Be precise, round twice

Recently after implementing a new feature in a software that outputs lots of floating point numbers, I realized that the last digits were off by one for about one in a hundred numbers. As you might suspect at this point, the culprit was floating point arithmetic. This post is about a solution, that turned out to surprisingly easy.

The code I was working on loads a couple of thousands numbers from a database, stores all the numbers as doubles, does some calculations with them and outputs some results rounded half-up to two decimal places. The new feature I had to implement involved adding constants to those numbers. For one value, 0.315, the constant in one of my test cases was 0.80. The original output was “0.32” and I expected to see “1.12” as the new rounded result, but what I saw instead was “1.11”.

What happened?

After the fact, nothing too surprising – I just hit decimals which do not have a finite representation as a binary floating point number. Let me explain, if you are not familiar with this phenomenon: 1/3 happens to be a fraction which does not have a finte representation as a decimal:

1/3=0.333333333333…

If a fraction has a finite representation or not, depends not only on the fraction, but also on the base of your numbersystem. And so it happens, that some innocent looking decimal like 0.8=4/5 has the following representation with base 2:

4/5=0.1100110011001100… (base 2)

So if you represent 4/5 as a double, it will turn out to be slightly less. In my example, both numbers, 0.315 and 0.8 do not have a finite binary representation and with those errors, their sum turns out to be slightly less than 1.115 which yields “1.11” after rounding. On a very rough count, in my case, this problem appeared for about one in a hundred numbers in the output.

What now?

The customer decided that the problem should be fixed, if it appears too often and it does not take to much time to fix it. When I started to think about some automated way to count the mistakes, I began to realize, that I actually have all the information I need to compute the correct output – I just had to round twice. Once say, at the fourth decimal place and a second time to the required second decimal place:

```(new BigDecimal(0.8d+0.315d))
.setScale(4, RoundingMode.HALF_UP)
.setScale(2, RoundingMode.HALF_UP)
```

Which produces the desired result “1.12”.

If doubles are used, the errors explained above can only make a difference of about $10^{-15}$, so as long as we just add a double to a number with a short decimal representation while staying in the same order of magnitude, we can reproduce the precise numbers from doubles by setting the scale (which amounts to rounding) of our double as a BigDecimal.

But of course, this can go wrong, if we use numbers, that do not have a short neat decimal representation like 0.315. In my case, I was lucky. First, I knew that all the input numbers have a precision of three decimal places. There are some calculations to be done with those numbers. But: All numbers are roughly in the same order of magnitude and there is only comparing, sorting, filtering and the only honest calculation is taking arithmetic means. And the latter only means I had to increase the scale from 4 to 8 to never see any error again.

So, this solution might look a bit sketchy, but in the end it solves the problem with the limited time budget, since the only change happens in the output function. And it can also be a valid first step of a migration to numbers with managed precision.

One thought on “Be precise, round twice”

1. Martin Grajcar

Nice trick. If you converted both operands to scale 4 BigDecimals, you’d work with exact numbers. Knowing that the error stays small, you can do the conversion afterwards and still get the exact result. Things like this work nearly always: https://stackoverflow.com/q/42871564/581205

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