Hello

I have no choice but use Grid:-Map and Grid:-Seq in my calculations due to the size of them.  Here is a very small example that is puzzling me (Perhaps I did something really silly and did not realize). 

ansa:=CodeTools:-Usage(Grid:-Map(w->CondswithOnesolutionTest(w,eqns,vars,newvars,tlim),conds5s)):

with the following result:

ansa:=set([{alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 1] = 0, alpha[3, 2] = 0, alpha[3, 3] = 0, alpha[3, 4] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}, {}, {}, {}, {}, {}], [{}, {}, {}, {}, {}, {alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 1] = 0, alpha[3, 2] = 0, alpha[3, 3] = 0, alpha[3, 5] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}], [{alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 1] = 0, alpha[3, 3] = 0, alpha[3, 4] = 0, alpha[3, 5] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}, {}, {}, {}, {}, {}], [{}, {}, {}, {}, {}, {alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 2] = 0, alpha[3, 3] = 0, alpha[3, 4] = 0, alpha[3, 5] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}])

The same thing but now using only map

ansb:=CodeTools:-Usage(map(w->CondswithOnesolutionTest(w,eqns,vars,newvars,tlim),conds5s)):

ansb:={[{}, {}, {}, {}, {}, {alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 1] = 0, alpha[3, 2] = 0, alpha[3, 3] = 0, alpha[3, 5] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}], [{}, {}, {}, {}, {}, {alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 2] = 0, alpha[3, 3] = 0, alpha[3, 4] = 0, alpha[3, 5] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}], [{alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 1] = 0, alpha[3, 2] = 0, alpha[3, 3] = 0, alpha[3, 4] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}, {}, {}, {}, {}, {}], [{alpha[1, 1] = 0, alpha[1, 2] = 0, alpha[1, 3] = 0, alpha[1, 4] = 0, alpha[1, 5] = 0, alpha[1, 6] = 0, alpha[1, 8] = 0, alpha[1, 9] = 0, alpha[2, 0] = 0, alpha[2, 1] = 0, alpha[2, 2] = 0, alpha[2, 4] = 0, alpha[2, 5] = 0, alpha[2, 7] = 0, alpha[2, 8] = 0, alpha[2, 9] = 0, alpha[3, 0] = 0, alpha[3, 1] = 0, alpha[3, 3] = 0, alpha[3, 4] = 0, alpha[3, 5] = 0, alpha[3, 6] = 0, alpha[3, 7] = 0, alpha[3, 8] = 0, alpha[3, 9] = 0}, {}, {}, {}, {}, {}]}

(This is what I expected as the result).

Why did Grid:-Map add set to the answer?  What am I missing?  

Many thanks

Here is the problem. I start Maple 2020 on windows 10. Run a script which takes 1-2 days to complete. 

During this time, I can't use that Maple at all, since it is busy. 

I could start Maple 2019, and that runs as completely separate process. But I want to use Maple 2020 since some things in my scripts do not work on Maple 2019 that work on Maple 2020.

If I start a new instance of Maple 2020, by doing Start->Maple 2020. it does seem to start it OK, but I noticed it seems to be somehow still connected to the one running somehow.  May be they are sharing the same interface?

I can use the new instance now and open new worksheet and use it. But it seems to become very slow, as if it is sharing something with the other Maple 2020 running the long script which uses lots of resources. It is not RAM issue, I have 64 GB RAM, and there is plenty of free RAM left. 

When I close the new Maple 2020 workseet I started, I get a message asking if I want to save the worksheet that I have open from the earlier instance which is still running ! 

I say no ofcourse, as I do not want to terminate that instance, I want to keep it running until the script is completed.

My question is: Could someone may be explain exactly what happens when one starts new Maple 2020, while one is allready running? Why it seems they are sharing either the interface or something else.  How to start completely separate Maple 2020 instance on same PC while one is allready running?

With Mathematica, this issue does not happen. I can start two instances of same version on same PC, and there is nothing shared between them at all.  This does not seem to be the case with Maple.

Maple 2020.1 on windows 10.

Is it possible to determine an analytic solution to the following system of two differential equations for $A$ and $B$ using Maple.  My suspicion is that trial and error would find an analytic solution in theory and so that Maple could find the solution.  M is a constant and \sigma is some arbitrary function of t and the spatial coordinates. 

\[ \Bigg( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} + \frac{1}{2} \Bigg( 1 + \frac{M}{2 \sqrt{x^2 + y^2 + z^2}} \Bigg) \Bigg( \frac{\partial \sigma}{\partial x }\frac{\partial}{\partial x} +\frac{\partial \sigma}{\partial y}\frac{\partial}{\partial y} +\frac{\partial \sigma}{\partial z}\frac{\partial}{\partial z} \Bigg) \Bigg)B=0, \]

\[\frac{d A}{dt} = AB.\]

Furthermore, the boundary conditions are 

\[B \rightarrow -1  \: \text{as}  \: \sqrt{x^2 + y^2 + z^2} \rightarrow \infty,\]

\[A \rightarrow e^{-t} \: \text{as} \: \sqrt{x^2 + y^2 + z^2} \rightarrow \infty \]

System_of_Equations.pdf

I do not quite understand why prof asks this question. Or I am doing right? Where can I improve? Or I understand this question completely wrong. To be honest, I did not get the point 

Hi there.

As we all know if we multiply two polynomials f(x) and g(x) of degrees m and n respectively we get polynomial h(x)= f(x)*g(x) of degree m+n and with m+n+1 coefficients in general. Function modp1(('Multiply')(...)) doing this very well. But sometimes we don't need full resulting h(x) - just subset of monomials and subset of coefficients of h(x) - so we don't need to calculate all m+n+1 coefficients of h(x) and waste time and resources for that.

I would request some additional rework of modp1 package: by adding to modp1(('Multiply')(...)) two optional parameters - degrees of first and last calculating coefficients of h(x).

For example:

h:=modp1(Multiply(f, g,n-1,n+1), p) could calculate only monomials with n-1, n and n+1 degrees and set other monomials to zero.

Or maybe it should be new function:

h:=modp1(Multiply_Truncate(f, g,n-1,n+1), p)

Is it possible?

It would be great and very efficient in many tasks.

Thank you.

Download spatial_1.mw

Hi all,

We want to find a curve fit for an integer sequence.

We have n such that n^2+n+17 is a prime number.

See oeis.org/A007635 and comments.

Use the Maple CurveFitting package.

I tried with(CurveFitting).

We do not know if this is best represented by a polynomial or exponential curve fit.

n2_and_n_and_17_in_OEIS_007635.mw

n2_and_n_and_17_in_OEIS_007635.pdf

Regards,

Matt

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

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

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.

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.

How to find sgn on maple?

signum.mw

When I select "oldest first", the first post shown is the newest.  Vice-versa when selecting "newest first".  Am I misunderstanding the meaning of the term?  I suppose it doesn't matter as both options are available but it's weird.

Maple Apps-Venn Diagrams does not work.  In box on the right there is an error message.

#include<iostream>
#include<vector>
#include<cmath>
#define NODE 8

using namespace std;
int graph[NODE][NODE] = {
   {0,1,1,0,0,0,0,0},
   {1,0,1,1,1,0,0,0},
   {1,1,0,1,0,1,0,0},
   {0,1,1,0,0,0,0,0},
   {0,1,0,0,0,1,1,1},
   {0,0,1,0,1,0,1,1},
   {0,0,0,0,1,1,0,0},
   {0,0,0,0,1,1,0,0}
};
int tempGraph[NODE][NODE];
int findStartVert() {
   for(int i = 0; i<NODE; i++) {
      int deg = 0;
      for(int j = 0; j<NODE; j++) {
         if(tempGraph[i][j])
            deg++; //increase degree, when connected edge found
      }
      if(deg % 2 != 0) //when degree of vertices are odd
      return i; //i is node with odd degree
   }
   return 0; //when all vertices have even degree, start from 0
}
int dfs(int prev, int start, bool visited[]){
   int count = 1;
   visited[start] = true;
   for(int u = 0; u<NODE; u++){
      if(prev != u){
         if(!visited[u]){
            if(tempGraph[start][u]){
               count += dfs(start, u, visited);
            }
         }
      }
   }
   return count;
}
bool isBridge(int u, int v) {
   int deg = 0;
   for(int i = 0; i<NODE; i++)
      if(tempGraph[v][i])
   deg++;
   if(deg>1) {
      return false; //the edge is not forming bridge
   }
   return true; //edge forming a bridge
}
int edgeCount() {
   int count = 0;
   for(int i = 0; i<NODE; i++)
      for(int j = i; j<NODE; j++)
         if(tempGraph[i][j])
   count++;
   return count;
}
void fleuryAlgorithm(int start) {
   static int edge = edgeCount();
   static int v_count = NODE;
   for(int v = 0; v<NODE; v++) {
      if(tempGraph[start][v]) {
         bool visited[NODE] = {false};
         if(isBridge(start, v)){
            v_count--;
         }
         int cnt = dfs(start, v, visited);
         if(abs(v_count-cnt) <= 2){
            cout << start << "--" << v << " ";
            if(isBridge(v, start)){
               v_count--;
            }
            tempGraph[start][v] = tempGraph[v][start] = 0; //remove edge from graph
            edge--;
            fleuryAlgorithm(v);
         }
      }
   }
}
int main() {
   for(int i = 0; i<NODE; i++) //copy main graph to tempGraph
   for(int j = 0; j<NODE; j++)
      tempGraph[i][j] = graph[i][j];
   cout << "Euler Path Or Circuit: ";
   fleuryAlgorithm(findStartVert());
}

Kind help 

Hi, I generated latex formate of an equation by using a command of maple but when I paste it into MathType, could not get the required equation, can anyone help me

${\frac {1}{51200\, \left( {x}^{2}+2 \right) ^{6}} \left( -187110\,

 \left( {x}^{2}+2 \right) ^{6}\sqrt {2} \left( {Q}^{3}+ \left( {\frac

{18\,k}{11}}-{\frac{18}{11}} \right) {Q}^{2}+ \left( {\frac {320\,{k}^

{2}}{297}}-{\frac {40\,k}{27}}+{\frac{320}{297}} \right) Q+{\frac {80

\,{k}^{3}}{297}}-{\frac {80\,{k}^{2}}{189}}+{\frac {80\,k}{189}}+{

\frac {640\,\lambda}{2079}}-{\frac{80}{297}} \right) \arctan \left( 1/

2\,x\sqrt {2} \right) -93555\, \left( {x}^{2}+2 \right) ^{6}\pi\,

 \left( {Q}^{3}+ \left( {\frac {18\,k}{11}}-{\frac{18}{11}} \right) {Q

}^{2}+ \left( {\frac {320\,{k}^{2}}{297}}-{\frac {40\,k}{27}}+{\frac{

320}{297}} \right) Q+{\frac {80\,{k}^{3}}{297}}-{\frac {80\,{k}^{2}}{

189}}+{\frac {80\,k}{189}}+{\frac {640\,\lambda}{2079}}-{\frac{80}{297

}} \right) \sqrt {2}-374220\, \left(  \left( {Q}^{3}+ \left( {\frac {

18\,k}{11}}-{\frac{18}{11}} \right) {Q}^{2}+ \left( {\frac {320\,{k}^{

2}}{297}}-{\frac {40\,k}{27}}+{\frac{320}{297}} \right) Q+{\frac {80\,

{k}^{3}}{297}}-{\frac {80\,{k}^{2}}{189}}+{\frac {80\,k}{189}}+{\frac

{640\,\lambda}{2079}}-{\frac{80}{297}} \right) {x}^{10}+ \left( {

\frac {34\,{Q}^{3}}{3}}+ \left( {\frac {204\,k}{11}}-{\frac{204}{11}}

 \right) {Q}^{2}+ \left( {\frac {10880\,{k}^{2}}{891}}-{\frac {1360\,k

}{81}}+{\frac{10880}{891}} \right) Q+{\frac {2720\,{k}^{3}}{891}}-{

\frac {2720\,{k}^{2}}{567}}+{\frac {2720\,k}{567}}+{\frac {21760\,

\lambda}{6237}}-{\frac{2720}{891}} \right) {x}^{8}+ \left( {\frac {264

\,{Q}^{3}}{5}}+ \left( {\frac {432\,k}{5}}-{\frac{432}{5}} \right) {Q}

^{2}+ \left( {\frac {512\,{k}^{2}}{9}}-{\frac {704\,k}{9}}+{\frac{512}

{9}} \right) Q+{\frac {128\,{k}^{3}}{9}}-{\frac {1408\,{k}^{2}}{63}}+{

\frac {1408\,k}{63}}+{\frac {97280\,\lambda}{6237}}-{\frac{128}{9}}

 \right) {x}^{6}+ \left( {\frac {4496\,{Q}^{3}}{35}}+ \left( {\frac {

80928\,k}{385}}-{\frac{80928}{385}} \right) {Q}^{2}+ \left( {\frac {

287744\,{k}^{2}}{2079}}-{\frac {35968\,k}{189}}+{\frac{287744}{2079}}

 \right) Q+{\frac {3328\,{k}^{3}}{99}}-{\frac {3328\,{k}^{2}}{63}}+{

\frac {3328\,k}{63}}+{\frac {10240\,\lambda}{297}}-{\frac{3328}{99}}

 \right) {x}^{4}+ \left( {\frac {10672\,{Q}^{3}}{63}}+ \left( {\frac {

21344\,k}{77}}-{\frac{21344}{77}} \right) {Q}^{2}+ \left( {\frac {

1094656\,{k}^{2}}{6237}}-{\frac {136832\,k}{567}}+{\frac{1094656}{6237

}} \right) Q+{\frac {35584\,{k}^{3}}{891}}-{\frac {35584\,{k}^{2}}{567

}}+{\frac {35584\,k}{567}}+{\frac {235520\,\lambda}{6237}}-{\frac{

35584}{891}} \right) {x}^{2}+{\frac {25376\,{Q}^{3}}{231}}+ \left( {

\frac {12352\,k}{77}}-{\frac{12352}{77}} \right) {Q}^{2}+ \left( -{

\frac {7936\,k}{63}}+{\frac {63488\,{k}^{2}}{693}}+{\frac{63488}{693}}

 \right) Q-{\frac{512}{27}}+{\frac {512\,{k}^{3}}{27}}-{\frac {5632\,{

k}^{2}}{189}}+{\frac {102400\,\lambda}{6237}}+{\frac {5632\,k}{189}}

 \right) x \right) }$