Maximum variable size allowed by the program is exceeded.

2020 4 10
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更新时间：2020 4 11
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Hi my code is below
clc
format long g
a=588545.909;
b=10167.688;
c=150;
ox=302;
oy=192;
for oz= 1800:1:2313
oev=a.*ox+b.*oy+c.*oz ;
oce=oev+500000;
less=round(oce.*0.9,3);
x1=250:.001:1550;
y2=150:.001:1450;
z3=1200:.001:2500;
x=perms(x1)
y=perms(x1)
z=perms(x1)
then the error come
Maximum variable size allowed by the program is exceeded for x=perms(x1)
need to overcome the error. How can I?

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Steven Lord
Steven Lord 2020-4-10

0 个投票

You can't.
The output of perms(x) has factorial(n) rows, where n is the number of elements in x. How big is that?
>> x1=250:.001:1550;
>> vpa(factorial(sym(numel(x1))))
ans =
1.2633683274814429271576127065521e+7383553
The number of hydrogen atoms in the observable universe is estimated to be around 1e80.

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MD.MASHRAVI SHAMS
MD.MASHRAVI SHAMS 2020-4-10
dear Steven,
actually my total code is given below-
clc
format long g
a=588545.909;
b=10167.688;
c=150;
ox=302;
oy=192;
for oz= 1800:1:2313
oev=a.*ox+b.*oy+c.*oz ;
oce=oev+500000;
less=round(oce.*0.9,3);
x=250:.001:1550;
y=150:.001:1450;
z=1200:.001:2500;
o=((a.*x+b.*y+c.*z));
t=o+500000;
v=0.9.*(t);
hvp=((t-less).*100)./t;
hvptrunc=round(hvp,3);
qp=round(t-(t.*(hvptrunc./100)),3);
qpwod= round(t,3);
difwod=round(abs(qpwod-less),3);
dif=round(abs(qp-less),3);
[m,ind]=mink(dif,1);
[mwod,indwod]=mink(difwod,3);
% disp([oz x(ind(1)) y(ind(1)) z(ind(1)) dif(ind(1))]);
T = table(oz,x(ind(1)),y(ind(1)),z(ind(1)),hvptrunc(ind(1)),less,qp(ind(1)),dif(ind(1)))
T1= table(oz,x(indwod(1)),y(indwod(1)),z(indwod(1)),less,qpwod(indwod(1)),dif(indwod(1)))
end
I want to know all the possible combination of the values of x, y and z for whose value will give qp-less= 0.
Steven Lord
Steven Lord 2020-4-10
Brute force isn't going to work.
Let's take a step back. What's the main problem you're trying to solve? If you explain your goal in words (not equations and not code) we may be able to offer suggestions for other ways to achieve your goal without trying to create such a huge array. Explain like you were explaining to someone with knowledge of mathematics but little or no knowledge of your specific application.
At the very least, once you have a fixed value for x and y that likely severely restricts the possible values z can have. Requiring the expression you round to form qp to be within a certain distance of less (so qp, the rounded form of that expression, is equal to less) will put some limits on what t can be and so what z can be.
MD.MASHRAVI SHAMS
MD.MASHRAVI SHAMS 2020-4-11
编辑：MD.MASHRAVI SHAMS 2020-4-11
thank you very much for your kind advice dear Steven.
ax+by+cz= T
i need to get solution for x,y and z where
  1. a,b and c is constant
  2. T is known
  3. i know the value ranges of x,y and z
  4. the value of the 3rd digit after point(.) can not be zero for x,y and z. for example x= 1200.001 or x=1200.005 or x= 1200.009 , but it can not be 1200.000.
  5. all the calculations has to be done upto 3rd digit after point(.)
  6. i can apply percentage on (ax+by+cz) to achive the value of T. for example you can apply 1%-99.999% to achive the exact value of T.
  7. i need to get all the possible solutions of x,y,and z toachive the value T.
dear brother is it a bit clear for you to understand my problem? i need the solution Steven,please help.
Steven Lord
Steven Lord 2020-4-11
If you multiply both sides of the equation by 1000 (so x, y, and z are integers) you may be able to use some of the techniques for Diophantine equations to obtain your solutions.
MD.MASHRAVI SHAMS
MD.MASHRAVI SHAMS 2020-4-11
dear steven, i have write another code for Diophantine equation. but it runs for last 7 hrs
a=588545.909;
b=10167.688;
c=150;
x=250:.001:1550;
y=150:.001:1450;
z=1200:.001:2500;
o=(a.*x+b.*y+c.*z);
t=o+500000==162486009.553;
[xSol, ySol, zsol] = solve(t,[x y z])
is there any wrong in this code?

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