Skip to main content

function construction - Another difference between Set and Setdelayed. Evaluation shortcut?



A little while ago I wondered why


f[x_] = f[x]

gives an infinite iteration. I ended up discovering the following difference in evaluation between Set and SetDelayed with Evaluate.


count = 0;
ClearAll@a
a /; (count++ < 20) = {a}
a // OwnValues
count = 0;
a


Output


{{{{{{{{{{{{{{{{{{{{{a}}}}}}}}}}}}}}}}}}}}}
{HoldPattern[a /; count++ < 20] :> {a}}
{a}

and


count = 0;
ClearAll@b
b /; (count++ < 20) := Evaluate@{b}

b // OwnValues
count = 0;
b

Output


{HoldPattern[b /; count++ < 20] :> {b}}
{{{{{{{{{{{{{{{{{{{{b}}}}}}}}}}}}}}}}}}}}

Can somebody explain the difference? Can we say that there is an evaluation shortcut at work here?


Related



This is a follow up question: Strange results of definitions using OwnValues


Why x = x doesn't cause an infinite loop, but f[x_] := f[x] does?


Does Set vs. SetDelayed have any effect after the definition was done?



Answer



I thought to give a bit more insight into why Update is needed, as pointed out in the other answers. Its documentation says Update may be needed when a change in 1 symbol changes another via a condition test.


In Jacob's example, setting count = 0 changes the condition test outcome, and thus a or b on the LHS. Consequently, a or b on the RHS is supposed to change. However, RHS a equals the old LHS a, which was undefined because count>=20, and needs Update to be changed. RHS b behaves the same, but was not evaluated in SetDelayed because Evaluate occurs before SetDelayed, so count is unchanged, and RHS b evaluates to LHS b with count<20. If we now reset count=0, evaluating b will return {b}.


To illustrate, I modify the example to separate LHS and RHS. MMA is clever enough to automatically update LHS declared as a variable, so I have to make a function:


count=0;
ClearAll[LHS,RHS];
LHS[]/;(count++<20)={RHS};

RHS=Unevaluated@LHS[];

count=0;
RHS (* Equals LHS[] with count >= 20 *)

(* Tell Wolfram Language about changes affecting RHS which depends on LHS *)
Update@Unevaluated@LHS;
RHS



LHS[]


{{{{{{{{{{{{{{{{{{{{LHS[]}}}}}}}}}}}}}}}}}}}}



Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

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}]