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for i = 1:5
b1(i,:)= round(((1860-0)*(rand(1)))+0)
end
b1
for i = 1:5
b2(i,:)= round(((1908-0)*(rand(1)))+0);
end
b2
for i = 1:5
b3(i)= round(((1316-0)*(rand(1)))+0);
end
b3
for i = 1:5
b4(i)= round(((2940-0)*(rand(1)))+0);
end
b4
for i=1:5
b5(i,:)= round(((1860-0)*(rand(1)))+0);
end
b5
for i=1:5
b6(i,:)= round(((1860-0)*(rand(1)))+0);
end
b6
In this code I want a condition that if b5+b6=b1 then only it should execute the result otherwise search for solutions until this criterion matches. Please help me with this.
3 Comments
Answers (1)
Guillaume
on 26 Feb 2015
Edited: Guillaume
on 26 Feb 2015
As per my comment, it seems very silly to generate random sets that have no way of working when you could just generate sets that always work, so this will get you there faster:
invalidset = true;
while invalidset
b1 = randi([0 1860], 5, 1);
%... same for b2, b3, ...
b5 = randi([0 1860], 5, 1);
b6 = b1 - b5;
invalidset = any(b6 <0 | b6 > 1860);
end
But if you really want to waste time generating a random bunch of 3 sets that have no chance of working, it's:
invalidset = true;
while invalidset
b1 = randi([0 1860], 5, 1);
%... same for b2, b3, ...
b5 = randi([0 1860], 5, 1);
b6 = randi([0 1860], 5, 1);
invalidset = ~isequal(b1, b5 + b6);
end
First option took 0.007044 seconds to generate a set on my machine. Second option is still running after a minute.
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