为Maple 15的for循环嵌套 [英] Nested for loops for Maple 15
问题描述
对于某些复数值的特定坐标点，我需要运行最里面的for循环.但是，我不知道特定的坐标点.我想猜测并检查得到答案.但是，那是tediuos.因此，我想提出嵌套的for循环来为我完成检查不同值的工作.当我尝试运行下面的代码时，我遇到了一个无限循环.我的代码有错误吗?
I need to run the inner most for loop for some particular coordinate points that are complex number valued. However, I do not know the particular coordinate points. I am suppose to guess and check to get an answer. However that is tediuos. Hence I wanted to come up with nested for loops to do the work for me of checking for different values. When I try to run the code below I get a unterminating loop. Is there an error in my code?
for i from -100.0 to 100.0 do
for j from -100.0 to 100.0 do
for m from -100.0 to 100.0 do
for n from -100.0 to 100.0 do
X__0:=[i+j*I, m+n*I]
for K from 1 to 20 while evalf(abs(X[K-1][1]-X[K-2][1]),25)<>0 and evalf(abs(X[K-1][2]-X[K-2][2]),25)<>0
do
X[K]:=evalf(G(X[K-1][1], X[K-1][2]), 25)
end do;
如果可能的话，是否可以在循环中添加条件以不打印与[16.19615242270663188058234，-1.928203230275509509174109109785]或[5.803847577293368119417661、11.928203230275509174174978]相匹配的X [k]解决方案?
Also if it is possible, Can I add a condition to the loop to not print solutions of X[k] which matches [16.19615242270663188058234, -1.928203230275509174109785] or [5.803847577293368119417661, 11.92820323027550917410978]?
推荐答案
Clearly this relates to your earlier question. So I'll try to address both here.
您可能会想到完全"的搜索网格可以更好地找到大多数(或全部)根，但是随机过程通常可以更快地找到较小的子集.但是，随着随机方法发现越来越多的根，那么每个新生成的随机起点将收敛到已经找到的根的可能性就增加了.
You may be get the idea that a "thorough" search grid does better at finding most (or sometimes all) roots, but that a random process can often be faster at finding a smaller subset. However as more and more roots are found by the random method then the chances grow are that each newly generated random starting point will converge to a root already found.
您在一个问题中提到了Maple 15，因此我在Windows的64位Maple 15.01中运行了以下代码.
You mentioned Maple 15 in one of your questions, so I ran the code below in 64bit Maple 15.01 for Windows.
我将您的基本步骤包装在proc中，以使事情更易于使用和(希望)理解.
I wrapped your basic steps inside a proc, to make things easier to use and (hopefully) understand.
我没有更改您的迭代方案(可能是家庭作业)，但是请注意，它需要一个初始点.请注意，X__0与X [0]类似地显示为漂亮打印的2D Math输出，但它们并不相同.这对于您在Question上的代码，对于在您其他Question上的一个示例中的转录代码来说都是一个问题.如果您不太清楚，建议您将1D Maple Notation用作输入模式.
I did not change your iteration scheme (which may be homework), but notice that it requires an initial point. Note that X__0 displays similarly to X[0] as pretty-printed 2D Math output, but they are not the same. This was a problem for your code on the Question, and also for the transcribed code on one of the examples in your other Question. I suggest you use 1D Maple Notation for your input mode if that is not entirely clear to you.
不需要使用不推荐使用的linalg [jacobian].我调整为使用VectorCalculus:-Jacobian.
There's no need to use linalg[jacobian] which is deprecated. I adjusted to use VectorCalculus:-Jacobian.
restart;
f := (x-7)^4+(y-2)^2-100:
g := (x-11)^2+(y-5)^2-75:
G := unapply(convert(Vector([x,y])
-1/VectorCalculus:-Jacobian([f, g],[x, y]).Vector([f,g]),
list),x,y):
p := proc(X0::[complex(numeric),complex(numeric)], G)
local X, K;
X[0]:=X0;
for K to 30 while evalf[Digits+5](abs(X[K-1][1]-X[K-2][1])) <> 0
and evalf[Digits+5](abs(X[K-1][2]-X[K-2][2])) <> 0 do
X[K] := evalf[Digits+5](G(X[K-1][1], X[K-1][2]));
end do;
if not type(X[K-1],[complex(numeric),complex(numeric)]) then
error "nonnumeric results"; end if;
if K<29 then map(simplify@fnormal, evalf(X[K-1]));
else error "did not converge"; end if;
end proc:
p( [17, -2], G );
[9.879819419, -3.587502283]
p( [5, 11], G );
[5.127257522, 11.36481703]
p( [-1.0-11.*I, -2.0*I], G );
[7.144643342 - 2.930435630 I, -3.398413328 + 1.345239163 I]
Digits := 20:
p( [-1.0-11.*I, -2.0*I], G );
[7.1446433421702820770 - 2.9304356302329792484 I,
-3.3984133281207314618 + 1.3452391631560967251 I]
Equate([x,y],%);
[x = 7.1446433421702820770 - 2.9304356302329792484 I,
y = -3.3984133281207314618 + 1.3452391631560967251 I]
eval( [f,g], % );
[ -18 -18 -18 ]
[1 10 + 2 10 I, -1 10 + 0. I]
Digits := 10:
NN:=2: # This attempts (NN+1)^4 iterates
incx,incy:=5.0,5.0:
Sols:={}:
count:=0:
st := time():
for a from -NN to NN do
for b from -NN to NN do
for c from -NN to NN do
for d from -NN to NN do
count:=count+1;
try
cand := p( [a*incx+b*incy*I, c*incx+d*incy*I], G );
if not member(cand,{Sols}) then
Sols := Sols union {cand}; end if;
catch:
end try;
end do;
end do;
end do;
end do;
(time() - st)*'seconds', count*'attempts';
34.695 seconds, 625 attempts
nops( Sols );
8
sort( Sols );
{[3.867005122, 0.08874923598], [5.127257522, 11.36481703],
[5.721021477, 11.86530303], [9.879819419, -3.587502283],
[7.144643342 - 2.930435630 I, -3.398413328 + 1.345239163 I],
[7.144643342 + 2.930435630 I, -3.398413328 - 1.345239163 I],
[8.557804888 - 1.867139097 I, 13.53272982 - 0.5344031829 I],
[8.557804888 + 1.867139097 I, 13.53272982 + 0.5344031829 I]}
seq( eval( max(abs(f),abs(g)), Equate([x,y],xypoint) ), xypoint=Sols );
-8 -7 -8 -8 -8 -8
3 10 , 1 10 , 9 10 , 4 10 , 3.162277660 10 , 3.162277660 10 ,
-7 -7
1.019803903 10 , 1.019803903 10
randomize():
fgen := proc(a::numeric,b::numeric,i::nonnegint:=1)
seq(RandomTools:-Generate(float('range'=a..b,'digits'=5)),
ii=1..i);
end proc:
fgen(-100.0, 100.0); # Usage example, a random point
-26.41
randSols := {}:
A,B := 15, 15:
numattempts:=100:
st := time():
for s from 1 to numattempts do
try
cand := p( [fgen(-A,A)+fgen(-B,B)*I, fgen(-A,A)+fgen(-B,B)*I], G );
if not member(cand,{randSols}) then
randSols := randSols union {cand}; end if;
catch:
end try;
end do;
(time() - st)*'seconds', numattempts*'attempts';
5.756 seconds, 100 attempts
nops(randSols);
5
sort( randSols );
{[3.867005122, 0.08874923598],
[7.144643342 - 2.930435630 I, -3.398413328 + 1.345239163 I],
[7.144643342 + 2.930435630 I, -3.398413328 - 1.345239163 I],
[8.557804888 - 1.867139097 I, 13.53272982 - 0.5344031829 I],
[8.557804888 + 1.867139097 I, 13.53272982 + 0.5344031829 I]}
seq( eval( max(abs(f),abs(g)), Equate([x,y],xypoint) ), xypoint=randSols );
-8 -8 -8 -7 -7
3 10 , 3.162277660 10 , 3.162277660 10 , 1.019803903 10 , 1.019803903 10
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