gepo

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16 years, 20 days

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These are answers submitted by gepo

I am writing a function in worksheet mode. I made a newline with "shift+enter". A new line indeed is made. like: MON_DEGREE :=proc(f,variables,weights) local deg, i; if not type(f,polynom) then ERROR(`ERROR: The First argument must be a monomial.`); end if; end But I found in other worksheet file, it is like: >MON_DEGREE :=proc(f,variables,weights) local deg, i; >if not type(f,polynom) then >ERROR(`ERROR: The First argument must be a monomial.`); >end if; >end there is a '>' at the beginning of each new line. How can I make it in this way? Only "shift+center" is not enough. thank you.

Just a quick question:

The string in my file "s.txt" is

p=a^c*m;

after parsing it, it became:

ac m;

 

is that weird?

 

Thanks

Sorry, I fixed it.

Before using cnt, first assigning some value to it.

T[cnt]:=parse(line);

and then

for cnt do

......

 

 

Thanks a lot.

I got an error message:

"error, (in unknown) invalid terms in product"

HandleGB :=proc(file) local line, T,L;
f:=fopen(file,READ);
try
T:=table();
for cnt do
 line:=readline(f);
if line=0 then break; end if;
T[cnt]:=parse(line);
end do;
finally
fclose(f);
end try; L:=convert(T,'list');
end;

 

 

Thanks

 

I changed the code in your way:

/***************code***********************/

Handle :=proc(f) local line,line2, L;
L:=[];
line:=readline("s.txt");
while line <> 0
do
L:=[op(L),line2];
line:=readline("s.txt");
line2:=parse(line);
end do;
L;
end proc;

/***************code***********************/

when I executed the function, I got a message:

invalid input: parse expects its 1st argument, st, to be of type string, but received 0

 

Could you explain it a little bit?

Thanks

another weird problem is when I created a file using document mode, the readline can work fine. readline("s.txt"); "1+1=2" the problem here is that i do not want the quotation marks ". How can I remove the quotation marks ". Thanks in advance.
Thank you. I just have a quick read and it is not easy to master the openmaple quickly. I just want to use Groebner Basis in my application. Does anyone have any ideas how to do that?
Thank you. It works.
Sorry, the second question should be: Why are there brackets in the result which is actually NOT a fully expanded expression.
it works. Thanks.
For a given galois field, suppose there is such a polynomial function mapping from GF(q) to GF(q): x^(q+m)+x^(q+m-1)+x^(q+m-2)+...+x^(1). I can use x^q-x=0 to "reduce" it to such the following form: x^(q-1)+x^(q-2)+x^(q-3)+...+x^(1) now, the problem is whether this is the only method (x^q-x=0) to reduce a polynomial function mapping from GF(q) to GF(q)? Thanks a lot.
From GF(2^m) to GF(2^m), there is a polynomial function: sum of f(a)(1-(x-a)^(q-1)), for all a belongs to GF(2^m), I wonder whether such a polynomial function is canonical? Thanks in advance , Alex.
alias(f=RootOf(_Z^2+_Z+1)); with calculations mod 2 defines GF(4) with elements 0, 1, f, f+1. -------------------------------------------------------------------------------- for this part, I can understand, but i cannot understand the following part: -------------------------------------------------------------------------------- alias(k=RootOf(_Z^2+_Z+f)); with calculations mod 2 defines GF(16) with elements 0, 1, f, f+1, k, k+1, k+f, k+f+1, kf, kf+1, kf+f, kf+f+1, kf+k, kf+k+1, kf+k+f, kf+k+f+1. ---------------------------------------------------------------------------- this is the form for each element in GF(16), and x is the root of _Z^2+_Z+f=0 ax^3+bx^2+cx+d substitute a, b,c,d from 0000 to 1111, we can get 16 different elements. How can we generate 'f' here? Thanks
yes, but do we have a polynomial function between them? For example, from GF(4) to GF(16)? Besides, whether there is a function from GF(16) to GF(4)? Thanks a lot, Alex.

Thank you for your explain. 

After further study, I do not think there is such a mapping exist. If such a mapping exist, I mean, if there is a polynomial function to map between  Z(2^m) and GF(2^m) to satisfy the addition and multiplication ,then, Z(2^m) will be a field which is impossible.

 

 

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