Found $x^8$ while calculating inverse of $(x^6+1)$ in finite field $GF(2^8)$. Help???Quadratic Extension of Finite fieldInverse of polynomial over $mathbb F_3$ finite field, quotient spaceExtended Euclidean Algorithm, what is our answer?Calculating the generator of a Finite FieldIs every finite extension of a finite field a finite field?Algebraic Field Extension of Finite FieldPrescribing norm and trace of elements in a finite field.Help determining if a field is finite?degree of finite extension of finite fieldHelp with a basic question on Finite Field characteristic
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Found $x^8$ while calculating inverse of $(x^6+1)$ in finite field $GF(2^8)$. Help???
Quadratic Extension of Finite fieldInverse of polynomial over $mathbb F_3$ finite field, quotient spaceExtended Euclidean Algorithm, what is our answer?Calculating the generator of a Finite FieldIs every finite extension of a finite field a finite field?Algebraic Field Extension of Finite FieldPrescribing norm and trace of elements in a finite field.Help determining if a field is finite?degree of finite extension of finite fieldHelp with a basic question on Finite Field characteristic
$begingroup$
So I was running the EEA (Extended Euclidean Algorithm) to find the multiplicative inverse of $(x^6+1)$ in the finite field $GF(2^8)$. Everything was going fine until the second last iteration where I was supposed to get my $t(x)$ auxiliary polynomial that was going to be the inverse. However, this is what I got:
$$1=(x+1)-1(x)
=(r_1+x^2 r_0+x^4 r_1+xr_0+x^3 r_1)+x(x^5 r_0+x^4 r_0+x^3 r_0+x^2 r_0+r_0+x^7 r_1+x^6 r_1+x^5 r_1+x^4 r_1+x^3 r_1+x^2 r_1+xr_1)$$
This equated to $$(x^6+x^5+x^4+x^3+x^2 ) r_0+(x^8+x^7+x^6+x^5+x^3+1)r_1$$
But as much as I know, there shouldn't be a value greater than x^7 in the polynomial, should there? Please Help I need to submit an assignment day after tomorrow...
(EDIT)
enter image description here
This is an image of the EEA calculations
finite-fields extension-field euclidean-algorithm
asked 2 days ago
hassan zaidihassan zaidi
11
New contributor
hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 2 more comments
$begingroup$
So I was running the EEA (Extended Euclidean Algorithm) to find the multiplicative inverse of $(x^6+1)$ in the finite field $GF(2^8)$. Everything was going fine until the second last iteration where I was supposed to get my $t(x)$ auxiliary polynomial that was going to be the inverse. However, this is what I got:
$$1=(x+1)-1(x)
=(r_1+x^2 r_0+x^4 r_1+xr_0+x^3 r_1)+x(x^5 r_0+x^4 r_0+x^3 r_0+x^2 r_0+r_0+x^7 r_1+x^6 r_1+x^5 r_1+x^4 r_1+x^3 r_1+x^2 r_1+xr_1)$$
This equated to $$(x^6+x^5+x^4+x^3+x^2 ) r_0+(x^8+x^7+x^6+x^5+x^3+1)r_1$$
But as much as I know, there shouldn't be a value greater than x^7 in the polynomial, should there? Please Help I need to submit an assignment day after tomorrow...
(EDIT)
enter image description here
This is an image of the EEA calculations
finite-fields extension-field euclidean-algorithm
asked 2 days ago
hassan zaidihassan zaidi
11
New contributor
hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Please use MathJax to make your question readable. Start by putting$signs around the math expressions.
$endgroup$
– saulspatz
2 days ago
$begingroup$
Changed it, sorry didnt realize that at first..
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
Welcome to Maths SX! What is $x$ here?
$endgroup$
– Bernard
2 days ago
$begingroup$
x is just a variable, that shows the bit value in an 8-bit vector. So, for example, the polynomial $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ in bit value is (11111110)
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
Btw in the look up table the inverse is actually equal to (11111110) meaning that the 4th iteration's $$r_1$$ is the correct answer
$endgroup$
– hassan zaidi
2 days ago
|
show 2 more comments
$begingroup$
So I was running the EEA (Extended Euclidean Algorithm) to find the multiplicative inverse of $(x^6+1)$ in the finite field $GF(2^8)$. Everything was going fine until the second last iteration where I was supposed to get my $t(x)$ auxiliary polynomial that was going to be the inverse. However, this is what I got:
$$1=(x+1)-1(x)
=(r_1+x^2 r_0+x^4 r_1+xr_0+x^3 r_1)+x(x^5 r_0+x^4 r_0+x^3 r_0+x^2 r_0+r_0+x^7 r_1+x^6 r_1+x^5 r_1+x^4 r_1+x^3 r_1+x^2 r_1+xr_1)$$
This equated to $$(x^6+x^5+x^4+x^3+x^2 ) r_0+(x^8+x^7+x^6+x^5+x^3+1)r_1$$
But as much as I know, there shouldn't be a value greater than x^7 in the polynomial, should there? Please Help I need to submit an assignment day after tomorrow...
(EDIT)
enter image description here
This is an image of the EEA calculations
finite-fields extension-field euclidean-algorithm
asked 2 days ago
hassan zaidihassan zaidi
11
New contributor
hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
So I was running the EEA (Extended Euclidean Algorithm) to find the multiplicative inverse of $(x^6+1)$ in the finite field $GF(2^8)$. Everything was going fine until the second last iteration where I was supposed to get my $t(x)$ auxiliary polynomial that was going to be the inverse. However, this is what I got:
$$1=(x+1)-1(x)
=(r_1+x^2 r_0+x^4 r_1+xr_0+x^3 r_1)+x(x^5 r_0+x^4 r_0+x^3 r_0+x^2 r_0+r_0+x^7 r_1+x^6 r_1+x^5 r_1+x^4 r_1+x^3 r_1+x^2 r_1+xr_1)$$
This equated to $$(x^6+x^5+x^4+x^3+x^2 ) r_0+(x^8+x^7+x^6+x^5+x^3+1)r_1$$
But as much as I know, there shouldn't be a value greater than x^7 in the polynomial, should there? Please Help I need to submit an assignment day after tomorrow...
(EDIT)
enter image description here
This is an image of the EEA calculations
finite-fields extension-field euclidean-algorithm
finite-fields extension-field euclidean-algorithm
asked 2 days ago
hassan zaidihassan zaidi
11
New contributor
hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
hassan zaidihassan zaidi
11
New contributor
hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
hassan zaidi
asked 2 days ago
hassan zaidihassan zaidi
11
New contributor
hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
hassan zaidihassan zaidi
11
asked 2 days ago
hassan zaidihassan zaidi
11
11
New contributor
hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
hassan zaidi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Please use MathJax to make your question readable. Start by putting$signs around the math expressions.
$endgroup$
– saulspatz
2 days ago
$begingroup$
Changed it, sorry didnt realize that at first..
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
Welcome to Maths SX! What is $x$ here?
$endgroup$
– Bernard
2 days ago
$begingroup$
x is just a variable, that shows the bit value in an 8-bit vector. So, for example, the polynomial $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ in bit value is (11111110)
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
Btw in the look up table the inverse is actually equal to (11111110) meaning that the 4th iteration's $$r_1$$ is the correct answer
$endgroup$
– hassan zaidi
2 days ago
|
show 2 more comments
1
$begingroup$
Please use MathJax to make your question readable. Start by putting$signs around the math expressions.
$endgroup$
– saulspatz
2 days ago
$begingroup$
Changed it, sorry didnt realize that at first..
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
Welcome to Maths SX! What is $x$ here?
$endgroup$
– Bernard
2 days ago
$begingroup$
x is just a variable, that shows the bit value in an 8-bit vector. So, for example, the polynomial $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ in bit value is (11111110)
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
Btw in the look up table the inverse is actually equal to (11111110) meaning that the 4th iteration's $$r_1$$ is the correct answer
$endgroup$
– hassan zaidi
2 days ago
1
1
$begingroup$
Please use MathJax to make your question readable. Start by putting
$ signs around the math expressions.$endgroup$
– saulspatz
2 days ago
$begingroup$
Please use MathJax to make your question readable. Start by putting
$ signs around the math expressions.$endgroup$
– saulspatz
2 days ago
$begingroup$
Changed it, sorry didnt realize that at first..
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
Changed it, sorry didnt realize that at first..
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
Welcome to Maths SX! What is $x$ here?
$endgroup$
– Bernard
2 days ago
$begingroup$
Welcome to Maths SX! What is $x$ here?
$endgroup$
– Bernard
2 days ago
$begingroup$
x is just a variable, that shows the bit value in an 8-bit vector. So, for example, the polynomial $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ in bit value is (11111110)
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
x is just a variable, that shows the bit value in an 8-bit vector. So, for example, the polynomial $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ in bit value is (11111110)
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
Btw in the look up table the inverse is actually equal to (11111110) meaning that the 4th iteration's $$r_1$$ is the correct answer
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
Btw in the look up table the inverse is actually equal to (11111110) meaning that the 4th iteration's $$r_1$$ is the correct answer
$endgroup$
– hassan zaidi
2 days ago
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Judging from the calculation at the link you provided, you're taking $ x $ to be a root of the polynomial $ x^8 + x^4 + x^3 + x + 1 $. There's nothing particulary wrong about having terms of degree $8$ or more in an expression for the inverse of an element of the field, but you can always replace them with a combination of terms of smaller degree by using the equation $ x^8 = x^4 + x^3 + x + 1 $. As it happens, when I multiplied your putative inverse $ x^8+x^7+x^6+x^5+x^3+1= x^7+x^6+x^5+x^4+x $ by $ x^6+1 $ I didn't get $1$. I got $ x^5 $ instead.
In fact, there appears to be an error on line $2$ of the calculation pointed to by your link. I believe the remainder on the right side of the equation should be $ x^4 + x^3 +x^2 + x + 1 $ rather than $ x^4 + x^3 + x + 1 $.
answered 2 days ago
lonza leggieralonza leggiera
1,03228
$endgroup$
$begingroup$
Yes you are right there was a problem in line 2, however, it was only a mistake of not writing the $$x^2$$ and I'm saying that because I rechecked my calculation twice after you pointed it out. And yes you are right $$x^7+x^6+x^5+x^4+x$$ is not the correct inverse of $$x^6+1$$ the correct inverse I got from a look table was actually $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ . And I get it now that we can write the >8 terms using smaller degrees, but still, I don't quite understand why I'm not getting the correct answer. And btw this is of tremendous help.
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
I suggest you check your calculation of the coefficient of $ r_1 $ in the second column of line 5. According to my calculations this should be $ left(x^4+x^3+1right)left(x^3+xright) + x^2 $, which gives me a different result from what you have.
$endgroup$
– lonza leggiera
yesterday
$begingroup$
Guys this was extremely helpful and I'd like to thank everyone that commented. I realized my mistake, it was in the 5th iteration. The coefficient that was next to $$r_0$$ was written as $$x^8+x^7+x^6+x^5+x^3+1$$ when it was actually $$x^8+x^7+x^6+x^5+x^2+1$$. Thank you all for the comments.
$endgroup$
– hassan zaidi
yesterday
add a comment |
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$begingroup$
Judging from the calculation at the link you provided, you're taking $ x $ to be a root of the polynomial $ x^8 + x^4 + x^3 + x + 1 $. There's nothing particulary wrong about having terms of degree $8$ or more in an expression for the inverse of an element of the field, but you can always replace them with a combination of terms of smaller degree by using the equation $ x^8 = x^4 + x^3 + x + 1 $. As it happens, when I multiplied your putative inverse $ x^8+x^7+x^6+x^5+x^3+1= x^7+x^6+x^5+x^4+x $ by $ x^6+1 $ I didn't get $1$. I got $ x^5 $ instead.
In fact, there appears to be an error on line $2$ of the calculation pointed to by your link. I believe the remainder on the right side of the equation should be $ x^4 + x^3 +x^2 + x + 1 $ rather than $ x^4 + x^3 + x + 1 $.
answered 2 days ago
lonza leggieralonza leggiera
1,03228
$endgroup$
$begingroup$
Yes you are right there was a problem in line 2, however, it was only a mistake of not writing the $$x^2$$ and I'm saying that because I rechecked my calculation twice after you pointed it out. And yes you are right $$x^7+x^6+x^5+x^4+x$$ is not the correct inverse of $$x^6+1$$ the correct inverse I got from a look table was actually $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ . And I get it now that we can write the >8 terms using smaller degrees, but still, I don't quite understand why I'm not getting the correct answer. And btw this is of tremendous help.
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
I suggest you check your calculation of the coefficient of $ r_1 $ in the second column of line 5. According to my calculations this should be $ left(x^4+x^3+1right)left(x^3+xright) + x^2 $, which gives me a different result from what you have.
$endgroup$
– lonza leggiera
yesterday
$begingroup$
Guys this was extremely helpful and I'd like to thank everyone that commented. I realized my mistake, it was in the 5th iteration. The coefficient that was next to $$r_0$$ was written as $$x^8+x^7+x^6+x^5+x^3+1$$ when it was actually $$x^8+x^7+x^6+x^5+x^2+1$$. Thank you all for the comments.
$endgroup$
– hassan zaidi
yesterday
add a comment |
$begingroup$
Judging from the calculation at the link you provided, you're taking $ x $ to be a root of the polynomial $ x^8 + x^4 + x^3 + x + 1 $. There's nothing particulary wrong about having terms of degree $8$ or more in an expression for the inverse of an element of the field, but you can always replace them with a combination of terms of smaller degree by using the equation $ x^8 = x^4 + x^3 + x + 1 $. As it happens, when I multiplied your putative inverse $ x^8+x^7+x^6+x^5+x^3+1= x^7+x^6+x^5+x^4+x $ by $ x^6+1 $ I didn't get $1$. I got $ x^5 $ instead.
In fact, there appears to be an error on line $2$ of the calculation pointed to by your link. I believe the remainder on the right side of the equation should be $ x^4 + x^3 +x^2 + x + 1 $ rather than $ x^4 + x^3 + x + 1 $.
answered 2 days ago
lonza leggieralonza leggiera
1,03228
$endgroup$
$begingroup$
Yes you are right there was a problem in line 2, however, it was only a mistake of not writing the $$x^2$$ and I'm saying that because I rechecked my calculation twice after you pointed it out. And yes you are right $$x^7+x^6+x^5+x^4+x$$ is not the correct inverse of $$x^6+1$$ the correct inverse I got from a look table was actually $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ . And I get it now that we can write the >8 terms using smaller degrees, but still, I don't quite understand why I'm not getting the correct answer. And btw this is of tremendous help.
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
I suggest you check your calculation of the coefficient of $ r_1 $ in the second column of line 5. According to my calculations this should be $ left(x^4+x^3+1right)left(x^3+xright) + x^2 $, which gives me a different result from what you have.
$endgroup$
– lonza leggiera
yesterday
$begingroup$
Guys this was extremely helpful and I'd like to thank everyone that commented. I realized my mistake, it was in the 5th iteration. The coefficient that was next to $$r_0$$ was written as $$x^8+x^7+x^6+x^5+x^3+1$$ when it was actually $$x^8+x^7+x^6+x^5+x^2+1$$. Thank you all for the comments.
$endgroup$
– hassan zaidi
yesterday
add a comment |
$begingroup$
Judging from the calculation at the link you provided, you're taking $ x $ to be a root of the polynomial $ x^8 + x^4 + x^3 + x + 1 $. There's nothing particulary wrong about having terms of degree $8$ or more in an expression for the inverse of an element of the field, but you can always replace them with a combination of terms of smaller degree by using the equation $ x^8 = x^4 + x^3 + x + 1 $. As it happens, when I multiplied your putative inverse $ x^8+x^7+x^6+x^5+x^3+1= x^7+x^6+x^5+x^4+x $ by $ x^6+1 $ I didn't get $1$. I got $ x^5 $ instead.
In fact, there appears to be an error on line $2$ of the calculation pointed to by your link. I believe the remainder on the right side of the equation should be $ x^4 + x^3 +x^2 + x + 1 $ rather than $ x^4 + x^3 + x + 1 $.
answered 2 days ago
lonza leggieralonza leggiera
1,03228
$endgroup$
Judging from the calculation at the link you provided, you're taking $ x $ to be a root of the polynomial $ x^8 + x^4 + x^3 + x + 1 $. There's nothing particulary wrong about having terms of degree $8$ or more in an expression for the inverse of an element of the field, but you can always replace them with a combination of terms of smaller degree by using the equation $ x^8 = x^4 + x^3 + x + 1 $. As it happens, when I multiplied your putative inverse $ x^8+x^7+x^6+x^5+x^3+1= x^7+x^6+x^5+x^4+x $ by $ x^6+1 $ I didn't get $1$. I got $ x^5 $ instead.
In fact, there appears to be an error on line $2$ of the calculation pointed to by your link. I believe the remainder on the right side of the equation should be $ x^4 + x^3 +x^2 + x + 1 $ rather than $ x^4 + x^3 + x + 1 $.
answered 2 days ago
lonza leggieralonza leggiera
1,03228
edited 2 days ago
answered 2 days ago
lonza leggieralonza leggiera
1,03228
answered 2 days ago
lonza leggieralonza leggiera
1,03228
answered 2 days ago
lonza leggieralonza leggiera
1,03228
1,03228
$begingroup$
Yes you are right there was a problem in line 2, however, it was only a mistake of not writing the $$x^2$$ and I'm saying that because I rechecked my calculation twice after you pointed it out. And yes you are right $$x^7+x^6+x^5+x^4+x$$ is not the correct inverse of $$x^6+1$$ the correct inverse I got from a look table was actually $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ . And I get it now that we can write the >8 terms using smaller degrees, but still, I don't quite understand why I'm not getting the correct answer. And btw this is of tremendous help.
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
I suggest you check your calculation of the coefficient of $ r_1 $ in the second column of line 5. According to my calculations this should be $ left(x^4+x^3+1right)left(x^3+xright) + x^2 $, which gives me a different result from what you have.
$endgroup$
– lonza leggiera
yesterday
$begingroup$
Guys this was extremely helpful and I'd like to thank everyone that commented. I realized my mistake, it was in the 5th iteration. The coefficient that was next to $$r_0$$ was written as $$x^8+x^7+x^6+x^5+x^3+1$$ when it was actually $$x^8+x^7+x^6+x^5+x^2+1$$. Thank you all for the comments.
$endgroup$
– hassan zaidi
yesterday
add a comment |
$begingroup$
Yes you are right there was a problem in line 2, however, it was only a mistake of not writing the $$x^2$$ and I'm saying that because I rechecked my calculation twice after you pointed it out. And yes you are right $$x^7+x^6+x^5+x^4+x$$ is not the correct inverse of $$x^6+1$$ the correct inverse I got from a look table was actually $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ . And I get it now that we can write the >8 terms using smaller degrees, but still, I don't quite understand why I'm not getting the correct answer. And btw this is of tremendous help.
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
I suggest you check your calculation of the coefficient of $ r_1 $ in the second column of line 5. According to my calculations this should be $ left(x^4+x^3+1right)left(x^3+xright) + x^2 $, which gives me a different result from what you have.
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– lonza leggiera
yesterday
$begingroup$
Guys this was extremely helpful and I'd like to thank everyone that commented. I realized my mistake, it was in the 5th iteration. The coefficient that was next to $$r_0$$ was written as $$x^8+x^7+x^6+x^5+x^3+1$$ when it was actually $$x^8+x^7+x^6+x^5+x^2+1$$. Thank you all for the comments.
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– hassan zaidi
yesterday
$begingroup$
Yes you are right there was a problem in line 2, however, it was only a mistake of not writing the $$x^2$$ and I'm saying that because I rechecked my calculation twice after you pointed it out. And yes you are right $$x^7+x^6+x^5+x^4+x$$ is not the correct inverse of $$x^6+1$$ the correct inverse I got from a look table was actually $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ . And I get it now that we can write the >8 terms using smaller degrees, but still, I don't quite understand why I'm not getting the correct answer. And btw this is of tremendous help.
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– hassan zaidi
2 days ago
$begingroup$
Yes you are right there was a problem in line 2, however, it was only a mistake of not writing the $$x^2$$ and I'm saying that because I rechecked my calculation twice after you pointed it out. And yes you are right $$x^7+x^6+x^5+x^4+x$$ is not the correct inverse of $$x^6+1$$ the correct inverse I got from a look table was actually $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ . And I get it now that we can write the >8 terms using smaller degrees, but still, I don't quite understand why I'm not getting the correct answer. And btw this is of tremendous help.
$endgroup$
– hassan zaidi
2 days ago
$begingroup$
I suggest you check your calculation of the coefficient of $ r_1 $ in the second column of line 5. According to my calculations this should be $ left(x^4+x^3+1right)left(x^3+xright) + x^2 $, which gives me a different result from what you have.
$endgroup$
– lonza leggiera
yesterday
$begingroup$
I suggest you check your calculation of the coefficient of $ r_1 $ in the second column of line 5. According to my calculations this should be $ left(x^4+x^3+1right)left(x^3+xright) + x^2 $, which gives me a different result from what you have.
$endgroup$
– lonza leggiera
yesterday
$begingroup$
Guys this was extremely helpful and I'd like to thank everyone that commented. I realized my mistake, it was in the 5th iteration. The coefficient that was next to $$r_0$$ was written as $$x^8+x^7+x^6+x^5+x^3+1$$ when it was actually $$x^8+x^7+x^6+x^5+x^2+1$$. Thank you all for the comments.
$endgroup$
– hassan zaidi
yesterday
$begingroup$
Guys this was extremely helpful and I'd like to thank everyone that commented. I realized my mistake, it was in the 5th iteration. The coefficient that was next to $$r_0$$ was written as $$x^8+x^7+x^6+x^5+x^3+1$$ when it was actually $$x^8+x^7+x^6+x^5+x^2+1$$. Thank you all for the comments.
$endgroup$
– hassan zaidi
yesterday
add a comment |
hassan zaidi is a new contributor. Be nice, and check out our Code of Conduct.
hassan zaidi is a new contributor. Be nice, and check out our Code of Conduct.
hassan zaidi is a new contributor. Be nice, and check out our Code of Conduct.
hassan zaidi is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Please use MathJax to make your question readable. Start by putting
$signs around the math expressions.$endgroup$
– saulspatz
2 days ago
$begingroup$
Changed it, sorry didnt realize that at first..
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– hassan zaidi
2 days ago
$begingroup$
Welcome to Maths SX! What is $x$ here?
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– Bernard
2 days ago
$begingroup$
x is just a variable, that shows the bit value in an 8-bit vector. So, for example, the polynomial $$x^7+x^6+x^5+x^4+x^3+x^2+x$$ in bit value is (11111110)
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– hassan zaidi
2 days ago
$begingroup$
Btw in the look up table the inverse is actually equal to (11111110) meaning that the 4th iteration's $$r_1$$ is the correct answer
$endgroup$
– hassan zaidi
2 days ago