Finding Prime numbers (NO isprime function)
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Hi, I am trying to return if the value is prime or not. And here is my code
vec = [3 4 88 5 1371];
for i = 1:length(vec)
b = mod(vec(i),2);
if vec(i)==1
disp('not prime');
elseif vec(i)==2
disp('prime');
elseif b==0
disp('not prime');
elseif b>0
for j = 3:2:(vec(i)-1)
b = mod(vec(i), j);
if b==0
disp('not prime');
end
end
disp('prime');
else
disp('not prime');
end
end
However, this code returned this
prime
not prime
not prime
prime
not prime
not prime
prime
Could you please tell me why its returning 7 outputs not 5? Also could you fix this code so it only returns 5 outputs?
2 个评论
Stephen23
2020-6-13
编辑:Stephen23
2020-6-13
"Could you please tell me why its returning 7 outputs not 5?"
Look at your j-loop: within that loop your code could call disp zero, one, or more times, depending on the vec(j) value.
And then after the j-loop, you call disp once more. Your code does nothing to ensure that you only call disp once for each vec value.
"Also could you fix this code so it only returns 5 outputs?"
Get rid of the j-loop and write vectorized code. It will be simpler.
Note that you do not need to check up to vec(i)-1, up to sqrt(vec(i)) is sufficient and more efficient.
回答(1 个)
Walter Roberson
2020-6-13
if b==0
disp('not prime');
end
You should exit the j loop as soon as you prove the number is not prime.
disp('prime');
You are doing that no matter what the outcome of your for j loop is.
2 个评论
Walter Roberson
2020-6-13
if b==0
disp('not prime');
break;
end
end
if b ~= 0
disp('prime')
end
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