Algorithm for equality condition.

Question

Abhinav 2015-2-25

0
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https://ww2.mathworks.cn/matlabcentral/answers/180271-algorithm-for-equality-condition

编辑： John D'Errico 2015-2-25

b5=zeros(5,1)
b6=zeros(5,1)
b1=[ 1417.1;
     1299;
     236.19;
     501.9;
     708.57]
while (~((b5+b6)==b1))
    restart = true;
while restart
   restart = false; %in case it succeeds
   for i = 1:5
      s5(i, :) = randi([0 1],1,6);
      xl5=0;
      xu5=4000;
      l5=6;
      si5=bi2de(s5);
      b5=(xl5+((xu5-xl5)/((2^l5)-1)))*si5;
      if all(s5(i, :) == 0)
          restart = true;%stop the for loop, thus restarting from scratch
      end
      end  
      end
end
disp([s5,si5,b5]);
restart = true;
while restart
   restart = false; %in case it succeeds
   for i = 1:5
      s6(i, :) = randi([0 1],1,6);
      xl6=0;
      xu6=4000;
      l6=6;
      si6=bi2de(s6);
      b6=(xl6+((xu6-xl6)/((2^l6)-1)))*si6;
       if all(s6(i, :) == 0)
           restart = true;%stop the for loop, thus restarting from scratch
           end
       end
end
disp([s6,si6,b6]);
b5+b6

Is there any fast method to satisfy the first constraint (~((b5+b6)==b1)) I am doing PSO optimization and this constraint has to be satisfied. Normally it is going on and going on and not MATLAB status is Busy

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John D'Errico 2015-2-25

编辑：John D'Errico 2015-2-25

Please use the code button "{} Code" when you post code. it is unreadable otherwise. I've done it for you this time.

John D'Errico 2015-2-25

编辑：John D'Errico 2015-2-25

在 MATLAB Online 中打开

I looked at your code once I could read it. A bunch of numbered variables, that mean absolutely nothing to anyone but you. I love this one, by the way:

l5=6;

Yes, it probably means l5, where l is a lower case L. But if your goal is to create unreadable code, you should recognize that this variable is virtually undistinguishable with the number 15. Depends on the font it is viewed with of course.

No hint as to what the code is supposed to do, so when you ask for a fast way to do it, we have no real idea what is the goal, so finding a faster way to do what we don't know what you are doing is impossible.

Your comments don't even make sense. For example, this one:

restart = true;%stop the for loop, thus restarting from scratch

restart has nothing to do with stopping any for loop it is contained in. Once that for loop does terminate, restart allows the while loop around it all to continue working.

Confusing spaghetti code, with no clue provided as to what it does. How can we possibly help you? My point is, if you want help, then help those who could possibly help you. If you can't write clear code, at least explain what it does, or what it should do!

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