Probability that arbitrary rank-$w$ matrix $mathbf{M} in mathbb{F}_2^{n times m}$ splits in two rank-$(w /...
For simplicity, assume $n$ and $w$ are even.
Let $mathbf{M} in mathbb{F}_2^{n times m}$ be an arbitrary matrix with $operatorname{rank} mathbf{M} = w$. What is the probability that we can write
$$
mathbf{M} = begin{bmatrix} mathbf{M}_1 \ mathbf{M}_2 end{bmatrix}text,
$$
where $mathbf{M}_1, mathbf{M}_2 in mathbb{F}_2^{(n / 2) times m}$ and $operatorname{rank} mathbf{M}_1 = operatorname{rank} mathbf{M}_2 = w / 2$?
What follows is what I have worked on so far.
There are (see 1)
$$
C := binom{n}{w}_2 prod_{i = 0}^{w - 1} left(2^m - 2^iright)
$$
choices for $mathbf{M}$, where $binom{n}{w}_2$ denotes the $2$-binomial coefficient.
Given some $mathbf{M}_2$ with $operatorname{rank} mathbf{M}_2 = w / 2$, the number of ways we can "extend" it (as per problem statement) by an $mathbf{M}_1$ and get a rank-$w$ matrix is (see Introduction in 2)
$$
C_1 := 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^{w / 2 + i}right)text.
$$
Analogously to $mathbf{M}$, there are
$$
C_2 := binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^iright)
$$
choices for $mathbf{M}_2$.
Therefore, the wanted probability is
$$
frac{C_1 C_2}{C} = 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2^2 binom{n}{w}_2^{-1}text.
$$
(The products cancel out.) Any hopes to verify this? Are you aware of related work that might help?
combinatorics matrices finite-fields
asked Nov 20 at 15:35
d125q
1,6471015
add a comment |
For simplicity, assume $n$ and $w$ are even.
Let $mathbf{M} in mathbb{F}_2^{n times m}$ be an arbitrary matrix with $operatorname{rank} mathbf{M} = w$. What is the probability that we can write
$$
mathbf{M} = begin{bmatrix} mathbf{M}_1 \ mathbf{M}_2 end{bmatrix}text,
$$
where $mathbf{M}_1, mathbf{M}_2 in mathbb{F}_2^{(n / 2) times m}$ and $operatorname{rank} mathbf{M}_1 = operatorname{rank} mathbf{M}_2 = w / 2$?
What follows is what I have worked on so far.
There are (see 1)
$$
C := binom{n}{w}_2 prod_{i = 0}^{w - 1} left(2^m - 2^iright)
$$
choices for $mathbf{M}$, where $binom{n}{w}_2$ denotes the $2$-binomial coefficient.
Given some $mathbf{M}_2$ with $operatorname{rank} mathbf{M}_2 = w / 2$, the number of ways we can "extend" it (as per problem statement) by an $mathbf{M}_1$ and get a rank-$w$ matrix is (see Introduction in 2)
$$
C_1 := 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^{w / 2 + i}right)text.
$$
Analogously to $mathbf{M}$, there are
$$
C_2 := binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^iright)
$$
choices for $mathbf{M}_2$.
Therefore, the wanted probability is
$$
frac{C_1 C_2}{C} = 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2^2 binom{n}{w}_2^{-1}text.
$$
(The products cancel out.) Any hopes to verify this? Are you aware of related work that might help?
combinatorics matrices finite-fields
asked Nov 20 at 15:35
d125q
1,6471015
add a comment |
For simplicity, assume $n$ and $w$ are even.
Let $mathbf{M} in mathbb{F}_2^{n times m}$ be an arbitrary matrix with $operatorname{rank} mathbf{M} = w$. What is the probability that we can write
$$
mathbf{M} = begin{bmatrix} mathbf{M}_1 \ mathbf{M}_2 end{bmatrix}text,
$$
where $mathbf{M}_1, mathbf{M}_2 in mathbb{F}_2^{(n / 2) times m}$ and $operatorname{rank} mathbf{M}_1 = operatorname{rank} mathbf{M}_2 = w / 2$?
What follows is what I have worked on so far.
There are (see 1)
$$
C := binom{n}{w}_2 prod_{i = 0}^{w - 1} left(2^m - 2^iright)
$$
choices for $mathbf{M}$, where $binom{n}{w}_2$ denotes the $2$-binomial coefficient.
Given some $mathbf{M}_2$ with $operatorname{rank} mathbf{M}_2 = w / 2$, the number of ways we can "extend" it (as per problem statement) by an $mathbf{M}_1$ and get a rank-$w$ matrix is (see Introduction in 2)
$$
C_1 := 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^{w / 2 + i}right)text.
$$
Analogously to $mathbf{M}$, there are
$$
C_2 := binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^iright)
$$
choices for $mathbf{M}_2$.
Therefore, the wanted probability is
$$
frac{C_1 C_2}{C} = 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2^2 binom{n}{w}_2^{-1}text.
$$
(The products cancel out.) Any hopes to verify this? Are you aware of related work that might help?
combinatorics matrices finite-fields
asked Nov 20 at 15:35
d125q
1,6471015
For simplicity, assume $n$ and $w$ are even.
Let $mathbf{M} in mathbb{F}_2^{n times m}$ be an arbitrary matrix with $operatorname{rank} mathbf{M} = w$. What is the probability that we can write
$$
mathbf{M} = begin{bmatrix} mathbf{M}_1 \ mathbf{M}_2 end{bmatrix}text,
$$
where $mathbf{M}_1, mathbf{M}_2 in mathbb{F}_2^{(n / 2) times m}$ and $operatorname{rank} mathbf{M}_1 = operatorname{rank} mathbf{M}_2 = w / 2$?
What follows is what I have worked on so far.
There are (see 1)
$$
C := binom{n}{w}_2 prod_{i = 0}^{w - 1} left(2^m - 2^iright)
$$
choices for $mathbf{M}$, where $binom{n}{w}_2$ denotes the $2$-binomial coefficient.
Given some $mathbf{M}_2$ with $operatorname{rank} mathbf{M}_2 = w / 2$, the number of ways we can "extend" it (as per problem statement) by an $mathbf{M}_1$ and get a rank-$w$ matrix is (see Introduction in 2)
$$
C_1 := 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^{w / 2 + i}right)text.
$$
Analogously to $mathbf{M}$, there are
$$
C_2 := binom{n / 2}{w / 2}_2 prod_{i = 0}^{w / 2 - 1} left(2^m - 2^iright)
$$
choices for $mathbf{M}_2$.
Therefore, the wanted probability is
$$
frac{C_1 C_2}{C} = 2^{w (n - w) / 4} binom{n / 2}{w / 2}_2^2 binom{n}{w}_2^{-1}text.
$$
(The products cancel out.) Any hopes to verify this? Are you aware of related work that might help?
combinatorics matrices finite-fields
combinatorics matrices finite-fields
asked Nov 20 at 15:35
d125q
1,6471015
asked Nov 20 at 15:35
d125q
1,6471015
edited Nov 20 at 19:34
asked Nov 20 at 15:35
d125q
1,6471015
asked Nov 20 at 15:35
d125q
1,6471015
asked Nov 20 at 15:35
d125q
1,6471015
1,6471015
add a comment |
add a comment |
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