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Hi! I'm trying to replicate, in Matlab, compiled algorithms implemented in C using single-precision arithmetic; I have the C source code, and an executable which is reputed to implement said C source, but no convenient means to compile and link said source myself. (Yes, I know I can download the Microsoft C compiler, incl. a no-frills version of Visual Studio, for free, but that's hardly "convenient.") I've implemented the algorithms and I'm 99% sure they're right because the "residuals" (the difference between my results and those of the original C implementations) are more-or-less randomly distributed around zero, and about 85% of them are within half of the precision of the C implementations' output (which varies, but, for example, when it's 1e-4, 85% of my residuals satisfy abs(res)<5e-5). It's the other 15% that leave me 1% uncertain, and my attempts to explain the discrepancy have brought me to the single/double precision issue: I have good cause to suspect that the C implementations are using single-precision arithmetic, and I've been led to believe that Matlab always performs double-precision arithmetic, even on single precision "data," yes? (I tried making 'single' all the data in my implementation, and confirmed after each "gross" calculation that the output remained so, but that only reduced the number of bad residuals by 3 out of 4118 total input points.) So, I'm wondering if there's either a "convenient" way to simulate single-precision arithmetic in Matlab (I searched File Exchange for relevant functions, but found none) or to estimate/bound the "error" introduced by using single-precision arithmetic (including power/root and exponential/log functions) instead of double. Thanks!
Look on undocumentedmatlab.com for the information about the features supported by system_dependent()
Matlab uses single precision arithmetic for single precision values. Therefore there is no reason to simulate it.
I'd suggest to use double arithmetic in the C-function instead for the intermediate results to increase the accuracy.
Downloading and installing the Windows-SDK needs several minutes only.
It is very difficult to verify what is actually happening behind the scenes for single precision calculations. E.g., on a PC it may be the case that certain calculations on float variables get converted to 80-bit extended precision for the actual calculations and then re-converted to float to store the result. This can be the case for your C code even though all the variables are float. I am not sure which functions of MATLAB are strictly single and which are not. Bottom line is I seriously doubt you will be able to get an exact match no matter how much time you spend on the problem, unless the code is very simple. Whether the residuals you are currently seeing are "small" or not depends on the algorithms you are implementing. I think Jan's advice is a pretty good one ... do it all in double and compare that result to the single result.
OK, points well-taken, so here's the nitty-gritty:
In what follows, P ("big" O(1-10)), T (O(1-10)), and K (O(0.1-10)) are inputs; in the C implementation, they come in w/ precisions of 1e-3, 1e-4, and 1e-6, resp., i.e., the precision is held constant, not the sig figs--yes, I know, bad science, but hey, it's not my software.
k = 0.0162; K35150 = 42.914; A = [2.070e-5 -6.370e-10 3.989e-15]; B = [3.426e-2 4.464e-4 4.215e-1 -3.107e-3]; G = [6.766097e-1 2.00564e-2 1.104259e-4 -6.9698e-7 1.0031e-9]; a = [0.0080 -0.1692 25.3851 14.0941 -7.0261 2.7081]; b = [0.0005 -0.0056 -0.0066 -0.0375 0.0636 -0.0144]; S = zeros(size(T)); RP = S; RT = S; % Begin calculation R = K / K35150; val = 1 + B(1)*T + B(2)*T.*T + B(3)*R + B(4)*R.* T; good = (val~=0); RP(good) = 1 + (P(good) .* (A(1) + P(good) .* (A(2) + A(3)*P(good)))) ./ val(good); val = RP .* (G(1) + (T .* (G(2) + T .* (G(3) + T .* (G(4) + G(5)*T))))); good = (val~=0); RT(good) = R(good) ./ val(good); RT(RT <= 0) = 1e-6; tmp = RT'; tmp = [ones(size(tmp)); tmp.^0.5; tmp; tmp.^1.5; tmp.^2; tmp.^2.5]; sum1 = a * tmp; sum2 = b * tmp; val = 1.0 + k * (T - 15.0); good = (val~=0); S(good) = sum1(good)' + sum2(good)'.*(T(good) - 15.0)./val(good);
So, as you can see, there are lots of additions, multiplications, and powers (incl. fractional), involving coefficients of widely different scales; I've posted a plot of the "residuals" = the (same) inputs processed by the compiled C app (output precision = 1e-4) - processed by my Matlab implementation. Note: for the inputs, I'm able to duplicate the C (implemented in double precision, I'm informed) results to within +/-0.5*output precision, uniformly for T and K, and to about 99% in the case of P. So, please opine: should I be concerned? Thanks!
Ouch, the C program's output format bit me; it's somewhat complicated, so suffice it to say I was comparing apples to oranges (though why the C program outputs both apples and oranges when one would expect it to output only one or the other remains in question)--once I figured this out and started comparing apples to apples, I found that my apples are just like theirs! Sorry for the noise.