What is a “n-valued function”? The Next CEO of Stack OverflowProve domain of partial...
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What is a “n-valued function”?
The Next CEO of Stack OverflowProve domain of partial computable function existsShowing that a function is not computable.What does 'real-valued' function mean in topology?What does it mean if a function is onto?Real (Valued) Functions in GermanWhy are function definitions not written with the := signWhat is the definition of a single valued functionOnline Encyclopedia of continuous and/or computable real valued functions?Are single-valued function also set-valued functions?Name of Property: The Order Of Parameters in a Function Does Not Matter
$begingroup$
Has $n$ parameters?
i.e.
0-valued function: $f(emptyset)=2$
1-valued function: $f(x)=x$
2-valued function: $f(x,y)=x+y$
3-valued function: $f(x,y,z)=x+y+z$
Not sure
terminology computability
asked Mar 18 at 0:28
A_for_ AbacusA_for_ Abacus
967925
$endgroup$
add a comment |
$begingroup$
Has $n$ parameters?
i.e.
0-valued function: $f(emptyset)=2$
1-valued function: $f(x)=x$
2-valued function: $f(x,y)=x+y$
3-valued function: $f(x,y,z)=x+y+z$
Not sure
terminology computability
asked Mar 18 at 0:28
A_for_ AbacusA_for_ Abacus
967925
$endgroup$
1
$begingroup$
Pretty sure that "0/1-valued" means that the output is always 0 or 1. But context is everything, so please let us know where you found this.
$endgroup$
– Morgan Rodgers
Mar 18 at 0:35
3
$begingroup$
Just the phrase "0/1-valued function" suggests that the function's codomain is the set ${0,1}$.
$endgroup$
– vadim123
Mar 18 at 0:36
$begingroup$
If I hear "n-valued function" I think the mean $f(x)$ may have multiple outputs (violating the standard definition of "function" which is practically writ in stone that a well-defined function has a single output for each point in the domain). For example $f:[-1,1]to [0,2pi)$ via $f(x) = theta$ if $sin theta x$ is a 2-value "function". But in this context I really actually have no idea.
$endgroup$
– fleablood
Mar 18 at 0:37
$begingroup$
"retty sure that "0/1-valued" means that the output is always 0 or 1." and "Just the phrase "0/1-valued function" suggests that the function's codomain is the set {0,1}". D'oh! That is almost certainly the case....
$endgroup$
– fleablood
Mar 18 at 0:37
$begingroup$
It's from a class on Theory of Computation, topic: Recursive and Recursively Enumerable Sets. "total" here means there's always an output (never undefined) so I think 1st commenter is right.
$endgroup$
– A_for_ Abacus
Mar 18 at 0:40
add a comment |
$begingroup$
Has $n$ parameters?
i.e.
0-valued function: $f(emptyset)=2$
1-valued function: $f(x)=x$
2-valued function: $f(x,y)=x+y$
3-valued function: $f(x,y,z)=x+y+z$
Not sure
terminology computability
asked Mar 18 at 0:28
A_for_ AbacusA_for_ Abacus
967925
$endgroup$
Has $n$ parameters?
i.e.
0-valued function: $f(emptyset)=2$
1-valued function: $f(x)=x$
2-valued function: $f(x,y)=x+y$
3-valued function: $f(x,y,z)=x+y+z$
Not sure
terminology computability
terminology computability
asked Mar 18 at 0:28
A_for_ AbacusA_for_ Abacus
967925
asked Mar 18 at 0:28
A_for_ AbacusA_for_ Abacus
967925
asked Mar 18 at 0:28
A_for_ AbacusA_for_ Abacus
967925
asked Mar 18 at 0:28
A_for_ AbacusA_for_ Abacus
967925
asked Mar 18 at 0:28
A_for_ AbacusA_for_ Abacus
967925
967925
1
$begingroup$
Pretty sure that "0/1-valued" means that the output is always 0 or 1. But context is everything, so please let us know where you found this.
$endgroup$
– Morgan Rodgers
Mar 18 at 0:35
3
$begingroup$
Just the phrase "0/1-valued function" suggests that the function's codomain is the set ${0,1}$.
$endgroup$
– vadim123
Mar 18 at 0:36
$begingroup$
If I hear "n-valued function" I think the mean $f(x)$ may have multiple outputs (violating the standard definition of "function" which is practically writ in stone that a well-defined function has a single output for each point in the domain). For example $f:[-1,1]to [0,2pi)$ via $f(x) = theta$ if $sin theta x$ is a 2-value "function". But in this context I really actually have no idea.
$endgroup$
– fleablood
Mar 18 at 0:37
$begingroup$
"retty sure that "0/1-valued" means that the output is always 0 or 1." and "Just the phrase "0/1-valued function" suggests that the function's codomain is the set {0,1}". D'oh! That is almost certainly the case....
$endgroup$
– fleablood
Mar 18 at 0:37
$begingroup$
It's from a class on Theory of Computation, topic: Recursive and Recursively Enumerable Sets. "total" here means there's always an output (never undefined) so I think 1st commenter is right.
$endgroup$
– A_for_ Abacus
Mar 18 at 0:40
add a comment |
1
$begingroup$
Pretty sure that "0/1-valued" means that the output is always 0 or 1. But context is everything, so please let us know where you found this.
$endgroup$
– Morgan Rodgers
Mar 18 at 0:35
3
$begingroup$
Just the phrase "0/1-valued function" suggests that the function's codomain is the set ${0,1}$.
$endgroup$
– vadim123
Mar 18 at 0:36
$begingroup$
If I hear "n-valued function" I think the mean $f(x)$ may have multiple outputs (violating the standard definition of "function" which is practically writ in stone that a well-defined function has a single output for each point in the domain). For example $f:[-1,1]to [0,2pi)$ via $f(x) = theta$ if $sin theta x$ is a 2-value "function". But in this context I really actually have no idea.
$endgroup$
– fleablood
Mar 18 at 0:37
$begingroup$
"retty sure that "0/1-valued" means that the output is always 0 or 1." and "Just the phrase "0/1-valued function" suggests that the function's codomain is the set {0,1}". D'oh! That is almost certainly the case....
$endgroup$
– fleablood
Mar 18 at 0:37
$begingroup$
It's from a class on Theory of Computation, topic: Recursive and Recursively Enumerable Sets. "total" here means there's always an output (never undefined) so I think 1st commenter is right.
$endgroup$
– A_for_ Abacus
Mar 18 at 0:40
1
1
$begingroup$
Pretty sure that "0/1-valued" means that the output is always 0 or 1. But context is everything, so please let us know where you found this.
$endgroup$
– Morgan Rodgers
Mar 18 at 0:35
$begingroup$
Pretty sure that "0/1-valued" means that the output is always 0 or 1. But context is everything, so please let us know where you found this.
$endgroup$
– Morgan Rodgers
Mar 18 at 0:35
3
3
$begingroup$
Just the phrase "0/1-valued function" suggests that the function's codomain is the set ${0,1}$.
$endgroup$
– vadim123
Mar 18 at 0:36
$begingroup$
Just the phrase "0/1-valued function" suggests that the function's codomain is the set ${0,1}$.
$endgroup$
– vadim123
Mar 18 at 0:36
$begingroup$
If I hear "n-valued function" I think the mean $f(x)$ may have multiple outputs (violating the standard definition of "function" which is practically writ in stone that a well-defined function has a single output for each point in the domain). For example $f:[-1,1]to [0,2pi)$ via $f(x) = theta$ if $sin theta x$ is a 2-value "function". But in this context I really actually have no idea.
$endgroup$
– fleablood
Mar 18 at 0:37
$begingroup$
If I hear "n-valued function" I think the mean $f(x)$ may have multiple outputs (violating the standard definition of "function" which is practically writ in stone that a well-defined function has a single output for each point in the domain). For example $f:[-1,1]to [0,2pi)$ via $f(x) = theta$ if $sin theta x$ is a 2-value "function". But in this context I really actually have no idea.
$endgroup$
– fleablood
Mar 18 at 0:37
$begingroup$
"retty sure that "0/1-valued" means that the output is always 0 or 1." and "Just the phrase "0/1-valued function" suggests that the function's codomain is the set {0,1}". D'oh! That is almost certainly the case....
$endgroup$
– fleablood
Mar 18 at 0:37
$begingroup$
"retty sure that "0/1-valued" means that the output is always 0 or 1." and "Just the phrase "0/1-valued function" suggests that the function's codomain is the set {0,1}". D'oh! That is almost certainly the case....
$endgroup$
– fleablood
Mar 18 at 0:37
$begingroup$
It's from a class on Theory of Computation, topic: Recursive and Recursively Enumerable Sets. "total" here means there's always an output (never undefined) so I think 1st commenter is right.
$endgroup$
– A_for_ Abacus
Mar 18 at 0:40
$begingroup$
It's from a class on Theory of Computation, topic: Recursive and Recursively Enumerable Sets. "total" here means there's always an output (never undefined) so I think 1st commenter is right.
$endgroup$
– A_for_ Abacus
Mar 18 at 0:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$phi_x:{Bbb N}_0^krightarrow {Bbb N}_0$ is a total 0/1-valued function means that the domain of $phi_x$ is ${Bbb N}_0^k$ and the range is $phi({Bbb N}_0^k)subseteq {0,1}$.
answered Mar 18 at 16:53
WuestenfuxWuestenfux
5,3331513
$endgroup$
add a comment |
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$begingroup$
$phi_x:{Bbb N}_0^krightarrow {Bbb N}_0$ is a total 0/1-valued function means that the domain of $phi_x$ is ${Bbb N}_0^k$ and the range is $phi({Bbb N}_0^k)subseteq {0,1}$.
answered Mar 18 at 16:53
WuestenfuxWuestenfux
5,3331513
$endgroup$
add a comment |
$begingroup$
$phi_x:{Bbb N}_0^krightarrow {Bbb N}_0$ is a total 0/1-valued function means that the domain of $phi_x$ is ${Bbb N}_0^k$ and the range is $phi({Bbb N}_0^k)subseteq {0,1}$.
answered Mar 18 at 16:53
WuestenfuxWuestenfux
5,3331513
$endgroup$
add a comment |
$begingroup$
$phi_x:{Bbb N}_0^krightarrow {Bbb N}_0$ is a total 0/1-valued function means that the domain of $phi_x$ is ${Bbb N}_0^k$ and the range is $phi({Bbb N}_0^k)subseteq {0,1}$.
answered Mar 18 at 16:53
WuestenfuxWuestenfux
5,3331513
$endgroup$
$phi_x:{Bbb N}_0^krightarrow {Bbb N}_0$ is a total 0/1-valued function means that the domain of $phi_x$ is ${Bbb N}_0^k$ and the range is $phi({Bbb N}_0^k)subseteq {0,1}$.
answered Mar 18 at 16:53
WuestenfuxWuestenfux
5,3331513
answered Mar 18 at 16:53
WuestenfuxWuestenfux
5,3331513
answered Mar 18 at 16:53
WuestenfuxWuestenfux
5,3331513
answered Mar 18 at 16:53
WuestenfuxWuestenfux
5,3331513
5,3331513
add a comment |
add a comment |
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1
$begingroup$
Pretty sure that "0/1-valued" means that the output is always 0 or 1. But context is everything, so please let us know where you found this.
$endgroup$
– Morgan Rodgers
Mar 18 at 0:35
3
$begingroup$
Just the phrase "0/1-valued function" suggests that the function's codomain is the set ${0,1}$.
$endgroup$
– vadim123
Mar 18 at 0:36
$begingroup$
If I hear "n-valued function" I think the mean $f(x)$ may have multiple outputs (violating the standard definition of "function" which is practically writ in stone that a well-defined function has a single output for each point in the domain). For example $f:[-1,1]to [0,2pi)$ via $f(x) = theta$ if $sin theta x$ is a 2-value "function". But in this context I really actually have no idea.
$endgroup$
– fleablood
Mar 18 at 0:37
$begingroup$
"retty sure that "0/1-valued" means that the output is always 0 or 1." and "Just the phrase "0/1-valued function" suggests that the function's codomain is the set {0,1}". D'oh! That is almost certainly the case....
$endgroup$
– fleablood
Mar 18 at 0:37
$begingroup$
It's from a class on Theory of Computation, topic: Recursive and Recursively Enumerable Sets. "total" here means there's always an output (never undefined) so I think 1st commenter is right.
$endgroup$
– A_for_ Abacus
Mar 18 at 0:40