Is it possible to get 0 by subtracting two unequal floating point numbers?

Question

Is it possible to get division by 0 (or infinity) in the following example?

public double calculation(double a, double          


        

Answer 1

You wouldn't get a division by zero regardless of the value of a - b, since floating point division by 0 doesn't throw an exception. It returns infinity.

Now, the only way a == b would return true is if a and b contain the exact same bits. If they differ by just the least significant bit, the difference between them will not be 0.

EDIT :

As Bathsheba correctly commented, there are some exceptions:

1. "Not a number compares" false with itself but will have identical bit patterns.

2. -0.0 is defined to compare true with +0.0, and their bit patterns are different.

So if both a and b are Double.NaN, you will reach the else clause, but since NaN - NaN also returns NaN, you will not be dividing by zero.

Answer 2

The supplied function can indeed return infinity:

public class Test {
    public static double calculation(double a, double b)
    {
         if (a == b)
         {
             return 0;
         }
         else
         {
             return 2 / (a - b);
         }
    }    

    /**
     * @param args
     */
    public static void main(String[] args) {
        double d1 = Double.MIN_VALUE;
        double d2 = 2.0 * Double.MIN_VALUE;
        System.out.println("Result: " + calculation(d1, d2)); 
    }
}

The output is Result: -Infinity.

When the result of the division is to big to be stored in a double, infinity is returned even if the denominator is non-zero.

Answer 3

There is no case where a division by zero can happen here.

The SMT Solver Z3 supports precise IEEE floating point arithmetic. Let's ask Z3 to find numbers a and b such that a != b && (a - b) == 0:

(set-info :status unknown)
(set-logic QF_FP)
(declare-fun b () (FloatingPoint 8 24))
(declare-fun a () (FloatingPoint 8 24))
(declare-fun rm () RoundingMode)
(assert
(and (not (fp.eq a b)) (fp.eq (fp.sub rm a b) +zero) true))
(check-sat)

The result is UNSAT. There are no such numbers.

The above SMTLIB string also allows Z3 to pick an arbitrary rounding mode (rm). This means that the result holds for all possible rounding modes (of which there are five). The result also includes the possibility that any of the variables in play might be NaN or infinity.

a == b is implemented as fp.eq quality so that +0f and -0f compare equal. The comparison with zero is implemented using fp.eq as well. Since the question is aimed at avoiding a division by zero this is the appropriate comparison.

If the equality test was implemented using bitwise equality, +0f and -0f would have been a way to make a - b zero. An incorrect previous version of this answer contains mode details about that case for the curious.

Z3 Online does not yet support the FPA theory. This result was obtained using the latest unstable branch. It can be reproduced using the .NET bindings as follows:

var fpSort = context.MkFPSort32();
var aExpr = (FPExpr)context.MkConst("a", fpSort);
var bExpr = (FPExpr)context.MkConst("b", fpSort);
var rmExpr = (FPRMExpr)context.MkConst("rm", context.MkFPRoundingModeSort());
var fpZero = context.MkFP(0f, fpSort);
var subExpr = context.MkFPSub(rmExpr, aExpr, bExpr);
var constraintExpr = context.MkAnd(
        context.MkNot(context.MkFPEq(aExpr, bExpr)),
        context.MkFPEq(subExpr, fpZero),
        context.MkTrue()
    );

var smtlibString = context.BenchmarkToSMTString(null, "QF_FP", null, null, new BoolExpr[0], constraintExpr);

var solver = context.MkSimpleSolver();
solver.Assert(constraintExpr);

var status = solver.Check();
Console.WriteLine(status);

Using Z3 to answer IEEE float questions is nice because it is hard to overlook cases (such as NaN, -0f, +-inf) and you can ask arbitrary questions. No need to interpret and cite specifications. You can even ask mixed float and integer questions such as "is this particular int log2(float) algorithm correct?".

Answer 4

In Java, a - b is never equal to 0 if a != b. This is because Java mandates IEEE 754 floating point operations which support denormalized numbers. From the spec:

In particular, the Java programming language requires support of IEEE 754 denormalized floating-point numbers and gradual underflow, which make it easier to prove desirable properties of particular numerical algorithms. Floating-point operations do not "flush to zero" if the calculated result is a denormalized number.

If an FPU works with denormalized numbers, subtracting unequal numbers can never produce zero (unlike multiplication), also see this question.

For other languages, it depends. In C or C++, for example, IEEE 754 support is optional.

That said, it is possible for the expression 2 / (a - b) to overflow, for example with a = 5e-308 and b = 4e-308.

Answer 5

In a floating-point implementation that conforms to IEEE-754, each floating-point type can hold numbers in two formats. One ("normalized") is used for most floating-point values, but the second-smallest number it can represent is only a tiny bit larger than the smallest, and so the difference between them is not representable in that same format. The other ("denormalized") format is used only for very small numbers that are not representable in the first format.

Circuitry to handle the denormalized floating-point format efficiently is expensive, and not all processors include it. Some processors offer a choice between either having operations on really small numbers be much slower than operations on other values, or having the processor simply regard numbers which are too small for normalized format as zero.

The Java specifications imply that implementations should support denormalized format, even on machines where doing so would make code run more slowly. On the other hand, it's possible that some implementations might offer options to allow code to run faster in exchange for slightly sloppy handling of values which would for most purposes be way too small to matter (in cases where values are too small to matter, it can be annoying having calculations with them take ten times as long as calculations that do matter, so in many practical situations flush-to-zero is more useful than slow-but-accurate arithmetic).

Answer 6

Division by zero is undefined, since the limit from positive numbers tend to infinity, the limited from negative numbers tend to negative infinity.

Not sure if this is C++ or Java since there is no language tag.

double calculation(double a, double b)
{
     if (a == b)
     {
         return nan(""); // C++

         return Double.NaN; // Java
     }
     else
     {
         return 2 / (a - b);
     }
}