Fake Proof: Finding mistakes in a “proof” of Menger's theorem The Next CEO of Stack OverflowProof for Menger's TheoremUnderstanding the Proof of Dirac's Theorem Regarding Graph ConnectivityFinding maximal set of vertex disjoint augmenting paths in Hopcroft KarpBetween any two vertices $u,v$ in a 3-connected graph, there are two internally disjoint $u$-$v$ paths of different lengths?A graph with maximum vertex degree $3$ can be divided into $2$ groups with simple structureSeparating a planar graph into two components containing shortest pathsFixing a proof on triangles in graphsProving this corollary to Menger's theorem in graph theoryProof of simple graph using pigeonhole theoremInductive Proof for Menger's Theorem
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Fake Proof: Finding mistakes in a “proof” of Menger's theorem
The Next CEO of Stack OverflowProof for Menger's TheoremUnderstanding the Proof of Dirac's Theorem Regarding Graph ConnectivityFinding maximal set of vertex disjoint augmenting paths in Hopcroft KarpBetween any two vertices $u,v$ in a 3-connected graph, there are two internally disjoint $u$-$v$ paths of different lengths?A graph with maximum vertex degree $3$ can be divided into $2$ groups with simple structureSeparating a planar graph into two components containing shortest pathsFixing a proof on triangles in graphsProving this corollary to Menger's theorem in graph theoryProof of simple graph using pigeonhole theoremInductive Proof for Menger's Theorem
$begingroup$
I am working on the following exercise:
Find all mistakes the following ‘proof’ of the set version of Menger’s
theorem. What statement(s) would be necessary to prove in order to complete the
proof?
Let S be a smallest $A-B$ separator. We say that a component $C$ of $G-S$ meets $A$ if $C$ contains a vertex of $A$. Denote by $G_A$ the graph that $G$ induces on the union of $S$ and all vertex sets of components of $G-S$ that meet $A$. Define $G_B$ analogously.
By the choice of $S$ and induction, $G_A$ contains $|S|$ disjoint $A-S$ paths, while $G_B$ contains $|S|$ disjoint $S-B$ paths. Joining these paths yields the desired set of $|S|$ disjoint $A-B$ paths.
In my opinion there is a huge gap in the part "By the choice of $S$ and induction, $G_A$ contains $|S|$ disjoint $A-S$ paths, while $G_B$ contains $|S|$ disjoint $S-B$ paths." I do not really understand what is happening here.
Do you think that this is a sufficient answer for this exercise?
graph-theory fake-proofs
asked Mar 18 at 11:19
3nondatur3nondatur
404111
$endgroup$
add a comment |
$begingroup$
I am working on the following exercise:
Find all mistakes the following ‘proof’ of the set version of Menger’s
theorem. What statement(s) would be necessary to prove in order to complete the
proof?
Let S be a smallest $A-B$ separator. We say that a component $C$ of $G-S$ meets $A$ if $C$ contains a vertex of $A$. Denote by $G_A$ the graph that $G$ induces on the union of $S$ and all vertex sets of components of $G-S$ that meet $A$. Define $G_B$ analogously.
By the choice of $S$ and induction, $G_A$ contains $|S|$ disjoint $A-S$ paths, while $G_B$ contains $|S|$ disjoint $S-B$ paths. Joining these paths yields the desired set of $|S|$ disjoint $A-B$ paths.
In my opinion there is a huge gap in the part "By the choice of $S$ and induction, $G_A$ contains $|S|$ disjoint $A-S$ paths, while $G_B$ contains $|S|$ disjoint $S-B$ paths." I do not really understand what is happening here.
Do you think that this is a sufficient answer for this exercise?
graph-theory fake-proofs
asked Mar 18 at 11:19
3nondatur3nondatur
404111
$endgroup$
$begingroup$
What is the induction hypothesis?
$endgroup$
– hbm
Mar 18 at 15:39
add a comment |
$begingroup$
I am working on the following exercise:
Find all mistakes the following ‘proof’ of the set version of Menger’s
theorem. What statement(s) would be necessary to prove in order to complete the
proof?
Let S be a smallest $A-B$ separator. We say that a component $C$ of $G-S$ meets $A$ if $C$ contains a vertex of $A$. Denote by $G_A$ the graph that $G$ induces on the union of $S$ and all vertex sets of components of $G-S$ that meet $A$. Define $G_B$ analogously.
By the choice of $S$ and induction, $G_A$ contains $|S|$ disjoint $A-S$ paths, while $G_B$ contains $|S|$ disjoint $S-B$ paths. Joining these paths yields the desired set of $|S|$ disjoint $A-B$ paths.
In my opinion there is a huge gap in the part "By the choice of $S$ and induction, $G_A$ contains $|S|$ disjoint $A-S$ paths, while $G_B$ contains $|S|$ disjoint $S-B$ paths." I do not really understand what is happening here.
Do you think that this is a sufficient answer for this exercise?
graph-theory fake-proofs
asked Mar 18 at 11:19
3nondatur3nondatur
404111
$endgroup$
I am working on the following exercise:
Find all mistakes the following ‘proof’ of the set version of Menger’s
theorem. What statement(s) would be necessary to prove in order to complete the
proof?
Let S be a smallest $A-B$ separator. We say that a component $C$ of $G-S$ meets $A$ if $C$ contains a vertex of $A$. Denote by $G_A$ the graph that $G$ induces on the union of $S$ and all vertex sets of components of $G-S$ that meet $A$. Define $G_B$ analogously.
By the choice of $S$ and induction, $G_A$ contains $|S|$ disjoint $A-S$ paths, while $G_B$ contains $|S|$ disjoint $S-B$ paths. Joining these paths yields the desired set of $|S|$ disjoint $A-B$ paths.
In my opinion there is a huge gap in the part "By the choice of $S$ and induction, $G_A$ contains $|S|$ disjoint $A-S$ paths, while $G_B$ contains $|S|$ disjoint $S-B$ paths." I do not really understand what is happening here.
Do you think that this is a sufficient answer for this exercise?
graph-theory fake-proofs
graph-theory fake-proofs
asked Mar 18 at 11:19
3nondatur3nondatur
404111
asked Mar 18 at 11:19
3nondatur3nondatur
404111
asked Mar 18 at 11:19
3nondatur3nondatur
404111
asked Mar 18 at 11:19
3nondatur3nondatur
404111
asked Mar 18 at 11:19
3nondatur3nondatur
404111
404111
$begingroup$
What is the induction hypothesis?
$endgroup$
– hbm
Mar 18 at 15:39
add a comment |
$begingroup$
What is the induction hypothesis?
$endgroup$
– hbm
Mar 18 at 15:39
$begingroup$
What is the induction hypothesis?
$endgroup$
– hbm
Mar 18 at 15:39
$begingroup$
What is the induction hypothesis?
$endgroup$
– hbm
Mar 18 at 15:39
add a comment |
0
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$begingroup$
What is the induction hypothesis?
$endgroup$
– hbm
Mar 18 at 15:39