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Finding set B when A and C is given along with relation .

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0

$begingroup$

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

share|cite|improve this question

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 | 

0

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

edited Mar 13 at 10:12

Floris Claassens

89116

asked Mar 13 at 10:01

Kshitij DhyaniKshitij Dhyani

104

$endgroup$

share|cite|improve this question

edited Mar 13 at 10:12

Floris Claassens

89116

asked Mar 13 at 10:01

Kshitij DhyaniKshitij Dhyani

104

share|cite|improve this question

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

asked Mar 13 at 10:01

Kshitij DhyaniKshitij Dhyani

104

asked Mar 13 at 10:01

Kshitij DhyaniKshitij Dhyani

104

1 Answer
1

active

oldest

votes

0

$begingroup$

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.

share|cite|improve this answer

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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0

$begingroup$

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.

share|cite|improve this answer

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 | 

0

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Mar 13 at 10:18

Floris ClaassensFloris Claassens

89116

$endgroup$

share|cite|improve this answer

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