@Carl Love

This error occurs in Maple 12 Classic Worksheet. 

@Carl Love 

ISC:=proc(T::list, Eq::`=`)

local f:= unapply((lhs-rhs)(Eq), indets(Eq,name)[]);

     `or`(seq(f(i[]) >= 0, i= T)) and `or`(seq(f(i[]) <= 0, i= T))

end proc:

ISC([[0,0],[10,10],[10,0]], x+y=-1);

Error, (in ISC) invalid input: or expects 2 arguments, but received 3

@Carl Love 

ISC:=proc(T::list, Eq::`=`)

local f:= unapply((lhs-rhs)(Eq), indets(Eq,name)[]);

     `or`(seq(f(i[]) >= 0, i= T)) and `or`(seq(f(i[]) <= 0, i= T))

end proc:

ISC([[0,0],[10,10],[10,0]], x+y=-1);

Error, (in ISC) invalid input: or expects 2 arguments, but received 3

M. Hirnyk, your code does not work in Maple 16:

trigsubs(sin(a+b)^2 - sin(a-b)^2)[];

   Error, (in trigsubs) expecting the sum or difference of two functions but got sin(a+b)^2-sin(a-b)^2

M. Hirnyk, your code does not work in Maple 16:

trigsubs(sin(a+b)^2 - sin(a-b)^2)[];

   Error, (in trigsubs) expecting the sum or difference of two functions but got sin(a+b)^2-sin(a-b)^2

If formally integrated in each interval, we get

int(abs(x-2), x)=-x^2/2+2*x+C1, x<=2

int(abs(x-2), x)=x^2/2-2*x+C2, x>=2

But the function  int(abs(x-2), x)   is continuous throughout the real axis and, in particular, at the point  x=2  because the function  abs(x-2)  is continuous. Therefore, at the point  x=2  we obtain  2+C1=-2+C2  or  C2=C1+4

@Carl Love 

Thanks for the explanation!

@Carl Love 

Thanks for the explanation!

It's interesting, why my simple version takes less time than your multithreaded version?

t:=time(): PP({i $ i=1..20}): time()-t;

t:=time(): PowerSet({i $ i=1..20}): time()-t;

                        10.640

                        14.766

It's interesting, why my simple version takes less time than your multithreaded version?

t:=time(): PP({i $ i=1..20}): time()-t;

t:=time(): PowerSet({i $ i=1..20}): time()-t;

                        10.640

                        14.766

@Carl Love

Regarding this example, you are right. But brute force method is more universal. Try to solve by  fsolve  the same question, for example,  for a series  

sum(ln(n)/n^2, n = 1 .. infinity);

@Carl Love

Regarding this example, you are right. But brute force method is more universal. Try to solve by  fsolve  the same question, for example,  for a series  

sum(ln(n)/n^2, n = 1 .. infinity);

If in the code of the procedure to replace the line  

L:=indets(A); m:=nops(L);  

by two lines

L0:=indets(A);
L:=remove(x->is(type(x,`^`)), L0);  m:=nops(L);

then operation of raising to a power is also supported. 

Note: the encoded numbers should not contain protected constants. For example, should be written  pi  rather than  PI or  Pi .

Attached file corrected.

Classic example (Knuth Volume 4 Fascicle 2):

Ksolve(pi * R ^ 2 = AREA);

          96*7^2 = 4704

Another example:

Ksolve(a^a=bcda);                                

             5^5  = 3125

Kryptarithms.mws

Acer, thank you very much!

Acer, thank you very much!