Skip to main content

parallelization - ParallelTable and Precision


I'm using ParallelTable[] to calculate a function over a range of my parameters , ($\omega,\ell$). This seems to be working well (in terms of speed increase) except for some strange Precision issues.


I set the $MinPrecision=40 at the start of the notebook as well as my working precision, precision goal and accuracy goal for some computations later in the notebook as


 wp=$MinPrecision;

ac=$MinPrecision-8;
pg=wp/2;

The rest of my code basically defines some helper functions before going on to the main function (that I feed into ParallelTable). This main function uses the helper functions along with NDSolve and NIntegrate to do some computations, and it's in these that I feed the Working Precision->wp ,AccuracyGoal->ac, PrecisionGoal->pg.


I have used DistributeDefinitions on all my variables, helper functions, and the main function, and even on $MinPrecision, also all my initial conditions for NDSolve have N[...,wp] wrapped around them, but still when I use ParallelTable[...] I get errors:


 Precision::precsm. Requested precision 39.153977328439204` is smaller than $MinPrecision. 

I don't get any such error when I simply use Do[...] or Table[..] on my main function, and indeed the results differ at the 33rd decimal place, which is the decimal the above number is precision too. I have no idea what this number is unfortunately, it doesn't look like any of my outputs or inputs.


I just don't understand how this could not crop up with the non-parallelized forms, but crop up with parallelized?


Minimal Example



This is much simpler than my code, but I think it still captures it and the problem:


Definitions:


M = 1;
$MinPrecision = 40;
wp = $MinPrecision;
ac = $MinPrecision - 8;
pg = wp/2;
rinf = 15000;

The main function to be parallelized:



dGenBessE[\[Omega]_?NumericQ, l_?IntegerQ] := 
Block[{\[CapitalPhi]out, init, dinit},

init = 0.00006630728036817007679447778124486601253323 +
6.913102762021489976135937610105907096265*10^-6 I;
dinit = -6.958226432502243329110910813935678705519*10^-7 +
6.631148430876236565520382557577187147081*10^-6 I;

\[CapitalPhi]out[\[Omega], l] = \[CapitalPhi] /.
Block[{$MaxExtraPrecision = 100},

NDSolve[{\[CapitalPhi]''[r] + (2 (r - M))/(
r (r - 2 M)) \[CapitalPhi]'[
r] + ((\[Omega]^2 r^2)/(r - 2 M)^2 - (l (l + 1))/(
r (r - 2 M))) \[CapitalPhi][r] ==
0, \[CapitalPhi][rinf] ==
N[init, wp], \[CapitalPhi]'[rinf] ==
N[dinit, wp]}, \[CapitalPhi], {r, 30, 40},
WorkingPrecision -> wp, AccuracyGoal -> ac,
PrecisionGoal -> pg, MaxSteps -> \[Infinity]]][[1]];
Print["For \[Omega]=", \[Omega], " and l=", l, ": "];

Print["Kernel ID: ", $KernelID];
Print["Precision of init: ", Precision[init]];
Print["Precision of dinit: ", Precision[dinit]];
Print["\[CapitalPhi]out at 35=" ,
N[\[CapitalPhi]out[\[Omega], l][35], wp] ];
Print["Precision of \[CapitalPhi]out: ",
Precision[\[CapitalPhi]out[\[Omega], l][35]]];
]

Do Paralleization prereqs:



LaunchKernels[4]
DistributeDefinitions[M, rinf, wp, ac, pg, $MinPrecision,dGenBessE];

Attempt to run it with Paralleize for two different values:


Parallelize[{dGenBessE[1/10, 0], dGenBessE[1/10, 1]}] // AbsoluteTiming

Result


For me this leads to a


Precision::precsm: Requested precision 38.95475956978393` is smaller than   $MinPrecision. Using $MinPrecision instead.

Answer




I think your problem arises because the value of $MinPrecision is not distributed correctly (If I remember correctly none of the variables in the System context are distributed automatically).


So we have to do this by hand


ParallelEvaluate[$MinPrecision = 40]

Parallelize[{dGenBessE[1/10, 0], dGenBessE[1/10, 1]}] // AbsoluteTiming

Should now work without problems.


Also, presonally I would write the last example as


ParallelMap[dGenBessE[1/10, #]&,{0,1}] // AbsoluteTiming


or


Parallelize@Scan[dGenBessE[1/10, #] &, {0, 1}]

depending on whether you need to return values or not.


Comments

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.