Domanda

Suppose I have an expression like (actually mine is much more complex, thousands of characters)

expr:a+b*c+b*c*d;

and I want to replace an internal sub-expression with a symbol (useful to avoid recomputation of common subexpressions), say k in place of b*c:

subst(b*c=k,expr);

returns

k+b*c*d+a

How I can make Maxima calculate the "right" substitution so to return (apart from obviuos simplification, here)

k+k*d+a

?

È stato utile?

Soluzione

Take a look at let and letsimp. E.g.:

(%i2) expr : a + b*c + b*c*d;
(%o2) b*c*d+b*c+a
(%i3) let (b*c, k);
(%o3) b*c --> k
(%i4) letsimp (expr);
(%o4) d*k+k+a

letsimp differs from subst and tellsimp or defrule in that those other functions make only formal substitutions, i.e., replacing subexpressions which are exactly the same as some pattern.

Altri suggerimenti

You can try optimize

http://maxima.sourceforge.net/docs/manual/en/maxima_6.html#IDX219

(%i14) example(optimize);

(%i15) diff(exp(y+x^2)/(y+x),x,2)
                        2            2              2            2
               2   y + x        y + x          y + x        y + x
            4 x  %e         2 %e         4 x %e         2 %e
(%o15)      ------------- + ---------- - ------------ + ----------
                y + x         y + x               2             3
                                           (y + x)       (y + x)
(%i16) optimize(%)
                                                 2         y + %2       1
(%o16) block([%1, %2, %3, %4], %1 : y + x, %2 : x , %3 : %e      , %4 : --, 
                                                                        %1
                                                                 4 x %3   2 %3
                                          4 %2 %4 %3 + 2 %4 %3 - ------ + ----)
                                                                    2       3
                                                                  %1      %1
Autorizzato sotto: CC-BY-SA insieme a attribuzione
Non affiliato a StackOverflow
scroll top