functions - Default behaviour of Hash[expr] and hashing in different versions of Mathematica

In different versions of Mathematica, Wolfram silently changed the behaviour of Hash when the algorithm is not specified explicitly

Hash[1]
(* 6568131406215528669 (Version 10.1) *)

Hash[1]

(* 4371187653775642860 (Version 9.0.1) *)

Hash[1]
(* 1742717557 (Version 8.0.4) *)

This is a serious violation of how a public function/API should be supported in such a large software as Mathematica. Especially, since the documentation of Hash suggests that the behaviour will be consistent:

Hash[expr,...] will always give the same result for the same expression expr.

More importantly, the default hashing algorithms seems to be completely detached from all available settings, as it seems impossible to recreate the default hash when explicitly choosing one.

Hash[1]
(* 6568131406215528669 *)

Hash[1,#]&/@{"Adler32","CRC32","MD2","MD5","SHA","SHA256","SHA384","SHA512"}//Column
(* 
3959688615
3017272578
277753940344783714340401450212752361952
68231128815270908652080701364659390939
846778260378026149058641558857036959755342858310

35440229038221092521327873929090932360118094198559938935455836283354874680335
32975197285603495667013724312557093882636446440150950620624270433328413926233671687205762131877578287401554708845055
12579926171497332473039920596952835386489858401292624452730263741969134739018228297640298179049647746066620814234742520593670116132355345543156774710409041
*)

Does someone know, 1. how to reproduce the default hashing behaviour when explicitly choosing a method and 2. re-create the default hashing behaviour in different versions?

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