Strange multiplication result

问题

In my code I have this multiplications in a C++ code with all variable types as double[]

f1[0] = (f1_rot[0] * xu[0]) + (f1_rot[1] * yu[0]); 
f1[1] = (f1_rot[0] * xu[1]) + (f1_rot[1] * yu[1]); 
f1[2] = (f1_rot[0] * xu[2]) + (f1_rot[1] * yu[2]); 

f2[0] = (f2_rot[0] * xu[0]) + (f2_rot[1] * yu[0]); 
f2[1] = (f2_rot[0] * xu[1]) + (f2_rot[1] * yu[1]);
f2[2] = (f2_rot[0] * xu[2]) + (f2_rot[1] * yu[2]);

corresponding to these values

Force Rot1 : -5.39155e-07, -3.66312e-07
Force Rot2 : 4.04383e-07, -1.51852e-08

xu: 0.786857, 0.561981, 0.255018
yu: 0.534605, -0.82715, 0.173264

F1: -6.2007e-07, -4.61782e-16, -2.00963e-07
F2: 3.10073e-07, 2.39816e-07, 1.00494e-07

this multiplication in particular produces a wrong value -4.61782e-16 instead of 1.04745e-13

f1[1] = (f1_rot[0] * xu[1]) + (f1_rot[1] * yu[1]);  

I hand verified the other multiplications on a calculator and they all seem to produce the correct values.

this is an open mpi compiled code and the above result is for running a single processor, there are different values when running multiple processors for example 40 processors produces 1.66967e-13 as result of F1[1] multiplication.

Is this some kind of mpi bug ? or a type precision problem ? and why does it work okay for the other multiplications ?

回答1:

Your problem is an obvious result of what is called catastrophic summations: As we know, a double precision float can handle numbers of around 16 significant decimals.

f1[1] = (f1_rot[0] * xu[1]) + (f1_rot[1] * yu[1])
      = -3.0299486605499998e-07 + 3.0299497080000003e-07
      = 1.0474500005332475e-13

This is what we obtain with the numbers you have given in your example. Notice that (-7) - (-13) = 6, which corresponds to the number of decimals in the float you give in your example: (ex: -5.39155e-07 -3.66312e-07, each mantissa is of a precision of 6 decimals). It means that you used here single precision floats.

I am sure that in your calculations, the precision of your numbers is bigger, that's why you find a more precise result.

Anyway, if you use single precision floats, you can't expect a better precision. With a double precision, you can find a precision up to 16. You shouldn't trust a difference between two numbers, unless it is bigger than the mantissa:

• Simple precision floats: (a - b) / b >= ~1e-7
• Double precision floats: (a - b) / b >= ~4e-16

For further information, see these examples ... or the table in this article ...

来源：https://stackoverflow.com/questions/24432648/strange-multiplication-result