Bipartite graph $G=(A,B)$ with $delta(A)=3n/2$ and no $C_4$ has a matching which saturate each vertex in $A$.
$begingroup$
Say $G$ 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 $frac32 n$ and $|A|leq n^2$, then $G$ has a matching which uses every vertex in $A$.
My proof goes like this
Use of Hall marriage theorem. Take any $Xsubseteq A$ and let $|X|=k$. Also let $Y=N(X)$ and $|Y|=l$. Let us prove that $lgeq k$. Assume that $l<k$.
Let $Y^*$ be a set of all pairs ${y_i,y_j}$, $ine j$ of elements in $Y$ and connect a pair with $xin X$ iff booth in pair are connected with $x$ in $G$. Then the degree of each pair in this new bipartite graph $G^*$ is at most $1$ (since there is no $C_4$ in $G$) and a degree of each $xin X$ is at least $displaystyle{{3nover 2}choose 2}$. So we have $$kcdot {{3nover 2}choose 2}leq {lchoose 2}$$
Since we assume $l<k$ we have $${{3nover 2}choose 2}< {k-1over 2}$$ so $${3n(3n-2)over 4} < k-1$$ Since $kleq n^2$ we have $$3n(3n-2) leq 4n^2-8$$so
$$5n^2leq 6n-8$$ which is obviously not true.
Comment: It seems that we can replace $3n/2$ with $n$?
Is this correct?
proof-verification graph-theory
asked May 25 '18 at 15:10
greedoidgreedoid
43.5k1154107
$endgroup$
add a comment |
$begingroup$
Say $G$ 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 $frac32 n$ and $|A|leq n^2$, then $G$ has a matching which uses every vertex in $A$.
My proof goes like this
Use of Hall marriage theorem. Take any $Xsubseteq A$ and let $|X|=k$. Also let $Y=N(X)$ and $|Y|=l$. Let us prove that $lgeq k$. Assume that $l<k$.
Let $Y^*$ be a set of all pairs ${y_i,y_j}$, $ine j$ of elements in $Y$ and connect a pair with $xin X$ iff booth in pair are connected with $x$ in $G$. Then the degree of each pair in this new bipartite graph $G^*$ is at most $1$ (since there is no $C_4$ in $G$) and a degree of each $xin X$ is at least $displaystyle{{3nover 2}choose 2}$. So we have $$kcdot {{3nover 2}choose 2}leq {lchoose 2}$$
Since we assume $l<k$ we have $${{3nover 2}choose 2}< {k-1over 2}$$ so $${3n(3n-2)over 4} < k-1$$ Since $kleq n^2$ we have $$3n(3n-2) leq 4n^2-8$$so
$$5n^2leq 6n-8$$ which is obviously not true.
Comment: It seems that we can replace $3n/2$ with $n$?
Is this correct?
proof-verification graph-theory
asked May 25 '18 at 15:10
greedoidgreedoid
43.5k1154107
$endgroup$
add a comment |
$begingroup$
Say $G$ 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 $frac32 n$ and $|A|leq n^2$, then $G$ has a matching which uses every vertex in $A$.
My proof goes like this
Use of Hall marriage theorem. Take any $Xsubseteq A$ and let $|X|=k$. Also let $Y=N(X)$ and $|Y|=l$. Let us prove that $lgeq k$. Assume that $l<k$.
Let $Y^*$ be a set of all pairs ${y_i,y_j}$, $ine j$ of elements in $Y$ and connect a pair with $xin X$ iff booth in pair are connected with $x$ in $G$. Then the degree of each pair in this new bipartite graph $G^*$ is at most $1$ (since there is no $C_4$ in $G$) and a degree of each $xin X$ is at least $displaystyle{{3nover 2}choose 2}$. So we have $$kcdot {{3nover 2}choose 2}leq {lchoose 2}$$
Since we assume $l<k$ we have $${{3nover 2}choose 2}< {k-1over 2}$$ so $${3n(3n-2)over 4} < k-1$$ Since $kleq n^2$ we have $$3n(3n-2) leq 4n^2-8$$so
$$5n^2leq 6n-8$$ which is obviously not true.
Comment: It seems that we can replace $3n/2$ with $n$?
Is this correct?
proof-verification graph-theory
asked May 25 '18 at 15:10
greedoidgreedoid
43.5k1154107
$endgroup$
Say $G$ 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 $frac32 n$ and $|A|leq n^2$, then $G$ has a matching which uses every vertex in $A$.
My proof goes like this
Use of Hall marriage theorem. Take any $Xsubseteq A$ and let $|X|=k$. Also let $Y=N(X)$ and $|Y|=l$. Let us prove that $lgeq k$. Assume that $l<k$.
Let $Y^*$ be a set of all pairs ${y_i,y_j}$, $ine j$ of elements in $Y$ and connect a pair with $xin X$ iff booth in pair are connected with $x$ in $G$. Then the degree of each pair in this new bipartite graph $G^*$ is at most $1$ (since there is no $C_4$ in $G$) and a degree of each $xin X$ is at least $displaystyle{{3nover 2}choose 2}$. So we have $$kcdot {{3nover 2}choose 2}leq {lchoose 2}$$
Since we assume $l<k$ we have $${{3nover 2}choose 2}< {k-1over 2}$$ so $${3n(3n-2)over 4} < k-1$$ Since $kleq n^2$ we have $$3n(3n-2) leq 4n^2-8$$so
$$5n^2leq 6n-8$$ which is obviously not true.
Comment: It seems that we can replace $3n/2$ with $n$?
Is this correct?
proof-verification graph-theory
proof-verification graph-theory
asked May 25 '18 at 15:10
greedoidgreedoid
43.5k1154107
asked May 25 '18 at 15:10
greedoidgreedoid
43.5k1154107
edited Dec 11 '18 at 21:55
greedoid
asked May 25 '18 at 15:10
greedoidgreedoid
43.5k1154107
asked May 25 '18 at 15:10
greedoidgreedoid
43.5k1154107
asked May 25 '18 at 15:10
greedoidgreedoid
43.5k1154107
43.5k1154107
add a comment |
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