Magma function for modulo irreducible polynomial
$begingroup$
So, I am trying to make a program in Magma which returns the value table of a given function F over a field $GF(2^n)$. To do so I need a irreducible polyomial. For example, I've considered $GF(2^3)$ and the irreducible polynomial $p(x)=x^3+x+1$.
My program started like this:
F<a>:=GF(2^3);
for i in F do
i mod a^3+a+1;
end for;
The 'mod' apperantly only works with integers, is there a polynomial version for this?
finite-fields irreducible-polynomials cryptography magma-cas magma
asked Dec 20 '18 at 12:40
zermelovaczermelovac
504212
$endgroup$
add a comment |
$begingroup$
So, I am trying to make a program in Magma which returns the value table of a given function F over a field $GF(2^n)$. To do so I need a irreducible polyomial. For example, I've considered $GF(2^3)$ and the irreducible polynomial $p(x)=x^3+x+1$.
My program started like this:
F<a>:=GF(2^3);
for i in F do
i mod a^3+a+1;
end for;
The 'mod' apperantly only works with integers, is there a polynomial version for this?
finite-fields irreducible-polynomials cryptography magma-cas magma
asked Dec 20 '18 at 12:40
zermelovaczermelovac
504212
$endgroup$
add a comment |
$begingroup$
So, I am trying to make a program in Magma which returns the value table of a given function F over a field $GF(2^n)$. To do so I need a irreducible polyomial. For example, I've considered $GF(2^3)$ and the irreducible polynomial $p(x)=x^3+x+1$.
My program started like this:
F<a>:=GF(2^3);
for i in F do
i mod a^3+a+1;
end for;
The 'mod' apperantly only works with integers, is there a polynomial version for this?
finite-fields irreducible-polynomials cryptography magma-cas magma
asked Dec 20 '18 at 12:40
zermelovaczermelovac
504212
$endgroup$
So, I am trying to make a program in Magma which returns the value table of a given function F over a field $GF(2^n)$. To do so I need a irreducible polyomial. For example, I've considered $GF(2^3)$ and the irreducible polynomial $p(x)=x^3+x+1$.
My program started like this:
F<a>:=GF(2^3);
for i in F do
i mod a^3+a+1;
end for;
The 'mod' apperantly only works with integers, is there a polynomial version for this?
finite-fields irreducible-polynomials cryptography magma-cas magma
finite-fields irreducible-polynomials cryptography magma-cas magma
asked Dec 20 '18 at 12:40
zermelovaczermelovac
504212
asked Dec 20 '18 at 12:40
zermelovaczermelovac
504212
asked Dec 20 '18 at 12:40
zermelovaczermelovac
504212
asked Dec 20 '18 at 12:40
zermelovaczermelovac
504212
asked Dec 20 '18 at 12:40
zermelovaczermelovac
504212
504212
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The mod function works for polynomials, provided they are recognized by Magma as being elements in a polynomial ring. For example, put F := GF(2); and P<a> := PolynomialRing(F); and you will get the results you want if you ask for a^i mod a^3+a+1;.
Alternately, specific for your finite field example, you can put F<a> := GF(2^3); which you can verify is constructed with $x^3+x+1$ as the minimal polynomial for a (ask for DefiningPolynomial(F);). By default, Magma will print elements of F as powers of a, but if you put in the command SetPowerPrinting(F,false); it will give you a reduced polynomial in a instead. So then you can just type a^i; and it will return this field element as the remainder when divided by $a^3+a+1$.
(Note that if you type both F<a> := GF(2^3); and P<a> := PolynomialRing(F); then you have introduced some confusion as to whether a is a finite field element, or an indeterminate in your polynomial ring. You should really avoid doing this.)
answered Jan 7 at 22:07
Morgan RodgersMorgan Rodgers
9,85021440
$endgroup$
add a comment |
$begingroup$
I managed to make the program.
F<a>:=GF(2^3);
P<a>:=PolynomialRing(F);
function f(x)
return x^5;
end function;
A:=;
for i in [1..6] do
Append(~A, (a^i mod (a^3+a+1)));
end for;
A:=Reverse(A);
A:=Append(A,1);
A:=Append(A,0);
A:=Reverse(A);
F:=;
for i in [1..6] do
Append(~F,((f(a))^i mod (a^3+a+1)));
end for;
F:=Reverse(F);
F:=Append(F,1);
F:=Append(F,0);
F:=Reverse(F);
BinA:=;
for i in A do
for j in [a^2,a,1] do
if j in Terms(i) then
Append(~BinA,1);
else
Append(~BinA,0);
end if;
end for;
end for;
BinA:=Matrix(3,BinA);
BinF:=;
for i in F do
for j in [a^2,a,1] do
if j in Terms(i) then
Append(~BinF,1);
else
Append(~BinF,0);
end if;
end for;
end for;
print BinF;
answered Dec 20 '18 at 18:54
zermelovaczermelovac
504212
$endgroup$
$begingroup$
You could just doSetPowerPrinting(F,false);and then typing ina^iwill automatically give you the reduced form modulo the defining polynomial of the field (which is coincidentally $x^{3}+x+1$).
$endgroup$
– Morgan Rodgers
Dec 21 '18 at 6:52
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
The mod function works for polynomials, provided they are recognized by Magma as being elements in a polynomial ring. For example, put F := GF(2); and P<a> := PolynomialRing(F); and you will get the results you want if you ask for a^i mod a^3+a+1;.
Alternately, specific for your finite field example, you can put F<a> := GF(2^3); which you can verify is constructed with $x^3+x+1$ as the minimal polynomial for a (ask for DefiningPolynomial(F);). By default, Magma will print elements of F as powers of a, but if you put in the command SetPowerPrinting(F,false); it will give you a reduced polynomial in a instead. So then you can just type a^i; and it will return this field element as the remainder when divided by $a^3+a+1$.
(Note that if you type both F<a> := GF(2^3); and P<a> := PolynomialRing(F); then you have introduced some confusion as to whether a is a finite field element, or an indeterminate in your polynomial ring. You should really avoid doing this.)
answered Jan 7 at 22:07
Morgan RodgersMorgan Rodgers
9,85021440
$endgroup$
add a comment |
$begingroup$
The mod function works for polynomials, provided they are recognized by Magma as being elements in a polynomial ring. For example, put F := GF(2); and P<a> := PolynomialRing(F); and you will get the results you want if you ask for a^i mod a^3+a+1;.
Alternately, specific for your finite field example, you can put F<a> := GF(2^3); which you can verify is constructed with $x^3+x+1$ as the minimal polynomial for a (ask for DefiningPolynomial(F);). By default, Magma will print elements of F as powers of a, but if you put in the command SetPowerPrinting(F,false); it will give you a reduced polynomial in a instead. So then you can just type a^i; and it will return this field element as the remainder when divided by $a^3+a+1$.
(Note that if you type both F<a> := GF(2^3); and P<a> := PolynomialRing(F); then you have introduced some confusion as to whether a is a finite field element, or an indeterminate in your polynomial ring. You should really avoid doing this.)
answered Jan 7 at 22:07
Morgan RodgersMorgan Rodgers
9,85021440
$endgroup$
add a comment |
$begingroup$
The mod function works for polynomials, provided they are recognized by Magma as being elements in a polynomial ring. For example, put F := GF(2); and P<a> := PolynomialRing(F); and you will get the results you want if you ask for a^i mod a^3+a+1;.
Alternately, specific for your finite field example, you can put F<a> := GF(2^3); which you can verify is constructed with $x^3+x+1$ as the minimal polynomial for a (ask for DefiningPolynomial(F);). By default, Magma will print elements of F as powers of a, but if you put in the command SetPowerPrinting(F,false); it will give you a reduced polynomial in a instead. So then you can just type a^i; and it will return this field element as the remainder when divided by $a^3+a+1$.
(Note that if you type both F<a> := GF(2^3); and P<a> := PolynomialRing(F); then you have introduced some confusion as to whether a is a finite field element, or an indeterminate in your polynomial ring. You should really avoid doing this.)
answered Jan 7 at 22:07
Morgan RodgersMorgan Rodgers
9,85021440
$endgroup$
The mod function works for polynomials, provided they are recognized by Magma as being elements in a polynomial ring. For example, put F := GF(2); and P<a> := PolynomialRing(F); and you will get the results you want if you ask for a^i mod a^3+a+1;.
Alternately, specific for your finite field example, you can put F<a> := GF(2^3); which you can verify is constructed with $x^3+x+1$ as the minimal polynomial for a (ask for DefiningPolynomial(F);). By default, Magma will print elements of F as powers of a, but if you put in the command SetPowerPrinting(F,false); it will give you a reduced polynomial in a instead. So then you can just type a^i; and it will return this field element as the remainder when divided by $a^3+a+1$.
(Note that if you type both F<a> := GF(2^3); and P<a> := PolynomialRing(F); then you have introduced some confusion as to whether a is a finite field element, or an indeterminate in your polynomial ring. You should really avoid doing this.)
answered Jan 7 at 22:07
Morgan RodgersMorgan Rodgers
9,85021440
answered Jan 7 at 22:07
Morgan RodgersMorgan Rodgers
9,85021440
answered Jan 7 at 22:07
Morgan RodgersMorgan Rodgers
9,85021440
answered Jan 7 at 22:07
Morgan RodgersMorgan Rodgers
9,85021440
9,85021440
add a comment |
add a comment |
$begingroup$
I managed to make the program.
F<a>:=GF(2^3);
P<a>:=PolynomialRing(F);
function f(x)
return x^5;
end function;
A:=;
for i in [1..6] do
Append(~A, (a^i mod (a^3+a+1)));
end for;
A:=Reverse(A);
A:=Append(A,1);
A:=Append(A,0);
A:=Reverse(A);
F:=;
for i in [1..6] do
Append(~F,((f(a))^i mod (a^3+a+1)));
end for;
F:=Reverse(F);
F:=Append(F,1);
F:=Append(F,0);
F:=Reverse(F);
BinA:=;
for i in A do
for j in [a^2,a,1] do
if j in Terms(i) then
Append(~BinA,1);
else
Append(~BinA,0);
end if;
end for;
end for;
BinA:=Matrix(3,BinA);
BinF:=;
for i in F do
for j in [a^2,a,1] do
if j in Terms(i) then
Append(~BinF,1);
else
Append(~BinF,0);
end if;
end for;
end for;
print BinF;
answered Dec 20 '18 at 18:54
zermelovaczermelovac
504212
$endgroup$
$begingroup$
You could just doSetPowerPrinting(F,false);and then typing ina^iwill automatically give you the reduced form modulo the defining polynomial of the field (which is coincidentally $x^{3}+x+1$).
$endgroup$
– Morgan Rodgers
Dec 21 '18 at 6:52
add a comment |
$begingroup$
I managed to make the program.
F<a>:=GF(2^3);
P<a>:=PolynomialRing(F);
function f(x)
return x^5;
end function;
A:=;
for i in [1..6] do
Append(~A, (a^i mod (a^3+a+1)));
end for;
A:=Reverse(A);
A:=Append(A,1);
A:=Append(A,0);
A:=Reverse(A);
F:=;
for i in [1..6] do
Append(~F,((f(a))^i mod (a^3+a+1)));
end for;
F:=Reverse(F);
F:=Append(F,1);
F:=Append(F,0);
F:=Reverse(F);
BinA:=;
for i in A do
for j in [a^2,a,1] do
if j in Terms(i) then
Append(~BinA,1);
else
Append(~BinA,0);
end if;
end for;
end for;
BinA:=Matrix(3,BinA);
BinF:=;
for i in F do
for j in [a^2,a,1] do
if j in Terms(i) then
Append(~BinF,1);
else
Append(~BinF,0);
end if;
end for;
end for;
print BinF;
answered Dec 20 '18 at 18:54
zermelovaczermelovac
504212
$endgroup$
$begingroup$
You could just doSetPowerPrinting(F,false);and then typing ina^iwill automatically give you the reduced form modulo the defining polynomial of the field (which is coincidentally $x^{3}+x+1$).
$endgroup$
– Morgan Rodgers
Dec 21 '18 at 6:52
add a comment |
$begingroup$
I managed to make the program.
F<a>:=GF(2^3);
P<a>:=PolynomialRing(F);
function f(x)
return x^5;
end function;
A:=;
for i in [1..6] do
Append(~A, (a^i mod (a^3+a+1)));
end for;
A:=Reverse(A);
A:=Append(A,1);
A:=Append(A,0);
A:=Reverse(A);
F:=;
for i in [1..6] do
Append(~F,((f(a))^i mod (a^3+a+1)));
end for;
F:=Reverse(F);
F:=Append(F,1);
F:=Append(F,0);
F:=Reverse(F);
BinA:=;
for i in A do
for j in [a^2,a,1] do
if j in Terms(i) then
Append(~BinA,1);
else
Append(~BinA,0);
end if;
end for;
end for;
BinA:=Matrix(3,BinA);
BinF:=;
for i in F do
for j in [a^2,a,1] do
if j in Terms(i) then
Append(~BinF,1);
else
Append(~BinF,0);
end if;
end for;
end for;
print BinF;
answered Dec 20 '18 at 18:54
zermelovaczermelovac
504212
$endgroup$
I managed to make the program.
F<a>:=GF(2^3);
P<a>:=PolynomialRing(F);
function f(x)
return x^5;
end function;
A:=;
for i in [1..6] do
Append(~A, (a^i mod (a^3+a+1)));
end for;
A:=Reverse(A);
A:=Append(A,1);
A:=Append(A,0);
A:=Reverse(A);
F:=;
for i in [1..6] do
Append(~F,((f(a))^i mod (a^3+a+1)));
end for;
F:=Reverse(F);
F:=Append(F,1);
F:=Append(F,0);
F:=Reverse(F);
BinA:=;
for i in A do
for j in [a^2,a,1] do
if j in Terms(i) then
Append(~BinA,1);
else
Append(~BinA,0);
end if;
end for;
end for;
BinA:=Matrix(3,BinA);
BinF:=;
for i in F do
for j in [a^2,a,1] do
if j in Terms(i) then
Append(~BinF,1);
else
Append(~BinF,0);
end if;
end for;
end for;
print BinF;
answered Dec 20 '18 at 18:54
zermelovaczermelovac
504212
answered Dec 20 '18 at 18:54
zermelovaczermelovac
504212
answered Dec 20 '18 at 18:54
zermelovaczermelovac
504212
answered Dec 20 '18 at 18:54
zermelovaczermelovac
504212
504212
$begingroup$
You could just doSetPowerPrinting(F,false);and then typing ina^iwill automatically give you the reduced form modulo the defining polynomial of the field (which is coincidentally $x^{3}+x+1$).
$endgroup$
– Morgan Rodgers
Dec 21 '18 at 6:52
add a comment |
$begingroup$
You could just doSetPowerPrinting(F,false);and then typing ina^iwill automatically give you the reduced form modulo the defining polynomial of the field (which is coincidentally $x^{3}+x+1$).
$endgroup$
– Morgan Rodgers
Dec 21 '18 at 6:52
$begingroup$
You could just do
SetPowerPrinting(F,false); and then typing in a^i will automatically give you the reduced form modulo the defining polynomial of the field (which is coincidentally $x^{3}+x+1$).$endgroup$
– Morgan Rodgers
Dec 21 '18 at 6:52
$begingroup$
You could just do
SetPowerPrinting(F,false); and then typing in a^i will automatically give you the reduced form modulo the defining polynomial of the field (which is coincidentally $x^{3}+x+1$).$endgroup$
– Morgan Rodgers
Dec 21 '18 at 6:52
add a comment |
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