Skip to main content

numerics - Different floating-point numbers equal?


Let's define two different numbers.


x = 1.

y = 1. + 2^-52 (* equivalently, 1 + $MachineEpsilon *)

Let's make sure they're different with FullForm:


x // FullForm (* 1.` *)
y // FullForm (* 1.0000000000000002` *)

Those look pretty close... let's make sure they're different. I'm not a wizard with the developer tools, but I can export them as IEEE double-precision floating point numbers (which I'd bet is their internal representation):


StringJoin @@ 
 IntegerString[Reverse@ToCharacterCode[ExportString[x, "Real64"]], 
  16, 2]

(* 3ff0000000000000 *)
StringJoin @@ 
 IntegerString[Reverse@ToCharacterCode[ExportString[y, "Real64"]], 
  16, 2]
(* 3ff0000000000001 *)

We can see that they are indeed different. They represent the two numbers:


$$ \begin{align} x &= (1.){\underbrace{000 \cdots 000}_\text{51 zeros}}0_2 \times 2^{01111111111_2 - 1023} \equiv 1 \\ y &= (1.){\underbrace{000 \cdots 000}_\text{51 zeros}}1_2 \times 2^{01111111111_2 - 1023} \equiv 1 + \frac{1}{2^{52}} \end{align} $$


That is, x is exactly one, and y is the smallest IEEE double greater than one. Ok, so they're different. Hey Mathematica, you know they're diff-


x == y (* True *)


Oh. What if we try-


x === y (* True *)

Hey Python, you use doubles, right? Are you seeing this?


>>> 1. == 1.0000000000000002
False

Maybe it's because you're using quads?


>>> 1. == 1.0000000000000001

True

Yeah, I didn't think so. Mathematica, are you sure? I mean, this doesn't seem right...


y - x (* 2.22045*10^-16 *)

Aha! I knew it! Now let's try this:


y - x == 0 (* False *)

Success! Now let's just double-check (pun intended):


1.0000000000000001 - 1. (* 0. *)

% == 0 (* True *)

So you are using double-precision...


My question is, Why do Equal and SameQ return True, even though these numbers are obviously different? SameQ ignores the last bit, and Equal ignores the last seven bits!



Answer



It seems I found my answer in OleksandrR's comment to this question. He says,



Bear in mind Equal applies an extra tolerance in Mathematica. The proper comparison is


Block[{Internal`$EqualTolerance = -Infinity}, 1 == 1 + $MachineEpsilon] (* False *)
Block[{Internal`$EqualTolerance = -Infinity}, 1 == 1 + $MachineEpsilon/2] (* True *)


In fact, the value of Internal`$EqualTolerance * Log2[10] is 7., meaning that it ignores the last seven bits, just as I discovered!


(Analogously, Internal`$SameQTolerance * Log2[10] is 1., i.e. it drops the last bit.)


Note that this is mentioned in the documentation for Equal, under Details:




  • Approximate numbers with machine precision or higher are considered equal if they differ in at most their last seven binary digits (roughly their last two decimal digits).

  • For numbers below machine precision the required tolerance is reduced in proportion to the precision of the numbers.




However, I never thought to look at it, since (thought) I knew what == means! Lesson learned, always check the documentation, especially if you don't think you need to.


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