pyfinite gives wrong result for multiplication in field GF(2^8)
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I am using pyfinte to calculate multiplication for AES over the field it uses which is F(2^8) but when I do as following:
from pyfinite import ffield
a = 0xbf
b = 0x03
F = ffield.FField(8)
c = F.Multiply(a, b)
print(hex(c))
the actual result is 0xdc but it should be 0xda here is how I solve it by hand:
0xbf*0x03 = 0xbf * 0x02 + 0xbf * 0x01 = 0b10111111 * 0x02 + 0xbf = 0b01111110 + 0x1B + 0xbf = 0b11011010 = 0xda
here is better format of my hand soloving:
so where is the wrong? is it on my solving by hand? or in pyfinite? how to solve this?
python aes finite-field
asked Jan 4 at 13:19
markmark
477
|
show 1 more comment
I am using pyfinte to calculate multiplication for AES over the field it uses which is F(2^8) but when I do as following:
from pyfinite import ffield
a = 0xbf
b = 0x03
F = ffield.FField(8)
c = F.Multiply(a, b)
print(hex(c))
the actual result is 0xdc but it should be 0xda here is how I solve it by hand:
0xbf*0x03 = 0xbf * 0x02 + 0xbf * 0x01 = 0b10111111 * 0x02 + 0xbf = 0b01111110 + 0x1B + 0xbf = 0b11011010 = 0xda
here is better format of my hand soloving:
so where is the wrong? is it on my solving by hand? or in pyfinite? how to solve this?
python aes finite-field
asked Jan 4 at 13:19
markmark
477
Could you format and/or annotate your hand math a little more? I'm having a hard time following it. It's been a while, but I followed Wikipedia's guidance and got 0xdc just like pyfinite did. (Edit: which I guess makes this more of a math question than a programming question.)
– glibdud
Jan 4 at 13:43
@glibdud just a minute to formate my hand hand solving better
– mark
Jan 4 at 13:50
@glibdud I did what you asked and updated my question to include latex detailed formatted solving by hand
– mark
Jan 4 at 14:18
How do you know that the modulus is 0x11B?
– Matt Timmermans
Jan 4 at 14:31
@MattTimmermans this is the way we learned in college couple days ago but I do not know if I misunderstood here is summary: when multiplying any number X by 03 in MixColumns stage in AES we multipy X by 02 and then Xor it with X . Multiplying X by 2 is as follow: we convert X to it is binary representation and we look at the leftmost bit if it is one or zero then we shift X to the left by one and Xor it with 0x1B if the leftmost bit was one. I do not know if I am confusing what we learn with what I read online about AES I think what we learned is shortcut for multiply over GF(2^8)
– mark
Jan 4 at 14:38
|
show 1 more comment
I am using pyfinte to calculate multiplication for AES over the field it uses which is F(2^8) but when I do as following:
from pyfinite import ffield
a = 0xbf
b = 0x03
F = ffield.FField(8)
c = F.Multiply(a, b)
print(hex(c))
the actual result is 0xdc but it should be 0xda here is how I solve it by hand:
0xbf*0x03 = 0xbf * 0x02 + 0xbf * 0x01 = 0b10111111 * 0x02 + 0xbf = 0b01111110 + 0x1B + 0xbf = 0b11011010 = 0xda
here is better format of my hand soloving:
so where is the wrong? is it on my solving by hand? or in pyfinite? how to solve this?
python aes finite-field
asked Jan 4 at 13:19
markmark
477
I am using pyfinte to calculate multiplication for AES over the field it uses which is F(2^8) but when I do as following:
from pyfinite import ffield
a = 0xbf
b = 0x03
F = ffield.FField(8)
c = F.Multiply(a, b)
print(hex(c))
the actual result is 0xdc but it should be 0xda here is how I solve it by hand:
0xbf*0x03 = 0xbf * 0x02 + 0xbf * 0x01 = 0b10111111 * 0x02 + 0xbf = 0b01111110 + 0x1B + 0xbf = 0b11011010 = 0xda
here is better format of my hand soloving:
so where is the wrong? is it on my solving by hand? or in pyfinite? how to solve this?
python aes finite-field
python aes finite-field
asked Jan 4 at 13:19
markmark
477
asked Jan 4 at 13:19
markmark
477
edited Jan 4 at 14:16
mark
asked Jan 4 at 13:19
markmark
477
asked Jan 4 at 13:19
markmark
477
asked Jan 4 at 13:19
markmark
477
477
Could you format and/or annotate your hand math a little more? I'm having a hard time following it. It's been a while, but I followed Wikipedia's guidance and got 0xdc just like pyfinite did. (Edit: which I guess makes this more of a math question than a programming question.)
– glibdud
Jan 4 at 13:43
@glibdud just a minute to formate my hand hand solving better
– mark
Jan 4 at 13:50
@glibdud I did what you asked and updated my question to include latex detailed formatted solving by hand
– mark
Jan 4 at 14:18
How do you know that the modulus is 0x11B?
– Matt Timmermans
Jan 4 at 14:31
@MattTimmermans this is the way we learned in college couple days ago but I do not know if I misunderstood here is summary: when multiplying any number X by 03 in MixColumns stage in AES we multipy X by 02 and then Xor it with X . Multiplying X by 2 is as follow: we convert X to it is binary representation and we look at the leftmost bit if it is one or zero then we shift X to the left by one and Xor it with 0x1B if the leftmost bit was one. I do not know if I am confusing what we learn with what I read online about AES I think what we learned is shortcut for multiply over GF(2^8)
– mark
Jan 4 at 14:38
|
show 1 more comment
Could you format and/or annotate your hand math a little more? I'm having a hard time following it. It's been a while, but I followed Wikipedia's guidance and got 0xdc just like pyfinite did. (Edit: which I guess makes this more of a math question than a programming question.)
– glibdud
Jan 4 at 13:43
@glibdud just a minute to formate my hand hand solving better
– mark
Jan 4 at 13:50
@glibdud I did what you asked and updated my question to include latex detailed formatted solving by hand
– mark
Jan 4 at 14:18
How do you know that the modulus is 0x11B?
– Matt Timmermans
Jan 4 at 14:31
@MattTimmermans this is the way we learned in college couple days ago but I do not know if I misunderstood here is summary: when multiplying any number X by 03 in MixColumns stage in AES we multipy X by 02 and then Xor it with X . Multiplying X by 2 is as follow: we convert X to it is binary representation and we look at the leftmost bit if it is one or zero then we shift X to the left by one and Xor it with 0x1B if the leftmost bit was one. I do not know if I am confusing what we learn with what I read online about AES I think what we learned is shortcut for multiply over GF(2^8)
– mark
Jan 4 at 14:38
Could you format and/or annotate your hand math a little more? I'm having a hard time following it. It's been a while, but I followed Wikipedia's guidance and got 0xdc just like pyfinite did. (Edit: which I guess makes this more of a math question than a programming question.)
– glibdud
Jan 4 at 13:43
Could you format and/or annotate your hand math a little more? I'm having a hard time following it. It's been a while, but I followed Wikipedia's guidance and got 0xdc just like pyfinite did. (Edit: which I guess makes this more of a math question than a programming question.)
– glibdud
Jan 4 at 13:43
@glibdud just a minute to formate my hand hand solving better
– mark
Jan 4 at 13:50
@glibdud just a minute to formate my hand hand solving better
– mark
Jan 4 at 13:50
@glibdud I did what you asked and updated my question to include latex detailed formatted solving by hand
– mark
Jan 4 at 14:18
@glibdud I did what you asked and updated my question to include latex detailed formatted solving by hand
– mark
Jan 4 at 14:18
How do you know that the modulus is 0x11B?
– Matt Timmermans
Jan 4 at 14:31
How do you know that the modulus is 0x11B?
– Matt Timmermans
Jan 4 at 14:31
@MattTimmermans this is the way we learned in college couple days ago but I do not know if I misunderstood here is summary: when multiplying any number X by 03 in MixColumns stage in AES we multipy X by 02 and then Xor it with X . Multiplying X by 2 is as follow: we convert X to it is binary representation and we look at the leftmost bit if it is one or zero then we shift X to the left by one and Xor it with 0x1B if the leftmost bit was one. I do not know if I am confusing what we learn with what I read online about AES I think what we learned is shortcut for multiply over GF(2^8)
– mark
Jan 4 at 14:38
@MattTimmermans this is the way we learned in college couple days ago but I do not know if I misunderstood here is summary: when multiplying any number X by 03 in MixColumns stage in AES we multipy X by 02 and then Xor it with X . Multiplying X by 2 is as follow: we convert X to it is binary representation and we look at the leftmost bit if it is one or zero then we shift X to the left by one and Xor it with 0x1B if the leftmost bit was one. I do not know if I am confusing what we learn with what I read online about AES I think what we learned is shortcut for multiply over GF(2^8)
– mark
Jan 4 at 14:38
|
show 1 more comment
1 Answer
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I'm not intimately familiar with AES, but from the link you provided it appears that the generator polynomial for the field is 0x11b. By default, pyfinite uses a different generator, which will produce different multiplication results:
>>> f = ffield.FField(8)
>>> hex(f.generator)
'0x11d'
If you specify the generator when you create the field object, you get the result you're expecting:
>>> f = ffield.FField(8, gen=0x11b, useLUT=0)
>>> hex(f.Multiply(0xbf, 0x03))
'0xda'
answered Jan 4 at 15:45
glibdudglibdud
5,71921731
I can confirm AES uses 0x11b, where all non-zero elements can be considered to be some power of 0x03. For 0x11d, all non-zero elements can be considered to be a power of 0x02. Most implementations involving finite fields will choose a polynomial where all non-zero elements are a power of 2. I don't know why AES choose 0x11b.
– rcgldr
Jan 4 at 16:50
add a comment |
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I'm not intimately familiar with AES, but from the link you provided it appears that the generator polynomial for the field is 0x11b. By default, pyfinite uses a different generator, which will produce different multiplication results:
>>> f = ffield.FField(8)
>>> hex(f.generator)
'0x11d'
If you specify the generator when you create the field object, you get the result you're expecting:
>>> f = ffield.FField(8, gen=0x11b, useLUT=0)
>>> hex(f.Multiply(0xbf, 0x03))
'0xda'
answered Jan 4 at 15:45
glibdudglibdud
5,71921731
I can confirm AES uses 0x11b, where all non-zero elements can be considered to be some power of 0x03. For 0x11d, all non-zero elements can be considered to be a power of 0x02. Most implementations involving finite fields will choose a polynomial where all non-zero elements are a power of 2. I don't know why AES choose 0x11b.
– rcgldr
Jan 4 at 16:50
add a comment |
I'm not intimately familiar with AES, but from the link you provided it appears that the generator polynomial for the field is 0x11b. By default, pyfinite uses a different generator, which will produce different multiplication results:
>>> f = ffield.FField(8)
>>> hex(f.generator)
'0x11d'
If you specify the generator when you create the field object, you get the result you're expecting:
>>> f = ffield.FField(8, gen=0x11b, useLUT=0)
>>> hex(f.Multiply(0xbf, 0x03))
'0xda'
answered Jan 4 at 15:45
glibdudglibdud
5,71921731
I can confirm AES uses 0x11b, where all non-zero elements can be considered to be some power of 0x03. For 0x11d, all non-zero elements can be considered to be a power of 0x02. Most implementations involving finite fields will choose a polynomial where all non-zero elements are a power of 2. I don't know why AES choose 0x11b.
– rcgldr
Jan 4 at 16:50
add a comment |
I'm not intimately familiar with AES, but from the link you provided it appears that the generator polynomial for the field is 0x11b. By default, pyfinite uses a different generator, which will produce different multiplication results:
>>> f = ffield.FField(8)
>>> hex(f.generator)
'0x11d'
If you specify the generator when you create the field object, you get the result you're expecting:
>>> f = ffield.FField(8, gen=0x11b, useLUT=0)
>>> hex(f.Multiply(0xbf, 0x03))
'0xda'
answered Jan 4 at 15:45
glibdudglibdud
5,71921731
I'm not intimately familiar with AES, but from the link you provided it appears that the generator polynomial for the field is 0x11b. By default, pyfinite uses a different generator, which will produce different multiplication results:
>>> f = ffield.FField(8)
>>> hex(f.generator)
'0x11d'
If you specify the generator when you create the field object, you get the result you're expecting:
>>> f = ffield.FField(8, gen=0x11b, useLUT=0)
>>> hex(f.Multiply(0xbf, 0x03))
'0xda'
answered Jan 4 at 15:45
glibdudglibdud
5,71921731
edited Jan 4 at 15:55
answered Jan 4 at 15:45
glibdudglibdud
5,71921731
answered Jan 4 at 15:45
glibdudglibdud
5,71921731
answered Jan 4 at 15:45
glibdudglibdud
5,71921731
5,71921731
I can confirm AES uses 0x11b, where all non-zero elements can be considered to be some power of 0x03. For 0x11d, all non-zero elements can be considered to be a power of 0x02. Most implementations involving finite fields will choose a polynomial where all non-zero elements are a power of 2. I don't know why AES choose 0x11b.
– rcgldr
Jan 4 at 16:50
add a comment |
I can confirm AES uses 0x11b, where all non-zero elements can be considered to be some power of 0x03. For 0x11d, all non-zero elements can be considered to be a power of 0x02. Most implementations involving finite fields will choose a polynomial where all non-zero elements are a power of 2. I don't know why AES choose 0x11b.
– rcgldr
Jan 4 at 16:50
I can confirm AES uses 0x11b, where all non-zero elements can be considered to be some power of 0x03. For 0x11d, all non-zero elements can be considered to be a power of 0x02. Most implementations involving finite fields will choose a polynomial where all non-zero elements are a power of 2. I don't know why AES choose 0x11b.
– rcgldr
Jan 4 at 16:50
I can confirm AES uses 0x11b, where all non-zero elements can be considered to be some power of 0x03. For 0x11d, all non-zero elements can be considered to be a power of 0x02. Most implementations involving finite fields will choose a polynomial where all non-zero elements are a power of 2. I don't know why AES choose 0x11b.
– rcgldr
Jan 4 at 16:50
add a comment |
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Could you format and/or annotate your hand math a little more? I'm having a hard time following it. It's been a while, but I followed Wikipedia's guidance and got 0xdc just like pyfinite did. (Edit: which I guess makes this more of a math question than a programming question.)
– glibdud
Jan 4 at 13:43
@glibdud just a minute to formate my hand hand solving better
– mark
Jan 4 at 13:50
@glibdud I did what you asked and updated my question to include latex detailed formatted solving by hand
– mark
Jan 4 at 14:18
How do you know that the modulus is 0x11B?
– Matt Timmermans
Jan 4 at 14:31
@MattTimmermans this is the way we learned in college couple days ago but I do not know if I misunderstood here is summary: when multiplying any number X by 03 in MixColumns stage in AES we multipy X by 02 and then Xor it with X . Multiplying X by 2 is as follow: we convert X to it is binary representation and we look at the leftmost bit if it is one or zero then we shift X to the left by one and Xor it with 0x1B if the leftmost bit was one. I do not know if I am confusing what we learn with what I read online about AES I think what we learned is shortcut for multiply over GF(2^8)
– mark
Jan 4 at 14:38