Posts - MaplePrimes

Numbrix Puzzle by the branch and bound method

by: Kitonum 21435 Product: Maple 2018

July 28 2019

3 3

In this post, the Numbrix Puzzle is solved by the branch and bound method (see the details of this puzzle in https://www.mapleprimes.com/posts/210643-Solving-A-Numbrix-Puzzle-With-Logic). The main difference from the solution using the Logic package is that here we get not one but all possible solutions. In the case of a unique solution, the NumbrixPuzzle procedure is faster than the Numbrix one (for convenience, I inserted the code for Numbrix procedure into the worksheet below). In the case of many solutions, the Numbrix procedure is usually faster (see all the examples below).

>

restart;

>

NumbrixPuzzle:=proc(A::Matrix)
local A1, L, N, S, MS, OneStepLeft, OneStepRight, F1, F2, m, L1, p, q, a, b, T, k, s1, s, H, n, L2, i, j, i1, j1, R;
uses ListTools;
S:=upperbound(A); N:=nops(op(A)[3]); MS:=`*`(S);
A1:=convert(A, listlist);
for i from 1 to S[1] do
for j from 1 to S[2] do
for i1 from i to S[1] do
for j1 from 1 to S[2] do
if A1[i,j]<>0 and A1[i1,j1]<>0 and abs(A1[i,j]-A1[i1,j1])<abs(i-i1)+abs(j-j1) then return `no solutions` fi;
od; od; od; od;
L:=sort(select(e->e<>0, Flatten(A1)));
L1:=[`if`(L[1]>1,seq(L[1]-k, k=0..L[1]-2),NULL)];
L2:=[seq(seq(`if`(L[i+1]-L[i]>1,L[i]+k,NULL),k=0..L[i+1]-L[i]-2), i=1..nops(L)-1), `if`(L[-1]<MS,seq(L[-1]+k,k=0..MS-L[-1]-1),NULL)];


OneStepLeft:=proc(A1::listlist)
local s, M, m, k, T;
uses ListTools;
s:=Search(a, Matrix(A1));
M:=[[s[1]-1,s[2]],[s[1]+1,s[2]],[s[1],s[2]-1],[s[1],s[2]+1]];
T:=table(); k:=0;
for m in M do
if m[1]>=1 and m[1]<=S[1] and m[2]>=1 and m[2]<=S[2] and A1[op(m)]=0 then k:=k+1; T[k]:=subsop(m=a-1,A1);
fi;
od;
convert(T, list);
end proc;

OneStepRight:=proc(A1::listlist)
local s, M, m, k, T, s1;
uses ListTools;
s:=Search(a, Matrix(A1)); s1:=Search(a+2, Matrix(A1));
M:=[[s[1]-1,s[2]],[s[1]+1,s[2]],[s[1],s[2]-1],[s[1],s[2]+1]];
T:=table(); k:=0;
for m in M do
if m[1]>=1 and m[1]<=S[1] and m[2]>=1 and m[2]<=S[2] and A1[op(m)]=0 and `if`(a+2 in L, `if`(is(abs(s1[1]-m[1])+abs(s1[2]-m[2])>1),false,true),true) then k:=k+1; T[k]:=subsop(m=a+1,A1);
fi;
od;
convert(T, list);
end proc;

F1:=LM->ListTools:-FlattenOnce(map(OneStepLeft, LM));
F2:=LM->ListTools:-FlattenOnce(map(OneStepRight, LM));

T:=[A1];
for a in L1 do
T:=F1(T);
od;

for a in L2 do
T:=F2(T);
od;

R:=map(t->convert(t,Matrix), T);
if nops(R)=0 then return `no solutions` else R[] fi;

end proc:

>

Numbrix := proc( M :: ~Matrix, { inline :: truefalse := false } )

local S, adjacent, eq, i, initial, j, k, kk, m, n, one, single, sol, unique, val, var, x;

    (m,n) := upperbound(M);

    initial := &and(seq(seq(ifelse(M[i,j] = 0
                                   , NULL
                                   , x[i,j,M[i,j]]
                                  )
                            , i = 1..m)
                        , j = 1..n));

    adjacent := &and(seq(seq(seq(x[i,j,k] &implies &or(NULL
                                                       , ifelse(i>1, x[i-1, j, k+1], NULL)
                                                       , ifelse(i<m, x[i+1, j, k+1], NULL)
                                                       , ifelse(j>1, x[i, j-1, k+1], NULL)
                                                       , ifelse(j<n, x[i, j+1, k+1], NULL)
                                                      )
                                 , i = 1..m)
                             , j = 1..n)
                         , k = 1 .. m*n-1));

    one := &or(seq(seq(x[i,j,1], i=1..m), j=1..n));

    single := &not(&or(seq(seq(seq(seq(x[i,j,k] &and x[i,j,kk], kk = k+1..m*n), k = 1..m*n-1)
                                , i = 1..m), j = 1..n)));

    sol := Logic:-Satisfy(&and(initial, adjacent, one, single));

    if sol = NULL then
        error "no solution";
    end if;
if inline then
        S := M;
     else
        S := Matrix(m,n);
    end if;

    for eq in sol do
        (var, val) := op(eq);
        if val then
            S[op(1..2, var)] := op(3,var);
        end if;
    end do;
    S;
end proc:

Two simple examples

>	A:=<0,0,5; 0,0,0; 0,0,9>; # The unique solution NumbrixPuzzle(A); A:=<0,0,5; 0,0,0; 0,8,0>; # 4 solutions NumbrixPuzzle(A);

Matrix(3, 3, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 5, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 9})

Matrix(3, 3, {(1, 1) = 3, (1, 2) = 4, (1, 3) = 5, (2, 1) = 2, (2, 2) = 7, (2, 3) = 6, (3, 1) = 1, (3, 2) = 8, (3, 3) = 9})

Matrix(3, 3, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 5, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (3, 1) = 0, (3, 2) = 8, (3, 3) = 0})

Matrix(%id = 18446746210121682686), Matrix(%id = 18446746210121682806), Matrix(%id = 18446746210121674750), Matrix(%id = 18446746210121674870)

(1)

Comparison with Numbrix procedure. The example is taken from
http://rosettacode.org/wiki/Solve_a_Numbrix_puzzle

>

A:=<0, 0, 0, 0, 0, 0, 0, 0, 0;
0, 0, 46, 45, 0, 55, 74, 0, 0;
0, 38, 0, 0, 43, 0, 0, 78, 0;
0, 35, 0, 0, 0, 0, 0, 71, 0;
0, 0, 33, 0, 0, 0, 59, 0, 0;
0, 17, 0, 0, 0, 0, 0, 67, 0;
0, 18, 0, 0, 11, 0, 0, 64, 0;
0, 0, 24, 21, 0, 1, 2, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0, 0>;
CodeTools:-Usage(NumbrixPuzzle(A));
CodeTools:-Usage(Numbrix(A));

memory used=7.85MiB, alloc change=-3.01MiB, cpu time=172.00ms, real time=212.00ms, gc time=93.75ms

memory used=1.21GiB, alloc change=307.02MiB, cpu time=37.00s, real time=31.88s, gc time=9.30s

Matrix(%id = 18446746210094669942)

(2)

In the example below, which has 104 solutions, the Numbrix procedure is faster.

>	C:=Matrix(5,{(1,1)=1,(5,5)=25}); CodeTools:-Usage(NumbrixPuzzle(C)): nops([%]); CodeTools:-Usage(Numbrix(C)):

Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (3, 5) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 0, (4, 5) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 0, (5, 5) = 25})

memory used=0.94GiB, alloc change=-22.96MiB, cpu time=12.72s, real time=11.42s, gc time=2.28s

memory used=34.74MiB, alloc change=0 bytes, cpu time=781.00ms, real time=783.00ms, gc time=0ns

>

Download NumbrixPuzzle.mw

Printing "Maplesoft" in handwritten cursive script...

by: Samir Khan 2116 Product: Maple

July 24 2019

4 4

For no particular reason at all, these are parametric equations that print "Maplesoft" in handwritten cursive script when plotted

restart:
X := -2.05*sin(-2.70 + 2.45*t) - 3.36*sin(1.12 + 1.43*t) - 4.82*sin(-2.19 + 2.03*t) - 2.02*sin(1.36 + 2.31*t) - 2.41*sin(1.08 + 2.59*t) - 14.2*sin(1.51 + 0.185*t) - 5.25*sin(-2.04 + 1.85*t) - 2.81*sin(0.984 + 2.36*t) - 3.01*sin(-2.04 + 1.80*t) - 1.80*sin(-2.61 + 2.73*t) - 0.712*sin(-3.94 + 1.89*t) - 6.90*sin(-1.90 + 1.52*t) - 0.600*sin(-3.39 + 2.26*t) - 0.631*sin(-4.65 + 2.68*t) - 3.10*sin(-2.22 + 2.17*t) - 2.95*sin(1.38 + 1.25*t) - 1.43*sin(0.383 + 2.40*t) - 8.25*sin(-1.66 + 0.323*t) - 1.39*sin(-3.08 + 2.63*t) - 0.743*sin(-2.43 + 0.647*t) - 6.25*sin(-1.73 + 0.832*t) - 273.*sin(-1.58 + 0.0462*t) - 4.58*sin(-2.00 + 1.48*t) - 5.70*sin(-1.80 + 1.20*t) - 2.30*sin(1.42 + 0.462*t) - 3.24*sin(1.51 + 0.277*t) - 16.0*sin(-1.64 + 0.231*t) - 1.58*sin(0.779 + 1.71*t) - 0.571*sin(-2.08 + 0.970*t) - 8.85*sin(-1.88 + 1.34*t) - 1.10*sin(-2.24 + 2.08*t) - 1.49*sin(-2.27 + 1.02*t) - 2.19*sin(-1.70 + 1.94*t) - 4.47*sin(-2.06 + 1.57*t) - 2.08*sin(-2.02 + 1.06*t) - 5.70*sin(-1.86 + 1.62*t) - 2.26*sin(-1.66 + 1.16*t) - 3.95*sin(-1.98 + 1.29*t) - 0.928*sin(-2.08 + 1.76*t) - 2.98*sin(1.36 + 1.11*t) - 0.390*sin(-2.33 + 2.22*t) - 3.81*sin(1.01 + 2.54*t) - 0.613*sin(-1.43 + 1.66*t) - 19.7*sin(-1.60 + 0.138*t) - 0.524*sin(-2.87 + 0.414*t) - 2.15*sin(-4.63 + 0.694*t) - 0.782*sin(-1.56 + 2.49*t) - 5.27*sin(-1.81 + 1.38*t) - 5.18*sin(1.51 + 0.0923*t) - 6.83*sin(1.37 + 0.923*t) - 0.814*sin(-1.72 + 0.600*t) - 2.98*sin(-1.82 + 0.738*t) - 5.49*sin(1.44 + 0.509*t) - 3.90*sin(-1.76 + 0.785*t) - 0.546*sin(-2.18 + 0.876*t) - 1.92*sin(0.755 + 1.98*t) - 8.16*sin(1.38 + 0.553*t) - 0.504*sin(-1.56 + 0.371*t) - 3.43*sin(1.14 + 2.12*t):
Y := -1.05*sin(-3.81 + 2.68*t) - 7.72*sin(-4.59 + 0.231*t) - 6.38*sin(1.37 + 1.11*t) - 4.24*sin(-2.36 + 2.31*t) - 7.06*sin(1.18 + 1.80*t) - 4.60*sin(1.28 + 2.03*t) - 0.626*sin(-0.285 + 2.45*t) - 0.738*sin(-1.89 + 2.26*t) - 1.45*sin(-1.73 + 1.57*t) - 2.30*sin(-4.51 + 2.59*t) - 9.58*sin(-2.07 + 1.71*t) - 0.792*sin(-0.578 + 0.647*t) - 4.55*sin(1.49 + 1.25*t) - 14.0*sin(-2.13 + 1.62*t) - 1.02*sin(0.410 + 0.277*t) - 19.2*sin(-1.54 + 0.0462*t) - 17.3*sin(-1.86 + 1.20*t) - 1.96*sin(-0.845 + 2.63*t) - 0.754*sin(-0.0904 + 2.73*t) - 4.74*sin(1.11 + 1.48*t) - 1.79*sin(0.860 + 2.17*t) - 25.2*sin(-1.77 + 0.832*t) - 3.88*sin(1.30 + 0.462*t) - 20.8*sin(-1.66 + 0.323*t) - 17.6*sin(1.20 + 1.29*t) - 4.83*sin(0.169 + 2.36*t) - 10.8*sin(-2.01 + 1.85*t) - 8.69*sin(-2.17 + 2.22*t) - 5.48*sin(-1.69 + 1.34*t) - 18.1*sin(1.18 + 1.43*t) - 4.71*sin(0.728 + 2.08*t) - 1.15*sin(-3.44 + 1.52*t) - 2.53*sin(-2.61 + 2.54*t) - 5.48*sin(-2.02 + 1.94*t) - 4.67*sin(1.30 + 1.66*t) - 9.10*sin(1.37 + 0.970*t) - 6.45*sin(1.31 + 1.02*t) - 5.18*sin(-2.09 + 1.76*t) - 18.3*sin(-1.77 + 1.06*t) - 27.3*sin(1.31 + 1.16*t) - 2.83*sin(-3.01 + 2.40*t) - 2.93*sin(-1.70 + 0.138*t) - 4.17*sin(-2.06 + 2.12*t) - 1.60*sin(-4.25 + 1.38*t) - 2.69*sin(-1.89 + 0.371*t) - 7.92*sin(-1.78 + 0.600*t) - 19.6*sin(-1.79 + 0.738*t) - 22.6*sin(1.48 + 0.509*t) - 13.5*sin(1.21 + 0.923*t) - 5.53*sin(-1.64 + 0.0923*t) - 1.20*sin(0.145 + 2.49*t) - 3.15*sin(-1.57 + 0.414*t) - 1.74*sin(0.655 + 1.98*t) - 3.98*sin(-2.14 + 0.876*t) - 11.3*sin(-1.82 + 0.694*t) - 10.4*sin(0.987 + 1.89*t) - 8.39*sin(-1.53 + 0.185*t) - 27.8*sin(-1.76 + 0.785*t) - 9.39*sin(1.38 + 0.553*t):
plot([X, Y, t = 0 .. 68], scaling = constrained, axes = boxed);

Aleksandrov Denis, 8th grade

by: Alsu 445

July 04 2019

4 0

Aleksandrov_3dnew.mws

Murtazin Shamil, 7th grade

by: Alsu 445 Product: Maple V

July 04 2019

3 2

Hare in the forest

Bilnov Yevgeniy, 7th grade

by: Alsu 445 Product: Maple V

July 04 2019

4 2

The rocket flies

Быльнов_raketa_letit.mws

Ibragimova Evelina, 9th grade

by: Alsu 445

July 04 2019

2 0

Plotting the function of a complex variable

Plotting_the_function_of_a_complex_variable.mws

Ibragimova Evelina, 9th grade

by: Alsu 445 Product: Maple V

July 04 2019

7 0

Animated 3-D cascade of dolls

3d_matryoshkas_en.mws

Pearson correlation

by: Lenin Araujo Castillo 1790 Product: Maple 2018

June 21 2019

4 1

With this application developed entirely in Maple using native syntax and embedded components for science and engineering students. Just replace your data and you're done.

Pearson_Coeficient.mw

Lenin Araujo Castillo

Ambassador of Maple

Foucault’s Pendulum Exploration Using MAPLE18

by: swanwane 20 Product: Maple 18

June 18 2019

4 2

Foucault’s Pendulum Exploration Using MAPLE18

https://www.ias.ac.in/describe/article/reso/024/06/0653-0659

In this article, we develop the traditional differential equation for Foucault’s pendulum from physical situation and solve it from
standard form. The sublimation of boundary condition eliminates the constants and choice of the local parameters (latitude, pendulum specifications) offers an equation that can be used for a plot followed by animation using MAPLE. The fundamental conceptual components involved in preparing differential equation viz; (i) rotating coordinate system, (ii) rotation of the plane of oscillation and its dependence on the latitude, (iii) effective gravity with latitude, etc., are discussed in detail. The accurate calculations offer quantities up to the sixth decimal point which are used for plotting and animation. This study offers a hands-on experience. Present article offers a know-how to devise a Foucault’s pendulum just by plugging in the latitude of reader’s choice. Students can develop a miniature working model/project of the pendulum.

Vector Force in space 3d

by: Lenin Araujo Castillo 1790 Product: Maple 2018

June 04 2019

2 0

Exercises solved online with Maple exclusively in space. I attach the explanation links on my YouTube channel.

Part # 01

https://www.youtube.com/watch?v=8Aa2xzU8LwQ

Part # 02

https://www.youtube.com/watch?v=qyGT28CeSz4

Part # 03

https://www.youtube.com/watch?v=yf8rjSPbv5g

Part # 04

https://www.youtube.com/watch?v=FwHPW7ncZTg

Part # 05

https://www.youtube.com/watch?v=bm3frpukb0I

Link for download the file:

Vector_Exercises-Force_in_space.mw

Lenin AC

Ambassador of Maple

Algorithms

by: vv 13822 Product: Maple

June 02 2019

2 4

@chandrashekhar

There are no efficient algorithms for this.
How would you simplify by hand the expression

512*b^9 + (2303*a + 2304)*b^8 + (4616*a^2 + 9216*a + 4608)*b^7 + (5348*a^3 + 16128*a^2 + 16128*a + 5376)*b^6
 + (4088*a^4 + 16128*a^3 + 24192*a^2 + 16128*a + 4032)*b^5 + (1946*a^5 + 10080*a^4 + 20160*a^3 + 20160*a^2 
+ 10080*a + 2016)*b^4 + (728*a^6 + 4032*a^5 + 10080*a^4 + 13440*a^3 + 10080*a^2 + 4032*a + 672)*b^3 
+ (116*a^7 + 1008*a^6 + 3024*a^5 + 5040*a^4 + 5040*a^3 + 3024*a^2 + 1008*a + 144)*b^2 
+ (26*a^8 + 144*a^7 + 504*a^6 + 1008*a^5 + 1260*a^4 + 1008*a^3 + 504*a^2 + 144*a + 18)*b + 9*a^8 
+ 36*a^7 + 84*a^6 + 126*a^5 + 126*a^4 + 84*a^3 + 36*a^2 + 9*a + 1

to (a+2*b+1)^9 - a*(a-b)^8 ?

Using the Lindstedt-Poincaré Method to Account...

by: TechnicalSupport 790 Product: Maple

May 29 2019

9 0

We occasionally get asked questions about methods of Perturbation Theory in Maple, including the Lindstedt-Poincaré Method. Presented here is the most famous application of this method.

Introduction

During the dawn of the 20th Century, one problem that bothered astronomers and astrophysicists was the precession of the perihelion of Mercury. Even when considering the gravity from other planets and objects in the solar system, the equations from Newtonian Mechanics could not account for the discrepancy between the observed and predicted precession.

One of the early successes of Einstein's General Theory of Relativity was that the new model was able to capture the precession of Mercury, in addition to the orbits of all the other planets. The Einsteinian model, when applied to the orbit of Mercury, was in fact a non-negligible perturbation of the old model. In this post, we show how to use Maple to compute the perturbation, and derive the formula for calculating the precession.

In polar coordinates, the Einsteinian model can be written in the following form, where u(theta)=a(1-e^2)/r(theta), with theta being the polar angle, r(theta) being the radial distance, a being the semi-major axis length, and e being the eccentricity of the orbit:

# Original system.
f := (u,epsilon) -> -1 - epsilon * u^2;
omega := 1;
u0, du0 := 1 + e, 0;
de1 := diff( u(theta), theta, theta ) + omega^2 * u(theta) + f( u(theta), epsilon );
ic1 := u(0) = u0, D(u)(0) = du0;

The small parameter epsilon (along with the amount of precession) can be found in terms of the physical constants, but for now we leave it arbitrary:

# Parameters.
P := [
    a = 5.7909050e10 * Unit(m),
    c = 2.99792458e8 * Unit(m/s),
    e = 0.205630,
    G = 6.6740831e-11 * Unit(N*m^2/kg^2), 
    M = 1.9885e30 * Unit(kg), 
    alpha = 206264.8062, 
    beta = 415.2030758 
];
epsilon = simplify( eval( 3 * G * M / a / ( 1 - e^2 ) / c^2, P ) ); # approximately 7.987552635e-8

Note that c is the speed of light, G is the gravitational constant, M is the mass of the sun, alpha is the number of arcseconds per radian, and beta is the number of revolutions per century for Mercury.

We will show that the radial distance, predicted by Einstein's model, is close to that for an ellipse, as predicted by Newton's model, but the perturbation accounts for the precession (42.98 arcseconds/century). During one revolution, the precession can be determined to be approximately

sigma = simplify( eval( 6 * Pi * G * M / a / ( 1 - e^2 ) / c^2, P ) ); # approximately 5.018727337e-7

and so, per century, it is alpha*beta*sigma, which is approximately 42.98 arcseconds/century.
It is worth checking out this question on Stack Exchange, which includes an animation generated numerically using Maple for a similar problem that has a more pronounced precession.

Lindstedt-Poincaré Method in Maple

In order to obtain a perturbation solution to the perturbed differential equation u'+omega^2*u=1+epsilon*u^2 which is periodic, we need to write both u and omega as a series in the small parameter epsilon. This is because otherwise, the solution would have unbounded oscillatory terms ("secular terms"). Using this Lindstedt-Poincaré Method, we substitute arbitrary series in epsilon for u and omega into the initial value problem, and then choose the coefficient constants/functions so that both the initial value problem is satisfied and there are no secular terms. Note that a first-order approximation provides plenty of agreement with the measured precession, but higher-order approximations can be obtained.

To perform this in Maple, we can do the following:

# Transformed system, with the new independent variable being the original times a series in epsilon.
de2 := op( PDEtools:-dchange( { theta = phi/b }, { de1 }, { phi }, params = { b, epsilon }, simplify = true ) );
ic2 := ic1;

# Order and series for the perturbation solutions of u(phi) and b. Here, n = 1 is sufficient.
n := 1;
U := unapply( add( p[k](phi) * epsilon^k, k = 0 .. n ), phi );
B := omega + add( q[k] * epsilon^k, k = 1 .. n );

# DE in terms of the series.
de3 := series( eval( de2, [ u = U, b = B ] ), epsilon = 0, n + 1 );

# Successively determine the coefficients p[k](phi) and q[k].
for k from 0 to n do

    # Specify the initial conditions for the kth DE, which involves p[k](phi).
    # The original initial conditions appear only in the coefficient functions with index k = 0,
    # and those for k > 1 are all zero.
    if k = 0 then
        ic3 := op( expand( eval[recurse]( [ ic2 ], [ u = U, epsilon = 0 ] ) ) );
    else
        ic3 := p[k](0), D(p[k])(0);
    end if:

    # Solve kth DE, which can be found from the coefficients of the powers of epsilon in de3, for p[k](phi).
    # Then, update de3 with the new information.
    soln := dsolve( { simplify( coeff( de3, epsilon, k ) ), ic3 } );
    p[k] := unapply( rhs( soln ), phi );
    de3 := eval( de3 );

    # Choose q[k] to eliminate secular terms. To do this, use the frontend() command to keep only the terms in p[k](phi)
    # which have powers of t, and then solve for the value of q[k] which makes the expression zero. 
    # Note that frontend() masks the t-dependence within the sine and cosine terms.
    # Note also that this method may need to be amended, based on the form of the terms in p[k](phi).
    if k > 0 then
        q[1] := solve( frontend( select, [ has, p[k](phi), phi ] ) = 0, q[1] );
        de3 := eval( de3 );
    end if;

end do:

# Final perturbation solution.
'u(theta)' = eval( eval( U(phi), phi = B * theta ) ) + O( epsilon^(n+1) );

# Angular precession in one revolution.
sigma := convert( series( 2 * Pi * (1/B-1), epsilon = 0, n + 1 ), polynom ):
epsilon := 3 * G * M / a / ( 1 - e^2 ) / c^2;
'sigma' = sigma;

# Precession per century.
xi := simplify( eval( sigma * alpha * beta, P ) ); # returns approximately 42.98

Maple Worksheet: Lindstedt-Poincare_Method.mw

Solving a Numbrix Puzzle with Logic

by: Joe Riel 9660 Product: Maple

May 29 2019

4 13

Solving a Numbrix Puzzle with Logic

Background

Parade magazine, a filler in the local Sunday newspaper, contains a Numbrix puzzle, the object of which is to find a serpentine path of consecutive integers, 1 through 81, through a nine by nine grid. The puzzle typically specifies the content of every other border cell.

The Maple Logic package has a procedure, Satisfy , that can be used to solve this puzzle. Satisfy is a SAT-solver; given a boolean expression it attempts to find a set of equations of the form ${x__1 = b__1, x__2 = b__2, () .. ()}$ , where are the boolean variables in the given expression and are boolean values (true or false) that satisfy the expression (cause it to evaluate true).

A general technique to solve this and other puzzles with Satisfy is to select boolean-values variables that encode the state of the puzzle (a trial solution, whether valid or not), generate a boolean-expression of these variables that is satisfied (true) if and only if the variables are given values that correspond to a solution, pass this expression to Satisfy, then translate the returned set of boolean values (if any) to the puzzle solution.

Setup

Assign a matrix that defines the grid and the initial position. Use zeros to indicate the cells that need values. To make it easy to inspect the expressions, a small 2 x 3 matrix is used for this demo---a full size example is given at the end.

>	M := Matrix(2,3, {(1,1) = 1, (1,3) = 5});

Matrix(2, 3, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 5, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0})

(2.1)

Extract the dimensions of the Matrix

>	(m,n) := upperbound(M);

(2.2)

	Boolean Variables
	Let the boolean variable x[i,j,k] mean that cell (i,j) has value k. For example, x[2,3,6] is true when cell (2,3) contains 6, otherwise it is false. There are boolean variables.

Initial Position

The initial position can be expressed as the following and-clause.

>	initial := &and(seq(seq(ifelse(M[i,j] = 0, NULL, x[i,j,M[i,j]]), i = 1..m), j = 1..n));

(4.1)

Adjacent Cells

The requirement that an interior cell with value k is adjacent to the cell with value k+1 can be expressed as the implication

x[i,j,k] &implies &or(x[i-1,j,k+1], x[i+1,j,k+1], x[i,j-1,k+1], x[i,j+1,k+1])

Extending this to handle all cells results in the following boolean expression.

>

adjacent := &and(seq(seq(seq(
         x[i,j,k] &implies &or(NULL
                               , ifelse(1<i, x[i-1, j, k+1], NULL)
                               , ifelse(i<m, x[i+1, j, k+1], NULL)
                               , ifelse(1<j, x[i, j-1, k+1], NULL)
                               , ifelse(j<n, x[i, j+1, k+1], NULL)
                               )
                            , i = 1..m)
                        , j = 1..n)
                    , k = 1 .. m*n-1));

(5.1)

All Values Used

The following expression is true when each integer , from 1 to , is assigned to one or more cells.

>	allvals := &and(seq(seq(&or(seq(x[i,j,k], k=1..m*n)), i=1..m), j=1..n));

(6.1)

Single Value

The following expression is satisfied when each cell has no more than one value.

>	single := &not &or(seq(seq(seq(seq(x[i,j,k] &and x[i,j,kk], kk = k+1..mn), k = 1..mn-1), i = 1..m), j = 1..n));

(7.1)

Solution

Combine the boolean expressions into a a single and-clause and pass it to Satisfy.

>	sol := Logic:-Satisfy(&and(initial, adjacent, allvals, single));

(8.1)

Select the equations whose right size is true

>	sol := select(rhs, sol);

(8.2)

Extract the lhs of the true equations

>	vars := map(lhs, sol);

(8.3)

Extract the result from the indices of the vars and assign to a new Matrix

>	S := Matrix(m,n):

>	for v in vars do S[op(1..2,v)] := op(3,v); end do:

>

S;

Matrix(2, 3, {(1, 1) = 1, (1, 2) = 6, (1, 3) = 5, (2, 1) = 2, (2, 2) = 3, (2, 3) = 4})

(8.4)

	Procedure
	We can now combine the manual steps into a procedure that takes an initialized Matrix and fills in a solution. Numbrix := proc( M :: ~Matrix, { inline :: truefalse := false } )

Example

Create the initial position for a 9 x 9 Numbrix and solve it.

>	P := Matrix(9, {(1,1)=11, (1,3)=7, (1,5)=3, (1,7)=81, (1,9)=77, (3,9)=75, (5,9)=65, (7,9)=55, (9,9)=53 , (9,7)=47, (9,5)=41, (9,3)=39, (9,1)=37, (7,1)=21, (5,1)=17, (3,1)=13});

(10.1)

>	CodeTools:-Usage(Numbrix(P));

memory used=0.77GiB, alloc change=220.03MiB, cpu time=15.55s, real time=12.78s, gc time=3.85s

(10.2)

>

numbrix.mw

Maple 2019.1 update

by: eithne 2072 Product: Maple

May 28 2019

6 0

We have just released an update to Maple, Maple 2019.1.

Maple 2019.1 includes corrections and improvement to the mathematics engine (including LPSolve, sum, statistics, and the Physics package), visualization (including annotations and the Plot Builder), export to LaTeX (exporting output) and PDF (freezing on export), network licensing, MATLAB connectivity, and more. We recommend that all Maple 2019 users install these updates.

This update is available through Tools>Check for Updates in Maple, and is also available from our website on the Maple 2019.1 download page, where you can also find more details.

Maple Conference – New Deadline for Submissions

by: Kat 232 Products: Maple , Maple Toolboxes

May 24 2019

2 2

I just wanted to let everyone know that the Call for Papers and Extended Abstracts deadline for the Maple Conference has been extended to June 14.

The papers and extended abstracts presented at the 2019 Maple Conference will be published in the Communications in Computer and Information Science Series from Springer. We welcome topics that fall into the following broad categories:

Maple in Education
Algorithms and Software
Applications of Maple

You can learn more about the conference or submit your paper or abstract here:

https://www.maplesoft.com/mapleconference/Papers-and-Presentations.aspx

Hope to hear from you soon!

E-Mail Address:
Password:
Remember Me:	Automatically sign in on future visits

E-Mail Address:
Password:
Remember Me:	Automatically sign in on future visits

Ask a Question

Create a Post

MaplePrimes Posts

MaplePrimes Posts are for sharing your experiences, techniques and opinions about Maple, MapleSim and related products, as well as general interests in math and computing.

Latest Post
Latest Posts Feed

Numbrix Puzzle by the branch and bound method

Printing "Maplesoft" in handwritten cursive script...

Aleksandrov Denis, 8th grade

Murtazin Shamil, 7th grade

Bilnov Yevgeniy, 7th grade

Ibragimova Evelina, 9th grade

Ibragimova Evelina, 9th grade

Pearson correlation

Foucault’s Pendulum Exploration Using MAPLE18

Vector Force in space 3d

Algorithms

Using the Lindstedt-Poincaré Method to Account...

Solving a Numbrix Puzzle with Logic

Maple 2019.1 update

Maple Conference – New Deadline for Submissions

Save this setting as your default sorting preference?

Ask a Question

Create a Post

Generating PDF…

Save this setting as your default sorting preference?
Note: You can change your preference any time in your account settings
Don't show this again

Ask a Question

Create a Post

MaplePrimes Posts

MaplePrimes Posts are for sharing your experiences, techniques and opinions about Maple, MapleSim and related products, as well as general interests in math and computing.

Latest Post Latest Posts Feed

Save this setting as your default sorting preference?

Ask a Question

Create a Post

Generating PDF…

Latest Post
Latest Posts Feed