Upper bound for certain number of colorings of $(2k,k^2)$-graph $G$
up vote
2
down vote
favorite
Let $G$ be a graph with $2k$ vertices and $k^2$ edges, $kgeq 1$, such that $G$ contains $K_{k-i, k+i}$ as a subgraph, where $0leq i leq k-1$.
Suppose that the complete bipartite subgraph $K_{k-i, k+i}$ has a bipartition $A_1cup A_2$, where $|A_1|=k-i$ and $|A_2|=k+i$. We define
$$P(G, A_1, A_2, 5)={ text{$5$-(vertex) colorings of $G$ such that every vertex of $A_1$ is color $1$ or $2$ and every vertex of $A_2$ is color $3,4$, or $5$} }.$$
I want to show that for any such graph $G$ we have
$$P(G, A_1, A_2, 5) leq 6^k.$$
***If for whatever reason the bound $6^k$ is too small, I am also content with obtaining an upper bound of $6^k+ o(6^k)$.
I thought that perhaps I could prove this claim by induction on $0leq ileq k-i$. One nice observation I have made is that the parameter $P(G, A_1, A_2, 5)$ satisfies a deletion-contraction identity, just as the chromatic polynomial does. That is,
$$P(G, A_1, A_2, 5)=P(G-e, A_1, A_2, 5)-P(G/e, A_1, A_2, 5),$$
for any $ein E(G)$.
Alright, so here is my base case: Suppose $i=1$. Then $K_{k-1,k+1}$ is a subgraph of $G$ which means that there is one edge $e$ in $G$ whose vertex ends are both in $A_1$ or both in $A_2$.
Case 1: Both ends of $e$ are in $A_1$.
We have
$$P(G, A_1, A_2, 5)=P(G-e, A_1, A_2, 5)-P(G/e, A_1, A_2, 5)$$
$$=2^{k-1}3^{k+1}-2^{k-2}3^{k+1} $$
$$=2^{k-2}3^{k+1}=6^k(3/4)<6^k.$$
Case 2: Both ends of $e$ are in $A_2$.
We have
$$P(G, A_1, A_2, 5)=P(G-e, A_1, A_2, 5)-P(G/e, A_1, A_2, 5)$$
$$=2^{k-1}3^{k+1}-2^{k-1}3^{k} $$
$$=2^{k-1}3^{k}(3-1)=6^k.$$
Thus, the bound holds when $i=1$.
I am struggling a lot with the inductive case. If I know that $K_{k-i, k+i}$ is a subgraph of $G$ then there are $i^2$ edges in $G$ that are not in $K_{k-i, k+i}$. Each of these edges must have both ends in $A_1$ or in $A_2$. I can't seem to get the inductive hypothesis to "pop out".
Any help is immensely appreciated. Thank you.
combinatorics discrete-mathematics graph-theory extremal-combinatorics
add a comment |
up vote
2
down vote
favorite
Let $G$ be a graph with $2k$ vertices and $k^2$ edges, $kgeq 1$, such that $G$ contains $K_{k-i, k+i}$ as a subgraph, where $0leq i leq k-1$.
Suppose that the complete bipartite subgraph $K_{k-i, k+i}$ has a bipartition $A_1cup A_2$, where $|A_1|=k-i$ and $|A_2|=k+i$. We define
$$P(G, A_1, A_2, 5)={ text{$5$-(vertex) colorings of $G$ such that every vertex of $A_1$ is color $1$ or $2$ and every vertex of $A_2$ is color $3,4$, or $5$} }.$$
I want to show that for any such graph $G$ we have
$$P(G, A_1, A_2, 5) leq 6^k.$$
***If for whatever reason the bound $6^k$ is too small, I am also content with obtaining an upper bound of $6^k+ o(6^k)$.
I thought that perhaps I could prove this claim by induction on $0leq ileq k-i$. One nice observation I have made is that the parameter $P(G, A_1, A_2, 5)$ satisfies a deletion-contraction identity, just as the chromatic polynomial does. That is,
$$P(G, A_1, A_2, 5)=P(G-e, A_1, A_2, 5)-P(G/e, A_1, A_2, 5),$$
for any $ein E(G)$.
Alright, so here is my base case: Suppose $i=1$. Then $K_{k-1,k+1}$ is a subgraph of $G$ which means that there is one edge $e$ in $G$ whose vertex ends are both in $A_1$ or both in $A_2$.
Case 1: Both ends of $e$ are in $A_1$.
We have
$$P(G, A_1, A_2, 5)=P(G-e, A_1, A_2, 5)-P(G/e, A_1, A_2, 5)$$
$$=2^{k-1}3^{k+1}-2^{k-2}3^{k+1} $$
$$=2^{k-2}3^{k+1}=6^k(3/4)<6^k.$$
Case 2: Both ends of $e$ are in $A_2$.
We have
$$P(G, A_1, A_2, 5)=P(G-e, A_1, A_2, 5)-P(G/e, A_1, A_2, 5)$$
$$=2^{k-1}3^{k+1}-2^{k-1}3^{k} $$
$$=2^{k-1}3^{k}(3-1)=6^k.$$
Thus, the bound holds when $i=1$.
I am struggling a lot with the inductive case. If I know that $K_{k-i, k+i}$ is a subgraph of $G$ then there are $i^2$ edges in $G$ that are not in $K_{k-i, k+i}$. Each of these edges must have both ends in $A_1$ or in $A_2$. I can't seem to get the inductive hypothesis to "pop out".
Any help is immensely appreciated. Thank you.
combinatorics discrete-mathematics graph-theory extremal-combinatorics
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $G$ be a graph with $2k$ vertices and $k^2$ edges, $kgeq 1$, such that $G$ contains $K_{k-i, k+i}$ as a subgraph, where $0leq i leq k-1$.
Suppose that the complete bipartite subgraph $K_{k-i, k+i}$ has a bipartition $A_1cup A_2$, where $|A_1|=k-i$ and $|A_2|=k+i$. We define
$$P(G, A_1, A_2, 5)={ text{$5$-(vertex) colorings of $G$ such that every vertex of $A_1$ is color $1$ or $2$ and every vertex of $A_2$ is color $3,4$, or $5$} }.$$
I want to show that for any such graph $G$ we have
$$P(G, A_1, A_2, 5) leq 6^k.$$
***If for whatever reason the bound $6^k$ is too small, I am also content with obtaining an upper bound of $6^k+ o(6^k)$.
I thought that perhaps I could prove this claim by induction on $0leq ileq k-i$. One nice observation I have made is that the parameter $P(G, A_1, A_2, 5)$ satisfies a deletion-contraction identity, just as the chromatic polynomial does. That is,
$$P(G, A_1, A_2, 5)=P(G-e, A_1, A_2, 5)-P(G/e, A_1, A_2, 5),$$
for any $ein E(G)$.
Alright, so here is my base case: Suppose $i=1$. Then $K_{k-1,k+1}$ is a subgraph of $G$ which means that there is one edge $e$ in $G$ whose vertex ends are both in $A_1$ or both in $A_2$.
Case 1: Both ends of $e$ are in $A_1$.
We have
$$P(G, A_1, A_2, 5)=P(G-e, A_1, A_2, 5)-P(G/e, A_1, A_2, 5)$$
$$=2^{k-1}3^{k+1}-2^{k-2}3^{k+1} $$
$$=2^{k-2}3^{k+1}=6^k(3/4)<6^k.$$
Case 2: Both ends of $e$ are in $A_2$.
We have
$$P(G, A_1, A_2, 5)=P(G-e, A_1, A_2, 5)-P(G/e, A_1, A_2, 5)$$
$$=2^{k-1}3^{k+1}-2^{k-1}3^{k} $$
$$=2^{k-1}3^{k}(3-1)=6^k.$$
Thus, the bound holds when $i=1$.
I am struggling a lot with the inductive case. If I know that $K_{k-i, k+i}$ is a subgraph of $G$ then there are $i^2$ edges in $G$ that are not in $K_{k-i, k+i}$. Each of these edges must have both ends in $A_1$ or in $A_2$. I can't seem to get the inductive hypothesis to "pop out".
Any help is immensely appreciated. Thank you.
combinatorics discrete-mathematics graph-theory extremal-combinatorics
Let $G$ be a graph with $2k$ vertices and $k^2$ edges, $kgeq 1$, such that $G$ contains $K_{k-i, k+i}$ as a subgraph, where $0leq i leq k-1$.
Suppose that the complete bipartite subgraph $K_{k-i, k+i}$ has a bipartition $A_1cup A_2$, where $|A_1|=k-i$ and $|A_2|=k+i$. We define
$$P(G, A_1, A_2, 5)={ text{$5$-(vertex) colorings of $G$ such that every vertex of $A_1$ is color $1$ or $2$ and every vertex of $A_2$ is color $3,4$, or $5$} }.$$
I want to show that for any such graph $G$ we have
$$P(G, A_1, A_2, 5) leq 6^k.$$
***If for whatever reason the bound $6^k$ is too small, I am also content with obtaining an upper bound of $6^k+ o(6^k)$.
I thought that perhaps I could prove this claim by induction on $0leq ileq k-i$. One nice observation I have made is that the parameter $P(G, A_1, A_2, 5)$ satisfies a deletion-contraction identity, just as the chromatic polynomial does. That is,
$$P(G, A_1, A_2, 5)=P(G-e, A_1, A_2, 5)-P(G/e, A_1, A_2, 5),$$
for any $ein E(G)$.
Alright, so here is my base case: Suppose $i=1$. Then $K_{k-1,k+1}$ is a subgraph of $G$ which means that there is one edge $e$ in $G$ whose vertex ends are both in $A_1$ or both in $A_2$.
Case 1: Both ends of $e$ are in $A_1$.
We have
$$P(G, A_1, A_2, 5)=P(G-e, A_1, A_2, 5)-P(G/e, A_1, A_2, 5)$$
$$=2^{k-1}3^{k+1}-2^{k-2}3^{k+1} $$
$$=2^{k-2}3^{k+1}=6^k(3/4)<6^k.$$
Case 2: Both ends of $e$ are in $A_2$.
We have
$$P(G, A_1, A_2, 5)=P(G-e, A_1, A_2, 5)-P(G/e, A_1, A_2, 5)$$
$$=2^{k-1}3^{k+1}-2^{k-1}3^{k} $$
$$=2^{k-1}3^{k}(3-1)=6^k.$$
Thus, the bound holds when $i=1$.
I am struggling a lot with the inductive case. If I know that $K_{k-i, k+i}$ is a subgraph of $G$ then there are $i^2$ edges in $G$ that are not in $K_{k-i, k+i}$. Each of these edges must have both ends in $A_1$ or in $A_2$. I can't seem to get the inductive hypothesis to "pop out".
Any help is immensely appreciated. Thank you.
combinatorics discrete-mathematics graph-theory extremal-combinatorics
combinatorics discrete-mathematics graph-theory extremal-combinatorics
edited Nov 20 at 22:33
asked Nov 20 at 20:10
Sarah
880820
880820
add a comment |
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006822%2fupper-bound-for-certain-number-of-colorings-of-2k-k2-graph-g%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown