@itsme   Thank you! Of cause, you are right.

@amrramadaneg   If you specify a function of two variables  x  and  y , defining the displacements, then the same can be done in automatic mode:

P:=[[0, 1], [2, 1], [2, -1], [0, -1]]:

A:=plottools[polygon](P, style=line,thickness = 2, color=red, numpoints=5000):

u:=(x,y)->[x+0.03*y, y+0.03*x];

B:=plottools[polygon](map(t->u(op(t)),P), style=line, thickness = 2, color=blue, numpoints=5000):

plots[display](A,B); 

If we have  n  different numbers passing in a certain order, the number of different ways  of parentheses setting  is equal to  C[n-1]=binomial(2*n-2, n-1)/n  (C[n] is nth Catalan number). Using this formula, we can easily calculate the number of of different variants in the exhaustive search of all variants. For example, if  n = 6 , using all four operations, the order of numbers is  arbitrary  and parentheses are used, the number of variants is

6!*4^5*eval(binomial(2*n-2, n-1)/n, n=6); 

                               30965760

Therefore, in such cases it is not recommended to take  n>6 , otherwise the time of calculation can be unacceptably large.

Consider a few more examples of the application  NumbersGame  procedure.

Example 4.  The problem: Is it possible to place between 14 numbers 1 .. 14  signs  "+"  and  "-"  so that the result is 0?

NumbersGame(0, [$ 1..14], ["+","-"]);

                          No solutions

Example 5.  The problem: What is the smallest number  n  such that placing between numbers  1..n  arithmetic operators we obtain the known mystical number 666 ?

for n from 6 do

NumbersGame(666, [$ 1..n]):

if nops(M)>0 then break fi;

od:

%;

 1 + 2 * 3 * 4 * 5 * 6 - 7 * 8 - 9 + 10 = 666,   1 + 2 * 3 * 4 * 5 * 6 + 7 - 8 * 9 + 10  =  666

Example 6. The problem: How smallest number of digits can write this number  666  using arithmetic operations and parentheses ?

for n from 3 do

T:=combinat[cartprod]([[$ 1..9] $ n]): 

while not T[finished] do

m:=T[nextvalue](): NumbersGame(666, m,  yes):

if nops(M)>0 then break  fi:

od:

if nops(M)>0 then break fi: 

od:

op(M);

        ( 2 + ( 8 * 9 ) ) * 9   =  666

@Markiyan Hirnyk  My code is correct, because there is not verified  a prime number  p  itself, but the next prime number is checked. Of course, you can write the equivalent code:

restart;

p:=41: n:=0:

while p<=107 do

sol:=msolve(y^2 + y - 11=0, p);

if  sol<>NULL then  n:=n+1; L[n]:=[p, sol]  fi;

p:=nextprime(p);

od:

L:=convert(L, list):

op(L);

@Vesnog   Read help to  name .

@Preben Alsholm  This system can be easily solved numerically, for example for  h1=0, h2=1.

restart;

Eq2 := diff(T(y), y, y)+(diff(sigma(y), y))*(diff(T(y), y))+(diff(T(y), y))^2+(exp(-y)+exp(y))^2 = 0;

Eq3 := diff(sigma(y), y, y)+diff(T(y), y, y) = 0;

bcs2 := T(0) = 0, T(1) = 1, sigma(0) = 0, sigma(1) = 1;

sol := dsolve({Eq2, Eq3, bcs2}, numeric);

plots[odeplot](sol, [[y, T(y)], [y, sigma(y)]], y = 0 .. 1, color = [red, blue]);

But if these values  h1  and  h2  are substituted into your explicit formulas, an error occurs "division by zero"

@Carl Love   Very ingenious way! Vote up.

@H-R   Read help to these commands. Everything is explained in detail.

@Dmitry Lyakhov

Examples:

A0:=(x,y)->x^3+y^2+1;  # or

A1:=unapply(x^3+y^2+1, x,y);

A0(2,3), A1(2,3);  

@Dmitry Lyakhov  If you speak in Russian, you can ask your questions here

@Markiyan Hirnyk  Only one inequality is allowed to multiply by a number. Should be

map(x->-3*x, `and`((1/3)*x-1 > 1, (1/3)*x-1 < 5));

                             -x < -6 and -18 < -x

@Markiyan Hirnyk  You are wrong. Just noticed that substitutions the inequality signs is not required. Maple does this automatically.

Example:

(-3)*(x < 2);

         -6 < -3*x

@Markiyan Hirnyk  You multiply by -3, not me. If I multiply by -3, of course I would have changed the inequality sign by reversed.

(1/3)*x-1 > 1 and (1/3)*x-1 < 5;

subs(`>` = `<`, `<` = `>`, map(x->-3*x, %));

solve(%);

@Markiyan Hirnyk  I do not understand the meaning of your comment.

@Preben Alsholm  Thank you. Very useful comment!

@Carl Love  Can we write this using  InertForm  package:

add(convert(i, symbol), i = 1 .. 20) = `+`($ 1 .. 20);