Bounds on Ramsey Numbers
$begingroup$
I'm working on a script with a section on Ramsey Theory. I know that $R(s,t) leq R(s-1,t) + R(s,t-1)$ and that you can add a -1 on the right side if both $R(s-1,t)$ and $R(s,t-1)$ are even. Using this inductively, one can prove $R(s,t) leq {s + t - 2 choose s-1}$, or $R(3,t) leq frac{1}{2} cdot (t^2 + t)$. Now, my script claims that this can be sharpened using the two upper inequalities to $R(3,t) leq frac{1}{2} cdot (t^2 + 3)$. How? I've been trying this for the past half hour and I can't find an answer. Thanks in advance for any help.
ramsey-theory
asked Jan 24 at 16:22
MateInTwoMateInTwo
207
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add a comment |
$begingroup$
I'm working on a script with a section on Ramsey Theory. I know that $R(s,t) leq R(s-1,t) + R(s,t-1)$ and that you can add a -1 on the right side if both $R(s-1,t)$ and $R(s,t-1)$ are even. Using this inductively, one can prove $R(s,t) leq {s + t - 2 choose s-1}$, or $R(3,t) leq frac{1}{2} cdot (t^2 + t)$. Now, my script claims that this can be sharpened using the two upper inequalities to $R(3,t) leq frac{1}{2} cdot (t^2 + 3)$. How? I've been trying this for the past half hour and I can't find an answer. Thanks in advance for any help.
ramsey-theory
asked Jan 24 at 16:22
MateInTwoMateInTwo
207
$endgroup$
add a comment |
$begingroup$
I'm working on a script with a section on Ramsey Theory. I know that $R(s,t) leq R(s-1,t) + R(s,t-1)$ and that you can add a -1 on the right side if both $R(s-1,t)$ and $R(s,t-1)$ are even. Using this inductively, one can prove $R(s,t) leq {s + t - 2 choose s-1}$, or $R(3,t) leq frac{1}{2} cdot (t^2 + t)$. Now, my script claims that this can be sharpened using the two upper inequalities to $R(3,t) leq frac{1}{2} cdot (t^2 + 3)$. How? I've been trying this for the past half hour and I can't find an answer. Thanks in advance for any help.
ramsey-theory
asked Jan 24 at 16:22
MateInTwoMateInTwo
207
$endgroup$
I'm working on a script with a section on Ramsey Theory. I know that $R(s,t) leq R(s-1,t) + R(s,t-1)$ and that you can add a -1 on the right side if both $R(s-1,t)$ and $R(s,t-1)$ are even. Using this inductively, one can prove $R(s,t) leq {s + t - 2 choose s-1}$, or $R(3,t) leq frac{1}{2} cdot (t^2 + t)$. Now, my script claims that this can be sharpened using the two upper inequalities to $R(3,t) leq frac{1}{2} cdot (t^2 + 3)$. How? I've been trying this for the past half hour and I can't find an answer. Thanks in advance for any help.
ramsey-theory
ramsey-theory
asked Jan 24 at 16:22
MateInTwoMateInTwo
207
asked Jan 24 at 16:22
MateInTwoMateInTwo
207
asked Jan 24 at 16:22
MateInTwoMateInTwo
207
asked Jan 24 at 16:22
MateInTwoMateInTwo
207
asked Jan 24 at 16:22
MateInTwoMateInTwo
207
207
add a comment |
add a comment |
1 Answer
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$begingroup$
Observe first that $R(2,t) = t$ since any $2$-coloring of $K_t$ either contains an edge of the first color, or else is a monochromatic $K_t$ of the second color.
This, together with your recurrence for $R(s,t)$, gives:
$$
R(3,t) le
begin{cases}
R(3,t-1) + t - 1 & text{if $R(3,t-1)$ and $t$ are both even,} \
R(3,t-1) + t & text{otherwise.}
end{cases}
$$
(Okay, technically the first case applies if our upper bound on $R(3,t-1)$ is even, as well as $t$; it doesn't matter if the true value of $R(3,t-1)$ is odd.)
Try this for small values. We have
$R(3,3) = 6$ as the base case.
$R(3,4) le 6 + 4 - 1 = 9$ since $6$ and $4$ are both even.
$R(3,5) le 9 + 5 = 14$.
$R(3,6) le 14 + 6 - 1 = 19$ since $14$ and $6$ are both even.
And so on. It becomes clear that our upper bound on $R(3,t)$ will be odd for even $t$ and even for odd $t$, so that we subtract $-1$ every other step. This may be shown by induction.
We subtract $1$ once for $t=5$, twice for $t=7$, three times for $t=9$, and so on, saving us a total of $frac{t-3}{2}$ for odd $t$. (For even $t$ it is $frac{t-2}{2}$, which is slightly better.) So the usual bound of $R(3,t) le frac{t^2+t}{2}$ improves to $R(3,t) le frac{t^2+t}{2} - frac{t-3}{2} = frac{t^2+3}{2}.$
answered Jan 26 at 2:22
Misha LavrovMisha Lavrov
47.3k657107
$endgroup$
$begingroup$
"Okay, technically the first case applies if our upper bound on R(3,t−1) is even, as well as t; it doesn't matter if the true value of R(3,t−1) is odd." That's what I hadn't understood. Thank you.
$endgroup$
– MateInTwo
Jan 26 at 12:48
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Observe first that $R(2,t) = t$ since any $2$-coloring of $K_t$ either contains an edge of the first color, or else is a monochromatic $K_t$ of the second color.
This, together with your recurrence for $R(s,t)$, gives:
$$
R(3,t) le
begin{cases}
R(3,t-1) + t - 1 & text{if $R(3,t-1)$ and $t$ are both even,} \
R(3,t-1) + t & text{otherwise.}
end{cases}
$$
(Okay, technically the first case applies if our upper bound on $R(3,t-1)$ is even, as well as $t$; it doesn't matter if the true value of $R(3,t-1)$ is odd.)
Try this for small values. We have
$R(3,3) = 6$ as the base case.
$R(3,4) le 6 + 4 - 1 = 9$ since $6$ and $4$ are both even.
$R(3,5) le 9 + 5 = 14$.
$R(3,6) le 14 + 6 - 1 = 19$ since $14$ and $6$ are both even.
And so on. It becomes clear that our upper bound on $R(3,t)$ will be odd for even $t$ and even for odd $t$, so that we subtract $-1$ every other step. This may be shown by induction.
We subtract $1$ once for $t=5$, twice for $t=7$, three times for $t=9$, and so on, saving us a total of $frac{t-3}{2}$ for odd $t$. (For even $t$ it is $frac{t-2}{2}$, which is slightly better.) So the usual bound of $R(3,t) le frac{t^2+t}{2}$ improves to $R(3,t) le frac{t^2+t}{2} - frac{t-3}{2} = frac{t^2+3}{2}.$
answered Jan 26 at 2:22
Misha LavrovMisha Lavrov
47.3k657107
$endgroup$
$begingroup$
"Okay, technically the first case applies if our upper bound on R(3,t−1) is even, as well as t; it doesn't matter if the true value of R(3,t−1) is odd." That's what I hadn't understood. Thank you.
$endgroup$
– MateInTwo
Jan 26 at 12:48
add a comment |
$begingroup$
Observe first that $R(2,t) = t$ since any $2$-coloring of $K_t$ either contains an edge of the first color, or else is a monochromatic $K_t$ of the second color.
This, together with your recurrence for $R(s,t)$, gives:
$$
R(3,t) le
begin{cases}
R(3,t-1) + t - 1 & text{if $R(3,t-1)$ and $t$ are both even,} \
R(3,t-1) + t & text{otherwise.}
end{cases}
$$
(Okay, technically the first case applies if our upper bound on $R(3,t-1)$ is even, as well as $t$; it doesn't matter if the true value of $R(3,t-1)$ is odd.)
Try this for small values. We have
$R(3,3) = 6$ as the base case.
$R(3,4) le 6 + 4 - 1 = 9$ since $6$ and $4$ are both even.
$R(3,5) le 9 + 5 = 14$.
$R(3,6) le 14 + 6 - 1 = 19$ since $14$ and $6$ are both even.
And so on. It becomes clear that our upper bound on $R(3,t)$ will be odd for even $t$ and even for odd $t$, so that we subtract $-1$ every other step. This may be shown by induction.
We subtract $1$ once for $t=5$, twice for $t=7$, three times for $t=9$, and so on, saving us a total of $frac{t-3}{2}$ for odd $t$. (For even $t$ it is $frac{t-2}{2}$, which is slightly better.) So the usual bound of $R(3,t) le frac{t^2+t}{2}$ improves to $R(3,t) le frac{t^2+t}{2} - frac{t-3}{2} = frac{t^2+3}{2}.$
answered Jan 26 at 2:22
Misha LavrovMisha Lavrov
47.3k657107
$endgroup$
$begingroup$
"Okay, technically the first case applies if our upper bound on R(3,t−1) is even, as well as t; it doesn't matter if the true value of R(3,t−1) is odd." That's what I hadn't understood. Thank you.
$endgroup$
– MateInTwo
Jan 26 at 12:48
add a comment |
$begingroup$
Observe first that $R(2,t) = t$ since any $2$-coloring of $K_t$ either contains an edge of the first color, or else is a monochromatic $K_t$ of the second color.
This, together with your recurrence for $R(s,t)$, gives:
$$
R(3,t) le
begin{cases}
R(3,t-1) + t - 1 & text{if $R(3,t-1)$ and $t$ are both even,} \
R(3,t-1) + t & text{otherwise.}
end{cases}
$$
(Okay, technically the first case applies if our upper bound on $R(3,t-1)$ is even, as well as $t$; it doesn't matter if the true value of $R(3,t-1)$ is odd.)
Try this for small values. We have
$R(3,3) = 6$ as the base case.
$R(3,4) le 6 + 4 - 1 = 9$ since $6$ and $4$ are both even.
$R(3,5) le 9 + 5 = 14$.
$R(3,6) le 14 + 6 - 1 = 19$ since $14$ and $6$ are both even.
And so on. It becomes clear that our upper bound on $R(3,t)$ will be odd for even $t$ and even for odd $t$, so that we subtract $-1$ every other step. This may be shown by induction.
We subtract $1$ once for $t=5$, twice for $t=7$, three times for $t=9$, and so on, saving us a total of $frac{t-3}{2}$ for odd $t$. (For even $t$ it is $frac{t-2}{2}$, which is slightly better.) So the usual bound of $R(3,t) le frac{t^2+t}{2}$ improves to $R(3,t) le frac{t^2+t}{2} - frac{t-3}{2} = frac{t^2+3}{2}.$
answered Jan 26 at 2:22
Misha LavrovMisha Lavrov
47.3k657107
$endgroup$
Observe first that $R(2,t) = t$ since any $2$-coloring of $K_t$ either contains an edge of the first color, or else is a monochromatic $K_t$ of the second color.
This, together with your recurrence for $R(s,t)$, gives:
$$
R(3,t) le
begin{cases}
R(3,t-1) + t - 1 & text{if $R(3,t-1)$ and $t$ are both even,} \
R(3,t-1) + t & text{otherwise.}
end{cases}
$$
(Okay, technically the first case applies if our upper bound on $R(3,t-1)$ is even, as well as $t$; it doesn't matter if the true value of $R(3,t-1)$ is odd.)
Try this for small values. We have
$R(3,3) = 6$ as the base case.
$R(3,4) le 6 + 4 - 1 = 9$ since $6$ and $4$ are both even.
$R(3,5) le 9 + 5 = 14$.
$R(3,6) le 14 + 6 - 1 = 19$ since $14$ and $6$ are both even.
And so on. It becomes clear that our upper bound on $R(3,t)$ will be odd for even $t$ and even for odd $t$, so that we subtract $-1$ every other step. This may be shown by induction.
We subtract $1$ once for $t=5$, twice for $t=7$, three times for $t=9$, and so on, saving us a total of $frac{t-3}{2}$ for odd $t$. (For even $t$ it is $frac{t-2}{2}$, which is slightly better.) So the usual bound of $R(3,t) le frac{t^2+t}{2}$ improves to $R(3,t) le frac{t^2+t}{2} - frac{t-3}{2} = frac{t^2+3}{2}.$
answered Jan 26 at 2:22
Misha LavrovMisha Lavrov
47.3k657107
answered Jan 26 at 2:22
Misha LavrovMisha Lavrov
47.3k657107
answered Jan 26 at 2:22
Misha LavrovMisha Lavrov
47.3k657107
answered Jan 26 at 2:22
Misha LavrovMisha Lavrov
47.3k657107
47.3k657107
$begingroup$
"Okay, technically the first case applies if our upper bound on R(3,t−1) is even, as well as t; it doesn't matter if the true value of R(3,t−1) is odd." That's what I hadn't understood. Thank you.
$endgroup$
– MateInTwo
Jan 26 at 12:48
add a comment |
$begingroup$
"Okay, technically the first case applies if our upper bound on R(3,t−1) is even, as well as t; it doesn't matter if the true value of R(3,t−1) is odd." That's what I hadn't understood. Thank you.
$endgroup$
– MateInTwo
Jan 26 at 12:48
$begingroup$
"Okay, technically the first case applies if our upper bound on R(3,t−1) is even, as well as t; it doesn't matter if the true value of R(3,t−1) is odd." That's what I hadn't understood. Thank you.
$endgroup$
– MateInTwo
Jan 26 at 12:48
$begingroup$
"Okay, technically the first case applies if our upper bound on R(3,t−1) is even, as well as t; it doesn't matter if the true value of R(3,t−1) is odd." That's what I hadn't understood. Thank you.
$endgroup$
– MateInTwo
Jan 26 at 12:48
add a comment |
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