Confusion about the proof of Menger's Theorem in “Introduction to Graph Theory” by Douglas West
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The proof of Menger's Theorem in the book "Introduction to Graph Theory" by Douglas West (2nd Edition; Page 167) has been divided into two cases.
The second case assumes that
"Every minimum $x,y$-cut is $N(x)$ or $N(y)$",
where $N(x)$ denote the set of neighbors of $x$.
However, it seems that the graph for Case 2 (see below) in the illustration does not satisfy this assumption. What is going on here?
graph-theory proof-explanation connectedness graph-connectivity
asked Nov 22 at 12:36
hengxin
1,5381427
add a comment |
up vote
3
down vote
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The proof of Menger's Theorem in the book "Introduction to Graph Theory" by Douglas West (2nd Edition; Page 167) has been divided into two cases.
The second case assumes that
"Every minimum $x,y$-cut is $N(x)$ or $N(y)$",
where $N(x)$ denote the set of neighbors of $x$.
However, it seems that the graph for Case 2 (see below) in the illustration does not satisfy this assumption. What is going on here?
graph-theory proof-explanation connectedness graph-connectivity
asked Nov 22 at 12:36
hengxin
1,5381427
Do you have the link to the book? I can only find solution manual: home.ku.edu.tr/mudogan/public_html/…
– mathnoob
Nov 22 at 12:48
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The proof of Menger's Theorem in the book "Introduction to Graph Theory" by Douglas West (2nd Edition; Page 167) has been divided into two cases.
The second case assumes that
"Every minimum $x,y$-cut is $N(x)$ or $N(y)$",
where $N(x)$ denote the set of neighbors of $x$.
However, it seems that the graph for Case 2 (see below) in the illustration does not satisfy this assumption. What is going on here?
graph-theory proof-explanation connectedness graph-connectivity
asked Nov 22 at 12:36
hengxin
1,5381427
The proof of Menger's Theorem in the book "Introduction to Graph Theory" by Douglas West (2nd Edition; Page 167) has been divided into two cases.
The second case assumes that
"Every minimum $x,y$-cut is $N(x)$ or $N(y)$",
where $N(x)$ denote the set of neighbors of $x$.
However, it seems that the graph for Case 2 (see below) in the illustration does not satisfy this assumption. What is going on here?
graph-theory proof-explanation connectedness graph-connectivity
graph-theory proof-explanation connectedness graph-connectivity
asked Nov 22 at 12:36
hengxin
1,5381427
asked Nov 22 at 12:36
hengxin
1,5381427
edited Nov 23 at 9:26
asked Nov 22 at 12:36
hengxin
1,5381427
asked Nov 22 at 12:36
hengxin
1,5381427
asked Nov 22 at 12:36
hengxin
1,5381427
1,5381427
Do you have the link to the book? I can only find solution manual: home.ku.edu.tr/mudogan/public_html/…
– mathnoob
Nov 22 at 12:48
add a comment |
Do you have the link to the book? I can only find solution manual: home.ku.edu.tr/mudogan/public_html/…
– mathnoob
Nov 22 at 12:48
Do you have the link to the book? I can only find solution manual: home.ku.edu.tr/mudogan/public_html/…
– mathnoob
Nov 22 at 12:48
Do you have the link to the book? I can only find solution manual: home.ku.edu.tr/mudogan/public_html/…
– mathnoob
Nov 22 at 12:48
add a comment |
1 Answer
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The illustrations will match up with the cases, if we change the descriptions of the cases to:
Case 1'. $G$ has a minimum $x,y$-cut $S$ not contained in $N(x) cup N(y)$.
Case 2'. Every minimum $x,y$-cut is contained in $N(x) cup N(y)$.
If we follow these descriptions, then the proof still works, because the only way we use the assumption in case 2 is to say that every vertex outside ${x} cup N(x) cup N(y) cup {y}$ is in no minimum $x,y$-cut, and this is still true in case 2'. (Since case 1' is a subcase of case 1, there is nothing to worry about there.)
In general, whenever $G$ falls under both case 1 and case 2' (that is, every minimum $x,y$-cut is contained in $N(x) cup N(y)$, but there is some minimum $x,y$-cut $S$ not equal to $N(x)$ or $N(y)$) then we can handle $G$ by the argument from either case, which is where this flexibility comes from.
Pedagogical note: when I taught this proof last year, I began by considering the case where $N(x) cap N(y) = varnothing$ and $V(G) = {x} cup N(x) cup N(y) cup {y}$, which falls under case 2' and is the case where we can apply König-Egerváry. Then I dealt with the three possibilities below:
$v in N(x) cap N(y)$, which is handled in the case 2 proof. (Delete $v$, reducing $kappa(x,y)$ by $1$.)
$v notin {x} cup N(x) cup N(y) cup {y}$, but $v$ is not part of any minimum $x,y$-cut, which is also handled in the case 2 proof. (Delete $v$, not changing $kappa(x,y)$.)- There is a minimum $x,y$-cut $S$ not contained in $N(x) cup N(y)$. (This is case 1', and we can apply the case 1 proof.)
In some sense, this is the logical progression: we apply König-Egerváry for some cases, and then show that all other cases can be reduced to smaller ones.
answered Nov 23 at 18:16
Misha Lavrov
43.1k555103
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The illustrations will match up with the cases, if we change the descriptions of the cases to:
Case 1'. $G$ has a minimum $x,y$-cut $S$ not contained in $N(x) cup N(y)$.
Case 2'. Every minimum $x,y$-cut is contained in $N(x) cup N(y)$.
If we follow these descriptions, then the proof still works, because the only way we use the assumption in case 2 is to say that every vertex outside ${x} cup N(x) cup N(y) cup {y}$ is in no minimum $x,y$-cut, and this is still true in case 2'. (Since case 1' is a subcase of case 1, there is nothing to worry about there.)
In general, whenever $G$ falls under both case 1 and case 2' (that is, every minimum $x,y$-cut is contained in $N(x) cup N(y)$, but there is some minimum $x,y$-cut $S$ not equal to $N(x)$ or $N(y)$) then we can handle $G$ by the argument from either case, which is where this flexibility comes from.
Pedagogical note: when I taught this proof last year, I began by considering the case where $N(x) cap N(y) = varnothing$ and $V(G) = {x} cup N(x) cup N(y) cup {y}$, which falls under case 2' and is the case where we can apply König-Egerváry. Then I dealt with the three possibilities below:
$v in N(x) cap N(y)$, which is handled in the case 2 proof. (Delete $v$, reducing $kappa(x,y)$ by $1$.)
$v notin {x} cup N(x) cup N(y) cup {y}$, but $v$ is not part of any minimum $x,y$-cut, which is also handled in the case 2 proof. (Delete $v$, not changing $kappa(x,y)$.)- There is a minimum $x,y$-cut $S$ not contained in $N(x) cup N(y)$. (This is case 1', and we can apply the case 1 proof.)
In some sense, this is the logical progression: we apply König-Egerváry for some cases, and then show that all other cases can be reduced to smaller ones.
answered Nov 23 at 18:16
Misha Lavrov
43.1k555103
add a comment |
up vote
1
down vote
accepted
The illustrations will match up with the cases, if we change the descriptions of the cases to:
Case 1'. $G$ has a minimum $x,y$-cut $S$ not contained in $N(x) cup N(y)$.
Case 2'. Every minimum $x,y$-cut is contained in $N(x) cup N(y)$.
If we follow these descriptions, then the proof still works, because the only way we use the assumption in case 2 is to say that every vertex outside ${x} cup N(x) cup N(y) cup {y}$ is in no minimum $x,y$-cut, and this is still true in case 2'. (Since case 1' is a subcase of case 1, there is nothing to worry about there.)
In general, whenever $G$ falls under both case 1 and case 2' (that is, every minimum $x,y$-cut is contained in $N(x) cup N(y)$, but there is some minimum $x,y$-cut $S$ not equal to $N(x)$ or $N(y)$) then we can handle $G$ by the argument from either case, which is where this flexibility comes from.
Pedagogical note: when I taught this proof last year, I began by considering the case where $N(x) cap N(y) = varnothing$ and $V(G) = {x} cup N(x) cup N(y) cup {y}$, which falls under case 2' and is the case where we can apply König-Egerváry. Then I dealt with the three possibilities below:
$v in N(x) cap N(y)$, which is handled in the case 2 proof. (Delete $v$, reducing $kappa(x,y)$ by $1$.)
$v notin {x} cup N(x) cup N(y) cup {y}$, but $v$ is not part of any minimum $x,y$-cut, which is also handled in the case 2 proof. (Delete $v$, not changing $kappa(x,y)$.)- There is a minimum $x,y$-cut $S$ not contained in $N(x) cup N(y)$. (This is case 1', and we can apply the case 1 proof.)
In some sense, this is the logical progression: we apply König-Egerváry for some cases, and then show that all other cases can be reduced to smaller ones.
answered Nov 23 at 18:16
Misha Lavrov
43.1k555103
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The illustrations will match up with the cases, if we change the descriptions of the cases to:
Case 1'. $G$ has a minimum $x,y$-cut $S$ not contained in $N(x) cup N(y)$.
Case 2'. Every minimum $x,y$-cut is contained in $N(x) cup N(y)$.
If we follow these descriptions, then the proof still works, because the only way we use the assumption in case 2 is to say that every vertex outside ${x} cup N(x) cup N(y) cup {y}$ is in no minimum $x,y$-cut, and this is still true in case 2'. (Since case 1' is a subcase of case 1, there is nothing to worry about there.)
In general, whenever $G$ falls under both case 1 and case 2' (that is, every minimum $x,y$-cut is contained in $N(x) cup N(y)$, but there is some minimum $x,y$-cut $S$ not equal to $N(x)$ or $N(y)$) then we can handle $G$ by the argument from either case, which is where this flexibility comes from.
Pedagogical note: when I taught this proof last year, I began by considering the case where $N(x) cap N(y) = varnothing$ and $V(G) = {x} cup N(x) cup N(y) cup {y}$, which falls under case 2' and is the case where we can apply König-Egerváry. Then I dealt with the three possibilities below:
$v in N(x) cap N(y)$, which is handled in the case 2 proof. (Delete $v$, reducing $kappa(x,y)$ by $1$.)
$v notin {x} cup N(x) cup N(y) cup {y}$, but $v$ is not part of any minimum $x,y$-cut, which is also handled in the case 2 proof. (Delete $v$, not changing $kappa(x,y)$.)- There is a minimum $x,y$-cut $S$ not contained in $N(x) cup N(y)$. (This is case 1', and we can apply the case 1 proof.)
In some sense, this is the logical progression: we apply König-Egerváry for some cases, and then show that all other cases can be reduced to smaller ones.
answered Nov 23 at 18:16
Misha Lavrov
43.1k555103
The illustrations will match up with the cases, if we change the descriptions of the cases to:
Case 1'. $G$ has a minimum $x,y$-cut $S$ not contained in $N(x) cup N(y)$.
Case 2'. Every minimum $x,y$-cut is contained in $N(x) cup N(y)$.
If we follow these descriptions, then the proof still works, because the only way we use the assumption in case 2 is to say that every vertex outside ${x} cup N(x) cup N(y) cup {y}$ is in no minimum $x,y$-cut, and this is still true in case 2'. (Since case 1' is a subcase of case 1, there is nothing to worry about there.)
In general, whenever $G$ falls under both case 1 and case 2' (that is, every minimum $x,y$-cut is contained in $N(x) cup N(y)$, but there is some minimum $x,y$-cut $S$ not equal to $N(x)$ or $N(y)$) then we can handle $G$ by the argument from either case, which is where this flexibility comes from.
Pedagogical note: when I taught this proof last year, I began by considering the case where $N(x) cap N(y) = varnothing$ and $V(G) = {x} cup N(x) cup N(y) cup {y}$, which falls under case 2' and is the case where we can apply König-Egerváry. Then I dealt with the three possibilities below:
$v in N(x) cap N(y)$, which is handled in the case 2 proof. (Delete $v$, reducing $kappa(x,y)$ by $1$.)
$v notin {x} cup N(x) cup N(y) cup {y}$, but $v$ is not part of any minimum $x,y$-cut, which is also handled in the case 2 proof. (Delete $v$, not changing $kappa(x,y)$.)- There is a minimum $x,y$-cut $S$ not contained in $N(x) cup N(y)$. (This is case 1', and we can apply the case 1 proof.)
In some sense, this is the logical progression: we apply König-Egerváry for some cases, and then show that all other cases can be reduced to smaller ones.
answered Nov 23 at 18:16
Misha Lavrov
43.1k555103
answered Nov 23 at 18:16
Misha Lavrov
43.1k555103
answered Nov 23 at 18:16
Misha Lavrov
43.1k555103
answered Nov 23 at 18:16
Misha Lavrov
43.1k555103
43.1k555103
add a comment |
add a comment |
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Do you have the link to the book? I can only find solution manual: home.ku.edu.tr/mudogan/public_html/…
– mathnoob
Nov 22 at 12:48