Mixing
What means a formula with ∞1?
$begingroup$
In the paper https://arxiv.org/abs/1602.04938 there is a formula
What does the formula for $Omega(g)$ mean? I understand that $w_g$ is a vector and that, using the zero norm $$||w_g||_0 > K$$ means "the number of non-zero elements of $w_g$ must be greater than $K$", but I don't get the rest.
notation
asked Jan 24 at 14:55
Make42Make42
19210
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add a comment |
$begingroup$
In the paper https://arxiv.org/abs/1602.04938 there is a formula
What does the formula for $Omega(g)$ mean? I understand that $w_g$ is a vector and that, using the zero norm $$||w_g||_0 > K$$ means "the number of non-zero elements of $w_g$ must be greater than $K$", but I don't get the rest.
notation
asked Jan 24 at 14:55
Make42Make42
19210
$endgroup$
add a comment |
$begingroup$
In the paper https://arxiv.org/abs/1602.04938 there is a formula
What does the formula for $Omega(g)$ mean? I understand that $w_g$ is a vector and that, using the zero norm $$||w_g||_0 > K$$ means "the number of non-zero elements of $w_g$ must be greater than $K$", but I don't get the rest.
notation
asked Jan 24 at 14:55
Make42Make42
19210
$endgroup$
In the paper https://arxiv.org/abs/1602.04938 there is a formula
What does the formula for $Omega(g)$ mean? I understand that $w_g$ is a vector and that, using the zero norm $$||w_g||_0 > K$$ means "the number of non-zero elements of $w_g$ must be greater than $K$", but I don't get the rest.
notation
notation
asked Jan 24 at 14:55
Make42Make42
19210
asked Jan 24 at 14:55
Make42Make42
19210
asked Jan 24 at 14:55
Make42Make42
19210
asked Jan 24 at 14:55
Make42Make42
19210
asked Jan 24 at 14:55
Make42Make42
19210
19210
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The $mathbb{1}$ means the indicator function, which is used here to say that it should take the value 0 when the condition inside is false, and the value 1 when the condition inside is true. That is, as a function of $g$, we have
$$
mathbb 1[|w_g|_0 > K] = begin{cases}
1 & text{if $|w_g|_0 > K$,}\
0 & text{otherwise.}
end{cases}
$$
The infinity symbol here just means "infinity" (some value larger than all real numbers), and the fact that they are written next to each other just means "multiplication", where it is supposed to be understood that $infty times 1 = infty$, and $infty times 0 = 0$. Thus the entire function is
$$
Omega(g) = infty times mathbb 1[|w_g|_0 > K] = begin{cases}
infty & text{if $|w_g|_0 > K$,}\
0 & text{otherwise.}
end{cases}
$$
(None of this is standard by the way, and it makes sense that you're confused.)
answered Jan 24 at 15:02
Mees de VriesMees de Vries
17.4k12957
$endgroup$
$begingroup$
Thanks, yeah, not too standard indeed - especially $infty cdot 0 = 0$ is debatable... Your explanation makes a lot of sense in the context though, so I believe you to be right.
$endgroup$
– Make42
Jan 24 at 15:08
add a comment |
$begingroup$
My guess
$$
mathbb 1 [cdots]
$$
means the indicator function of $[cdots]$, so that formula for $Omega(g)$ means: $infty$ on the set $[cdots]$ and $0$ otherwise.
answered Jan 24 at 15:02
GEdgarGEdgar
63.1k267171
$endgroup$
add a comment |
$begingroup$
I believe it means $Omega(g)$ is infinity if $w_g $ exceeds $K$ and zero otherwise. That would be consistent with implementing a hard constraint as a cost function.
answered Jan 24 at 15:04
MarkMark
2,08732450
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The $mathbb{1}$ means the indicator function, which is used here to say that it should take the value 0 when the condition inside is false, and the value 1 when the condition inside is true. That is, as a function of $g$, we have
$$
mathbb 1[|w_g|_0 > K] = begin{cases}
1 & text{if $|w_g|_0 > K$,}\
0 & text{otherwise.}
end{cases}
$$
The infinity symbol here just means "infinity" (some value larger than all real numbers), and the fact that they are written next to each other just means "multiplication", where it is supposed to be understood that $infty times 1 = infty$, and $infty times 0 = 0$. Thus the entire function is
$$
Omega(g) = infty times mathbb 1[|w_g|_0 > K] = begin{cases}
infty & text{if $|w_g|_0 > K$,}\
0 & text{otherwise.}
end{cases}
$$
(None of this is standard by the way, and it makes sense that you're confused.)
answered Jan 24 at 15:02
Mees de VriesMees de Vries
17.4k12957
$endgroup$
$begingroup$
Thanks, yeah, not too standard indeed - especially $infty cdot 0 = 0$ is debatable... Your explanation makes a lot of sense in the context though, so I believe you to be right.
$endgroup$
– Make42
Jan 24 at 15:08
add a comment |
$begingroup$
The $mathbb{1}$ means the indicator function, which is used here to say that it should take the value 0 when the condition inside is false, and the value 1 when the condition inside is true. That is, as a function of $g$, we have
$$
mathbb 1[|w_g|_0 > K] = begin{cases}
1 & text{if $|w_g|_0 > K$,}\
0 & text{otherwise.}
end{cases}
$$
The infinity symbol here just means "infinity" (some value larger than all real numbers), and the fact that they are written next to each other just means "multiplication", where it is supposed to be understood that $infty times 1 = infty$, and $infty times 0 = 0$. Thus the entire function is
$$
Omega(g) = infty times mathbb 1[|w_g|_0 > K] = begin{cases}
infty & text{if $|w_g|_0 > K$,}\
0 & text{otherwise.}
end{cases}
$$
(None of this is standard by the way, and it makes sense that you're confused.)
answered Jan 24 at 15:02
Mees de VriesMees de Vries
17.4k12957
$endgroup$
$begingroup$
Thanks, yeah, not too standard indeed - especially $infty cdot 0 = 0$ is debatable... Your explanation makes a lot of sense in the context though, so I believe you to be right.
$endgroup$
– Make42
Jan 24 at 15:08
add a comment |
$begingroup$
The $mathbb{1}$ means the indicator function, which is used here to say that it should take the value 0 when the condition inside is false, and the value 1 when the condition inside is true. That is, as a function of $g$, we have
$$
mathbb 1[|w_g|_0 > K] = begin{cases}
1 & text{if $|w_g|_0 > K$,}\
0 & text{otherwise.}
end{cases}
$$
The infinity symbol here just means "infinity" (some value larger than all real numbers), and the fact that they are written next to each other just means "multiplication", where it is supposed to be understood that $infty times 1 = infty$, and $infty times 0 = 0$. Thus the entire function is
$$
Omega(g) = infty times mathbb 1[|w_g|_0 > K] = begin{cases}
infty & text{if $|w_g|_0 > K$,}\
0 & text{otherwise.}
end{cases}
$$
(None of this is standard by the way, and it makes sense that you're confused.)
answered Jan 24 at 15:02
Mees de VriesMees de Vries
17.4k12957
$endgroup$
The $mathbb{1}$ means the indicator function, which is used here to say that it should take the value 0 when the condition inside is false, and the value 1 when the condition inside is true. That is, as a function of $g$, we have
$$
mathbb 1[|w_g|_0 > K] = begin{cases}
1 & text{if $|w_g|_0 > K$,}\
0 & text{otherwise.}
end{cases}
$$
The infinity symbol here just means "infinity" (some value larger than all real numbers), and the fact that they are written next to each other just means "multiplication", where it is supposed to be understood that $infty times 1 = infty$, and $infty times 0 = 0$. Thus the entire function is
$$
Omega(g) = infty times mathbb 1[|w_g|_0 > K] = begin{cases}
infty & text{if $|w_g|_0 > K$,}\
0 & text{otherwise.}
end{cases}
$$
(None of this is standard by the way, and it makes sense that you're confused.)
answered Jan 24 at 15:02
Mees de VriesMees de Vries
17.4k12957
answered Jan 24 at 15:02
Mees de VriesMees de Vries
17.4k12957
answered Jan 24 at 15:02
Mees de VriesMees de Vries
17.4k12957
answered Jan 24 at 15:02
Mees de VriesMees de Vries
17.4k12957
17.4k12957
$begingroup$
Thanks, yeah, not too standard indeed - especially $infty cdot 0 = 0$ is debatable... Your explanation makes a lot of sense in the context though, so I believe you to be right.
$endgroup$
– Make42
Jan 24 at 15:08
add a comment |
$begingroup$
Thanks, yeah, not too standard indeed - especially $infty cdot 0 = 0$ is debatable... Your explanation makes a lot of sense in the context though, so I believe you to be right.
$endgroup$
– Make42
Jan 24 at 15:08
$begingroup$
Thanks, yeah, not too standard indeed - especially $infty cdot 0 = 0$ is debatable... Your explanation makes a lot of sense in the context though, so I believe you to be right.
$endgroup$
– Make42
Jan 24 at 15:08
$begingroup$
Thanks, yeah, not too standard indeed - especially $infty cdot 0 = 0$ is debatable... Your explanation makes a lot of sense in the context though, so I believe you to be right.
$endgroup$
– Make42
Jan 24 at 15:08
add a comment |
$begingroup$
My guess
$$
mathbb 1 [cdots]
$$
means the indicator function of $[cdots]$, so that formula for $Omega(g)$ means: $infty$ on the set $[cdots]$ and $0$ otherwise.
answered Jan 24 at 15:02
GEdgarGEdgar
63.1k267171
$endgroup$
add a comment |
$begingroup$
My guess
$$
mathbb 1 [cdots]
$$
means the indicator function of $[cdots]$, so that formula for $Omega(g)$ means: $infty$ on the set $[cdots]$ and $0$ otherwise.
answered Jan 24 at 15:02
GEdgarGEdgar
63.1k267171
$endgroup$
add a comment |
$begingroup$
My guess
$$
mathbb 1 [cdots]
$$
means the indicator function of $[cdots]$, so that formula for $Omega(g)$ means: $infty$ on the set $[cdots]$ and $0$ otherwise.
answered Jan 24 at 15:02
GEdgarGEdgar
63.1k267171
$endgroup$
My guess
$$
mathbb 1 [cdots]
$$
means the indicator function of $[cdots]$, so that formula for $Omega(g)$ means: $infty$ on the set $[cdots]$ and $0$ otherwise.
answered Jan 24 at 15:02
GEdgarGEdgar
63.1k267171
answered Jan 24 at 15:02
GEdgarGEdgar
63.1k267171
answered Jan 24 at 15:02
GEdgarGEdgar
63.1k267171
answered Jan 24 at 15:02
GEdgarGEdgar
63.1k267171
63.1k267171
add a comment |
add a comment |
$begingroup$
I believe it means $Omega(g)$ is infinity if $w_g $ exceeds $K$ and zero otherwise. That would be consistent with implementing a hard constraint as a cost function.
answered Jan 24 at 15:04
MarkMark
2,08732450
$endgroup$
add a comment |
$begingroup$
I believe it means $Omega(g)$ is infinity if $w_g $ exceeds $K$ and zero otherwise. That would be consistent with implementing a hard constraint as a cost function.
answered Jan 24 at 15:04
MarkMark
2,08732450
$endgroup$
add a comment |
$begingroup$
I believe it means $Omega(g)$ is infinity if $w_g $ exceeds $K$ and zero otherwise. That would be consistent with implementing a hard constraint as a cost function.
answered Jan 24 at 15:04
MarkMark
2,08732450
$endgroup$
I believe it means $Omega(g)$ is infinity if $w_g $ exceeds $K$ and zero otherwise. That would be consistent with implementing a hard constraint as a cost function.
answered Jan 24 at 15:04
MarkMark
2,08732450
answered Jan 24 at 15:04
MarkMark
2,08732450
answered Jan 24 at 15:04
MarkMark
2,08732450
answered Jan 24 at 15:04
MarkMark
2,08732450
2,08732450
add a comment |
add a comment |
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