Also available:

## Remove the permutation...

Asked by:

Hi all, i have a problem someone help me.

Define per(f(a,b,c)) = {f(a,b,c), f(b,c,a), f(c,a,b)}. I want to write a function removePer() that removes the permutations, example:

ds := {a, b, a^2, b^2, c, c + 2a, a - b, c^2, a + 2b, b + 2c}

then removePer(ds) return {a, a^2, c + 2a, a - b} because per(a) = {a, b, c}, per(a^2)  = {a^2, b^2, c^2} and per(c+2a) ={a+2b, b+2c, c+a}. Note that removePer(ds) can return {b, a^2, c + 2a, a - b} or {c, a^2, c + 2a, a - b}, ...  but just one result.

ds := {ab, bc, a - b^2, b - c^2, a^2, c - a^2},

then removePer(ds) return {ab, a - b^2, a^2}.

Thank you very much.

## Convert {} to [] and [] to {}....

Asked by:

Hi all, I have a problem someone can help me

F := {a^2, b^2, c^2, ab, bc, ca}

G := [a^2, b^2, c^2, ab, bc, ca]

How to convert F to G and G to F ?

Thanks you very much.

## The opery awards

by: Maple

most effective built in operator code award goes to ppl that wrote the code for the union and intercect set operations for maple. Very important simple example below of  one of its applications.

When i work with algorithms, probably one of my most primary ports of enquiry (figuratively jeez skynet)  is to set up and if statement triggered to terminate the loop once the operations performed for any further cycles is INDEMPOTENT. this doesnt always mean your output is convergent in every case but it allows you to minimize the amount of time the cpu needs to collect data( ie the point at which it would produce that same set as it did in the last most loop)

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## Set Theory, how to create 'set functions' in Maple...

Asked by:

Typically sets are created like:

A:={a,b,c};

B:={c,q,w,e};

and then you can carryout A union B or B\A

what if  you wanted to create the set as

A:={values in some three dimensional space};

B:={volume, based on values taken from A};

Can these relationships be set up in Maple? If so, how? If there are commands that specifically handle these types of sets, what does maple call them?  I've seen the term 'set function' but what might Maple call them?

Note: I am not even sure i 'tagged' this correctly because it I am not sure the proper terms for these functions/sets.

Thanks in advance for any help.

## Rearrange 7 equations ...

Asked by:

Hello,

I tried to rearrange below set of equations to have the equations in term of P[0], P[1], P[2], P[3], P[4], P[5] and P[6]. I used the symbol := for function definition for all Ps except one of them. Thus maple will rearrange that excepted one. However, I got error massage stating "Error, (in P[3]) too many levels of recursion" when I tried to rearranged equations for P[0].

Can I get help to rearranged them.

P[0](s) = (P[1](s)*mu[1]+P[2](s)*mu[2]+1)/(s+3*lambda+3*sigma)

P[1](s) = (3*P[0](s)*lambda+3*P[3](s)*mu[1]+P[4](s)*mu[2])/(s+mu[1]+2*lambda+2*sigma)

P[2](s) = (3*P[0](s)*sigma+P[4](s)*mu[2]+P[3](s)*mu[1])/(s+mu[2]+2*lambda+2*sigma)

P[3](s) = 2*lambda(P[1](s)+P[2](s))/(s+2*mu[1]+lambda)

P[4](s) = 2*sigma(P[1](s)+P[2](s))/(s+2*mu[2]+sigma)

P[5](s) = lambda(P[3](s)+P[4](s))/s

P[6](s) = sigma(P[3](s)+P[4](s))/s

Thank you for your help

## Command for product of sets?...

Asked by:

Is there any Maple command for computing the product of two set A and B?

For example if A={a,b,c} and B={x,y} then we have A*B={ax,ay,bx,by,cx,cy}.

## how to shift a list or set in maple...

Asked by:

though swap can do this,

hope to know a simple way to shift a list or set to the left in cycle in maple

for example

1,2,3

2,3,1

3,1,2

## Is that element of array?...

Asked by:

Hi,

Suppose I have an array

A[1]:=a;

A[2]:=b;

A[3]:=c;

I would like to check if an element, e.g. "c", belong to this array. I can contrsuct a "for... end do" command to compare each element and/or construct a subroutine.

Is there any build-in function in maple can do that? Perferably return true/false

Thank you!

## Subtract member from set...

Asked by:

I know how we can add a member to a mathematical set ,for example :

C := NULL;

C := C, V, B, X;

print(C)

V, B, X

but, my problem is that how we can subtract a member from the defined set ?

Thanks,

## how to get the supplementary set ...

Asked by:

assume x,y is a real number and {x>4,y>4}, how can i get the supplementary set ({x<=4,y<=4}) of x,y in the real domain by maple?

## how can i get a unique list...

Asked by:

eq:=[Vector[row](4, {(1) = 0., (2) = 0., (3) = 0., (4) = 0.}), Vector[row](4, {(1) = -0.3381778526e-2, (2) = -.3381778526, (3) = -.3185010532, (4) = 0.}), Vector[row](4, {(1) = 0., (2) = 0., (3) = 0., (4) = 0.}), Vector[row](4, {(1) = 0., (2) = 0., (3) = 0., (4) = 0.}), Vector[row](4, {(1) = 0., (2) = 0., (3) = 0., (4) = 0.})]

how i can get a unique results:[Vector[row](4, {(1) = 0., (2) = 0., (3) = 0., (4) = 0.}), Vector[row](4, {(1) = -0.3381778526e-2, (2) = -.3381778526, (3) = -.3185010532, (4) = 0.})]

## Partition of a set into parts of any sizes

by:

The work contains two procedures called  SetPartition  and  NumbPart .

The first procedure  SetPartition  generates all the partitions of a set  S  into disjoint subsets of specific sizes defined by a list  Q. The third optional parameter is the symbol  `ordered`  or  `nonordered`  (by default) . It determines whether the order of subsets in a partition is significant.

The second procedure  NumbPart  returns the number of all partitions of the sizes indicated by  Q .

Codes of these procedures:

SetPartition:=proc(S::set, Q::list(posint), K::symbol:=nonordered)  # Procedure finds all partitions of the set  S  into subsets of the specific size given Q.

local L, P, T, S1, S2, n, i, j, m, k, M;

uses ListTools,combinat;

if `+`(op(Q))<>nops(S) then error "Should be `+`(op(Q))=nops(S)" fi;

L:=convert(Q,set);

T:=[seq([L[i],Occurrences(L[i],Q)], i=1..nops(L))];

P:=`if`(K=ordered,convert(T,list),convert(T,set));

S1:=S;  S2:={`if`(K=ordered,[],{})}; n:=nops(P);

for i to n do

m:=P[i,1]; k:=P[i,2];

for j to k do

S2:={seq(seq(`if`(K=ordered,[op(s),t],{op(s),t}), t=choose(S1 minus `union`(op(s)),m)), s=S2)};

od; od;

if K=ordered then {map(op@permute,S2)[]} else S2 fi;

end proc:

NumbPart:=proc(Q::list(posint), K::symbol:=nonordered)  # Procedure finds the number of all partitions of a set into subsets of the specific size given  Q

local L, T, P, n, S, N;

uses ListTools;

L:=convert(Q,set);

T:=[seq([L[i],Occurrences(L[i],Q)], i=1..nops(L))];

P:=`if`(K=ordered,convert(T,list),convert(T,set));

n:=nops(P);  N:=add(P[i,2], i=1..n);

S:=add(P[i,1]*P[i,2],i=1..n)!/mul(P[i,1]!^P[i,2],i=1..n)/mul(P[i,2]!,i=1..n);

if K=nonordered then return S else  S*N! fi;

end proc:

Examples of use:

SetPartition({a,b,c,d,e}, [1,1,3]);  nops(%);  # Nonordered partitions and their number

SetPartition({a,b,c,d,e}, [1,1,3], ordered);  nops(%);  # Ordered partitions and their number

Here's a more interesting example. 5 fruits  {apple, pear, orange, mango, peach}  must be put on three plates so that  on each of two plates there are 2  fruits, and there is one fruit  on one plate. Two variants to solve: 1) plates are indistinguishable and 2) different plates. In how many ways can this be done?

SetPartition({apple,pear,orange,mango,peach}, [1,2,2]);  nops(%);  # plates are indistinguishable

NumbPart([1,2,2], ordered);  # Number of ways for  different plates

90

Another example - how many ways can be divided  100 objects into 10 equal parts?

NumbPart([10\$10]);

64954656894649578274066349293466217242333450230560675312538868633528911487364888307200

## Creating two lists from a set with pairs using loo...

Asked by:

I paste below a simple code illustrating what I want to create: two lists from a set with pairs. I wonder it is a very simple task but I my lists aren't create in the end.

> restart;
>
> lista:=[[1,10],[2,20],[3,30]]:
> x:=[]:
> y:=[]:
>
> for i from 1 to nops(lista) do
>     for j from 1 to 2 do
>
>         if j = 1 then
>             x[i,j]:=op(1,op(i,lista));
>        fi;
>        if j = 2 then
>             y[i,j]:=op(2,op(i,lista));
>       fi;
>    od;
> od;
>
> x;
> y;

In resume then after the for loop is terminated I want to be left with two lists:

x:= [1,2,3] and y:=[10,20,30]. I'm sure it has a quick fix but I'm stuck and would appreciate any help/advice.

Thanks in advance!

## why the length of this set is always equal to 9?...

Asked by:

I tried

nops({x, y, z,1+x, 2+y, -3+z,-2+x, 3+y, -1+z });

and Maple out 9.

If x = 1, y = 2, z = 3, we have

nops({1, 2, 3, 2, 4, 0, -1, 5, 2});

equal to 7.

Edit.

I edited my question. The word "list" into "set".

My question means, there is a triple (x, y, z) = (1, 2, 3) so that the number of element of the set {x, y, z,1+x, 2+y, -3+z,-2+x, 3+y, -1+z } is 7, not 9. Why we can confirm the number of elements of the set {x, y, z, 1+x, 2+y, -3+z,-2+x, 3+y, -1+z } is 9? And, how must select the integer numbers x, y, z (0 < x <10, 0< y <10, 0< z < 10) so that the number of elements of the set {x, y, z, 1+x, 2+y, -3+z,-2+x, 3+y, -1+z } is 9?

I tried

restart:

ListTools[Categorize]:

L:=[]:

for x from -2 to 5  do

for y from -2 to 5  do

for z from -2 to 5  do

a:=[x,y,z]:

b:=[1+x,2+y,-3+z]:

c:=[-2+x,3+y,-1+z]:

if op(1,a)*op(2,a)*op(3,a)*op(1,b)*op(2,b)*op(3,b)*op(1,c)*op(2,c)*op(3,c)<> 0

and op(1,a)<>op(2,a) and  op(1,a)<>op(3,a) and  op(1,a)<>op(2,b) and  op(1,a)<>op(3,b) and  op(1,a)<>op(2,c) and  op(1,a)<>op(3,c) and op(2,a)<>op(3,a) and op(2,a)<>op(1,b) and

op(2,a)<>op(3,b) and

op(2,a)<>op(1,c) and

op(2,a)<>op(3,c) and

op(3,a)<>op(1,b) and

op(3,a)<>op(2,b) and

op(3,a)<>op(1,c) and

op(3,a)<>op(2,c) and

op(1,b)<>op(2,b) and

op(1,b)<>op(3,b) and

op(1,b)<>op(2,c) and

op(1,b)<>op(3,c) and

op(2,b)<>op(3,b) and

op(2,b)<>op(1,c) and

op(2,b)<>op(3,c) and

op(3,b)<>op(1,c) and

op(3,b)<>op(2,c) and

op(1,c)<>op(2,c) and

op(1,c)<>op(3,c) and op(2,c)<>op(3,c) then L:=[op(L), {a,b,c}] fi; od: od: od:

nops(L);

L;

## Very strange behavior of set of vectors...

Asked by:

I experienced strange operation of "union" for sets of vectors.

Mt1:=Matrix(2, 4, [[ 0,1, 0, 0], [ 0,  0,  1, 1]]); Ms := Vector[column](4, [8,4,2,1]); St1 := {}:

St1:= `union`(St1, {Mt1 . Ms});

I am surprised, because each execution of union adds new and the same vector <4 | 3> to set St1:

But after copying any set in the clipboard and pasting the set St1 has only one instance of vector <4 | 3>:

What does it mean?

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