I want to write a function which takes as input three integers a, b and c, and generates and returns a random integer that is equally likely to take any value between a and b (inclusive), but will not return the value given by c.

This is what I have so far:

    function f = randnum_reject(a,b,c)

    randnum_reject = (b-a)*rand + a; 
    f = randnum_reject;
    if f == c
        f = (b-a)*rand + a;
    else
        f = randnum_reject;
    end
end

I know I probably made the mistake right after I assigned the output to f. I just really don't have any ideas as to how exclude c when I call random numbers from a to b

[Walter Roberson]

Is c always a single value or could it be a vector of values to avoid?

[Alex Benavides]

C is always a variable it can't be a vector

Answer 1

[Roger Stafford]

You don't need a for-loop for this problem.

 p = [a:c-1,c+1:b];
 f = p(ceil((b-a)*rand));

[Image Analyst]

You can also use randi() to pick out one of the possible values:

f2 = int32(p(randi(length(p))))

I also added int32() to make the type of variable a true integer instead of a double that just happened to have an integer for a value.

[Alex Benavides]

So I don't need an 'If' statement?

[Walter Roberson]

Yes, It is possible to program it without an "if" statement.

[Alex Benavides]

This gives me an error, any other suggestions?

[Walter Roberson]

What error does it give, with what inputs ?

[Walter Roberson]

p( randperm(length(p),1) )

if your MATLAB is a newer one.

Answer 2

[Walter Roberson]

Do not use a variable with the same name as your function. Inside your function rename the variable randum_reject to something else.

[Walter Roberson]

Infinite loop. Try something. Was the result acceptable? If it was, return it. Otherwise loop again.

[Alex Benavides]

It's still not working :S, can you give me some more tips, how would you write this program? thanks a lot, I'm really stuck on this one!

[Walter Roberson]

What was your trial code along these lines?

[Alex Benavides]

When I test it, it doesn't skip the value it's supposed to skip

[Image Analyst]

Are you not wanting to try any of the working solutions under Roger's answer for some reason, and are continuing to use your code (which you admit doesn't work)? If so, why?

[Walter Roberson]

What code was it that you tested that did not skip the value it was supposed to?