Kitonum

14034 Reputation

24 Badges

11 years, 195 days

MaplePrimes Activity


These are Posts that have been published by Kitonum

We assume that the radius of the outer stationary circle is  1. If we set the radius  x  of the inner stationary circle, all the other circles are uniquely determined by solving the system Sys.  Should be  x<=1/3 . If  x=1/3  then all the inner circles have a radius  1/3 . The following picture explains the meaning of symbols in the procedure Circles:

                                   

 

 

Circles:=proc(x)

local OO, O1, O2, O3, O4, O2x, O2y, O3x, O3y, OT, T1, T2, T3, s, t, dist, Sys, Sol, sol, y, u, v, z, C0, R0, P;

uses plottools, plots;

OO:=[0,0]: O1:=[x+y,0]: O2:=[O2x,O2y]: O3:=[O3x,O3y]: O4:=[-x-z,0]: OT:=[x+2*y-1,0]:

T1:=(O2*~y+O1*~u)/~(y+u): T2:=(O3*~u+O2*~v)/~(u+v): T3:=(O4*~v+O3*~z)/~(v+z):

solve({(T2-T1)[1]*(s-((T1+T2)/2)[1])+(T2-T1)[2]*(t-((T1+T2)/2)[2])=0, (T3-T2)[1]*(s-((T2+T3)/2)[1])+(T3-T2)[2]*(t-((T3+T2)/2)[2])=0}, {s,t}):

assign(%);

dist:=(A,B)->sqrt((B[1]-A[1])^2+(B[2]-A[2])^2):

Sys:={dist(O1,O2)^2=(y+u)^2, dist(OO,O2)^2=(x+u)^2, dist(O2,O3)^2=(u+v)^2, dist(OO,O3)^2=(x+v)^2, dist(O3,O4)^2=(z+v)^2, x+y+z=1, dist(O2,OT)^2=(1-u)^2, dist(O3,OT)^2=(1-v)^2};

Sol:=op~([allvalues([solve(Sys)])]);

sol:=select(i->is(eval(convert([y>0,u>0,v>0,z>0,O2y>0,x<=y,u<=y,v<=u,z<=v],`and`),i)), Sol)[];

assign(sol);

O1:=[x+y,0]: O2:=[O2x,O2y]: O3:=[O3x,O3y]: O4:=[-x-z,0]: OT:=[x+2*y-1,0]:

C0:=eval([s,t],sol);

R0:=eval(dist(T1,C0),sol):

P:=proc(phi)

local eq, r1, r, R, Ot, El, i, S, s, t, P1, P2;

uses plots,plottools;

eq:=1-dist([r*cos(s),r*sin(s)],OT)=r-x;

r1:=solve(eq,r);

r:=eval(r1,s=phi);

R[1]:=evalf(r-x);

Ot[1]:=evalf([r*cos(phi),r*sin(phi)]);

El:=plot([r1*cos(s),r1*sin(s),s=0..2*Pi],color="Green",thickness=3);

for i from 2 to 6 do

S:=[solve({1-dist(OT,[s,t])=dist(Ot[i-1],[s,t])-R[i-1], 1-dist(OT,[s,t])=dist(OO,[s,t])-x})];

P1:=eval([s,t],S[1]); P2:=eval([s,t],S[2]);

Ot[i]:=`if`(evalf(Ot[i-1][1]*P1[2]-Ot[i-1][2]*P1[1])>0,P1,P2);

R[i]:=dist(Ot[i],OO)-x;

od;

display(El,seq(disk(Ot[k],0.012),k=1..6),circle(C0,R0,color=gold,thickness=3),circle([x+2*y-1,0],1, color=blue,thickness=4), circle(OO,x, color=red,thickness=4), seq(circle(Ot[k],R[k], thickness=3),k=1..6), scaling=constrained, axes=none);

end proc:

animate(P,[phi], phi=0..Pi, frames=120);

end proc:  

 

Example of use (I got  x=0.22  just by measuring the ruler displayed original animation):

Circles(0.22);

                               

 

 

The curve on the following animation is an astroid (a special case of hypocycloid). See wiki for details. Hypocycloid procedure creates animation for any hypocycloid.  Parameters of the procedure: R is the radius of the outer circle, r is the radius of the inner circle.

Hypocycloid:=proc(R,r)

local A, B, f, g, F;

uses plots,plottools;

A:=circle(R,color=green,thickness=4):

B:=display(circle([R-r,0],r,color=red,thickness=4),line([R-r,0],[R,0],color=red,thickness=4)):

f:=t->plot([(R-r)*cos(s)+r*cos((R-r)/r*s),(R-r)*sin(s)-r*sin((R-r)/r*s),s=0..t],color=blue,thickness=4):

g:=t->rotate(rotate(B,-R/r*t,[R-r,0]),t):

F:=t->display(A,f(t),g(t),scaling=constrained):

animate(F,[t], t=0..2*Pi*denom(R/r), frames=90);

end proc:

 

Examples of use:

Hypocycloid(4,1); 

                                      

 

 

Hypocycloid(5,3);

                                      

 

 

 Круги.mw

IntegerPoints2  procedure generalizes  IntegerPoints1  procedure and finds all the integer points inside a bounded curved region of arbitrary dimension.  We also use a brute force method, but to find the ranges for each variable  Optimization[Minimize]  and   Optimization[Maximize]  is used instead of  simplex[minimize]  or  simplex[minimize] .

Required parameters of the procedure: SN is a set or a list of  inequalities and/or equations with any number of variables, the Var is the list of variables. Bound   is an optional parameter - list of ranges for each variable in the event, if  Optimization[Minimize/Maximize]  fails. By default  Bound  is NULL.

If all constraints are linear, then in this case it is recommended to use  IntegerPoints1  procedure, as it is better to monitor specific cases (no solutions or an infinite number of solutions for an unbounded region).

Code of the procedure:

IntegerPoints2 := proc (SN::{list, set}, Var::(list(symbol)), Bound::(list(range)) := NULL)

local SN1, sn, n, i, p, q, xl, xr, Xl, Xr, X, T, k, t, S;

uses Optimization, combinat;

n := nops(Var);

if Bound = NULL then

SN1 := SN;

for sn in SN1 do

if type(sn, `<`) then

SN1 := subs(sn = (`<=`(op(sn))), SN1) fi od;

for i to n do

p := Minimize(Var[i], SN1); q := Maximize(Var[i], SN1);

xl[i] := eval(Var[i], p[2]); xr[i] := eval(Var[i], q[2]) od else

assign(seq(xl[i] = lhs(Bound[i]), i = 1 .. n));

assign(seq(xr[i] = rhs(Bound[i]), i = 1 .. n)) fi;

Xl := map(floor, convert(xl, list)); Xr := map(ceil, convert(xr, list));

X := [seq([$ Xl[i] .. Xr[i]], i = 1 .. n)];

T := cartprod(X); S := table();

for k while not T[finished] do

t := T[nextvalue]();

if convert(eval(SN, zip(`=`, Var, t)), `and`) then

S[k] := t fi od;

convert(S, set);

end proc:

 

In the first example, we find all the integer points in the four-dimensional ball of radius 10:

Ball := IntegerPoints2({x1^2+x2^2+x3^2+x4^2 < 10^2}, [x1, x2, x3, x4]):  # All the integer points

nops(Ball);  # The total number of the integer points

seq(Ball[1000*n], n = 1 .. 10);  # Some points

                                                                    48945

                  [-8, 2, 0, -1], [-7, 0, 1, -3], [-6, -4, -6, 2], [-6, 1, 1, 1], [-5, -6, -2, 4], [-5, -1, 2, 0],

                                [-5, 4, -6, -2], [-4, -5, 1, 5], [-4, -1, 6, 1], [-4, 3, 5, 6]

 

 

In the second example, with the visualization we find all the integer points in the inside intersection of  a cone and a cylinder:

A := <1, 0, 0; 0, (1/2)*sqrt(3), -1/2; 0, 1/2, (1/2)*sqrt(3)>:  # Matrix of rotation around x-axis at Pi/6 radians

f := unapply(A^(-1) . <x, y, z-4>, x, y, z):  

S0 := {4*x^2+4*y^2 < z^2}:  # The inner of the cone

S1 := {x^2+z^2 < 4}:  # The inner of the cylinder

S2 := evalf(eval(S1, {x = f(x, y, z)[1], y = f(x, y, z)[2], z = f(x, y, z)[3]})):

S := IntegerPoints2(`union`(S0, S2), [x, y, z]);  # The integer points inside of the intersection of the cone and the rotated cylinder

Points := plots[pointplot3d](S, color = red, symbol = solidsphere, symbolsize = 8):

Sp := plot3d([r*cos(phi), r*sin(phi), 2*r], phi = 0 .. 2*Pi, r = 0 .. 5, style = surface, color = "LightBlue", transparency = 0.7):

F := plottools[transform]((x, y, z)->convert(A . <x, y, z>+<0, 0, 4>, list)):

S11 := plot3d([2*cos(t), y, 2*sin(t)], t = 0 .. 2*Pi, y = -4 .. 7, style = surface, color = "LightBlue", transparency = 0.7):

plots[display]([F(S11), Sp, Points], scaling = constrained, orientation = [25, 75], axes = normal);

      

 

 

In the third example, we are looking for the integer points in a non-convex area between two parabolas. Here we have to specify ourselves the ranges to enumeration (Optimization[Minimize] command fails for this example):

P := IntegerPoints2([y > (-x^2)*(1/2)+2, y < -x^2+8], [x, y], [-4 .. 4, -4 .. 8]);

A := plots[pointplot](P, color = red, symbol = solidcircle, symbolsize = 10):

B := plot([(-x^2)*(1/2)+2, -x^2+8], x = -4 .. 4, -5 .. 9, color = blue):

plots[display](A, B, scaling = constrained);

     

 

 IntegerPoints2.mw

 

This post is my attempt to answer the question from   here : how to find all integer points (all points with integer coordinates) in the intersection of two cubes. The following procedure  IntegerPoints  solves a more general problem: it finds all the integer points of a bounded polyhedral region of arbitrary dimension, defined by a system of linear inequalities and / or equations.

Required parameters of the procedure: SN is a set or a list of linear inequalities and/or equations with any number of variables, the Var is the list of variables. The procedure returns the set of all integer points, satisfying the conditions  SN .

Code of the procedure:

restart;

IntegerPoints := proc (SN::{list, set}, Var::list)

local SN1, sn, n, Sol, k, i, s, S, R;

uses PolyhedralSets, SolveTools[Inequality];

SN1 := convert(evalf(SN), fraction);

for sn in SN1 do

if type(sn, `<`) then SN1 := subs(sn = (`<=`(op(sn))), SN1)

end if; end do;

if IsBounded(PolyhedralSet(SN1)) = false then error "The region should be bounded" end if;

n := nops(Var);

Sol := LinearMultivariateSystem(SN, Var);

if Sol = {} then return {} else

k := 0;

for s in Sol do if nops(indets(s[1])) = 1 then

S[0] := [[]];

for i to n do

S[i] := [seq(seq([op(j1), op(j2)], j2 = [isolve(eval(s[i], j1))]), j1 = S[i-1])] end do;

k := k+1; R[k] := op(S[n]);

end if; end do;

convert(R, set);

map(t->rhs~(t), %);

end if;

end proc:

 

Examples of use:

IntegerPoints({x > 0, y > 0, z > 0, 2*x+3*y+z < 12}, [x, y, z]);

       

  {[1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 1, 4], [1, 1, 5], [1, 1, 6], [1, 2, 1], [1, 2, 2], [1, 2, 3], [2, 1, 1], [2, 1, 2],

                                   [2, 1, 3], [2, 1, 4], [2, 2, 1], [3, 1, 1], [3, 1, 2]}

 

IntegerPoints({x > 0, y > 0, z > 0, 2*x+3*y+z = 12}, [x, y, z]);

                                    {[1, 1, 7], [1, 2, 4], [1, 3, 1], [2, 1, 5], [2, 2, 2], [3, 1, 3], [4, 1, 1]}

 

IntegerPoints([x > 0, y > 0, z > 0, 2*x+3*y+z = 12, x+y+z <= 6], [x, y, z]);

                                                           {[1, 3, 1], [2, 2, 2], [4, 1, 1]}

isolve({x > 0, y > 0, z > 0, 2*x+3*y+z < 12});  #  isolve fails with these examples

              Warning, solutions may have been lost

isolve({x > 0, y > 0, z > 0, 2*x+3*y+z = 12});

              Warning, solutions may have been lost

 

In the following example (with a visualization) we find all integer point in the intersection of a square and a triangle:

S1 := {x > 0, y > 0, x < 13/2, y < 13/2}:

S2 := {y > (1/4)*x+1, y < 2*x, y+x < 12}:

S := IntegerPoints(`union`(S1, S2), [x, y]):

Region := plots[inequal](`union`(S1, S2), x = 0 .. 7, y = 0 .. 7, color = "LightGreen", nolines):

Points := plot([op(S)], style = point, color = red, symbol = solidcircle):

Square := plottools[curve]([[0, 0], [13/2, 0], [13/2, 13/2], [0, 13/2], [0, 0]], color = blue, thickness = 3):

Triangle := plottools[curve]([[4/7, 8/7], [4, 8], [44/5, 16/5], [4/7, 8/7]], color = blue, thickness = 3):

plots[display](Square, Triangle, Points, Region, scaling = constrained);

                                           

 

 

In the following example (with a visualization) we find all integer point in the intersection of two cubes. The second cube is obtained from the first cube by rotation with orthogonal matrix  A  and by a translation:

A := <1/3, 2/3, 2/3; -2/3, 2/3, -1/3; -2/3, -1/3, 2/3>:

f := unapply(A^(-1).<x+5, y-4, z-7>, x, y, z):

S1 := {x > 0, y > 0, z > 0, x < 6, y < 6, z < 6}:

S2 := eval(S1, {x = f(x, y, z)[1], y = f(x, y, z)[2], z = f(x, y, z)[3]}):

S := IntegerPoints(`union`(S1, S2), [x, y, z]);

Points := plots[pointplot3d](S, color = red, symbol = box):

Cube := plottools[cuboid]([0, 0, 0], [6, 6, 6], color = blue, linestyle = solid):

F := plottools[transform]((x, y, z)->convert(A.<x, y, z>+<-5, 4, 7>, list)):

plots[display](Cube,  F(Cube), Points, scaling = constrained, linestyle = solid, transparency = 0.7, orientation = [25, 75], axes = normal);

 

 

 

In the example below, all the ways to exchange $ 1 coins of 1, 5, 10, 25 and 50 cents, if the number of coins no more than 8, there is no pennies and there is at least one 50-cent coin:

IntegerPoints({x1 = 0, x2 >= 0, x3 >= 0, x4 >= 0, x5 >= 1,  x1+5*x2+10*x3+25*x4+50*x5 = 100, x1+x2+x3+x4+x5 <= 8}, [x1, x2, x3, x4, x5]);

nops(%);

                              {[0, 0, 0, 0, 2], [0, 0, 0, 2, 1], [0, 0, 5, 0, 1], [0, 1, 2, 1, 1], [0, 2, 4, 0, 1],

                                                 [0, 3, 1, 1, 1], [0, 4, 3, 0, 1], [0, 5, 0, 1, 1]}

                                                                                    8

 

Integer_points.mw

 

Addition: Below in my comments another procedure  IntegerPoints1  is presented that solves the same problem.

Consider the well-known Euler's formula  

 eix = cos x + i sin x   

When we calculate that for  x = π  we get:

eiπ = cos π + i sin π   or

eiπ = −1 + i × 0   (because cos π = −1 and sin π = 0)  or

eiπ = −1  or  eiπ + 1 = 0

It seems absolutely magical that such a neat equation combines  5  fundamental constants: e ,  i ,  π , 1 , 0

The purpose of this post - to give a simple visualization of equality  eiπ = −1  (statical and animated) by expanding  eiπ  in a series of complex numbers. These numbers we represent as vectors in the plane. We will see that the partial sums of this series are broken lines like a spiral, twisting around the point -1 steadily approaching to it.

Euler procedure has one required parameter  n is positive integer - the number of displayed terms of the series  for  eiπ  

Optional parameter  a  is any symbol (by default  a=NULL). We use this option if  instead of a static spiral want to see an animated spiral. 

Procedure code can be found in the attached file  Euler.mw

 

Examples of use.

The first example shows  8 terms of the series (broken line of 8 units):

Euler(8);

                

 

 

The terms of the series where  n> = 10  on the same plot can not be seen as very small. In this case, we use  the second plot with magnification of  100 : 1 .  

The second example:

Euler(14);

 

 

 

In the third example, we see an animated broken line. It's  first 9 units represented  on the left plot, and then for n> = 10 on the right plot:

Euler(13, a);

  

 

Euler.mw

This post is my attempt to answer the question from here .  

The procedure  ContoursWithLabels  has 2 required parameters: Expr  is an expression in  x  and  y  variables,  Range1  and  Range2  are ranges for  x  and  y . In this case, the output is the list of floats for the contours and 8 black contours (with labels) (the axis of coordinates as a box). 

The optional parameters: Number is positive integer - the number of contours (by default Number=8),  S is a set of real numbers  C  for contours (for which Expr=C) (by default  S={}),  GraphicOptions  is a list of graphic options for plotting (by default  GraphicOptions=[color = black, axes = box]),  Coloring  is an equality  Coloring=list of color options for  plots[dencityplot]  command (by default Coloring=NULL). 

The code of the procedure:

restart;

ContoursWithLabels := proc (Expr, Range1::(range(realcons)), Range2::(range(realcons)), Number::posint := 8, S::(set(realcons)) := {}, GraphicOptions::list := [color = black, axes = box], Coloring::`=` := NULL)

local r1, r2, L, f, L1, h, S1, P, P1, r, M, C, T, p, p1, m, n, A, B, E;

uses plots, plottools;

f := unapply(Expr, x, y);

if S = {} then r1 := rand(convert(Range1, float)); r2 := rand(convert(Range2, float));

L := [seq([r1(), r2()], i = 1 .. 205)];

L1 := convert(sort(select(a->type(a, realcons), [seq(f(op(t)), t = L)]), (a, b) ->is(abs(a) < abs(b))), set);

h := (L1[-6]-L1[1])/Number;

S1 := [seq(L1[1]+(1/2)*h+h*(n-1), n = 1 .. Number)] else

S1 := convert(S, list)  fi;

print(Contours = evalf[2](S1));

r := k->rand(20 .. k-20); M := []; T := [];

for C in S1 do

P := implicitplot(Expr = C, x = Range1, y = Range2, op(GraphicOptions), gridrefine = 3);

P1 := [getdata(P)];

for p in P1 do

p1 := convert(p[3], listlist); n := nops(p1);

if n < 500 then m := `if`(40 < n, (r(n))(), round((1/2)*n)); M := `if`(40 < n, [op(M), p1[1 .. m-11], p1[m+11 .. n]], [op(M), p1]); T := [op(T), [op(p1[m]), evalf[2](C)]] else

if 500 <= n then h := floor((1/2)*n); m := (r(h))(); M := [op(M), p1[1 .. m-11], p1[m+11 .. m+h-11], p1[m+h+11 .. n]]; T := [op(T), [op(p1[m]), evalf[2](C)], [op(p1[m+h]), evalf[2](C)]]

fi; fi; od; od;

A := plot(M, op(GraphicOptions));

B := plots:-textplot(T);

if Coloring = NULL then E := NULL else E := ([plots:-densityplot])(Expr, x = Range1, y = Range2, op(rhs(Coloring)))  fi;

display(E, A, B);

end proc:

 

Examples of use:

ContoursWithLabels(x^2+y^2, -3 .. 3, -3 .. 3);

                             

 

 

ContoursWithLabels(x^2-y^2, -5 .. 5, -5 .. 5, {-20, -15, -10, -5, 0, 5, 10, 15, 20}, [color = black, thickness = 2, axes = box], Coloring = [colorstyle = HUE, colorscheme = ["White", "Red"], style = surface]);

                           

 

 

The next example, I took from here:

ContoursWithLabels(sin(1.3*x)*cos(.9*y)+cos(.8*x)*sin(1.9*y)+cos(.2*x*y), -5 .. 0, 2 .. 5, {seq(-2 .. 2, 0.5)}, [color = black, axes = box], Coloring = [colorstyle = HUE, colorscheme = ["Cyan", "Red"], style = surface]);

                                 

 

There are many more examples can be found in the attached file. 

ContoursWithLabels1.mw

 

Edit. The attached file has been corrected.

2 3 4 5 6 7 8 Page 4 of 10