The graph we draw in the graph theory package appear to be vectorial. So the export pdf is also clear. 

with(GraphTheory):
DrawGraph(PathGraph(2), stylesheet=[vertexborder=false,
vertexpadding=5,edgecolor = blue,vertexcolor=black,edgethickness=2],size=[250,250])

draw1.pdf

We know that DrawGraph only offers straight forms for drawn edges.  Sometimes we need to draw graphs with curves.  But when I try to use maple, I find that the vertices (by pointplot) are blurred after zooming in (or even without zooming out).

with(plottools):    
with(plots):
Bezier:=proc(x0,y0,x1,y1,x2,y2,x3,y3)
local f,g,c,l1,l2,l3,p0,p1,p2,p3;
f:=t->x0*(1-t)^3+3*x1*t*(1-t)^2+3*x2*t^2*(1-t)+x3*t^3:
g:=t->y0*(1-t)^3+3*y1*t*(1-t)^2+3*y2*t^2*(1-t)+y3*t^3:
c:=plot([f(t),g(t),t=0..1],thickness=2,scaling=constrained, color=blue,axes=none):
p0:=pointplot([x0,y0],symbol= solidcircle,symbolsize = 50,color=black):
p1:=pointplot([x3,y3],symbol= solidcircle,symbolsize = 50,color=black):
display(c,p0,p1,size=[300,300]);
end:

x0:=0:   y0:=0:
x1:=-0.2:  y1:=0.2:
x2:=0.5:  y2:=0.2:
x3:=0.3:  y3:=0:
p:=Bezier(x0,y0,x1,y1,x2,y2,x3,y3)

 draw2.pdf 

How to make these vertices vectorially drawn (like DrawGraph)？

In logic and computer science, the Boolean satisfiability problem (sometimes called propositional satisfiability problem and abbreviated SAT) is the problem of determining if there exists an interpretation that satisfies a given Boolean formula.

We can see that some graph problems such as the coloring problem and the problem of finding k-cliques can be transformed into SAT problem.  I am not familiar with how maple uses the sat solver. I am trying to use maple to solve the following game for a solution. The game is a good choice for understanding the SAT.

• http://www.cril.univ-artois.fr/~roussel/satgame/satgame.php?level=1&lang=eng

The game is won when at least one cell on each line is green. 

Clicking on a number will color each cell with the same number in green, and each cell with the opposite number in red. 

For example:

If we choose ``1" as green(i.e. be chosen), then -1 is red (not be chosen) in any cell.  I guss that the maple function Satisfy can do it. 

That is, we are looking for a set of solutions such that at least one number in any line is selected as green. 

Here is a solution for winning this game:

How to use the SAT solver Satisfy in maple to find a solution from a sat game?

I open the command-line of maple 2022, and I can run the code alone. But when I save it in a text named ``6conn.txt", I don't know how to run it.

with(GraphTheory):
g:=Graph({{0 ,1},{3 ,4},{4 ,1},{1 ,3},{3 ,0},{0 ,4},
{2 ,1},{4 ,5},{4 ,2},{3 ,6},{4 ,6},{6 ,5},{1 ,6},{6 ,2},
{5 ,0},{0 ,2},{2 ,5},{7 ,8},{12 ,10},{10 ,11},{11 ,8},{8 ,10},
{10 ,7},{7 ,11},{9 ,8},{11 ,12},{11 ,9},{10 ,13},
{11 ,13},{13 ,12},{8 ,13},{13 ,9},{12 ,7},{7 ,9},{9 ,12},
{8 ,2},{13 ,0},{10 ,5},{12 ,3},{1 ,9},{7 ,6}
});

I tried  .\E:\\6conn.txt or !E:\\6conn.txt. Neither seems to work.

6conn.txt

It is easy to simplify the following expression (to 4), but maple's ceil function does not seem to be interested in simplifying it.

where n is greater than or equal to 4. 

simplify(ceil((3*n-8)/(n-3))) assuming n>=4, n::positive # As-is output

We have to rewrite "(3*n-8)/(n-3)" in this form "3 + 1/(n - 3)" to recognize.

simplify(ceil(3 + 1/(n - 3))) assuming  n>=4, n::positive

4

• My first question is: How to transform (3*n-8)/(n-3) into 3 + 1/(n - 3) by maple?
• My second question: Can we see the steps of execution of the simplification involving  ceil)?

I have an expression and I want to find its maximum value.

expr:=sin(sqrt(3)*t)*cos(sqrt(3)*t)*(sqrt(3)*cos(sqrt(3)*t) - sin(sqrt(3)*t))/3

It is easy to find its maximum value in a numerical form.

Optimization:-Maximize(sin(sqrt(3)*t)*cos(sqrt(3)*t)*(sqrt(3)*cos(sqrt(3)*t) - sin(sqrt(3)*t))/3)

[0.324263244248023330, [t = 1.39084587050767]]

The images of the expression is as follows.

But does it exist an acceptable maximum value in symbolic form?  As the function maximize seems to take a lot of time, I don't see any hope so far. Perhaps the expression is indeed complex.

maximize(expr)# it runs long time.

We try to find the derivative of expr and get some points where thier derivatives are 0.

s:=[solve(diff(expr,t),t)]
evalf~(s) # Some solutions seem to have been left out.

[0.7607468963 + 0.*I, -0.4229534936 - 0.*I, 0.2668063857 + 0.*I]

ex:=convert(expr,exp):
s:=[solve(diff(ex,t)=0,t)]:
s1:=evalf~(s);# choose the 6th item: 1.390845877 + (4.655829150*10^(-9))*I
fexpr := unapply(ex, t);
evalf(fexpr(s1[6]));
fexpr (s[6]); # a very long expression that is not quite acceptable.

-I/6*(-sqrt(-(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)*((1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3)*sqrt(3)*I - 2*I*sqrt(3)*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) + (1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3) + 36*I*sqrt(3) + 2*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) - 44))/(6*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)) + 6*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)/sqrt(-(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)*((1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3)*sqrt(3)*I - 2*I*sqrt(3)*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) + (1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3) + 36*I*sqrt(3) + 2*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) - 44)))*(-sqrt(-(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)*((1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3)*sqrt(3)*I - 2*I*sqrt(3)*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) + (1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3) + 36*I*sqrt(3) + 2*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) - 44))/(12*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)) - 3*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)/sqrt(-(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)*((1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3)*sqrt(3)*I - 2*I*sqrt(3)*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) + (1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3) + 36*I*sqrt(3) + 2*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) - 44)))*(sqrt(3)*(-sqrt(-(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)*((1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3)*sqrt(3)*I - 2*I*sqrt(3)*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) + (1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3) + 36*I*sqrt(3) + 2*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) - 44))/(12*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)) - 3*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)/sqrt(-(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)*((1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3)*sqrt(3)*I - 2*I*sqrt(3)*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) + (1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3) + 36*I*sqrt(3) + 2*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) - 44))) + (-sqrt(-(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)*((1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3)*sqrt(3)*I - 2*I*sqrt(3)*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) + (1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3) + 36*I*sqrt(3) + 2*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) - 44))/(6*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)) + 6*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)/sqrt(-(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3)*((1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3)*sqrt(3)*I - 2*I*sqrt(3)*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) + (1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(2/3) + 36*I*sqrt(3) + 2*(1404*I*sqrt(3) - 1396 + 36*sqrt(-3018 - 3018*I*sqrt(3)))^(1/3) - 44)))*I/2)

An interesting problem is that an acceptably concise expression (although it is very subjective) for the maximum value may not exist mathematically. But knowing this in advance is difficult.

Download compute_zlc.mw