Replacing dot by *?

Question

I am unable to replace dot by *.

Replace[a.a, .-> *]

Is there any solution to get this done?

Answer 1

You can use ReplaceAll (/.):

a.a /. Dot -> Times

a^2

Alternatively, you can temporarily redefine Dot as Times using Block:

Block[{Dot = Times}, a.a]

a^2

Answer 2

I think there are several reasonons pointed out by Szabolcs in the comments why the approach failed, and they are not all supper basic by my standard. I would like to point them out.

First the basics

Everything is an expression.

foo -> bar is the same thing as Rule[foo,bar] and the same thing as foo~Rule~bar.

This underlying expression is often hidden and not important, but checking the underlying description is very helpful when debugging. Check out FullForm and TreeForm.

The second argument .-> * is not a valid syntax

When parsing this input, the interpreter expects ., -> and * to be infix operators. Having two or more infix operators next to each other can not be interpreted and it is indicated by codehighlighting of the cell:

It effectivly reads:

(~Dot~) (~ReplaceAll~) (~Times~)

which can not be interpreted, as left and right operands are missing.

Also ToExpression[".->*"] returns a message

ToExpression::sntx: Invalid syntax in or before ".->*".

Repalce checks only the top level

Repalce will not replace a part of an expression. Compare:

In[1953]:= 
Replace[Dot, Dot -> Times]
Replace[Dot[a.a], Dot -> Times]
Replace[Dot[a.a], Dot[a.a] -> Times[a a]]

Out[1953]= Times

Out[1954]= a.a

Out[1955]= a^2

Working solutions

Either use more general replacement rule:

In[1939]:= Replace[a.b, (f_).(g_) :> (f*g)]

Out[1939]= a b

or instead of Replace use a ReplaceAll (/.) , which tries to apply the pattern also to subexpressions:

In[1943]:= 
FullForm[a.b]
ReplaceAll[a.b, Dot -> Times]
FullForm[%]

Out[1943]//FullForm= 
Dot[a,b]

Out[1944]= 
a b

Out[1950]//FullForm=
Times[a, b]