Bounds on Ramsey Numbers












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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







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    $begingroup$


    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







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    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






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      asked Jan 24 at 16:22









      MateInTwoMateInTwo

      207




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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 $R(3,t) le frac{t^2+t}{2} - frac{t-3}{2} = frac{t^2+3}{2}.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            "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













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          1 Answer
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          1 Answer
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          active

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          1












          $begingroup$

          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 $R(3,t) le frac{t^2+t}{2} - frac{t-3}{2} = frac{t^2+3}{2}.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            "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


















          1












          $begingroup$

          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 $R(3,t) le frac{t^2+t}{2} - frac{t-3}{2} = frac{t^2+3}{2}.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            "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
















          1












          1








          1





          $begingroup$

          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 $R(3,t) le frac{t^2+t}{2} - frac{t-3}{2} = frac{t^2+3}{2}.$






          share|cite|improve this answer









          $endgroup$



          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 $R(3,t) le frac{t^2+t}{2} - frac{t-3}{2} = frac{t^2+3}{2}.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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




















          • $begingroup$
            "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




















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