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Matching in a bipartite graph

bipartite graph matchingbipartite graph matching exerciseMatching in $n$ by $b$ bipartite graphProof bipartite graph matchingComplete matching in bipartite graphMaximum matching for bipartite graphHow would we prove that the following bipartite graph has a perfect matching?Proofs involving a $d$-regular bipartite graphGraph Theory - MatchingBipartite graph $G=(A,B)$ with $delta(A)=3n/2$ and no $C_4$ has a matching which saturate each vertex in $A$.

2

0

$begingroup$

Suppose that $G$ is a bipartite graph with bipartition $(A,B)$ and $G$ is $C_4$-free. Prove that if every vertex in $A$ has degree at least $frac32 x$ and $|A|leq x^2$, then $G$ has a matching which uses every vertex in $A$.

Would I have to use Hall's Theorem in some way because there is matching or Tuttes?

graph-theory matching-theory

share|cite|improve this question

edited Mar 21 '18 at 21:17

Maria Mazur

47k1260120

asked May 2 '15 at 1:55

LindseyLindsey

978

$endgroup$

add a comment | 

2

0

$begingroup$

Suppose that $G$ is a bipartite graph with bipartition $(A,B)$ and $G$ is $C_4$-free. Prove that if every vertex in $A$ has degree at least $frac32 x$ and $|A|leq x^2$, then $G$ has a matching which uses every vertex in $A$.

Would I have to use Hall's Theorem in some way because there is matching or Tuttes?

graph-theory matching-theory

share|cite|improve this question

edited Mar 21 '18 at 21:17

Maria Mazur

47k1260120

asked May 2 '15 at 1:55

LindseyLindsey

978

$endgroup$

add a comment | 

$begingroup$

Suppose that $G$ is a bipartite graph with bipartition $(A,B)$ and $G$ is $C_4$-free. Prove that if every vertex in $A$ has degree at least $frac32 x$ and $|A|leq x^2$, then $G$ has a matching which uses every vertex in $A$.

Would I have to use Hall's Theorem in some way because there is matching or Tuttes?

graph-theory matching-theory

share|cite|improve this question

edited Mar 21 '18 at 21:17

Maria Mazur

47k1260120

asked May 2 '15 at 1:55

LindseyLindsey

978

$endgroup$

share|cite|improve this question

edited Mar 21 '18 at 21:17

Maria Mazur

47k1260120

asked May 2 '15 at 1:55

LindseyLindsey

978

share|cite|improve this question

edited Mar 21 '18 at 21:17

Maria Mazur

47k1260120

asked May 2 '15 at 1:55

LindseyLindsey

978

edited Mar 21 '18 at 21:17

Maria Mazur

47k1260120

edited Mar 21 '18 at 21:17

Maria Mazur

47k1260120

asked May 2 '15 at 1:55

LindseyLindsey

978

asked May 2 '15 at 1:55

LindseyLindsey

978

1 Answer
1

active

oldest

votes

0

$begingroup$

Assume the statement is not true and let $M$ be a maximum matching, $v$ an unmatched vertex of $A$.
Note that the requirement that $G$ is $C_4$-free means, that two vertices cannot have 2 common neighbours.
Let $t=lceilfrac32xrceil$.

$v$ has at least $t$ neighbours in $B$ that are all matched by $M$, call them $b_1,ldots,b_t$
and let $a_1,ldots,a_t$ be the vertices of $A$ such that $a_ib_i$ is an edge of $M$.
Each of the $a_i$ has at least $t-1$ neighbours in $B$ that must be different from all $b_i$
and each of them must be matched by $M$ (why?).

So $a_1$ has at least $t-1$ neighbours different from the $b_i$, call this set $A_1$.

$a_2$ has at least $t-1$ neighbours different from the $b_i$, and it can only have one neighbour in common
with $a_1$, so there is a set $A_2$ of size at least $t-2$ this is disjoint from both the $b_i$ and $A_1$.

Continuing this way (make a nice induction argument if you like), we find
$t+(t-1)+ldots+1$ different elements of $B$, all matched by $M$, so $A$ also must have at least
this number of elements.

Since $t+(t-1)+ldots+1=fract(t+1)2geqfracfrac94x^22>x^2$ we found a contradiction.

share|cite|improve this answer

answered May 3 '15 at 9:17

Leen DroogendijkLeen Droogendijk

6,1551716

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why must each be matched by M?
  $endgroup$
  – Lindsey
  May 4 '15 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $a_i$ would have a neighbor $c$ that is not matched by $M$, then $vb_ia_ic$ is an $M$-augmenting path, contradicting the maximality of $M$.
  $endgroup$
  – Leen Droogendijk
  May 5 '15 at 6:14
  

  
  
  
  

add a comment | 

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0

$begingroup$

Assume the statement is not true and let $M$ be a maximum matching, $v$ an unmatched vertex of $A$.
Note that the requirement that $G$ is $C_4$-free means, that two vertices cannot have 2 common neighbours.
Let $t=lceilfrac32xrceil$.

$v$ has at least $t$ neighbours in $B$ that are all matched by $M$, call them $b_1,ldots,b_t$
and let $a_1,ldots,a_t$ be the vertices of $A$ such that $a_ib_i$ is an edge of $M$.
Each of the $a_i$ has at least $t-1$ neighbours in $B$ that must be different from all $b_i$
and each of them must be matched by $M$ (why?).

So $a_1$ has at least $t-1$ neighbours different from the $b_i$, call this set $A_1$.

$a_2$ has at least $t-1$ neighbours different from the $b_i$, and it can only have one neighbour in common
with $a_1$, so there is a set $A_2$ of size at least $t-2$ this is disjoint from both the $b_i$ and $A_1$.

Continuing this way (make a nice induction argument if you like), we find
$t+(t-1)+ldots+1$ different elements of $B$, all matched by $M$, so $A$ also must have at least
this number of elements.

Since $t+(t-1)+ldots+1=fract(t+1)2geqfracfrac94x^22>x^2$ we found a contradiction.

share|cite|improve this answer

answered May 3 '15 at 9:17

Leen DroogendijkLeen Droogendijk

6,1551716

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why must each be matched by M?
  $endgroup$
  – Lindsey
  May 4 '15 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $a_i$ would have a neighbor $c$ that is not matched by $M$, then $vb_ia_ic$ is an $M$-augmenting path, contradicting the maximality of $M$.
  $endgroup$
  – Leen Droogendijk
  May 5 '15 at 6:14
  

  
  
  
  

add a comment | 

0

$begingroup$

Assume the statement is not true and let $M$ be a maximum matching, $v$ an unmatched vertex of $A$.
Note that the requirement that $G$ is $C_4$-free means, that two vertices cannot have 2 common neighbours.
Let $t=lceilfrac32xrceil$.

$v$ has at least $t$ neighbours in $B$ that are all matched by $M$, call them $b_1,ldots,b_t$
and let $a_1,ldots,a_t$ be the vertices of $A$ such that $a_ib_i$ is an edge of $M$.
Each of the $a_i$ has at least $t-1$ neighbours in $B$ that must be different from all $b_i$
and each of them must be matched by $M$ (why?).

So $a_1$ has at least $t-1$ neighbours different from the $b_i$, call this set $A_1$.

$a_2$ has at least $t-1$ neighbours different from the $b_i$, and it can only have one neighbour in common
with $a_1$, so there is a set $A_2$ of size at least $t-2$ this is disjoint from both the $b_i$ and $A_1$.

Continuing this way (make a nice induction argument if you like), we find
$t+(t-1)+ldots+1$ different elements of $B$, all matched by $M$, so $A$ also must have at least
this number of elements.

Since $t+(t-1)+ldots+1=fract(t+1)2geqfracfrac94x^22>x^2$ we found a contradiction.

share|cite|improve this answer

answered May 3 '15 at 9:17

Leen DroogendijkLeen Droogendijk

6,1551716

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why must each be matched by M?
  $endgroup$
  – Lindsey
  May 4 '15 at 18:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $a_i$ would have a neighbor $c$ that is not matched by $M$, then $vb_ia_ic$ is an $M$-augmenting path, contradicting the maximality of $M$.
  $endgroup$
  – Leen Droogendijk
  May 5 '15 at 6:14
  

  
  
  
  

add a comment | 

$begingroup$

Assume the statement is not true and let $M$ be a maximum matching, $v$ an unmatched vertex of $A$.
Note that the requirement that $G$ is $C_4$-free means, that two vertices cannot have 2 common neighbours.
Let $t=lceilfrac32xrceil$.

$v$ has at least $t$ neighbours in $B$ that are all matched by $M$, call them $b_1,ldots,b_t$
and let $a_1,ldots,a_t$ be the vertices of $A$ such that $a_ib_i$ is an edge of $M$.
Each of the $a_i$ has at least $t-1$ neighbours in $B$ that must be different from all $b_i$
and each of them must be matched by $M$ (why?).

So $a_1$ has at least $t-1$ neighbours different from the $b_i$, call this set $A_1$.

$a_2$ has at least $t-1$ neighbours different from the $b_i$, and it can only have one neighbour in common
with $a_1$, so there is a set $A_2$ of size at least $t-2$ this is disjoint from both the $b_i$ and $A_1$.

Continuing this way (make a nice induction argument if you like), we find
$t+(t-1)+ldots+1$ different elements of $B$, all matched by $M$, so $A$ also must have at least
this number of elements.

Since $t+(t-1)+ldots+1=fract(t+1)2geqfracfrac94x^22>x^2$ we found a contradiction.

share|cite|improve this answer

answered May 3 '15 at 9:17

Leen DroogendijkLeen Droogendijk

6,1551716

$endgroup$

share|cite|improve this answer

answered May 3 '15 at 9:17

Leen DroogendijkLeen Droogendijk

6,1551716

answered May 3 '15 at 9:17

Leen DroogendijkLeen Droogendijk

6,1551716

answered May 3 '15 at 9:17

Leen DroogendijkLeen Droogendijk

6,1551716