Skip to main content

documentation - How has Hash changed in 11.3?


There are already few topic related to Hash[_String]:


How does Hash calculate hash for strings?


Incorrect calculating Hash SHA256


But it looks like changes are more severe:


Hash[{}] returns the same in V11.2 and V11.3, but e.g. Hash[{}, "MD5"] does not.


enter image description here


And I don't see an explanation in documentation:


enter image description here


What is the complete list of changes? How to make old code compatible with those changes?




Answer



There is already discussion about String and ByteArray in the linked previous Q & A, so I'll comment a bit about general expressions.


This only concerns the named hash algorithms like "MD5" or "SHA" etc.


The single argument form Hash[expr], which is equivalent to Hash[expr, "Expression"] is completely separate and based on the internal representation of expr. It has not been changed for 11.3.


Using @Kuba's example of a very simple expression, in 11.3 we have the following new hash value


Hash[{}, "MD5"]

(* 68244457821771821570522625853545795031 *)

The previous hash value can still be obtained using



Developer`LegacyHash[{}, "MD5"]

(* 272934427398090264974473461931457450337 *)

Why the change? The previous scheme for converting expressions (to strings) for hashing had a number of severe problems.


For example, it did not take into account the contexts of symbols so it could happen that the same expression would hash to a different value because of a different $ContextPath, it had issues with evaluation leaks etc.


These have now been addressed, but the fixes mean the hash values would inevitably change. Developer`LegacyHash is provided for people who in some way depend on the old hash values.


The wording in the documentation isn't fully accurate, because ToString[FullForm[expr]] is not used literally.


What is actually true is that the input given to the hashing algorithm is based on the bytes of ToString[Unevaluated[FullForm[expr]]], where all symbols in expr are qualified with their full contexts.


Furthermore, a constant 32-byte sequence prefix is added for (non-String and non-ByteArray) expressions to avoid collisions -- this ensures that the number 2 and the string "2" do not end up having the same hash value. This is because ToString[Unevaluated[FullForm[2]]] is the same as the string "2" but 2 and "2" are different expressions.



Below is a mock-up example (not the actual implementation) that could be used to replicate the 11.3 hash value even on earlier versions. It uses the byteHash utility defined in my previous answer.


prefix = {209, 74, 9, 190, 254, 30, 81, 99, 147, 98, 22, 44, 107, 239, 77, 113, 
          23, 185, 9, 18, 189, 28, 97, 183, 43, 63, 221, 103, 61, 127, 201, 101};

byteHash[Join[prefix, ToCharacterCode["System`List[]"]], "MD5"]

(* 68244457821771821570522625853545795031 *)

While the documentation ideally should give some idea of what serialization is used for general expressions, I would not hold the expectation that it must go into any deep level of detail or provide sufficient information to actually write an alternative implementation. Besides, the serialization could conceivably change some day again.


I think the moral is, if people want full control, they should themselves create a sequence of bytes to give as input to the hashing method in whatever way they see as appropriate. Then, what a named algorithm like "MD5" or "SHA" must return is fully determined, and a result different from that would certainly be a bug.



Comments

Post a Comment

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}]
Post a Comment
Read more

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 - 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