Firstly, are created a list of all pairs of diagonals for such squares, and then in a loop filled with the remaining entries. The total number of such  Latin squares is equal to 960. For illustration  3 squares are displayed .

restart:

L:=combinat[permute](5):

ND:=0: Diag:=[]:

for i in L do

A:=convert(i, set) minus {i[3]}:

M:=combinat[permute](A):

for j in M do

if i[1]<>j[1] and i[2]<>j[2] and i[4]<>j[3] and i[5]<>j[4] and i[1]<>j[4] and i[2]<>j[3] and i[4]<>j[2] and i[5]<>j[1] then ND:=ND+1: Diag:=[op(Diag),[[seq(i[k], k=1..5)], [j[1], j[2], i[3], j[3], j[4]]]]: fi:

od: od:

ND: # The number of pairs of diagonals

Diag: # The list of pairs of diagonals

B:={1, 2, 3, 4, 5}:

N:=0: List:=[]:

for i from 1 to nops(Diag) do

for a in B minus {Diag[i,1,1], Diag[i,2,5], Diag[i,1,2], Diag[i,2,2]} do

for b in B minus {Diag[i,1,1], Diag[i,2,5], Diag[i,1,3]} do

for c in B minus {Diag[i,1,1], Diag[i,2,5], Diag[i,2,4], Diag[i,1,4]} do

for d in B minus {Diag[i,1,2], Diag[i,2,4], Diag[i,1,1], Diag[i,2,1]} do

for e in B minus {Diag[i,1,2], Diag[i,2,4], Diag[i,1,3]} do

for f in B minus {Diag[i,1,2], Diag[i,2,4], Diag[i,2,5], Diag[i,1,5]} do

for g in B minus {Diag[i,1,3], Diag[i,1,1], Diag[i,2,1]} do

for h in B minus {Diag[i,1,3], Diag[i,1,2], Diag[i,2,2]} do

for l in B minus {Diag[i,1,3], Diag[i,2,4], Diag[i,1,4]} do

for m in B minus {Diag[i,1,3], Diag[i,2,5], Diag[i,1,5]} do

for p in B minus {Diag[i,2,2], Diag[i,1,4], Diag[i,1,1], Diag[i,2,1]} do

for q in B minus {Diag[i,2,2], Diag[i,1,4], Diag[i,1,3]} do

for r in B minus {Diag[i,2,2], Diag[i,1,4], Diag[i,2,5], Diag[i,1,5]} do

for u in B minus {Diag[i,2,1], Diag[i,1,5], Diag[i,1,2], Diag[i,2,2]} do

for v in B minus {Diag[i,2,1], Diag[i,1,5], Diag[i,1,3]} do

for w in B minus {Diag[i,2,1], Diag[i,1,5], Diag[i,2,4], Diag[i,1,4]} do

if nops({Diag[i,1,1],a,b,c,Diag[i,2,5]})=5 and nops({d,Diag[i,1,2],e,Diag[i,2,4],f})=5 and nops({g,h,Diag[i,1,3],l,m})=5 and nops({p,Diag[i,2,2],q,Diag[i,1,4],r})=5 and nops({Diag[i,2,1],u,v,w,Diag[i,1,5]})=5 and nops({Diag[i,1,1],d,g,p,Diag[i,2,1]})=5 and nops({a,Diag[i,1,2],h,Diag[i,2,2],u})=5 and nops({b,e,Diag[i,1,3],q,v})=5 and nops({c,Diag[i,2,4],l,Diag[i,1,4],w})=5 and nops({Diag[i,2,5],f,m,r,Diag[i,1,5]})=5 then N:=N+1: List:=[op(List), Matrix([[Diag[i,1,1],a,b,c,Diag[i,2,5]], [d,Diag[i,1,2],e,Diag[i,2,4],f], [g,h,Diag[i,1,3],l,m], [p,Diag[i,2,2],q,Diag[i,1,4],r], [Diag[i,2,1],u,v,w,Diag[i,1,5]]])]: fi:

od: od: od: od: od: od: od: od: od: od: od: od: od: od: od: od: od:

N;  # N -  the total number of squares,  List - the list of all squares

List[1], List[100], List[960];

A possible way :

with(LinearAlgebra):

M:=RandomMatrix(6);

G:=DiagonalMatrix(RandomVector(6));

M - DiagonalMatrix([seq(M[i,i], i=1..6)])+DiagonalMatrix([a$6]);

M - DiagonalMatrix([seq(M[i,i], i=1..6)])+G;

a:=2: b:=4: c:=7: d:=2:

if b>3 then z:=a:  y:=c elif b<3 then z:=d:  y:=a end if:

z;  y;

2

7

Add  numpoints  option:

plot3d(sin(sqrt(x^2+y^2))/sqrt(x^2+y^2), x=-3..3, y=-3..3, axes=normal, numpoints=3000);

Use

Optimization[Minimize]  command.

The same as in Mathematica (based on Markiyan's idea):

A := seq(plots[tubeplot]([x, y, x^2-y^2, y = -1 .. 1, radius = 0.02], scaling = constrained, color = grey, style = surface), x = -1 .. 1, .2):

B := seq(plots[tubeplot]([x, y, x^2-y^2, x = -1 .. 1, radius = 0.02], scaling = constrained, color = grey, style = surface), y = -1 .. 1, .2):

plots[display](A, B, orientation = [120, 60], lightmodel = light4);

It is easy to write a procedure that makes the same for any surface.

See this link  http://www.maplesoft.com/applications/view.aspx?SID=33406! The installation package includes detailed instructions for installation!

Example:

A:=Matrix([[1, 0.00009], [2, 1]]):

map(fnormal,A, 10, 1e-4);    

       [ 1   0. ]

       [ 2   1  ]

Of course, the constant a should not be an integer, otherwise the denominator would be 0. Unfortunately, Maple does not convert the sum of trigonometric functions in the product, so the denominator should be simplified by hand:

simplify(algsubs(1-cos(2*Pi*a)=2*sin(Pi*a)^2,-2*Pi*sin(Pi*a)/(-1+cos(2*Pi*a))));

                                                             Pi/sin(Pi*a)

The parameter  a  affects only the size of the curve and not its form. Therefore, for the plotting you can take a = 1.

To find the parametric equation of the curve make the substitution y=x*(tan(t))^3, substitute in the equation of the curve. First, express  x  through  t, then  y. Finally, you obtain

x=a*(cos(t))^3,  y=a*(sin(t))^3,   t=0..2*Pi

EM:=unapply([E01$a,E02$b,E03$c], a,b,c):

EM(1,3,2);

EM(2,2,1);

evalf(Int(sqrt(x^3+1), x=1..2));

2.129861293

I think I understand what you need. Use a double loop as follows:

L:=[0,1,2]:

M:=[]:

for i to nops(L) do

for j to nops(L) do

M:=[op(M), [L[i], L[j]]]:

od: od:

M;

If your polynomial is a quadratic form, then use the Sylvester criterion. For an arbitrary polynomial  the appropriate command in Maple is unknown to me.

Repetitive elements can easily remove from the list, making the set from the list.

L:=[0, 0, 1, 1, 2, 2]:

M:={op(L)}:

combinat[choose](M,2);

{{0,1},{0,2},{1,2}}