Items tagged with programming

for example

func1 := proc(system1)

for i from 1 to 100 do

solve([system1[1], system1[2]],[x,y]);

od:

end proc:

 

func1([diff(y,t) = data[i+t+1], diff(x,t) = data[i+t+1]])

i is depend on the for loop inside a function, but woud like to pass this system into a function with i

this will cause error

how to write better for passing a system as parameter using variable inside a function?

Hello, everyone. I have a group project where we have to explain the Josephus problem and use Maple to solve the problem. I am trying to solve the problem in multiple ways (because why not), but I am struggling with my third procedure. I understand the logic behind it and how its supposed to achieve O(k*logn), but the code that I wrote for it doesn't seem to produce the correct result.

JosephusImproved := proc (n, k)
local count, result:
if n = 1 then
return 0:
elif 1 < n < k then
return JosephusImproved(n - 1, k) + k + 1 mod n:
else
count := floor(n / k):
result := JosephusImproved(n - count, k):
result := result - n mod k:
if result < 0 then
result := result + n
else
result := floor(result /(k - 1)):
return result:
end if:
end if:
end proc:

Note: The regular recursive expression [Josephus(n - 1, k) + k + 1 mod n] has a "+ 1" since that was the only way I could make Maple do the calculation correctly. Proven with a Cyclic procedure I already made.
Note 2: I am using Maple 2016 and 2D Math.

I would like some insight as to how I could fix this so that it works, just like the regular recursive procedure and cyclic list that I have.

Cheers.

I want to make the blue output my procedure spits out a another color, and align it to the right, is this even possible? Or something like it?

I basically want to make a Maple procedure that does certain calculations and writes the explanation for each calculation. I do however want Maple to write these explanations as a text field like in a normal Maple worksheet, instead of the blue output in the middle. Is this possible? Or is there any alternative ideas you have that I could try? Would really appreciate any kind of help, thanks.

Hi I have this code

 psi:=proc(n,x);
 (1/sqrt(sqrt(pi)*2^n*factorial(n))*exp(-x^2/2)*HermiteH(n,x))
 end proc;
 psi := proc(n, x) exp(-1/2*x^2)*HermiteH(n, x)/sqrt(sqrt(pi)*2^n*n!) end proc

 psi2=proc(a,x);
 psi(a,x):=(1/sqrt(sqrt(pi)*2^a*factorial(a))*exp(-x^2/2)*HermiteH(a,x))
 end proc;
psi2 = (proc(a, x)
    psi(a, x) := exp(-1/2*x^2)*HermiteH(a, x)/sqrt(sqrt(pi)*2^a*a!)

end proc)
 for n from 0 to 2 do;
 for a from 0 to 2 do;
 result=proc(n,a);
 result(n,a)=psi*psi2
 end proc;
print(evalf(int(result(n,a),x=0..infinity)));
od;
od;


it returns

 Float(infinity) signum(result(0, 0))

                     Float(infinity) signum(result(0, 1))

                     Float(infinity) signum(result(0, 2))

                     Float(infinity) signum(result(1, 0))

                     Float(infinity) signum(result(1, 1))

                     Float(infinity) signum(result(1, 2))

                     Float(infinity) signum(result(2, 0))

                     Float(infinity) signum(result(2, 1))

                     Float(infinity) signum(result(2, 2))

I know the results for (0,0), (1,1) and (2,2) should be 1 and the rest should be 0.

Can anybody help fix this please

My code

checkInSet := proc(L, x)
    local i;
    if nops(L) = 0 then
        return 0;
    else    
        for i from 1 to nops(L) do
            if x = L[i] then
                return i;
            fi;
        end;
    end;
end proc:

 

Let

M := {(a-2*b-2*c)^2*(2*a-b+2*c)^2, (a-2*b-c)^2*(a-b+2*c)^2, (a-2*b+c)^2*(a+b-2*c)^2, (a-2*b+2*c)^2*(2*a+b-2*c)^2, (a-b-2*c)^2*(2*a-b+c)^2, (a-b-c)^2*(a-b+c)^2, (a-b+c)^2*(a+b-c)^2, (a-b+2*c)^2*(2*a+b-c)^2, (a+b-2*c)^2*(2*a-b-c)^2};

 

But checkInSet(M, (a-2*b-2*c)^2*(2*a-b+2*c)^2) not work? Someone please help me.

Thanks you very much.

Hello!

I was creating this code for the "Müller Method" for Numerical Analysis. Everything works fine till the if (for getting the x3) goes on. The problem is, the Maple doesn't detects the if.
I tried checking every variable and every variable is calculated except "x3"; even "disc".

I would like to know what happened. Anyways, here's the code:
http://imgur.com/a/DGlgl
 

                     
x0 := 0; x1 := 8.4; x2 := 10; iter := 1000; tol := 10^(-8); f := proc (x) options operator, arrow; 3*x^3+7*x^2+x+2 end proc; f(x); plot(f(x), color = green); printf(" n          x0               x1               x2              x3                Error \n"); for i to iter do d0 := evalf((f(x1)-f(x0))/(x1-x0)); d1 := evalf((f(x2)-f(x1))/(x2-x1)); h1 := evalf(x2-x1); h0 := evalf(x1-x0); a := evalf((d1-d0)/(h1-h0)); b := evalf(a*h1+d1); c := evalf(f(x2)); disc := sqrt(-4*a*c+b^2); if abs(b+disc) > abs(b-disc) then x3 := x2+(-2*c)*(1/(b+sqrt(-4*a*c+b^2))); erry := abs((x3-x2)/x3) else x3 := x2+(-2*c)*(1/(b-sqrt(-4*a*c+b^2))); erry := abs((x3-x2)/x3) end if; if erry > tol then x0 := x1; x1 := x2; x2 := x3; printf("%2d     %2.8f      %2.8f       %2.8f         %2.5 f     %2.8 f \n", i, a, b, c, x3, erry) else printf("una raiz es: %2.8f ", x3); break end if end do;

I put the image (as it looks on my maple) and the "code" so you can copy-paste it in Maple.

Let us consider the linear integer programming problem:

A := Matrix([[1, 7, 1, 3], [1, 6, 4, 6], [17, 1, 5, 1], [1, 6, 10, 4]]):
 n := 4; z := add(add(A[i, j]*x[i, j], j = 1 .. n), i = 1 .. n):
restr := {seq(add(x[i, j], i = 1 .. n) = 1, j = 1 .. n), seq(add(x[i, j], j = 1 .. n) = 1, i = 1 .. n)}:
 sol := Optimization[LPSolve](z, restr, assume = binary);

Error, (in Optimization:-LPSolve) no feasible integer point found; 
use feasibilitytolerance option to adjust tolerance

sol1 := Optimization[LPSolve](z, restr, assume = binary, feasibilitytolerance = 100, integertolerance = 1);

Error, (in Optimization:-LPSolve) no feasible integer point found;
 use feasibilitytolerance option to adjust tolerance

That was OK in Maple 16, outputting

.

The bug in one of the principal Maple commands lasts since Maple 2015, where the above code causes "Kernel connection has been lost". The SCRs about it were submitted three times (see http://www.mapleprimes.com/questions/204750-Bug-In-LPSolve-In-Maple-20151).

I found 4 ways so far to call a Maple sin function (to return numerical value).

Can you find more ways?

evalf(sin(Pi/2));
evalf(`sin`(Pi/2));
evalf('sin'(Pi/2));
`evalf/sin`(Pi/2);

 

 

 

 

 

I am maple newbie. Sometimes when I look at Maple code to try to understand the algorithm (which is hard, since I do not know Maple well), I see the code puts ` ` around some keywords. And sometimes it does not. For example, sometimes I see something like (these are random samples) from Maple code shown using showstat()

return `if`(assigned(r),r,{})
r := `union`(r,{solve(op(1,expr) = 1,vs)})
v2 := `intersect`(vs,indets(op(1,expr),'name'));

But sometimes, they do not put ` ` around functions name or keywords, like this:

if nops(v2) = 1 then
t1 := remove(a -> has(a,RootOf),t1);
for t in expr do

and so on.

Can some expert please give what is the rule thumb to use? Should user adopt this method also? When to put ` ` and when not to put ` `?   I understand that ` ` prevents one-time evaluation (or rather, holds off immediate evaluation) and ``  `` prevents two times evaluation and so on. But when to use ` ` is what is confusing me.

 

how to convert a nested for loop to iterative version with stack

I was trying to write a procedure that would compute a simple linear equation using the Extended Euclidean Algorithm. I was thinking of a procedure like the following:

solveEeaMatrix := proc (a::list, b::list) 
 local c::list;  
 c := a -iquo(a[1],b[1])*b;  
 print(c);  
 while (c[1] <> gcd(a[1],b[1]) do 
 ...

I am basically stuck at this part as

1) I don't know how to setup a multi-dimensional array that could dynamically grow(as a possible solution).

2) I can't come up with a recursive function that could possibly take care of this.

In short, if I am given for example an equation like: 84*x+203*y = 14

I will transform it into 2 linear equation as follow:
row0 := [203, 0, 1] row1 := [84, 1, 0] Subsequently, I will perform the following:

c := a -iquo(a[1],b[1])*b;  

Where aand b are both lists and arguments of the procedure and cbeing another list and a local variable.

But I don't know how to do the following programmatically:

row3 := row1-iquo(row1[1], row2[1])*row2;
row4 := row2-iquo(row2[1], row3[1])*row3;
row5 := row3-iquo(row3[1], row4[1])*row4;
and so on ...                  

Any hint would be appreciated.

Please Sir/Ma, I'm trying to generate a recurrent relations of this series and I try to use "if" "else" condition but I didn't get it right. Any one with useful suggestions. Appreciate 

 

restart;
Y[0] := A; Y[1] := B;
if k = a then delta(k-a) := 1 else 0 end if;
                               0
for k from 0 to 10 do Y[k+2] := solve(add(delta(i-1)*(k-i+1)*(k-i+2)*Y[k-i+2], i = 0 .. k)+add((delta(i)-delta(i-1))*(k-i+1)*Y[k-i+1], i = 0 .. k)+lambda*Y[k] = 0, Y[k+2]) end do;
y := sum(Y[j]*x^j, j = 0 .. 10);
 

Hello,

So as to build a function which gives several outputs, i have made a code with this manner :

Input:=[i1, ..., in]

Output:=proc(Input);

      export o1, o2, o3
      o1=f1(Input);
      o2=f2(Input);
      o3=f3(Input);
      end module

End proc;

By doing, output1:=Output(Input):-o1; , i obtained my result. However, as my structure begins with the procedure (proc End proc), I can't export my result in package.

For this reason, I'm thinking about changing the structure of my code by starting with the creation of the module and put all the functions inside.

Input:=[i1, ..., in]

Output:=module():

export o1,o2,o3;
option package;
o1:=proc(Input)
f1(Input);
end proc;

o2:=proc(Input)
f2(Input);
end proc;

o3:=proc(Input)
f3(Input);
end proc;
   
End module;

Can you give me your feedback on the two structures ? Do you think that the second choice Module->Functions is more appropriated ?

Thanks a lot of your help and feedback.

So below is a calculation im trying to create, but it just tells me there is an unexpected IF error, and returns me to the third For line. I have tried it straight and as a procedure but i just cant seem to get it to work. Any ideas or tweaks to make this work would be much appreciated. All of the other variables I've named work in their respective lines of code, which I have left out as it is long as it is!


ComlexAlgorithm:= proc(L)
local N, j, kf, o, DeltaMag, DeltaBond;
global NConfig, Configz1, Col, Row, OldConfig, NewConfig, NewTable;

# Set Delta values to zero.
DeltaMag:=0:
DeltaBond:=0:

# Creates loops to loop through the correct changes in values i.e. Add -1, then 1 to first point, move across each point one by one, repeat from the beginning for each configuration then repeat whole process for the amount of rows.
printlevel:=4:
for N from -1 to 1 by 2 do
   for j from 1 to Col do
      for w from 0 to NConfig-1 do   
         for o from 1 to Row do

# Calculate DeltaMag.
if N=-1 then DeltaMag:=DeltaMag+1 else DeltaMag:=DeltaMag-1
end if:

# Calculate all relevant bonds i.e. 3 values for DeltaMag, except at "corner" lattice points where there is only 2 bonds.
if N= Configz1[w](o, j) then DeltaBond:= DeltaBond-1 else DeltaBond:= DeltaBond+1
end if:

if j-1>1 then
   if N <> Configz1[w](o, j+1) then DeltaBond:= DeltaBond+1
      if N = Configz1[w](o, j+1) then DeltaBond:= DeltaBond-1
else DeltaBond:= DeltaBond
end if:
end if:
end if:

if j+1<Col then
   if N <> Configz1[w](o, j+1) then DeltaBond:= DeltaBond+1
      if N = Configz1[w](o, j+1) then DeltaBond:= DeltaBond-1
else DeltaBond:= DeltaBond
end if:
end if:
end if:

# Calculate if, and by what degree, conifguration number changes using 2^(j-1) which is the arithmetric series for binary. Changes only occur when "incoming" spin is different.
if N= -1 then
   if Configz1[w](o, j) <> N then NewConfig[w]:= OldConfig[w-(2^(j-1))] 
else NewConfig[w]:= OldConfig[w]
end if:
end if:

if N= 1 then
   if Configz1[kf](o, j) <> N then NewConfig[w]:= OldConfig[w+(2^(j-1))] 
else NewConfig[kf]:= OldConfig[w]
end if:
end if:

NewTable[kf]:= [NewConfig[w], BondEnergy(w)+DeltaBond, MagEnergy(w)+DeltaMag];

# Set Delta values to zero to finsh.
DeltaMag:=0:
DeltaBond:=0:

end do:
end do:
end do:
end do:
end proc:

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