Skip to main content

numerical integration - Whether NIntegrate evaluation is multithreaded or not?


About MultiThread evaluation I am talking about here, see this post


I am NIntegrate the function grarm[e, 3, dimension, 1, 1, 0.005]. There is matrix operation in function grarm, the dimension set the dim of the matrix.


I give the code of gram in the end.( take notice that you will have to manually change the dimension to get different gram with different dimension variable before integration)


OK, First I set dimension equals 7, and integrate the following


In:=NIntegrate[Im[grarm[e, 3, 7, 1, 1, 0.005]^2], {e, -0.3, 0}, 
  MaxRecursion -> 200] // AbsoluteTiming

Out:={7.454151, 0.0581345}


and If I use linux command top to monitor the cpu situation, I see it is single threaded. See enter image description here


now, I set dimension to 9, and integrate the following


In[55]:= NIntegrate[Im[grarm[e, 3, 9, 1, 1, 0.005]^2], {e, -0.3, 0}, 
  MaxRecursion -> 200] // AbsoluteTiming

Out[55]= {83.592780, 0.0271489}

and If I use linux command top to monitor the cpu situation, I see it is multi-threaded. See enter image description here


And Now I force the same integration to use one thread only, integrate the following



In[56]:= 
ParallelMap[
  NIntegrate[
    Im[grarm[e, #[[1]], 9, #[[2]], 1, 0.005]^2], {e, -0.3, 0}, 
    MaxRecursion -> 200] &, 
  Tuples[{Range[3, 3], Range[1, 1]}]] // AbsoluteTiming

Out[56]= {16.575363, {0.0271489}}

Notice it is much faster, that is odd.



In summary, There are two peculiar things waited to be explained



  1. why dimension=7 is single thread while dimension=9 is multi-thread? Can NIntegrate actually naturally multi-threaded or not? Or the multi-thread is actually due to the matrix operation in the integrand(well, I personally think this is not likely)?

  2. why multi-thread evaluation is much slower than single thread evaluation in the case of dimension 9?



In the end, the code for grarm.(don't forget to change the dimension, before integration)


dimension = 7;

Clear[armchairibbonmat]

armchairibbonmat[kx_, n_] :=
 (
  h = Table[0, {i, 1, 2 (n + 2)}, {j, 1, 2 (n + 2)}];
  a[m_] := 2 m - 1;
  b[m_] := 2 m;
  t1 = 1;
  aa = 1;
  Do[
   h[[b[i], a[i]]] = t1 E^(I kx aa);
   h[[a[i], b[i]]] = t1 E^(-I kx aa);

   h[[b[i + 1], a[i]]] = t1 E^(-I kx aa/2);
   h[[b[i - 1], a[i]]] = t1 E^(-I kx aa/2);
   h[[a[i - 1], b[i]]] = t1 E^(I kx aa/2);
   h[[a[i + 1], b[i]]] = t1 E^(I kx aa/2);
   , {i, 2, n + 1}];
  h = ArrayPad[h, -2]
  )

mat = Compile[{kx, n}, Evaluate@armchairibbonmat[kx, dimension]];
Clear[armchairibbonmat];

armchairibbonmat = mat;

Clear[gkarm]
gkarm[kx_?NumericQ, e_, xx_, width_, ii_, label_, \[Eta]_] := 
 Inverse[(e - I \[Eta]) IdentityMatrix[2*width] - 
    armchairibbonmat[kx, width]][[2*(ii - 1) + label, 
   2*(ii - 1) + label]]

Clear[fkx]
fkx[kx_?NumericQ, xx_] := E^(I kx xx)


Clear[grarm]
grarm[e_?NumericQ, xx_, width_, ii_, label_, \[Eta]_] := 
 NIntegrate[
  gkarm[kx, e, xx, width, ii, label, \[Eta]]*
   fkx[kx, xx], {kx, -\[Pi]/3, \[Pi]/3}, MaxRecursion -> 200]


Comments

Post a Comment

Popular posts from this blog

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}]
Post a Comment
Read more

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}]
Post a Comment
Read more

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....
Post a Comment
Read more