# Floating Point Numbers and Roundoff

CS 301 Lecture, Dr. Lawlor

## Normalized Numbers

In C++, "float" and "double" store numbers in an odd way--internally they're really storing the number in scientific notation, like
x = + 3.785746 * 105
Note that:
• You only need one bit to represent the sign--plus or minus.
• The exponent's just an integer, so you can store it as an integer.
• The 3.785746 part, called the "mantissa" or just "fraction" part, can be stored as the integer 3785746 (at least as long as you can figure out where the decimal point goes!).  Of course, we store the fraction part as a binary number.
Scientific notation is designed to be compatible with slide rules (here's a circular slide rule demo); slide rules are basically a log table starting at 1.  This works because log(1) = 0, and log(a) + log(b) = log(ab).  But slide rules only give you the mantissa; you need to figure out the exponent yourself.  The "order of magnitude" guess that engineers (and I) like so much is just a calculation using zero significant digits--no mantissa, all exponent.

One problem is scientific notation can represent the same number in several different ways:
x = + 3.785746 * 105  = + 0.3785746 * 106 = + 0.03785746 * 107 = + 37.85746 * 104

It's common to "normalize" a number in scientific notation so that:
1. There's exactly one digit to the left of the decimal point.
2. And that digit ain't zero.
This means the 105 version above is the "normal" way to write the number above.

In binary, a "normalized" number *always* has a 1 at the left of the decimal point (if it ain't zero, it's gotta be one).  So there's no reason to even store the 1; you just know it's there!

## Roundoff in Arithmetic

They're funny old things, floats.  The fraction part (mantissa) only stores so much precision; further bits are lost.  For example, in reality,
1.234* 104 + 7.654* 100 = 1.2347654 * 104
But if we only keep three decimal places,
1.234* 104 + 7.654* 100 = 1.234 * 104
which is to say, adding a tiny value to a great big value might not change the great big value at all, because the tiny value gets lost when rounding off to 3 places.   To avoid this "roundoff error", when you're doing arithmetic by hand, people recommend keeping lots of digits, and only rounding once, at the end.  But for a given value of "lots of digits", did you keep enough?

For example, on a real computer adding one to a float repeatedly will eventually stop changing the float!
`float f=0.73;while (1) {	volatile float g=f+1;	if (g==f) {		std::cout<<"f+1 == f  at f="<< f <<", or 2^"<< log(f)/log(2.0) <<std::endl;		return 0;	}	else f=g;}`

(Try this in NetRun now!)

For "double", you can add one more times, but eventually the double will stop changing despite your additions.  Recall that for integers, adding one repeatedly will *never* give you the same value--eventually the integer will wrap around, but it won't just stop moving like floats!

This has really weird effects.  For example, floating-point arithmetic isn't "associative"--if you change the order of operations, you change the result due to accumulated roundoff.  In exact arithmetic:
1.2355308 * 104 = 1.234* 104 + (7.654* 100 + 7.654* 100)
1.2355308 * 104 = (1.234* 104 + 7.654* 100) + 7.654* 100
In other words, parenthesis don't matter if you're computing the exact result.  But to three decimal places,
1.235 * 104 = 1.234* 104 + (7.654* 100 + 7.654* 100)
1.234 * 104 = (1.234* 104 + 7.654* 100) + 7.654* 100
In the first line, the small values get added together, and together they're enough to move the big value.  But separately, they splat like bugs against the windshield of the big value, and don't affect it at all!
`double lil=1.0;double big=pow(2.0,53); //<- carefully chosen for IEEE 64-bit float (52 bits of fraction + implicit 1)std::cout<<" big+(lil+lil) -big = "<<  big+(lil+lil) -big <<std::endl;std::cout<<"(big+lil)+lil  -big = "<< (big+lil)+lil  -big <<std::endl;`

(Try this in NetRun now!)

`float gnats=1.0;volatile float windshield=1<<24;float orig=windshield;for (int i=0;i<1000;i++)	windshield += gnats;if (windshield==orig) std::cout<<"You puny bugs can't harm me!\n";else std::cout<<"Gnats added "<<windshield-orig<<" to the windshield\n";`

In fact, if you've got a bunch of small values to add to a big value, it's more roundoff-friendly to add all the small values together first, then add them all to the big value:
`float gnats=1.0;volatile float windshield=1<<24;float orig=windshield;volatile float gnatcup=0.0;for (int i=0;i<1000;i++)	gnatcup += gnats;windshield+=gnatcup; /* add all gnats to the windshield at once */if (windshield==orig) std::cout<<"You puny bugs can't harm me!\n";else std::cout<<"Gnats added "<<windshield-orig<<" to the windshield\n";`

Roundoff can be very annoying.  But it's not the end of the world if you don't care about exact answers, like in computer games, and even in many simulations (where "exact" is unmeasureable anyway).  You just need to be able to estimate the amount of roundoff, and make sure it's not too much.

However, the amount of roundoff depends on the precision you keep in your numbers.  This, in turn, depends on the size of the numbers.  For example, a "float" is just 4 bytes, so it's not very precise.  A "double" is 8 bytes, and so more precise.  A "long double" is 12 bytes (or more!), using more memory, but it's got tons of precision.  There's often a serious tradeoff between precision and space (and time), so just using long double for everything isn't a good idea: your program may get bigger and slower, and you still might not have enough precision.

## Roundoff in Representation

Sadly, 0.1 decimal is an infinitely repeating pattern in binary: 0.0(0011), with 0011 repeating.  This means multiplying by some *finite* pattern to approximate 0.1 is only an approximation of really dividing by the integer 10.0.  The exact difference is proportional to the precision of the numbers and the size of the input data:
`for (int i=1;i<1000000000;i*=10) {	double mul01=i*0.1;	double div10=i/10.0;	double diff=mul01-div10;	std::cout<<"i="<<i<<"  diff="<<diff<<"\n";}`

In a perfect world, multiplying by 0.1 and dividing by 10 would give the exact same result.  But in reality, 0.1 has to be approximated by a finite series of binary digits, while the integer 10 can be stored exactly, so on NetRun's Pentium4 CPU, this code gives:
`i=1  diff=5.54976e-18i=10  diff=5.55112e-17i=100  diff=5.55112e-16i=1000  diff=5.55112e-15i=10000  diff=5.55112e-14i=100000  diff=5.55112e-13i=1000000  diff=5.54934e-12i=10000000  diff=5.5536e-11i=100000000  diff=5.54792e-10Program complete.  Return 0 (0x0)`
That is, there's a factor of 10^-18 difference between double-precision 0.1 and the true 1/10!  This can add up over time.

## Roundoff Taking Over Control

One place roundoff is very annoying is in your control structures.  For example, this loop will execute *seven* times, even though it looks like it should only execute *six* times:
`std::cout.precision(20);for (double k=0.0;k<1.0;k+=1.0/6.0) {	std::cout<<"k="<<k<<"\n";}`

(Try this in NetRun now!)

Notice that the very last iteration isn't quite 1.0, it's k=0.99999999999999988898.  Sigh.

The trouble is of course that 1/6 can't be represented exactly in floating-point, so if we add our approximation for 1/6 six times, we haven't quite hit 1.0, so the loop executes one additional time.  There are several possible fixes for this:

• Don't use floating-point as your loop variable.  Loop over an integer i (without roundoff), and divide by six to get k.  This is the recommended approach if you care about the exact number of times around the loop.  Many coding standards mandate this approach.
• Or, only increment the loop by a value with a short binary representation, like 0.125 (==0.001 in binary).  This clearly deserves a big obvious comment explaining what's subtle here.
• Or you could adjust the loop termination condition so it's "k<1.0-0.00001", where the "0.00001" provides some safety margin for roundoff.  This sort of "epsilon" value is common along floating-point boundaries, although too small and you can still get roundoff, and too big and you've screwed up the computation.
• Or you could use a lower-precision comparison, like "(float)k<1.0f".  This also provides roundoff margin, because the comparison is taking place at the lower "float" precision.
Any of these fixes will work, but you do have to realize this is a potential problem, and put the precision-compensation code in there!