Skip to main content

simplifying expressions - Counting multiplications (complexity function)


I'm trying to count the number of times multiplication appears in an expression (for the purpose of constructing a ComplexityFunction that would help me simplify a terrible algebraic expression for later numerical crunching in matlab).


For example, I consider the following expression to involve 3 multiplications:


a b c + d e


two in "abc" and one in "de".


I've tried Level, but unfortunately mathematica sees the term "abc" as one "Times" with three arguments:


 In:= Level[a b c + d e, {-1}, Heads->True]
Out:= {Plus, Times, a, b, c, Times, c, d}

In principle I could reverse-engineer the multiplicity of arguments of "Times" from the above, but that seems like an unnecessarily cumbersome way.


i'd greatly appreciate any help.



Answer



Here's one way to count the number of multiplications in an expression (equal to or greater than the number of Times in the expression). It should also work for several other binary operators, Listable or not (although I haven't tested it on them).



t[x_, oper_: Times] := Tr @ ((Length[#] - 1) & /@
                       (Extract[x, {Sequence @@ Drop[#, -1]}] & /@ Position[x, oper]))

Usage


t[a b c + d e + f[a b] - 1/f[f[c d]]]


6



Under the hood.


Let's examine a b c + d e + f[a b] - 1/f[f[c d]], using Position and Extract and TreeForm.


The tree structure of the expression...


ClearAll[a, b, c, d, e, f]
(x = a b c + d e + f[a b] - 1/f[f[c d]]) // TreeForm

times1


Then the positions of the head, Times.


Position[x, Times]



{{1, 0}, {2, 0}, {3, 1, 0}, {4, 0}, {4, 2, 1, 1, 1, 0}}

By dropping the final zero from each position, we obtain the instances of Times, including the arguments.


Extract[x, {Sequence @@ Drop[#, -1]}] & /@ %


{a b c, d e, a b, -(1/f[f[c d]]), c d}

...the number of items.



Length[%]


5

[t[] goes a step further to count the (arguments-1) for each Times, namely, 6.]


...a look at the structure of each of those instances of Times.


TreeForm /@ %%

times2



Comments

Post a Comment

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...
Post a Comment
Read more

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...
Post a Comment
Read more

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...
Post a Comment
Read more