MaplePrimes Questions

Hi,
I'm surprised by the error Maple returns when asked to find the JordanBlockMatrix or the JordanForm of a matrix whose some entries are floating points.
I don't understand why these operations are valid only with entries of the form a+I*b with a and b are algebraic numbers or with transcendent numbers such as Pi.
Is this a theoretical result or a technical limitation of Maple?
Jordan.mw

 

Download Jordan.mw

 

As the title states, I want to have an equation f(x), and f(x) = 0 if x < 0, f(x) = x if x >= 0. How could I accomplish this?

I'm actually trying to generating a differential equation something like y'(x) + k*h(y) = sin(x) where h(y) is what I described above. Is there any convenient way to do this?

I want to construct a list/set of square symmetric matrices (or Toeplitz etc.) matrices with entries from a certain finite field, GF(2) or GF(5).  

Thank you in adavnce.

Here is a division-by-zero bug in a solution produced by pdsolve.  Admittedly, this sort of problem can be difficult to avoid in a CAS, but here it is, in case there is a chance to get it fixed somehow.

restart;

kernelopts(version);

`Maple 2020.1, X86 64 LINUX, Jul 30 2020, Build ID 1482634`

pde := diff(u(x,y,t),t,t) = diff(u(x,y,t),x,x) + diff(u(x,y,t),y,y);

diff(diff(u(x, y, t), t), t) = diff(diff(u(x, y, t), x), x)+diff(diff(u(x, y, t), y), y)

bc := u(x,0,t)=0, u(x,1,t)=0, u(0,y,t)=0, u(1,y,t)=0;

u(x, 0, t) = 0, u(x, 1, t) = 0, u(0, y, t) = 0, u(1, y, t) = 0

ic := u(x,y,0) = x*y*sin(Pi*x)*sin(Pi*y),  D[3](u)(x,y,0)=0;

u(x, y, 0) = x*y*sin(Pi*x)*sin(Pi*y), (D[3](u))(x, y, 0) = 0

pdsol := pdsolve({pde, bc, ic});

"pdsol:=u(x,y,t)=(&sum;) (&sum;)-(2 sin(n Pi x) sin(n1 Pi y) cos(Pi sqrt(n^2+n1^2) t) ({[[Pi^2,n=1],[-(8 n ((-1)^n+1))/((n-1)^2 (n+1)^2),1<n]]) n1 ((-1)^n1+1))/(Pi^4 (-1+n1)^2 (n1+1)^2)"

eval(pdsol, infinity=4);
value(%);

"u(x,y,t)=(&sum;) (&sum;)-(2 sin(n Pi x) sin(n1 Pi y) cos(Pi sqrt(n^2+n1^2) t) ({[[Pi^2,n=1],[-(8 n ((-1)^n+1))/((n-1)^2 (n+1)^2),1<n]]) n1 ((-1)^n1+1))/(Pi^4 (-1+n1)^2 (n1+1)^2)"

Error, (in SumTools:-DefiniteSum:-ClosedForm) summand is singular in the interval of summation

 

 

 

Download pdsolve-bug.mw

 

Since I am a mathematician, I am wondering how Maple goes about solving an identity for 3 functions.
Let's say we have af1(t)+bf_2(t)+cf_3(t) = 0 for all t. How does maple actually find a triplet a,b,c that works for all real t?
It does with solve(identity( ),[a,b,c]). But what is the theory behind it?
We know, of course, a priori, that such a triplet exists.

Thank you!

mapleatha

 



 

I was saying to myself today, that one nice thing in Maple, is that it has good type system so one can check type of arguments aginst wrong types being used in the call.

So I was surprised when I called ArrayTools:-AllNonZero  using list as argument, where this function is supposed to accept only Matrix,Vector or Array. And it worked.  But it gave wrong answer at the same time.

Am I doing something wrong here? Should not have this call failed to go through?

restart;
interface(warnlevel=4);
kernelopts('assertlevel'=2):

A:=[0,0,0];
ArrayTools:-AllNonZero(A)

Maple replied 

                       true

But 

A:=Array([0,0,0]);
ArrayTools:-AllNonZero(A)

Now Maple gives the expected result  false

The question is why Maple did not detect the wrong type? Should it have detected wrong type?

And why it even gave wrong answer (but this is not as important, since the call should not be allowed in first place).

btw, this applied to all the other API's listed

 

 

I found a new strange thing with odetest that I have not seen before.

Would you say that these two solutions are the same or not?

They look the same to me.  One just have all the terms on one side, that is all.

Why would Maple odetest verify the first one but not the second? I did not know that all terms have to be on one side before. Is this documented somewhere?

Please see worksheet below.

restart;

interface(version);

`Standard Worksheet Interface, Maple 2020.1, Windows 10, July 30 2020 Build ID 1482634`

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 847 and is the same as the version installed in this computer, created 2020, October 17, 17:3 hours Pacific Time.`

ode:=diff(y(x), x) - 2*y(x) = 2*sqrt(y(x));
ic:=[y(0)=1];
maple_sol:=dsolve([ode,op(ic)],implicit);
odetest(maple_sol,[ode,op(ic)]) assuming x>0

diff(y(x), x)-2*y(x) = 2*y(x)^(1/2)

[y(0) = 1]

1-2*exp(x)+y(x)^(1/2) = 0

[0, 0]

#now move some terms to one side, and odetest no longer verifies it
my_sol:=y(x)^(1/2) = -1+2*exp(x);
odetest(my_sol,[ode,op(ic)]) assuming x>0

y(x)^(1/2) = -1+2*exp(x)

[diff(y(x), x)-2*y(x)+2-4*exp(x), 0]

#the solution is to put all terms on one side
my_sol:=y(x)^(1/2) = -1+2*exp(x);
odetest(rhs(my_sol)-lhs(my_sol)=0,[ode,op(ic)]) assuming x>0

y(x)^(1/2) = -1+2*exp(x)

[0, 0]

 


 

Download odetest_issue_oct_18_2020.mw

 

How do we use on the axes Pi, Pi/2, ...,  in Maple 13 plots without evaluating Pi? When I just put enter Pi, Maple gives me 3.1415... . "piticks" is not available for Maple 13. I do have access to the university ocmputer's Maple 18, but, recently, it has decided that Citrix iwill not serve Windows 7.

Thank you.

mapleatha

Hello, 

im looking to see if there is a way to read in RGB values of the pixels in images and put those values into matrices.

any help or suggestions are appreciated.

Hi,

I'm new to Maple and was wondering if anyone could help me with how to put a discrete distribution such as Pr(X=1)=0.25, Pr(X=2)=0.65, Pr(X=3)=0.1, into Maple.

Thanks.


Dear Colleagues,

Apologies for the generic question below.

I am trying to obtain the Nash equilibrium solutions for a two-person game. I am not sure of any in-built packages that can help me in obtaining the solutions computationally. The algorithms that I created do not seem to give good solutions that are meaningful in my application. Any suggestion would be much appreciated. 

Regards,

Omkar

 

 

Hi guys

I want to solve the following non-linear differential equation but by using dsolve(), the computer cannot solve it, so please guide me.

Q:=2*diff(a(t), t, t)*a(t)^3 - 3*diff(a(t), t)^4 + diff(a(t), t)^2*a(t)^2

with the best regards

i have two nonlinear functions from which one is exicuted properly but 2nd funnction not risponde properly. i can not understan how to overcome it
 

restart

with(LinearAlgebra):

F := proc (x) options operator, arrow; x^2 end proc:

F(u(t));

u(t)^2

(1)

G := proc (w) options operator, arrow; w*dw/dt end proc:

G(h(t));

h(t)*dw/dt

(2)

for n from 0 while n <= 6 do V[n] := (diff(F(sum(t^i*u[i], i = 0 .. n)), [`$`(t, n)]))/factorial(n); U[n] := (diff(G(sum(t^i*h[i], i = 0 .. n)), [`$`(t, n)]))/factorial(n) end do:

t := 0;

0

(3)

for i from 0 while i <= n-1 do A[i] := V[i]; B[i] := U[i] end do;

u[0]^2

 

h[0]*dw/dt

 

2*u[0]*u[1]

 

h[1]*dw/dt

 

2*u[0]*u[2]+u[1]^2

 

h[2]*dw/dt

 

2*u[0]*u[3]+2*u[1]*u[2]

 

h[3]*dw/dt

 

2*u[0]*u[4]+2*u[1]*u[3]+u[2]^2

 

h[4]*dw/dt

 

2*u[0]*u[5]+2*u[1]*u[4]+2*u[2]*u[3]

 

h[5]*dw/dt

 

2*u[0]*u[6]+2*u[1]*u[5]+2*u[2]*u[4]+u[3]^2

 

h[6]*dw/dt

(4)

for j from 0 while j <= n-1 do u[0] := 1; u[j+1] := int(x*B[j], x)+int(A[j], x) end do;

1

 

(1/2)*x^2*h[0]*dw/dt+x

 

1

 

(1/2)*x^2*h[1]*dw/dt+(1/3)*x^3*h[0]*dw/dt+x^2

 

1

 

(1/2)*x^2*h[2]*dw/dt+(1/3)*x^3*h[1]*dw/dt+(5/12)*x^4*h[0]*dw/dt+x^3+(1/20)*h[0]^2*dw^2*x^5/dt^2

 

1

 

(1/2)*x^2*h[3]*dw/dt+(1/3)*x^3*h[2]*dw/dt+(1/6)*x^4*h[1]*dw/dt+(1/6)*x^5*h[0]*dw/dt+(1/2)*x^4+(13/180)*h[0]^2*dw^2*x^6/dt^2+(2/5)*((1/3)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)*x^5+(1/2)*((1/2)*h[1]*dw/dt+1)*x^4

 

1

 

(1/2)*x^2*h[4]*dw/dt+(1/3)*x^3*h[3]*dw/dt+(5/12)*x^4*h[2]*dw/dt+(1/15)*x^5*h[1]*dw/dt+(1/18)*x^6*h[0]*dw/dt+(1/5)*x^5+(139/1260)*h[0]^2*dw^2*x^7/dt^2+(2/15)*((1/3)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)*x^6+(1/5)*((1/2)*h[1]*dw/dt+1)*x^5+(1/160)*h[0]^3*dw^3*x^8/dt^3+(1/3)*((5/12)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^6+(2/5)*((1/3)*h[1]*dw/dt+1+(1/4)*h[0]*dw^2*h[2]/dt^2)*x^5+(1/9)*h[0]*dw*((1/2)*h[1]*dw/dt+1)*x^6/dt+(1/5)*((1/2)*h[1]*dw/dt+1)^2*x^5

 

1

 

(1/2)*x^2*h[5]*dw/dt+(1/3)*x^3*h[4]*dw/dt+(5/12)*x^4*h[3]*dw/dt+(1/6)*x^5*h[2]*dw/dt+(1/45)*x^6*h[1]*dw/dt+(1/63)*x^7*h[0]*dw/dt+(139/5040)*h[0]^2*dw^2*x^8/dt^2+(17/1296)*h[0]^3*dw^3*x^9/dt^3+(1/15)*x^6+(4/105)*((1/3)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)*x^7+(1/15)*((1/2)*h[1]*dw/dt+1)*x^6+(2/21)*((5/12)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^7+(2/15)*((1/3)*h[1]*dw/dt+1+(1/4)*h[0]*dw^2*h[2]/dt^2)*x^6+(1/15)*((1/2)*h[1]*dw/dt+1)^2*x^6+(1/4)*((13/180)*h[0]^2*dw^2/dt^2+(1/2)*h[0]*dw*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)/dt)*x^8+(2/7)*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/2)*h[0]*dw*((5/12)*h[1]*dw/dt+1)/dt)*x^7+(1/3)*((5/12)*h[1]*dw/dt+1+(1/6)*h[0]*dw^2*h[2]/dt^2)*x^6+(2/5)*((1/3)*h[2]*dw/dt+(1/4)*h[0]*dw^2*h[3]/dt^2)*x^5+(1/4)*((1/20)*((1/2)*h[1]*dw/dt+1)*h[0]^2*dw^2/dt^2+(5/36)*h[0]^2*dw^2/dt^2)*x^8+(2/7)*((5/12)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/3)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^7+(1/3)*(((1/2)*h[1]*dw/dt+1)*((1/3)*h[1]*dw/dt+1)+(1/6)*h[0]*dw^2*h[2]/dt^2)*x^6+(1/5)*((1/2)*h[1]*dw/dt+1)*h[2]*dw*x^5/dt+(2/63)*h[0]*dw*((1/2)*h[1]*dw/dt+1)*x^7/dt

 

1

 

(1/20)*h[2]^2*dw^2*x^5/dt^2+(1/3)*x^3*h[5]*dw/dt+(5/12)*x^4*h[4]*dw/dt+(1/6)*x^5*h[3]*dw/dt+(1/18)*x^6*h[2]*dw/dt+(2/315)*x^7*h[1]*dw/dt+(1/252)*x^8*h[0]*dw/dt+(139/22680)*h[0]^2*dw^2*x^9/dt^2+(2167/90720)*h[0]^3*dw^3*x^10/dt^3+(7/8800)*h[0]^4*dw^4*x^11/dt^4+(1/2)*x^2*h[6]*dw/dt+(2/105)*x^7+(2/9)*((13/180)*((1/2)*h[1]*dw/dt+1)*h[0]^2*dw^2/dt^2+(1/3)*h[0]*dw*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)/dt)*x^9+(1/4)*(((1/2)*h[1]*dw/dt+1)*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)+(1/3)*h[0]*dw*((5/12)*h[1]*dw/dt+1)/dt)*x^8+(2/7)*(((1/2)*h[1]*dw/dt+1)*((5/12)*h[1]*dw/dt+1)+(1/9)*h[0]*dw^2*h[2]/dt^2)*x^7+(1/3)*((1/3)*((1/2)*h[1]*dw/dt+1)*h[2]*dw/dt+(1/6)*h[0]*dw^2*h[3]/dt^2)*x^6+(1/105)*((1/3)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)*x^8+(2/105)*((1/2)*h[1]*dw/dt+1)*x^7+(1/42)*((5/12)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^8+(4/105)*((1/3)*h[1]*dw/dt+1+(1/4)*h[0]*dw^2*h[2]/dt^2)*x^7+(2/105)*((1/2)*h[1]*dw/dt+1)^2*x^7+(1/18)*((13/180)*h[0]^2*dw^2/dt^2+(1/2)*h[0]*dw*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)/dt)*x^9+(1/14)*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/2)*h[0]*dw*((5/12)*h[1]*dw/dt+1)/dt)*x^8+(2/21)*((5/12)*h[1]*dw/dt+1+(1/6)*h[0]*dw^2*h[2]/dt^2)*x^7+(2/15)*((1/3)*h[2]*dw/dt+(1/4)*h[0]*dw^2*h[3]/dt^2)*x^6+(1/18)*((1/20)*((1/2)*h[1]*dw/dt+1)*h[0]^2*dw^2/dt^2+(5/36)*h[0]^2*dw^2/dt^2)*x^9+(1/14)*((5/12)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/3)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^8+(2/21)*(((1/2)*h[1]*dw/dt+1)*((1/3)*h[1]*dw/dt+1)+(1/6)*h[0]*dw^2*h[2]/dt^2)*x^7+(1/9)*((1/10)*((1/3)*h[1]*dw/dt+1)*h[0]^2*dw^2/dt^2+(25/144)*h[0]^2*dw^2/dt^2)*x^9+(1/8)*((1/20)*h[2]*dw^3*h[0]^2/dt^3+(5/6)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^8+(1/7)*((5/12)*h[0]*dw^2*h[2]/dt^2+((1/3)*h[1]*dw/dt+1)^2)*x^7+(2/9)*((139/1260)*h[0]^2*dw^2/dt^2+(1/2)*h[0]*dw*((43/180)*h[0]*dw/dt+(8/45)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/6)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)/dt)*x^9+(1/4)*((43/180)*h[0]*dw/dt+(8/45)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/6)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt+(1/2)*h[0]*dw*((3/10)*h[1]*dw/dt+4/5+(1/10)*h[0]*dw^2*h[2]/dt^2+(1/5)*((1/2)*h[1]*dw/dt+1)^2)/dt)*x^8+(2/7)*((3/10)*h[1]*dw/dt+4/5+(37/120)*h[0]*dw^2*h[2]/dt^2+(1/5)*((1/2)*h[1]*dw/dt+1)^2)*x^7+(1/3)*((5/12)*h[2]*dw/dt+(1/6)*h[0]*dw^2*h[3]/dt^2)*x^6+(2/5)*((1/3)*h[3]*dw/dt+(1/4)*h[0]*dw^2*h[4]/dt^2)*x^5+(1/5)*((1/2)*h[1]*dw/dt+1)*h[3]*dw*x^5/dt+(1/15)*((1/2)*h[1]*dw/dt+1)*h[2]*dw*x^6/dt+(1/126)*h[0]*dw*((1/2)*h[1]*dw/dt+1)*x^8/dt+(1/6)*h[2]*dw*((1/3)*h[1]*dw/dt+1)*x^6/dt

(5)

y := sum(u[l], l = 0 .. n-1);

1+(1/2)*x^4+(1/5)*x^5+(1/2)*x^2*h[0]*dw/dt+(1/2)*x^2*h[1]*dw/dt+(1/3)*x^3*h[0]*dw/dt+(1/2)*x^2*h[2]*dw/dt+(1/3)*x^3*h[1]*dw/dt+(5/12)*x^4*h[0]*dw/dt+(1/20)*h[0]^2*dw^2*x^5/dt^2+(1/2)*x^2*h[3]*dw/dt+(1/3)*x^3*h[2]*dw/dt+(1/6)*x^4*h[1]*dw/dt+(1/6)*x^5*h[0]*dw/dt+(13/180)*h[0]^2*dw^2*x^6/dt^2+(1/2)*x^2*h[4]*dw/dt+(1/3)*x^3*h[3]*dw/dt+(5/12)*x^4*h[2]*dw/dt+(1/15)*x^5*h[1]*dw/dt+(1/18)*x^6*h[0]*dw/dt+(139/1260)*h[0]^2*dw^2*x^7/dt^2+(1/160)*h[0]^3*dw^3*x^8/dt^3+(1/2)*x^2*h[5]*dw/dt+(1/3)*x^3*h[4]*dw/dt+(5/12)*x^4*h[3]*dw/dt+(1/6)*x^5*h[2]*dw/dt+(1/45)*x^6*h[1]*dw/dt+(1/63)*x^7*h[0]*dw/dt+(139/5040)*h[0]^2*dw^2*x^8/dt^2+(17/1296)*h[0]^3*dw^3*x^9/dt^3+x+(1/15)*x^6+x^3+(2/15)*((1/3)*h[1]*dw/dt+1+(1/4)*h[0]*dw^2*h[2]/dt^2)*x^6+(1/15)*((1/2)*h[1]*dw/dt+1)^2*x^6+(1/4)*((13/180)*h[0]^2*dw^2/dt^2+(1/2)*h[0]*dw*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)/dt)*x^8+(2/7)*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/2)*h[0]*dw*((5/12)*h[1]*dw/dt+1)/dt)*x^7+(1/3)*((5/12)*h[1]*dw/dt+1+(1/6)*h[0]*dw^2*h[2]/dt^2)*x^6+(2/5)*((1/3)*h[2]*dw/dt+(1/4)*h[0]*dw^2*h[3]/dt^2)*x^5+(1/4)*((1/20)*((1/2)*h[1]*dw/dt+1)*h[0]^2*dw^2/dt^2+(5/36)*h[0]^2*dw^2/dt^2)*x^8+(2/7)*((5/12)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/3)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^7+(1/3)*(((1/2)*h[1]*dw/dt+1)*((1/3)*h[1]*dw/dt+1)+(1/6)*h[0]*dw^2*h[2]/dt^2)*x^6+(4/105)*((1/3)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)*x^7+(1/15)*((1/2)*h[1]*dw/dt+1)*x^6+(2/21)*((5/12)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^7+(2/15)*((1/3)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)*x^6+(1/5)*((1/2)*h[1]*dw/dt+1)*x^5+(1/3)*((5/12)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^6+(2/5)*((1/3)*h[1]*dw/dt+1+(1/4)*h[0]*dw^2*h[2]/dt^2)*x^5+(1/5)*((1/2)*h[1]*dw/dt+1)^2*x^5+(2/5)*((1/3)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)*x^5+(1/2)*((1/2)*h[1]*dw/dt+1)*x^4+(1/5)*((1/2)*h[1]*dw/dt+1)*h[2]*dw*x^5/dt+(2/63)*h[0]*dw*((1/2)*h[1]*dw/dt+1)*x^7/dt+(1/9)*h[0]*dw*((1/2)*h[1]*dw/dt+1)*x^6/dt+x^2

(6)

``

NULL


 

Download incomplete_example.mw

I've been studying the  drawing  of graph lately .    One of the themes is  1-planar graph .

A 1-planar graph is a graph that can be drawn in the Euclidean plane in such a way that each edge has at most one crossing point,  where it crosses a single additional edge. If a 1-planar graph, one of the most natural generalizations of planar graphs, is drawn that way, the drawing is called a 1-plane graph or 1-planar embedding of the graph.

 

 

 

 

 

I know it is NP hard to determine whether a graph is a 1-planar . My idea is to take advantage of some mathematical software to provide some roughly and  intuitive understanding before determining .

Now,  the layout of vertices or edges becomes important.  The drawing of a plane graph is a good example.

G1:=AddEdge( CycleGraph([v__1,v__2,v__3,v__4]),{{v__2,v__4},{v__1,v__3}}):
DrawGraph(G1)
DrawGraph(G1,style=planar)

K5 := CompleteGraph(5);
DrawGraph(K5);
vp:=[[-1,0],[1,0],[-0.2,0.5],[0.2,0.5],[0,1]];
SetVertexPositions(K5,vp);  #modified the vertex position

DrawGraph(K5);

My problem is that I see that  Maple2020 has updated a lot of layouts about DrawGraph  graph theory backpack , and I don’t know which ones are working towards the least possible number of crossing of  each edges of graph . 

Some links that may be useful:

https://de.maplesoft.com/products/maple/new_features/Maple2020/graphtheory.aspx

https://de.maplesoft.com/support/help/Maple/view.aspx?path=GraphTheory/SetVertexPositions

I think the software can improve some calculations related to topological graph theory, such as crossing number of graph, etc.

 

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