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.
Prashant Kumar answered .
2025-11-20
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.