Why isn't MATLAB counting this as equal to 1?

I've been writing a code, a small part of which checks whether some of the inputs sum to 1 (as they're all proportions of a whole). If they do, the code continues. If not, the code stops and throws up an error message.

a = 0.4;
b = 0.3;
c = 0.2;
d = 0.1;
total = [a,b,c,d];

if sum(total) ~= 1
  warning('Sum of a,b,c,d must be 1.')
  return
end

The example I've given here DOES NOT WORK, even though a+b+c+d = 1. It throws up the error message and stops the script.

To be clear, this works fine for a,b,c and d = 0.25. It works for a = 0.5, b = 0.4, c and d = 0.05. It doesn't work for the example I've given, and it doesn't work for a,b,c = 0.3, d = 0.1. I haven't checked other combinations exhaustively.

Any help would be greatly appreciated, thanks.

[Adam]

(most) Floating point numbers cannot be represented with 100% accuracy so even though they may appear to be exact they are not and when you do maths with these those minor errors accumulate. You should never test equality of floating point numbers though unless there is some very clear purpose in doing so (e.g. two places that implement the same maths to get to a result)

[BecomeAnAstronaut]

Hi Adam, thanks for your response. I'm just trying to write a bit of code that stops if the values a,b,c,d don't sum to 1, as they're all supposed to proportions of the work done in a compressor, so the sum must be 1. I get that, with floating numbers, errors can build up, but surely adding four numbers together and checking if they're equal to 1 should work. I don't understand how that much error can accumulate, especially given that I tried madhan ravi's answer of including a tolerance.

Answer 1

[Steven Lord]

None of 0.4, 0.3, 0.2, or 0.1 can be exactly represented in IEEE double precision. See the "One-tenth" and "Hexadecimal format" sections of this post on Cleve Moler's blog for a bit more of an explanation. So those are not exactly equal to four-tenths, three-tenths, two-tenths, or one-tenth. This is not a MATLAB problem (other languages will behave the exact same way if they use double precision), this is a "double precision numbers aren't infinitely precise" situation.

To give a similar scenario that doesn't involve computers, take out a piece of paper and a pen or pencil (or anything you can use to write and something to write on.) Divide 1 by 3 to as many decimal places as you want and can write down. You're not allowed to just say that it's 0.3 repeating, you have to write out all the decimal places you want to use in the next step.

Now multiply the answer you computed in the first step by 3. Symbolically, (one third) * 3 is exactly 1. Practically, (the approximation you computed to one third) * 3 is not exactly 1.

The original code posted by madhan ravi has a mistake in it. [I saw it was edited, but I'm not sure if it was edited to correct the mistake.] Instead of this line:

if (sum(total) == 1) < 1e-4

you need to use:

if abs(sum(total) - 1) < 1e-4

This second line computes the distance between the sum of your vector and 1 (calling abs means we don't care which one is larger, we're just getting the absolute distance) and checks if that distance is "close enough" (which in this code is defined as a distance smaller than 1e-4.) If it is "close enough" the if statement's body will execute.

Added: You may wonder why it worked with 0.25, why sum([0.25 0.25 0.25 0.25]) was exactly 1. Unlike 0.4, 0.3, 0.2, and 0.1 the number 0.25 can be exactly represented in IEEE double precision. Using the fraction/mantissa and exponent notation from the "Single and Double Precision" section in Cleve's blog post, f is 0 and e is -2.

[madhan ravi]

I'm not sure if it was edited to correct the mistake

yes it was edited to correct the mistake @Steven

but wouldn't have managed it to explain clearly like you did

+1

Answer 2

[madhan ravi]

if abs(sum(total-1)) < 1e-4    %edited

The above is equivalent to

if sum(total)==1

[madhan ravi]

add tolerance see https://www.mathworks.com/help/matlab/ref/eq.html#bt2klek-3 for further explanation

[BecomeAnAstronaut]

Thank you!

[madhan ravi]

Anytime :)

[BecomeAnAstronaut]

Ok sorry, but I just added that code in, and the issue is now worse. Now no combinations work. I'm very confused because I used your exact code. It's especially weird because abs(sum(total)) works perfectly well and gives exactly 1 in the command window.

Ignoring the abs and just having sum(total) ~= 1 < 1e-4 doesn't work either, sorry

[madhan ravi]

what are you trying to do and what doesn't work ? did you see Adams comment?

[BecomeAnAstronaut]

I just want to stop the code if the four numbers a,b,c,d don't sum to 1. It's just to stop users putting in values that make no sense and then thinking the answers are correct.

I get that floating numbers can be an issue, and your tolerance solution makes a lot of sense, but for some reason trying

if abs(sum(total)~=1)<1e-4
  warning
end

doesn't work, with or without the abs. I'm just confused as to why this won't work.

[madhan ravi]

I just want to stop the code if the four numbers a,b,c,d don't sum to 1.

if (sum(total)~=1)<1e-4
  break       %use break to stop 
end

[BecomeAnAstronaut]

The stopping I'm fine with, I use return because break only works in a for or while loop. Regardless, the issue is that MATLAB, for whatever reason, considers the sum(total) not equal to 1. It's fine if total = 0.25,0.25,0.25,0.25, but not 0.4,0.3,0.2,0.1 for example. The tolerance solution also doesn't work. No idea if it's just my MATLAB or if it's something I'm missing.

[Matt J]

It should be

    if abs( sum(total)-1 ) < 1e-4

[madhan ravi]

Thank you @Matt edited already

[BecomeAnAstronaut]

Hi both, There was clearly a hardware issue on my end. I came home to test the code on my laptop and a couple of the suggested solutions work fine. No idea what I'll do about my computer... I'll have to debug it or something. Sorry for the inconvenience and thanks for your help.

[madhan ravi]

There was a mistake in my part (corrected it)

see https://www.mathworks.com/matlabcentral/answers/428577-very-odd-error-with-simple-multiplication-ok-not-so-odd#answer_345716 for further explanation on float numbers

make sure to accept the answer if it answered your question so people know the question is solved