Maple 2015 Questions and Posts

These are Posts and Questions associated with the product, Maple 2015
I have to crypt and decrypt with vigenere.

(procedures need lists)

"In a Caesar cipher, each letter of the alphabet is shifted along some number of places; for example, in a Caesar cipher of shift 3, A would become D, B would become E, Y would become B and so on. The Vigenère cipher consists of several Caesar ciphers in sequence with different shift values.

To encrypt, a table of alphabets can be used, termed a tabula recta, Vigenère square, or Vigenère table. It consists of the alphabet written out 26 times in different rows, each alphabet shifted cyclically to the left compared to the previous alphabet, corresponding to the 26 possible Caesar ciphers. At different points in the encryption process, the cipher uses a different alphabet from one of the rows. The alphabet used at each point depends on a repeating keyword.[citation needed]

For example, suppose that the plaintext to be encrypted is:

ATTACKATDAWN
The person sending the message chooses a keyword and repeats it until it matches the length of the plaintext, for example, the keyword "LEMON":

LEMON
Each row starts with a key letter. The remainder of the row holds the letters A to Z (in shifted order). Although there are 26 key rows shown, you will only use as many keys (different alphabets) as there are unique letters in the key string, here just 5 keys, {L, E, M, O, N}. For successive letters of the message, we are going to take successive letters of the key string, and encipher each message letter using its corresponding key row. Choose the next letter of the key, go along that row to find the column heading that matches the message character; the letter at the intersection of [key-row, msg-col] is the enciphered letter.

For example, the first letter of the plaintext, A, is paired with L, the first letter of the key. So use row L and column A of the Vigenère square, namely L. Similarly, for the second letter of the plaintext, the second letter of the key is used; the letter at row E and column T is X. The rest of the plaintext is enciphered in a similar fashion:

Plaintext:	ATTACKATDAWN
Key:	LEMON
Ciphertext:	LXFOPVEFRNHR
Decryption is performed by going to the row in the table corresponding to the key, finding the position of the ciphertext letter in this row, and then using the column's label as the plaintext. For example, in row L (from LEMON), the ciphertext L appears in column A, which is the first plaintext letter. Next we go to row E (from LEMON), locate the ciphertext X which is found in column T, thus T is the second plaintext letter."

I think that it can be done with a for loop but I do not know where to start.

Thanks in advance!

Dear friends:

I am facing to search the command which zoom a sactor of graph, I know it can be done by using maple tools whose present tool bar but I need a command for zooming please help me to fix this problem (attached) I want to see th sactor eta=0.8 to eta=1.2.

graph_phi_varies.mw

Please see the problem and correct as soon as possible. I am waiting your positive respone.

Muhammad Usman

School of Mathematical Sciences 
Peking University, Beijing, China

 

 

 

A man walks to different points.

I have to find the point that has a minimal length.
Perhaps through the Repetition Statement (for...while...do)

 

 

how to read a jpg picture and draw mesh on it?

mesh generator in maple

how to get all points and lines from this mesh on jpeg picture?

after solve a set of equations, i got a set of solutions, [A,B,C], if i remove one of solution [C]

is it possible to find this removed solution from solutions set [A,B]

how to check the dependency of groebner basis or a set of polynomials?

emal := proc(zipeq,abc1param,abc2param,abc3param,abc4param,abc5param)
source1 := Prefix(zipeq);
print(source1);
source2 := subs(s=abc5param,subs(v=abc4param,subs(u=abc3param,subs(t=abc2param,subs(a=abc1param,source1))))):
print(source2);
zipplus := proc(mm, pp)
 return zip((x,y) -> x+y, mm, pp)
end proc:
zipstar := proc(mm, pp)
 return zip((x,y) -> x*y, mm, pp)
end proc:
return eval(source2);
end proc:
 

would like to do the Prefix first and then eval the prefix later but do not expect do define zipstar and zippplus in global

just want to eval this Prefix which is formula of zipstar and zipplus function

 

above result same as print(source2), but eval in return , it have not evaluate, why?

if just expect to find whether equation is like

(a-b)*(c-d) + (e-f)*(g-h) = i

if use equal , it will care the variable name, is it possible treat the variable name as a wildcard to match the equation structure only 

the structure i mean is *, +, -, =, ()

Dear hope you will be fine I want to represent the data in term of solid line not dotted, asterik etc of the following code. Please help me to fix this problem

 

restart; epsilon := 0; Pr := 1; beta := .1; Sc := 1; S := 0; L := 15;
                               15
for i from -L while i <= L do a[i] := 1.0*i/L end do;
for i2 from -L while i2 <= L do fw := a[i2]; Eq1[i2] := eval(diff(F(eta), eta, eta, eta)+F(eta)*(diff(F(eta), eta, eta))-(diff(F(eta), eta))^2+S*(epsilon-(diff(F(eta), eta)))+epsilon^2); Eq2[i2] := eval((diff(G(eta), eta, eta))/Pr-G(eta)*(diff(F(eta), eta))+F(eta)*(diff(G(eta), eta))); Eq3[i2] := eval(diff(H(eta), eta, eta)+Sc*(F(eta)*(diff(H(eta), eta))-beta*H(eta))); IC[i2] := F(0) = a[i2], (D(F))(0) = 1, (D(F))(L) = epsilon, G(0) = 1, G(L) = 0, H(0) = 1, H(L) = 0; dsys1[i2] := {Eq1[i2], Eq2[i2], Eq3[i2], IC[i2]}; dsol1[i2] := dsolve(dsys1[i2], numeric, output = listprocedure, range = 0 .. L); dsol1x[i2] := subs(dsol1[i2], diff(F(eta), eta, eta)); dsol1y[i2] := subs(dsol1[i2], G(eta)); dsol1z[i2] := subs(dsol1[i2], H(eta)) end do;
for j from -L while j <= L do g[j] := eval(-dsol1x[j](0)) end do;
with(plots);

g1 := pointplot({seq([n/L, g[n]], n = -L .. L)}, symbol = asterisk, symbolsize = 15, color = blue);
display(g1);
 

how to convert a^2*b+c to func2(func1(func1(abc[1],abc[1]),abc[2]),abc[3])

when i use custom function func2 to represent plus, func1 to represent multiply

input 3 parameters,

one is a^2*b + c one is [func1, func2] and second is [abc[1],abc[2],abc[3]] corresponding to a, b, c
a^2*b + c = func2(func1(func1(abc[1],abc[1]),abc[2]),abc[3]);

I need two buttons.
one should remove a row from DataTable0
and the second should add a row.

I'm doing an interactive component and i need to plot a function and some points (which can change).
My question is if is it possible display the graphic with a value of automatic Plot Ranges.
For example the graphic would started with the first point and finisched with the last, like in this image (but i did it manually).

I want a sort of autozoom.

I am looking for persons with experience in setting up maple for the blind. The setting is a 21-year old blind
 student who wants to do math. He uses Mac OS/X and the built-in VoiceOver screen reader, plus refreshable Braille display (one-line), a Braille printer, and Nemeth, the Braille extension for math symbols.

Initially, we are working from no experience whatsoever, trying ideas, mostly failure. Getting some advice from persons who are blind and using maple would be wonderful. If anyone knows of an instance, a person who is blind and regularly uses maple, then we would both would be delighted to receive advice on what to install, how to set it up, and how to use maple on a daily basis.

One trouble for the VoiceOver screen reader is verbosity of maple output. Any methods to reduce the amount of output would be welcome. To uderstand the issue, look at the output of these two commands when interface(prettyprint=0) has been set to help out the VoiceOver screen reader:

matrix([[1,2],6,7]]); Matrix([[1,2],[6,7]]);

The first outputs the same text, which VoiceOver will read accurately. The second dumps out two lines of extraneous and confusing information, only 20% of which is useful. The verbosity is not restricted to matrix packages linalg or LinearAlgebra. It is a problem wtih nearly all packages.

-Grant Gustafson, Salt Lake City, Univ of Utah Math Dept

 

 

I'm doing this interactive component and for coordinates of points I use one TextArea for each coordinate.

a:= Do(%CB);

x1:=Do (%x1);
y1:=Do (%y1);
x2:=Do (%x2);
y2:=Do (%y2);
x3:=Do (%x3);
y3:=Do (%y3);
x4:=Do (%x4);
y4:=Do (%y4);
x5:=Do (%x5);
y5:=Do (%y5);

#x6:=Do (%x6);
#y6:=Do (%y6);
#x7:=Do (%x7);
#y7:=Do (%y7);
#x8:=Do (%x8);
#y8:=Do (%y8);
#x9:=Do (%x9);
#y9:=Do (%y9);
#x10:=Do (%x10);
#y10:=Do (%y10);




vx2:=[x1,x2];
vy2:=[y1,y2];

vx3:=[x1,x2,x3];
vy3:=[y1,y2,y3];

vx4:=[x1,x2,x3,x4];
vy4:=[y1,y2,y3,y4];

vx5:=[x1,x2,x3,x4,x5];
vy5:=[y1,y2,y3,y4,y5];



if a=2 then F2:=y=LinearFit([1, x], vx2, vy2, x);
end if:

if a=3 then F3:=evalf[2](LinearFit([1, x], vx3, vy3, x));
end if:


if a=4 then F4:=LinearFit([1, x], vx4, vy4, x);
end if:


if a=5 then F5:=LinearFit([1, x], vx5, vy5, x);
end if:

I was browsing on maple and i sow this:

I would use a table to insert coordinates of point like this image, but i don't know how can I do.

Can you help me plese?

Thanks in advance
Enrico

Dear friends:

I am facing two problems

1. one is to get solution of the below system of ODE for L=100 (highlited as red) and

2. the other is I want the graph in the form of solid line not poit, asterisk etc.

 

restart; epsilon := .1; Pr := 1; beta := .1; Sc := 1; S := 1; L := 20;
for i from -L while i <= L do;
a[i] := 1.0*i/L;
end do;
for i2 from -L while i2 <= L do;

fw := a[i2]; 

Eq1[i2] := eval(diff(F(eta), eta, eta, eta)+F(eta)*(diff(F(eta), eta, eta))-(diff(F(eta), eta))^2+S*(epsilon-(diff(F(eta), eta)))+epsilon^2);
Eq2[i2] := eval((diff(G(eta), eta, eta))/Pr-G(eta)*(diff(F(eta), eta))+F(eta)*(diff(G(eta), eta))); 
Eq3[i2] := eval(diff(H(eta), eta, eta)+Sc*(F(eta)*(diff(H(eta), eta))-beta*H(eta)));
IC[i2] := F(0) = a[i2], (D(F))(0) = 1, (D(F))(L) = epsilon, G(0) = 1, G(L) = 0, H(0) = 1, H(L) = 0;
dsys1[i2] := {Eq1[i2], Eq2[i2], Eq3[i2], IC[i2]};
dsol1[i2] := dsolve(dsys1[i2], numeric, output = listprocedure, range = 0 .. L);
dsol1x[i2] := subs(dsol1[i2], diff(F(eta), eta, eta));
dsol1y[i2] := subs(dsol1[i2], G(eta));
dsol1z[i2] := subs(dsol1[i2], H(eta)) end do;

for j from -L while j <= L do; 
g[j] := eval(-dsol1x[j](0)) end do;
with(plots); 

g6 := pointplot({seq([n/L, g[n]], n = -L .. L)}, symbol = asterisk, symbolsize = 15, color = red);
display(g6);


Please see the problem and correct as soon as possible. I am waiting your positive respone.

Muhammad Usman

School of Mathematical Sciences 
Peking University, Beijing, China

 

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