Why am I receiving 1 full integer less when dividing using fix()

So I'm working on Cody, and am trying to make a dollar amount (a) split into an array (change) in paper and coin money. I have it set up to divide the given amount (a) by each of the elements one x one in (change) using the function fix() to eliminate remainders. I then take the remainder after the division and rerun the for loop over the remainder.

This works perfectly until the final loop, done on the 0.01 element in (change). In this example the final return should be 4, but I repeatedly receive 3. Can anyone help me figure why this is the case?

change = [100 50 20 10 5 2 1 0.5 0.25 0.1 0.05 0.01];
a = 10035.99;
for i = 1:size(change,2)
    b(1,i) = fix(a/(change(i)))
    a = mod(a,change(i))
end
b(2,:) = change

Answer 1

[Azzi Abdelmalek]

To understand why the problem occurs, read this http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

In your case, let us run the code until a=0.04

change = [100 50 20 10 5 2 1 0.5 0.25 0.1 0.05 0.01];
b=zeros(2,numel(change))
a = 10035.99;
for i = 1:numel(change)-1
  b(1,i) = fix(a/(change(i)))
  a = mod(a,change(i))
end

Then test

a==0.04

You see that a is not equal to 0.04, you can check that it's less then 0.04, that's why you get the result 3. This is du to calculations errors.

To fix the problem use round function instead of fix function

change = [100 50 20 10 5 2 1 0.5 0.25 0.1 0.05 0.01];
a = 10035.99;
b=zeros(2,numel(change))
for i = 1:numel(change)
  b(1,i) = round(a/(change(i)))
  a = mod(a,change(i))+1e-6
end
b(2,:) = change

Answer 2

[John BG]

Hi Jonathan

Let me comment your lines

1.-

..
  for i = 1:size(change,2)

in this case you don't need size(change,2), size(change) is enough because MATLAB recognises that there is only one dimension >1.

2.- The same applies to the variable you intend to use to collect the results, b, you are indexing b(1,i) but with indexing b(i) would be enough.

3.- Azzi Abdelmalek is right, the main error is using fix(), instead use round(), or you can also use floor()

4.- And

..
b(2,:) = change
don't see use either.

So, you write 'it works perfectly until the final loop' (?) yet when I run your lines the variable you intend to use to collect the result, b, is empty, so, it's not that 'doesn't work perfectly', simply, it doesn't work.

The following works, I renamed your 'b' to 'change', seems reasonable, doesn't it? and your 'change' I call it 'base', as you are actually decomposing the amount into a base, aren't you have a look:

a=randi([1 2000000],1,1);a=a/100

    base = [100 50 20 10 5 2 1 0.5 0.25 0.1 0.05 0.01]; 
    change=zeros(1,length(base))

    a2=a;
    % for k=1:1:size(base)
    k=1
    while k<length(base)+1
        while a2/base(k)>1
            a2=a2-base(k);
            change(k)=change(k)+1;
        end
        k=k+1
    end
    change

So, Jonathan, having commented where your code doesn't work,

would you please mark my answer as ACCEPTED ANSWER

or at least click on the thumbs-up vote

Thanks in advance

John BG

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