Want to compute the permutations of {1, 2, ..., 11} with only 3 GB of memory

There is another way to calculate

Permutations [{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}]

without triggering an error, I have 3 GB of RAM with WIN 7

Edit:

This short code is the one that broke my head for a while, are 11 variables that must meet a very specific condition, their difference must be 1. either can take the 11 values, hence all permutations, this code is an adaptation of another code I saw here, which helps me to what I need.

juan[{a_, b_, c_, d_, e_, f_, g_, h_, i_, j_, k_}] := 
 Abs[Differences /@ ({{a, b}, {a, c}, {b, c}, {b, f}, {b, e}, {c, 
   e}, {c, f}, {c, g}, {d, f}, {d, g}, {e, b}, {e, a}, {e, f}, {e,
    i}, {e, h}, {f, g}, {f, j}, {f, i}, {f, h}, {g, i}, {g, 
   j}, {h, i}, {h, k}, {i, j}, {i, k}, {j, k}})] // Flatten
*(*per = Permutations[Range@11]*) (this line is calculated as 799    consecutive files in HD thanks to the collaboration of  rasher)

(*per=Import["C:\\Users\\M\\Desktop\\per.txt"]*)(as I upload the files   sequentially and that its securities are passing the variable "per" and will be prosecuted.?)
Select[per, FreeQ[juan@#, 1] &]

Answer

This will write the permutations to permutations.txt in list blocks of ~50,000 each.

Quiet@Block[{$ContextPath}, Needs["Combinatorica`"]]

len = 11
numchunk = 1000

chunks = Partition[Clip[FindDivisions[{0, len! - 1, 1}, numchunk],

                  {0, len! - 1}, {0,len! - 1}], 2, 1] // 
                  (# + Join[{{0, 0}}, ConstantArray[{1, 0}, Length@# - 1]]) &;

Monitor[(chunk = #; (Combinatorica`UnrankPermutation[#, 11] & /@ 
        Range @@ chunk) >> 
      "permutations-" <> ToString[First@#] <> "-" <> 
       ToString[Last@#] <> ".txt") & /@ chunks;, chunk]

Will take about an hour, I'd ventue...

If you must have equal sized files, you'll want to create your own chunks, since FindDivisions uses a heuristic that usually won't meet that criteria, e.g. in your case for 11 length:

p = Partition[Range[1, 102089*392, 102089] - 1, 2, 1];
p[[1]] = p[[1]] - {0, 1};
p[[2 ;;, 1]] = p[[2 ;;, 1]] + 1;
chunks = p;

Will create files all with same # of permutations (about 100k).

Comments

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 - 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 - 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....

Blog

Search This Blog