Exhaustion method with a condition

Dear :

I have a set of weight number [weight1 weight2 weight3 weight4],I want to use a exhaustion method with a condition find all of them satisfied the condition:

Both of their value are 0:0.01:0.28 condition is 0.5*(weight1^2+weight2^2+weight3^2+weight4^2)==0.125 (or (weight1^2+weight2^2+weight3^2+weight4^2)==0.25 )

Following is my code :

close all clc

%%wieght constraining
cont=0;

weight=zeros(1,4);
for weight1=0:0.01:0.28
    for weight2=0:0.01:0.28
        for weight3=0:0.01:0.28
            for weight4=0:0.01:0.28
                check =(weight1^2+weight2^2+weight3^2+weight4^2);
                if check == 0.25
                    cont= cont+1;
                    weight(cont,:)=[weight1,weight2,weight3,weight4];

                  end
              end
          end
      end
  end

But the value only show : [0.16 0.28 0.28 0.26] [0.25 0.25 0.25 0.25] and [0.28 0.16 0.28 0.26] ,but it should have more value for sure:[0.16 0.26 0.28 0.28] [0.16 0.28 0.26 0.28]......

What's wrong is my code?or could someone help me,maybe I ignored some parts.

Thanks a lot!!

[John D'Errico]

NEVER test for exact equality between floating point numbers. Always use a tolerance on something like this.

Double precision numbers use a BINARY mantissa to store the number, but values like 0.01 are not exactly representable in binary.

[Zoe Lin]

Dear John D'Errico:

Got it,Thx.

Answer 1

[Ameer Hamza]

As pointed out by @John in the comments, you are getting this error because of finite precision of floating point numbers in the computer systems. A way around this is to compare numbers with tolerance. To do that here is one option, replace

if check == 0.25

with

if ismembertol(check, 0.25, 1e-6)

the third number to ismembertol() is the tolerance value i.e. how much difference from 0.25 on the left side of comparison will be considered equal to 0.25. For your code, this gives following a total of 13 combinations of weights

weight =

    0.1600    0.2600    0.2800    0.2800
    0.1600    0.2800    0.2600    0.2800
    0.1600    0.2800    0.2800    0.2600
    0.2500    0.2500    0.2500    0.2500
    0.2600    0.1600    0.2800    0.2800
    0.2600    0.2800    0.1600    0.2800
    0.2600    0.2800    0.2800    0.1600
    0.2800    0.1600    0.2600    0.2800
    0.2800    0.1600    0.2800    0.2600
    0.2800    0.2600    0.1600    0.2800
    0.2800    0.2600    0.2800    0.1600
    0.2800    0.2800    0.1600    0.2600
    0.2800    0.2800    0.2600    0.1600

If you increase the tolerance

if ismembertol(check, 0.25, 1e-3)

then even more elements will satisfy the equality. 1e-3 gives 61 points

weight =

    0.1600    0.2600    0.2800    0.2800
    0.1600    0.2700    0.2700    0.2800
    0.1600    0.2700    0.2800    0.2700
    0.1600    0.2800    0.2600    0.2800
    0.1600    0.2800    0.2700    0.2700
    0.1600    0.2800    0.2800    0.2600
    0.1900    0.2500    0.2700    0.2800
    0.1900    0.2500    0.2800    0.2700
    0.1900    0.2700    0.2500    0.2800
    0.1900    0.2700    0.2800    0.2500
    0.1900    0.2800    0.2500    0.2700
    0.1900    0.2800    0.2700    0.2500
    0.2400    0.2500    0.2500    0.2600
    0.2400    0.2500    0.2600    0.2500
    0.2400    0.2600    0.2500    0.2500
    0.2500    0.1900    0.2700    0.2800
    0.2500    0.1900    0.2800    0.2700
    0.2500    0.2400    0.2500    0.2600
    0.2500    0.2400    0.2600    0.2500
    0.2500    0.2500    0.2400    0.2600
    0.2500    0.2500    0.2500    0.2500
    0.2500    0.2500    0.2600    0.2400
    0.2500    0.2600    0.2400    0.2500
    0.2500    0.2600    0.2500    0.2400
    0.2500    0.2700    0.1900    0.2800
    0.2500    0.2700    0.2800    0.1900
    0.2500    0.2800    0.1900    0.2700
    0.2500    0.2800    0.2700    0.1900
    0.2600    0.1600    0.2800    0.2800
    0.2600    0.2400    0.2500    0.2500
    0.2600    0.2500    0.2400    0.2500
    0.2600    0.2500    0.2500    0.2400
    0.2600    0.2800    0.1600    0.2800
    0.2600    0.2800    0.2800    0.1600
    0.2700    0.1600    0.2700    0.2800
    0.2700    0.1600    0.2800    0.2700
    0.2700    0.1900    0.2500    0.2800
    0.2700    0.1900    0.2800    0.2500
    0.2700    0.2500    0.1900    0.2800
    0.2700    0.2500    0.2800    0.1900
    0.2700    0.2700    0.1600    0.2800
    0.2700    0.2700    0.2800    0.1600
    0.2700    0.2800    0.1600    0.2700
    0.2700    0.2800    0.1900    0.2500
    0.2700    0.2800    0.2500    0.1900
    0.2700    0.2800    0.2700    0.1600
    0.2800    0.1600    0.2600    0.2800
    0.2800    0.1600    0.2700    0.2700
    0.2800    0.1600    0.2800    0.2600
    0.2800    0.1900    0.2500    0.2700
    0.2800    0.1900    0.2700    0.2500
    0.2800    0.2500    0.1900    0.2700
    0.2800    0.2500    0.2700    0.1900
    0.2800    0.2600    0.1600    0.2800
    0.2800    0.2600    0.2800    0.1600
    0.2800    0.2700    0.1600    0.2700
    0.2800    0.2700    0.1900    0.2500
    0.2800    0.2700    0.2500    0.1900
    0.2800    0.2700    0.2700    0.1600
    0.2800    0.2800    0.1600    0.2600
    0.2800    0.2800    0.2600    0.1600

So it depends on how much tolerance is acceptable for your problem.

[Zoe Lin]

Dear Ameer Hamza:

Thank a lot,very detailed explanation,I 'll try it later.

[Ameer Hamza]

You are welcome.

[Jan]

Be careful with ismembertol. I think it uses a strange definition of "tolerance":

abs(u-v) <= TOL * max(abs([A(:); B(:)]))

Using the maximum values of all inputs is less intuitive in my opinion. An elementwise application of the tolerance would be smarter:

abs(u-v) <= TOL * max(abs(A), abs(B))

In the case of this thread, this does not matter, because it is a scalar comparison. Then this might be easier:

if abs(check - 0.25) < 0.0001

[Zoe Lin]

Dear Jan :

Thanks,learned a lot!!