Finding set B when A and C is given along with relation .Set notation: subtracting elements with given cardinality from the powersetPower Set MathematicsIn a transitive relation does x and z have to be the same element?Can this relation be transitive but not symmetric and reflexive?How to find the number of subsets of any given set that contain a particular numberCardinality of the set $[a,b]$ where $0 < a leq b$ and $a,b in mathbbN$.can a set have elements and sets?Naive set theory, empty set as subset of mathematical objectsSubset of a family of setsFind the number of the elements for each set. $emptyset$ , $emptyset$ , $emptyset$, $emptyset, emptyset$
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Finding set B when A and C is given along with relation .
Set notation: subtracting elements with given cardinality from the powersetPower Set MathematicsIn a transitive relation does x and z have to be the same element?Can this relation be transitive but not symmetric and reflexive?How to find the number of subsets of any given set that contain a particular numberCardinality of the set $[a,b]$ where $0 < a leq b$ and $a,b in mathbbN$.can a set have elements and sets?Naive set theory, empty set as subset of mathematical objectsSubset of a family of setsFind the number of the elements for each set. $emptyset$ , $emptyset$ , $emptyset$, $emptyset, emptyset$
$begingroup$
Set $C$ contains all the elements that can be represented using sum of an element of $A$ and an element of $B$.
In such a case if we have the values of set $A$ and set $C$, how can we find out the set $B$.
For eg :
If set $A=1,2$
and set $C=3,4,5$,
then set $B$ would be $B=2,3$.
I can work this around for small sets, but I fail to understand this for bigger sets, is there a global solution for this ?
elementary-set-theory
edited Mar 13 at 10:12
Floris Claassens
89116
asked Mar 13 at 10:01
Kshitij DhyaniKshitij Dhyani
104
$endgroup$
add a comment |
$begingroup$
Set $C$ contains all the elements that can be represented using sum of an element of $A$ and an element of $B$.
In such a case if we have the values of set $A$ and set $C$, how can we find out the set $B$.
For eg :
If set $A=1,2$
and set $C=3,4,5$,
then set $B$ would be $B=2,3$.
I can work this around for small sets, but I fail to understand this for bigger sets, is there a global solution for this ?
elementary-set-theory
edited Mar 13 at 10:12
Floris Claassens
89116
asked Mar 13 at 10:01
Kshitij DhyaniKshitij Dhyani
104
$endgroup$
$begingroup$
Please, use "{" to list elements of sets.
$endgroup$
– Mauro ALLEGRANZA
Mar 13 at 10:05
$begingroup$
@MauroALLEGRANZA no context is correct .
$endgroup$
– Kshitij Dhyani
Mar 13 at 10:09
$begingroup$
@MauroALLEGRANZA : OP is trying to find $B$, not $C$
$endgroup$
– MPW
Mar 13 at 10:09
add a comment |
$begingroup$
Set $C$ contains all the elements that can be represented using sum of an element of $A$ and an element of $B$.
In such a case if we have the values of set $A$ and set $C$, how can we find out the set $B$.
For eg :
If set $A=1,2$
and set $C=3,4,5$,
then set $B$ would be $B=2,3$.
I can work this around for small sets, but I fail to understand this for bigger sets, is there a global solution for this ?
elementary-set-theory
edited Mar 13 at 10:12
Floris Claassens
89116
asked Mar 13 at 10:01
Kshitij DhyaniKshitij Dhyani
104
$endgroup$
Set $C$ contains all the elements that can be represented using sum of an element of $A$ and an element of $B$.
In such a case if we have the values of set $A$ and set $C$, how can we find out the set $B$.
For eg :
If set $A=1,2$
and set $C=3,4,5$,
then set $B$ would be $B=2,3$.
I can work this around for small sets, but I fail to understand this for bigger sets, is there a global solution for this ?
elementary-set-theory
elementary-set-theory
edited Mar 13 at 10:12
Floris Claassens
89116
asked Mar 13 at 10:01
Kshitij DhyaniKshitij Dhyani
104
edited Mar 13 at 10:12
Floris Claassens
89116
asked Mar 13 at 10:01
Kshitij DhyaniKshitij Dhyani
104
edited Mar 13 at 10:12
Floris Claassens
89116
edited Mar 13 at 10:12
Floris Claassens
89116
edited Mar 13 at 10:12
Floris Claassens
89116
89116
asked Mar 13 at 10:01
Kshitij DhyaniKshitij Dhyani
104
asked Mar 13 at 10:01
Kshitij DhyaniKshitij Dhyani
104
asked Mar 13 at 10:01
Kshitij DhyaniKshitij Dhyani
104
104
$begingroup$
Please, use "{" to list elements of sets.
$endgroup$
– Mauro ALLEGRANZA
Mar 13 at 10:05
$begingroup$
@MauroALLEGRANZA no context is correct .
$endgroup$
– Kshitij Dhyani
Mar 13 at 10:09
$begingroup$
@MauroALLEGRANZA : OP is trying to find $B$, not $C$
$endgroup$
– MPW
Mar 13 at 10:09
add a comment |
$begingroup$
Please, use "{" to list elements of sets.
$endgroup$
– Mauro ALLEGRANZA
Mar 13 at 10:05
$begingroup$
@MauroALLEGRANZA no context is correct .
$endgroup$
– Kshitij Dhyani
Mar 13 at 10:09
$begingroup$
@MauroALLEGRANZA : OP is trying to find $B$, not $C$
$endgroup$
– MPW
Mar 13 at 10:09
$begingroup$
Please, use "{" to list elements of sets.
$endgroup$
– Mauro ALLEGRANZA
Mar 13 at 10:05
$begingroup$
Please, use "{" to list elements of sets.
$endgroup$
– Mauro ALLEGRANZA
Mar 13 at 10:05
$begingroup$
@MauroALLEGRANZA no context is correct .
$endgroup$
– Kshitij Dhyani
Mar 13 at 10:09
$begingroup$
@MauroALLEGRANZA no context is correct .
$endgroup$
– Kshitij Dhyani
Mar 13 at 10:09
$begingroup$
@MauroALLEGRANZA : OP is trying to find $B$, not $C$
$endgroup$
– MPW
Mar 13 at 10:09
$begingroup$
@MauroALLEGRANZA : OP is trying to find $B$, not $C$
$endgroup$
– MPW
Mar 13 at 10:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm afraid you might have to go with a brute force solution.
You can easily find $min(B)$ and $max(B)$ by taking $min(B)=min(C)-min(A)$ and $max(B)=max(C)-max(A)$. After that I see no other way but looking at all $min(B)<n<max(B)$ to see whether $A+nsubset C$. If $A+nsubset C$ then $nin B$, if $A+nsubsetneq C$ then $nnotin B$. Note that this gives the biggest possible $B$.
There might be a more elegant solution out there, and there might be much better solutions out there for "nice" sets, until then I would just program a quick for loop.
answered Mar 13 at 10:18
Floris ClaassensFloris Claassens
89116
$endgroup$
$begingroup$
what is A here in A+n⊂C
$endgroup$
– Kshitij Dhyani
Mar 13 at 10:36
$begingroup$
So we have initial sets $A$ and $C$, and we try to find set $B$ such that $A+B=C$, so $A$ is already given. $A+n$ is just shorthand for $A+n$.
$endgroup$
– Floris Claassens
Mar 13 at 10:56
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
I'm afraid you might have to go with a brute force solution.
You can easily find $min(B)$ and $max(B)$ by taking $min(B)=min(C)-min(A)$ and $max(B)=max(C)-max(A)$. After that I see no other way but looking at all $min(B)<n<max(B)$ to see whether $A+nsubset C$. If $A+nsubset C$ then $nin B$, if $A+nsubsetneq C$ then $nnotin B$. Note that this gives the biggest possible $B$.
There might be a more elegant solution out there, and there might be much better solutions out there for "nice" sets, until then I would just program a quick for loop.
answered Mar 13 at 10:18
Floris ClaassensFloris Claassens
89116
$endgroup$
$begingroup$
what is A here in A+n⊂C
$endgroup$
– Kshitij Dhyani
Mar 13 at 10:36
$begingroup$
So we have initial sets $A$ and $C$, and we try to find set $B$ such that $A+B=C$, so $A$ is already given. $A+n$ is just shorthand for $A+n$.
$endgroup$
– Floris Claassens
Mar 13 at 10:56
add a comment |
$begingroup$
I'm afraid you might have to go with a brute force solution.
You can easily find $min(B)$ and $max(B)$ by taking $min(B)=min(C)-min(A)$ and $max(B)=max(C)-max(A)$. After that I see no other way but looking at all $min(B)<n<max(B)$ to see whether $A+nsubset C$. If $A+nsubset C$ then $nin B$, if $A+nsubsetneq C$ then $nnotin B$. Note that this gives the biggest possible $B$.
There might be a more elegant solution out there, and there might be much better solutions out there for "nice" sets, until then I would just program a quick for loop.
answered Mar 13 at 10:18
Floris ClaassensFloris Claassens
89116
$endgroup$
$begingroup$
what is A here in A+n⊂C
$endgroup$
– Kshitij Dhyani
Mar 13 at 10:36
$begingroup$
So we have initial sets $A$ and $C$, and we try to find set $B$ such that $A+B=C$, so $A$ is already given. $A+n$ is just shorthand for $A+n$.
$endgroup$
– Floris Claassens
Mar 13 at 10:56
add a comment |
$begingroup$
I'm afraid you might have to go with a brute force solution.
You can easily find $min(B)$ and $max(B)$ by taking $min(B)=min(C)-min(A)$ and $max(B)=max(C)-max(A)$. After that I see no other way but looking at all $min(B)<n<max(B)$ to see whether $A+nsubset C$. If $A+nsubset C$ then $nin B$, if $A+nsubsetneq C$ then $nnotin B$. Note that this gives the biggest possible $B$.
There might be a more elegant solution out there, and there might be much better solutions out there for "nice" sets, until then I would just program a quick for loop.
answered Mar 13 at 10:18
Floris ClaassensFloris Claassens
89116
$endgroup$
I'm afraid you might have to go with a brute force solution.
You can easily find $min(B)$ and $max(B)$ by taking $min(B)=min(C)-min(A)$ and $max(B)=max(C)-max(A)$. After that I see no other way but looking at all $min(B)<n<max(B)$ to see whether $A+nsubset C$. If $A+nsubset C$ then $nin B$, if $A+nsubsetneq C$ then $nnotin B$. Note that this gives the biggest possible $B$.
There might be a more elegant solution out there, and there might be much better solutions out there for "nice" sets, until then I would just program a quick for loop.
answered Mar 13 at 10:18
Floris ClaassensFloris Claassens
89116
answered Mar 13 at 10:18
Floris ClaassensFloris Claassens
89116
answered Mar 13 at 10:18
Floris ClaassensFloris Claassens
89116
answered Mar 13 at 10:18
Floris ClaassensFloris Claassens
89116
89116
$begingroup$
what is A here in A+n⊂C
$endgroup$
– Kshitij Dhyani
Mar 13 at 10:36
$begingroup$
So we have initial sets $A$ and $C$, and we try to find set $B$ such that $A+B=C$, so $A$ is already given. $A+n$ is just shorthand for $A+n$.
$endgroup$
– Floris Claassens
Mar 13 at 10:56
add a comment |
$begingroup$
what is A here in A+n⊂C
$endgroup$
– Kshitij Dhyani
Mar 13 at 10:36
$begingroup$
So we have initial sets $A$ and $C$, and we try to find set $B$ such that $A+B=C$, so $A$ is already given. $A+n$ is just shorthand for $A+n$.
$endgroup$
– Floris Claassens
Mar 13 at 10:56
$begingroup$
what is A here in A+n⊂C
$endgroup$
– Kshitij Dhyani
Mar 13 at 10:36
$begingroup$
what is A here in A+n⊂C
$endgroup$
– Kshitij Dhyani
Mar 13 at 10:36
$begingroup$
So we have initial sets $A$ and $C$, and we try to find set $B$ such that $A+B=C$, so $A$ is already given. $A+n$ is just shorthand for $A+n$.
$endgroup$
– Floris Claassens
Mar 13 at 10:56
$begingroup$
So we have initial sets $A$ and $C$, and we try to find set $B$ such that $A+B=C$, so $A$ is already given. $A+n$ is just shorthand for $A+n$.
$endgroup$
– Floris Claassens
Mar 13 at 10:56
add a comment |
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$begingroup$
Please, use "{" to list elements of sets.
$endgroup$
– Mauro ALLEGRANZA
Mar 13 at 10:05
$begingroup$
@MauroALLEGRANZA no context is correct .
$endgroup$
– Kshitij Dhyani
Mar 13 at 10:09
$begingroup$
@MauroALLEGRANZA : OP is trying to find $B$, not $C$
$endgroup$
– MPW
Mar 13 at 10:09