4 ms·

Decimal floating point just changes how cancellation/rounding appears, it doesn't do anything to increase accuracy. E.g. suppose you're using a decimal float wi

by dbaupp 11y ago

Decimal floating point just changes how cancellation/rounding appears, it doesn't do anything to increase accuracy. E.g. suppose you're using a decimal float with 1 digit of precision (i.e. it can have 1.0, 1.1, ... 9.9 as the mantissa), and compute 100 + 1 - 100 (i.e. 1.0e2 + 1.0e0 + -1.0e2): 100 + 1 is 101, which rounds to 100 (1.0e2), and 100 + -100 is of course 0. Complete catastrophic cancellation. In fact, binary floats use their storage better than decimal floats, so a binary float of a given size will have slightly higher accuracy than a decimal one of the same size. For example, there's 90 mantissae to encode in the example above, requiring a minimum of 7 bits, with a machine epsilon of 0.01. Using those 7 bits for a binary float gives an smaller epsilon: 1/128 ≈ 0.0078. Of course, decimal floats do have the advantage of matching a common input/output format, and their rounding artifacts better match the intuition of us base-10 accustomed humans.

legulere 11y ago
> In fact, binary floats use their storage better than decimal floats, so a binary float of a given size will have slightly higher accuracy than a decimal one of the same size. Not necessarily. The inefficiency is coped with by using less data for the exponent (which thus offers a smaller range). 64 bit decimal floats actually even have more data for the mantissa than their 64 bit binary counterpart. > Decimal floating point just changes how cancellation/rounding appears, it doesn't do anything to increase accuracy. E.g. suppose you're using a decimal float with 1 digit of precision (i.e. it can have 1.0, 1.1, ... 9.9 as the mantissa), and compute 100 + 1 - 100 (i.e. 1.0e2 + 1.0e0 + -1.0e2): 100 + 1 is 101, which rounds to 100 (1.0e2), and 100 + -100 is of course 0. Complete catastrophic cancellation. It also changes how often it occurs. With binary floats it already happens when you do 0.1 + 1.0 (I don't mean the rounding error of converting 0.1 to binary). When you enter 1.0 + 0.1 - 1.0 e.g. in python you will get 0.10000000000000009 as a result which is not equal to what you get when you enter 0.1. With decimal floats this doesn't happen. With every calculation you do the amount of precision needed can rise. With decimals converted to binary floats you use all of the precision right from the beginning. With decimal floats you can often stay easily within the precision offered by the data type.

by dbaupp 11y ago

legulere 11y ago

> In fact, binary floats use their storage better than decimal floats, so a binary float of a given size will have slightly higher accuracy than a decimal one of the same size. Not necessarily. The inefficiency is coped with by using less data for the exponent (which thus offers a smaller range). 64 bit decimal floats actually even have more data for the mantissa than their 64 bit binary counterpart. > Decimal floating point just changes how cancellation/rounding appears, it doesn't do anything to increase accuracy. E.g. suppose you're using a decimal float with 1 digit of precision (i.e. it can have 1.0, 1.1, ... 9.9 as the mantissa), and compute 100 + 1 - 100 (i.e. 1.0e2 + 1.0e0 + -1.0e2): 100 + 1 is 101, which rounds to 100 (1.0e2), and 100 + -100 is of course 0. Complete catastrophic cancellation. It also changes how often it occurs. With binary floats it already happens when you do 0.1 + 1.0 (I don't mean the rounding error of converting 0.1 to binary). When you enter 1.0 + 0.1 - 1.0 e.g. in python you will get 0.10000000000000009 as a result which is not equal to what you get when you enter 0.1. With decimal floats this doesn't happen. With every calculation you do the amount of precision needed can rise. With decimals converted to binary floats you use all of the precision right from the beginning. With decimal floats you can often stay easily within the precision offered by the data type.