Mixing
Linear constraint equivalent to one-to-oneness
$begingroup$
Suppose that $A$ is a finite set with cardinality $n$. Let $phi$ be a one-to-one function from $A$ to ${1,2,dots,n}$. I am wondering if it is possible to write the one-to-oneness condition on $phi$ to an equivalent linear equality (or inequality) constraint on $phi$. Any help is greatly appreciated.
real-analysis calculus optimization
asked Jan 17 at 6:40
KumaraKumara
219118
$endgroup$
add a comment |
$begingroup$
Suppose that $A$ is a finite set with cardinality $n$. Let $phi$ be a one-to-one function from $A$ to ${1,2,dots,n}$. I am wondering if it is possible to write the one-to-oneness condition on $phi$ to an equivalent linear equality (or inequality) constraint on $phi$. Any help is greatly appreciated.
real-analysis calculus optimization
asked Jan 17 at 6:40
KumaraKumara
219118
$endgroup$
1
$begingroup$
Are binary variables acceptable?
$endgroup$
– LinAlg
Jan 20 at 22:26
add a comment |
$begingroup$
Suppose that $A$ is a finite set with cardinality $n$. Let $phi$ be a one-to-one function from $A$ to ${1,2,dots,n}$. I am wondering if it is possible to write the one-to-oneness condition on $phi$ to an equivalent linear equality (or inequality) constraint on $phi$. Any help is greatly appreciated.
real-analysis calculus optimization
asked Jan 17 at 6:40
KumaraKumara
219118
$endgroup$
Suppose that $A$ is a finite set with cardinality $n$. Let $phi$ be a one-to-one function from $A$ to ${1,2,dots,n}$. I am wondering if it is possible to write the one-to-oneness condition on $phi$ to an equivalent linear equality (or inequality) constraint on $phi$. Any help is greatly appreciated.
real-analysis calculus optimization
real-analysis calculus optimization
asked Jan 17 at 6:40
KumaraKumara
219118
asked Jan 17 at 6:40
KumaraKumara
219118
asked Jan 17 at 6:40
KumaraKumara
219118
asked Jan 17 at 6:40
KumaraKumara
219118
asked Jan 17 at 6:40
KumaraKumara
219118
219118
1
$begingroup$
Are binary variables acceptable?
$endgroup$
– LinAlg
Jan 20 at 22:26
add a comment |
1
$begingroup$
Are binary variables acceptable?
$endgroup$
– LinAlg
Jan 20 at 22:26
1
1
$begingroup$
Are binary variables acceptable?
$endgroup$
– LinAlg
Jan 20 at 22:26
$begingroup$
Are binary variables acceptable?
$endgroup$
– LinAlg
Jan 20 at 22:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The answer to your question is NO. In fact it already fails for $n=2$. In this case, $A=lbrace a_1,a_2 rbrace $ and there are exactly two one-to-one functions $A to lbrace 1,2 rbrace $, namely $phi_1$ defined by $phi_1(a_1)=1,phi_1(a_2)=2$ and $phi_2$ defined by $phi_2(a_1)=2,phi_2(a_2)=1$.
Now, suppose a set of linear constraints is satisfied by both $phi_1$ and $phi_2$. Because the constraints are linear, they will still be satisfied by any linear combination of $phi_1$ and $phi_2$ with nonnegative coefficients. In particular, they will be satisfied by the mean $frac{phi_1+phi_2}{2}$ which is constant equal to $1$ and therefore not one-to-one.
Similarly, for arbitrary $n$ there are $n!$ different one-to-one functions $A to lbrace 1,2,ldots,n rbrace $ and their (scaled) mean, i.e. their sum divided by $n$, is constant equal to $1$.
answered Jan 20 at 6:17
Ewan DelanoyEwan Delanoy
41.8k442104
$endgroup$
add a comment |
$begingroup$
A bijection from A to S = { 1,2,.. n } usually it isn't linear.
How could it be if A, for example, were a set of open sets?
Even if A were a set of integers,
how could there be a linear bijection from { x$^2$ : x in S } to S?
answered Jan 17 at 7:43
William ElliotWilliam Elliot
8,2872720
$endgroup$
$begingroup$
I am not asking if $phi$ is linear; some linear constraint on $phi$. Something like $sum_x F(phi(x)) = k$ for some function $F$ and some constant $k$.
$endgroup$
– Kumara
Jan 17 at 9:19
$begingroup$
@Kumara. Yes. Let k = n and F is constant function 1 or k = 0 and F the constant function 0.
$endgroup$
– William Elliot
Jan 18 at 0:01
$begingroup$
Please understand the question properly. I am asking a condition on $phi$ $equivalent$ to one-to-oneness.
$endgroup$
– Kumara
Jan 18 at 4:10
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer to your question is NO. In fact it already fails for $n=2$. In this case, $A=lbrace a_1,a_2 rbrace $ and there are exactly two one-to-one functions $A to lbrace 1,2 rbrace $, namely $phi_1$ defined by $phi_1(a_1)=1,phi_1(a_2)=2$ and $phi_2$ defined by $phi_2(a_1)=2,phi_2(a_2)=1$.
Now, suppose a set of linear constraints is satisfied by both $phi_1$ and $phi_2$. Because the constraints are linear, they will still be satisfied by any linear combination of $phi_1$ and $phi_2$ with nonnegative coefficients. In particular, they will be satisfied by the mean $frac{phi_1+phi_2}{2}$ which is constant equal to $1$ and therefore not one-to-one.
Similarly, for arbitrary $n$ there are $n!$ different one-to-one functions $A to lbrace 1,2,ldots,n rbrace $ and their (scaled) mean, i.e. their sum divided by $n$, is constant equal to $1$.
answered Jan 20 at 6:17
Ewan DelanoyEwan Delanoy
41.8k442104
$endgroup$
add a comment |
$begingroup$
The answer to your question is NO. In fact it already fails for $n=2$. In this case, $A=lbrace a_1,a_2 rbrace $ and there are exactly two one-to-one functions $A to lbrace 1,2 rbrace $, namely $phi_1$ defined by $phi_1(a_1)=1,phi_1(a_2)=2$ and $phi_2$ defined by $phi_2(a_1)=2,phi_2(a_2)=1$.
Now, suppose a set of linear constraints is satisfied by both $phi_1$ and $phi_2$. Because the constraints are linear, they will still be satisfied by any linear combination of $phi_1$ and $phi_2$ with nonnegative coefficients. In particular, they will be satisfied by the mean $frac{phi_1+phi_2}{2}$ which is constant equal to $1$ and therefore not one-to-one.
Similarly, for arbitrary $n$ there are $n!$ different one-to-one functions $A to lbrace 1,2,ldots,n rbrace $ and their (scaled) mean, i.e. their sum divided by $n$, is constant equal to $1$.
answered Jan 20 at 6:17
Ewan DelanoyEwan Delanoy
41.8k442104
$endgroup$
add a comment |
$begingroup$
The answer to your question is NO. In fact it already fails for $n=2$. In this case, $A=lbrace a_1,a_2 rbrace $ and there are exactly two one-to-one functions $A to lbrace 1,2 rbrace $, namely $phi_1$ defined by $phi_1(a_1)=1,phi_1(a_2)=2$ and $phi_2$ defined by $phi_2(a_1)=2,phi_2(a_2)=1$.
Now, suppose a set of linear constraints is satisfied by both $phi_1$ and $phi_2$. Because the constraints are linear, they will still be satisfied by any linear combination of $phi_1$ and $phi_2$ with nonnegative coefficients. In particular, they will be satisfied by the mean $frac{phi_1+phi_2}{2}$ which is constant equal to $1$ and therefore not one-to-one.
Similarly, for arbitrary $n$ there are $n!$ different one-to-one functions $A to lbrace 1,2,ldots,n rbrace $ and their (scaled) mean, i.e. their sum divided by $n$, is constant equal to $1$.
answered Jan 20 at 6:17
Ewan DelanoyEwan Delanoy
41.8k442104
$endgroup$
The answer to your question is NO. In fact it already fails for $n=2$. In this case, $A=lbrace a_1,a_2 rbrace $ and there are exactly two one-to-one functions $A to lbrace 1,2 rbrace $, namely $phi_1$ defined by $phi_1(a_1)=1,phi_1(a_2)=2$ and $phi_2$ defined by $phi_2(a_1)=2,phi_2(a_2)=1$.
Now, suppose a set of linear constraints is satisfied by both $phi_1$ and $phi_2$. Because the constraints are linear, they will still be satisfied by any linear combination of $phi_1$ and $phi_2$ with nonnegative coefficients. In particular, they will be satisfied by the mean $frac{phi_1+phi_2}{2}$ which is constant equal to $1$ and therefore not one-to-one.
Similarly, for arbitrary $n$ there are $n!$ different one-to-one functions $A to lbrace 1,2,ldots,n rbrace $ and their (scaled) mean, i.e. their sum divided by $n$, is constant equal to $1$.
answered Jan 20 at 6:17
Ewan DelanoyEwan Delanoy
41.8k442104
answered Jan 20 at 6:17
Ewan DelanoyEwan Delanoy
41.8k442104
answered Jan 20 at 6:17
Ewan DelanoyEwan Delanoy
41.8k442104
answered Jan 20 at 6:17
Ewan DelanoyEwan Delanoy
41.8k442104
41.8k442104
add a comment |
add a comment |
$begingroup$
A bijection from A to S = { 1,2,.. n } usually it isn't linear.
How could it be if A, for example, were a set of open sets?
Even if A were a set of integers,
how could there be a linear bijection from { x$^2$ : x in S } to S?
answered Jan 17 at 7:43
William ElliotWilliam Elliot
8,2872720
$endgroup$
$begingroup$
I am not asking if $phi$ is linear; some linear constraint on $phi$. Something like $sum_x F(phi(x)) = k$ for some function $F$ and some constant $k$.
$endgroup$
– Kumara
Jan 17 at 9:19
$begingroup$
@Kumara. Yes. Let k = n and F is constant function 1 or k = 0 and F the constant function 0.
$endgroup$
– William Elliot
Jan 18 at 0:01
$begingroup$
Please understand the question properly. I am asking a condition on $phi$ $equivalent$ to one-to-oneness.
$endgroup$
– Kumara
Jan 18 at 4:10
add a comment |
$begingroup$
A bijection from A to S = { 1,2,.. n } usually it isn't linear.
How could it be if A, for example, were a set of open sets?
Even if A were a set of integers,
how could there be a linear bijection from { x$^2$ : x in S } to S?
answered Jan 17 at 7:43
William ElliotWilliam Elliot
8,2872720
$endgroup$
$begingroup$
I am not asking if $phi$ is linear; some linear constraint on $phi$. Something like $sum_x F(phi(x)) = k$ for some function $F$ and some constant $k$.
$endgroup$
– Kumara
Jan 17 at 9:19
$begingroup$
@Kumara. Yes. Let k = n and F is constant function 1 or k = 0 and F the constant function 0.
$endgroup$
– William Elliot
Jan 18 at 0:01
$begingroup$
Please understand the question properly. I am asking a condition on $phi$ $equivalent$ to one-to-oneness.
$endgroup$
– Kumara
Jan 18 at 4:10
add a comment |
$begingroup$
A bijection from A to S = { 1,2,.. n } usually it isn't linear.
How could it be if A, for example, were a set of open sets?
Even if A were a set of integers,
how could there be a linear bijection from { x$^2$ : x in S } to S?
answered Jan 17 at 7:43
William ElliotWilliam Elliot
8,2872720
$endgroup$
A bijection from A to S = { 1,2,.. n } usually it isn't linear.
How could it be if A, for example, were a set of open sets?
Even if A were a set of integers,
how could there be a linear bijection from { x$^2$ : x in S } to S?
answered Jan 17 at 7:43
William ElliotWilliam Elliot
8,2872720
answered Jan 17 at 7:43
William ElliotWilliam Elliot
8,2872720
answered Jan 17 at 7:43
William ElliotWilliam Elliot
8,2872720
answered Jan 17 at 7:43
William ElliotWilliam Elliot
8,2872720
8,2872720
$begingroup$
I am not asking if $phi$ is linear; some linear constraint on $phi$. Something like $sum_x F(phi(x)) = k$ for some function $F$ and some constant $k$.
$endgroup$
– Kumara
Jan 17 at 9:19
$begingroup$
@Kumara. Yes. Let k = n and F is constant function 1 or k = 0 and F the constant function 0.
$endgroup$
– William Elliot
Jan 18 at 0:01
$begingroup$
Please understand the question properly. I am asking a condition on $phi$ $equivalent$ to one-to-oneness.
$endgroup$
– Kumara
Jan 18 at 4:10
add a comment |
$begingroup$
I am not asking if $phi$ is linear; some linear constraint on $phi$. Something like $sum_x F(phi(x)) = k$ for some function $F$ and some constant $k$.
$endgroup$
– Kumara
Jan 17 at 9:19
$begingroup$
@Kumara. Yes. Let k = n and F is constant function 1 or k = 0 and F the constant function 0.
$endgroup$
– William Elliot
Jan 18 at 0:01
$begingroup$
Please understand the question properly. I am asking a condition on $phi$ $equivalent$ to one-to-oneness.
$endgroup$
– Kumara
Jan 18 at 4:10
$begingroup$
I am not asking if $phi$ is linear; some linear constraint on $phi$. Something like $sum_x F(phi(x)) = k$ for some function $F$ and some constant $k$.
$endgroup$
– Kumara
Jan 17 at 9:19
$begingroup$
I am not asking if $phi$ is linear; some linear constraint on $phi$. Something like $sum_x F(phi(x)) = k$ for some function $F$ and some constant $k$.
$endgroup$
– Kumara
Jan 17 at 9:19
$begingroup$
@Kumara. Yes. Let k = n and F is constant function 1 or k = 0 and F the constant function 0.
$endgroup$
– William Elliot
Jan 18 at 0:01
$begingroup$
@Kumara. Yes. Let k = n and F is constant function 1 or k = 0 and F the constant function 0.
$endgroup$
– William Elliot
Jan 18 at 0:01
$begingroup$
Please understand the question properly. I am asking a condition on $phi$ $equivalent$ to one-to-oneness.
$endgroup$
– Kumara
Jan 18 at 4:10
$begingroup$
Please understand the question properly. I am asking a condition on $phi$ $equivalent$ to one-to-oneness.
$endgroup$
– Kumara
Jan 18 at 4:10
add a comment |
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1
$begingroup$
Are binary variables acceptable?
$endgroup$
– LinAlg
Jan 20 at 22:26