f δ i ε ≤≤ n 1 ; //参见算法 4
_
_
7: end for

8: D ⇐ ∑ 1 n f ,0,0,true) ;
(
i= 1 i
9: else if f = f 2 f then
1
10: if f 1 =0 and f 2 >0 then

11: D ⇐ (0,0,0,true) ;
12: end if f 1 >0 then

13: D ⇐ AccurateEnough ({( , ) |1y ε i ≤≤ n },e f 2 ln( f× 1 ) , )ε
i
i
14: end if f 1 <0,f 2 =n′/n″(不可约),n′是偶数,and n″是奇数 then

15: D ⇐ AccurateEnough ({( , ) |1y ε ≤≤ n },e f 2 ln( f× − 1 ) , )ε ;
i

i i
16: end if f 1 <0,f 2 =n′/n″(不可约),and n′与 n″均是奇数 then
17: D ⇐ AccurateEnough ({( , ) |1y ε ≤≤ n }, e− f 2 ln( f× − 1 ) , )ε ;

i
i i
18: end if
19: end if f=式(34)中的表达式 then