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2018-03-19 03:56:04 +0200 commented answer solving matrix over GF(2)

As you said, we can get the $M\in S$ whose first column is $y-b$, but i don't know how to generate the set $S$ by Sage code.

2018-03-16 10:12:11 +0200 commented answer solving matrix over GF(2)

There exist other cases such as y = vector(GF(2), [1, 0, 1, 0, 1, 0, 0, 0]) and x = vector(GF(2), [0, 1, 1, 1, 1, 1, 0, 0]). The values of x and y are variable.

2018-03-16 02:59:15 +0200 commented question solving matrix over GF(2)

Yes, we need all possible solutions of $A$, since we have many equations like $Ax+b=y$, we will solve each equation and take the intersection to get the final $A$.

2018-03-16 02:06:28 +0200 asked a question solving matrix over GF(2)
A = matrix(GF(2), 8, 8, [])
b = vector(GF(2), [0, 1, 1, 0, 1, 0, 1, 1])
y = vector(GF(2), [0, 0, 0, 0, 1, 0, 1, 1])
x = vector(GF(2), [1, 0, 0, 0, 0, 0, 0, 0])

If the matrix $A$ is unkown, we have $Ax+b = y$.

How can we solve the matrix $A$?

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2018-03-10 18:22:06 +0200 asked a question irreducible polynomial defining the finite field

X = GF(2).polynomial_ring().gen()

F = GF(2^8, name="a", modulus=X^8 + X^6 + X^5 + X + 1)

As example above, the polynomial X^8 + X^6 + X^5 + X + 1 is one of irreducible polynomial defining the finite field GF(2^8).

How can I get all the irreducible polynomial which can define the finite field GF(2^8)?

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2018-02-07 08:38:44 +0200 commented question How to generate a cyclic matrix

A cyclic matrix means a circulant matrix. We need the first row the property that the sum of entries is not zero so that this cyclic matrix could be a invertible matrix.

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2018-02-06 19:22:26 +0200 asked a question How to generate a cyclic matrix

How to generate all 8×8 cyclic matrix $A$ with the first row $(a_0, a_1, \ldots, a_7)$,

the matrix $A\in GL(8, 2)$ and the first row $a_0 + a_1 + \ldots + a_7 \neq 0$

2017-07-18 14:52:58 +0200 asked a question How to check whether a permutation is linear?

Suppose that there is a permutation from (1, 2, 3, 4, 5, 6) to (2, 3, 5, 4, 6, 1), how could I check whether it is linear or not?


Now I have many permutations from 256 integers to 256 integers, how could i check?


For example, this is a permutation, from (0, 1, ..., 256) to (0, 29, 142, 147, 199, 218, 73, 84, 227, 254, 109, 112, 36, 57, 170, 183, 113, 108, 255, 226, 182, 171, 56, 37, 146, 143, 28, 1, 85, 72, 219, 198, 184, 165, 54, 43, 127, 98, 241, 236, 91, 70, 213, 200, 156, 129, 18, 15, 201, 212, 71, 90, 14, 19, 128, 157, 42, 55, 164, 185, 237, 240, 99, 126, 220, 193, 82, 79, 27, 6, 149, 136, 63, 34, 177, 172, 248, 229, 118, 107, 173, 176, 35, 62, 106, 119, 228, 249, 78, 83, 192, 221, 137, 148, 7, 26, 100, 121, 234, 247, 163, 190, 45, 48, 135, 154, 9, 20, 64, 93, 206, 211, 21, 8, 155, 134, 210, 207, 92, 65, 246, 235, 120, 101, 49, 44, 191, 162, 238, 243, 96, 125, 41, 52, 167, 186, 13, 16, 131, 158, 202, 215, 68, 89, 159, 130, 17, 12, 88, 69, 214, 203, 124, 97, 242, 239, 187, 166, 53, 40, 86, 75, 216, 197, 145, 140, 31, 2, 181, 168, 59, 38, 114, 111, 252, 225, 39, 58, 169, 180, 224, 253, 110, 115, 196, 217, 74, 87, 3, 30, 141, 144, 50, 47, 188, 161, 245, 232, 123, 102, 209, 204, 95, 66, 22, 11, 152, 133, 67, 94, 205, 208, 132, 153, 10, 23, 160, 189, 46, 51, 103, 122, 233, 244, 138, 151, 4, 25, 77, 80, 195, 222, 105, 116, 231, 250, 174, 179, 32, 61, 251, 230, 117, 104, 60, 33, 178, 175, 24, 5, 150, 139, 223, 194, 81, 76)